Leakage Inductance, Please Explain?
Christopher R. Carlen
I am interested in making a measurement of leakage inductance of a power
supply
transformer secondary. As with my post about capacitor ripple current,
I don't
know for sure what leakage inductance is either. But I suspect that it
appears
as a series inductance in the (simplified) equivalent circuit for the
transformer secondary:
(I have shown some component values measured below:)
O------Rs-----Ls-----+
|
|
|
E(t) Rl
|
|
|
O--------------------+
E(t) is the open circuit transformer secondary voltage = 14.2V (RMS) .
Rs is the transformer secondary resistance = 2.1 ohm .
Rl is the load resistance = 25.2 ohm .
Ls is the leakage inductance, which appears as a series inductance (I'm
really guessing, so correct me if I am wrong). This is unknown.
With the load applied, I measured 12.2V across Rl giving 0.484A in the
circuit.
The total circuit impedance that appears to E(t) is thus
Z = 14.2 V / 0.484 A = 29.3 ohm
and Ls should be
Ls = Sqrt( Z^2 - (Rs^2 + Rl^2) ) / (2 Pi f)
where of course, f = 60/s and so
Ls = 29 mH .
This seems a bit large. So I would appreciate any advice on this
approach, and
some explanations of leakage inductance.
Thanks.
_______________________
Christopher R. Carlen
cr...@epix.net <--- Reply here, please.
car...@cs.moravian.edu
My OS is Linux v2
Roy McCammon
This is about right, but Rs and Rl are both real so they add,
giving you Ls = Sqrt( Z^2 - (Rs + Rl)^2 ) / (2 Pi f)
which barely makes any difference.
Your value for Ls is not unbelievable but others may be
able to tell you if it is typical.
But the L and R you see from the secondary would include
any impedences in the primary divided by N^2 (N:1 transformer)
so you really have
O-Rs+Rp/N^2--Ls+Lp/N^2-----+
|
|
|
E(t) Rl
|
|
|
O--------------------------+
Rp and Lp include any resistence and inductance in the mains wiring
also. You can measue Rp directly to put it into your equation and
solve for Ls+Lp/N^2.
Lets say you find Rp = 80 ohms and let N = 9, now you get
Ls+Lp/N^2 = Sqrt( Z^2 - (80/9^2 + Rs + Rl)^2 ) / (2 Pi f) = 20.25 mH.
Now, that may be sufficient for your needs. That is you don't always
need to know the primary leakage separately from the secondary leakage,
rather that knowing Ls+Lp/N^2 is in fact what you need for your circuit
analysis.
But lets just guess that the primary leakage is N^2 times the secondary.
LS + (N^2Ls/N^2) = Ls(1+1) = 20.25 implies Ls = 10 mH.
Opinions expressed herein are my own and may not represent those of my employer.
Christopher R. Carlen
Christopher R. Carlen wrote:
> Ls = Sqrt( Z^2 - (Rs^2 + Rl^2) ) / (2 Pi f)
Oops, I meant to say
Ls = Sqrt( Z^2 - (Rs + Rl)^2 ) / (2 Pi f)
_______________________
Winfield Hill
Christopher R. Carlen, <cr...@epix.net> said...
>
>Christopher R. Carlen wrote:
>> Ls = Sqrt( Z^2 - (Rs^2 + Rl^2) ) / (2 Pi f)
>
>Oops, I meant to say
>
>Ls = Sqrt( Z^2 - (Rs + Rl)^2 ) / (2 Pi f)
Christopher, make life easy on yourself, and short the
primary before taking the impedance measurement.
Then you'll just be measuring R + Ls, both referred
to the secondary. R should be about Rs + Rp/N^2.
Incidentally, leakage-inductance measurements of this
type on power transformers are fairly consistant, unlike
magnetizing-inductance (open circuit) measurments.
--
Winfield Hill hi...@rowland.org _/_/_/ _/_/_/_/
The Rowland Institute for Science _/ _/ _/_/ _/
Cambridge, MA USA 02142-1297 _/_/_/_/ _/ _/ _/_/_/
_/ _/ _/ _/ _/
http://www.artofelectronics.com/ _/ _/ _/_/ _/_/_/_/
Christopher R. Carlen
Winfield Hill wrote:
> Christopher, make life easy on yourself, and short the
> primary before taking the impedance measurement.
> Then you'll just be measuring R + Ls, both referred
> to the secondary. R should be about Rs + Rp/N^2.
>
> Incidentally, leakage-inductance measurements of this
> type on power transformers are fairly consistant, unlike
> magnetizing-inductance (open circuit) measurments.
You mean I can short the primary, and just apply voltage to the
secondary to measure impedance? Yeah, I guess this makes sense, and
accounts for all the reflected stuff. Of course, it just makes me
curious to see if the different methods will give the same results!
Ultimately, I will put these values into a mathematical model for a full
wave rectifier to predict capacitor ripple current, peak charging
current, and all that.
Don't worry, I'm just doing this for fun, it's not my job.
Stay tuned...
John Woodgate
In article <34AA51CD...@epix.net>, "Christopher R. Carlen"
<cr...@epix.net> writes
>I am interested in making a measurement of leakage inductance of a power
>supply
>transformer secondary. As with my post about capacitor ripple current,
>I don't
>know for sure what leakage inductance is either. But I suspect that it
>appears
>as a series inductance in the (simplified) equivalent circuit for the
>transformer secondary:
>
>(I have shown some component values measured below:)
>
>
>
> O------Rs-----Ls-----+
> |
> |
> |
> E(t) Rl
> |
> |
> |
> O--------------------+
>
>
>
>E(t) is the open circuit transformer secondary voltage = 14.2V (RMS) .
>
>Rs is the transformer secondary resistance = 2.1 ohm .
>
>Rl is the load resistance = 25.2 ohm .
>
>Ls is the leakage inductance, which appears as a series inductance (I'm
>really guessing, so correct me if I am wrong). This is unknown.
Quite right.
>
>With the load applied, I measured 12.2V across Rl giving 0.484A in the
>circuit.
>The total circuit impedance that appears to E(t) is thus
>
>Z = 14.2 V / 0.484 A = 29.3 ohm
>
>and Ls should be
>
>Ls = Sqrt( Z^2 - (Rs^2 + Rl^2) ) / (2 Pi f)
>
>where of course, f = 60/s and so
>
>Ls = 29 mH .
>
>
>This seems a bit large. So I would appreciate any advice on this
>approach, and
>some explanations of leakage inductance.
You don't know what leakage inductance is, but you know 29 mH is 'a bit
large'?
You have everything right in principle, except:
1. The two resistances add directly, not as the sum of squares. i.e. (R1
+ Rs)^2 in the equation.
2. You have forgotten the effect of the resistance of the primary
winding, which also appears as a reflected resistsnce in series with Rs.
For a good transformer. the reflected resistance should be nearly equal
to Rs (equal winding areas for primary and secondary) . You then have Z
= 29.3 and Rtotal = 29.4!!! That does not mean that you have a negative
leakage inductance: it means that it is so small that the measurements
are not accurate enough to determine it.
--
Regards, John Woodgate, Phone +44 (0)1268 747839 Fax +44 (0)1268 777124.
OOO - Own Opinions Only. The first glass for thirst, the second for
nourishment, the third for pleasure and the fourth for madness.
John Woodgate
In article <34AA5A...@mmm.com>, Roy McCammon <rbmcc...@mmm.com>
writes
>Christopher R. Carlen wrote:
>>
>> With the load applied, I measured 12.2V across Rl giving 0.484A in the
>> circuit.
>> The total circuit impedance that appears to E(t) is thus
>>
>> Z = 14.2 V / 0.484 A = 29.3 ohm
>>
>> and Ls should be
>>
>> Ls = Sqrt( Z^2 - (Rs^2 + Rl^2) ) / (2 Pi f)
>>
>> where of course, f = 60/s and so
>>
>> Ls = 29 mH .
>
>
>giving you Ls = Sqrt( Z^2 - (Rs + Rl)^2 ) / (2 Pi f)
>which barely makes any difference.
>
>Your value for Ls is not unbelievable but others may be
>able to tell you if it is typical.
>
>But the L and R you see from the secondary would include
>any impedences in the primary divided by N^2 (N:1 transformer)
>so you really have
>
> O-Rs+Rp/N^2--Ls+Lp/N^2-----+
> |
> |
> |
> E(t) Rl
> |
> |
> |
> O--------------------------+
>
>Rp and Lp include any resistence and inductance in the mains wiring
>also. You can measue Rp directly to put it into your equation and
>solve for Ls+Lp/N^2.
>
>Lets say you find Rp = 80 ohms and let N = 9, now you get
That would not be a typical transformer, because Rs = 2.1 ohms but
Rp/N^2 = 1 ohm. If there are only two windings on the transformer the
refelcted primary resistance should be nearly equal to the secondary
resistance. That is why you are still getting too high a value from the
original figures.
>
>Now, that may be sufficient for your needs. That is you don't always
>need to know the primary leakage separately from the secondary leakage,
>rather that knowing Ls+Lp/N^2 is in fact what you need for your circuit
>analysis.
>
>But lets just guess that the primary leakage is N^2 times the secondary.
>
>LS + (N^2Ls/N^2) = Ls(1+1) = 20.25 implies Ls = 10 mH.
>
>
>
>Opinions expressed herein are my own and may not represent those of my
>employer.
>
--
John Woodgate
In article <68ebhk$2...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
>>
>>
>
> primary before taking the impedance measurement.
> Then you'll just be measuring R + Ls, both referred
> to the secondary. R should be about Rs + Rp/N^2.
>
> Incidentally, leakage-inductance measurements of this
> type on power transformers are fairly consistant, unlike
> magnetizing-inductance (open circuit) measurments.
>
energised transformer. He wouldn't want to short the primary in that
case! (;-).
Yes, leakage inductance measurements are reasonably constant because
most of the flux path is air. That is how the flux forms closed loops
around one winding only, and does not couple to the other one.
John Woodgate
In article <34AAE8FB...@epix.net>, "Christopher R. Carlen"
<cr...@epix.net> writes
>Winfield Hill wrote:
>
>> Christopher, make life easy on yourself, and short the
>> primary before taking the impedance measurement.
>> Then you'll just be measuring R + Ls, both referred
>> to the secondary. R should be about Rs + Rp/N^2.
>>
>> Incidentally, leakage-inductance measurements of this
>> type on power transformers are fairly consistant, unlike
>> magnetizing-inductance (open circuit) measurments.
>
>
>secondary to measure impedance? Yeah, I guess this makes sense, and
>accounts for all the reflected stuff. Of course, it just makes me
>curious to see if the different methods will give the same results!
>
>Ultimately, I will put these values into a mathematical model for a full
>wave rectifier to predict capacitor ripple current, peak charging
>current, and all that.
>
>Don't worry, I'm just doing this for fun, it's not my job.
>
>Stay tuned...
>
>_______________________
>Christopher R. Carlen
>cr...@epix.net <--- Reply here, please.
>car...@cs.moravian.edu
>My OS is Linux v2
Well, it's already been done at least five times. Look, for example, at
K L Smith, 'DC supplies from AC sources - Part 3', Electronics and
Wireless World, February 1995, p.24 -27. This gives 10 references to
earlier work, including mathematical analyses by Stout, Waidelich,
Schade (often cited) and Lieders (also cited because it is extensive and
dates only from 1979). The other references include practical
information.
Christopher R. Carlen
John Woodgate wrote:
> You don't know what leakage inductance is, but you know 29 mH is 'a bit
> large'?
I really didn't know what leakage inductance is before picking up some
hints from other posts. The reason I knew 29 mH was too large, is
because I have a mathematical model that I'm working on to simulate the
capacitor current waveforms, and when I put in 29 mH, or something in
that range, I got waveforms that were very different from what I saw on
the scope. When I took the inductance down to about 1 mH, then my scope
traces look almost exactly like the simulated waveforms. Hence, I
concluded that there must be something about leakage inductance I didn't
know yet. What that turns out to be is all this reflected stuff.
Something I had heard about many times, but hadn't studied yet, well,
because I'm a chemist not an EE.
The following information really straightens out my puzzle. I will try
to get the most accurate correlation possible between my simulated
waveforms and the measured ones before I post the model. Thanks for the
help and stay tuned...
> You have everything right in principle, except:
>
> 1. The two resistances add directly, not as the sum of squares. i.e. (R1
> + Rs)^2 in the equation.
>
> 2. You have forgotten the effect of the resistance of the primary
> winding, which also appears as a reflected resistsnce in series with Rs.
> For a good transformer. the reflected resistance should be nearly equal
> to Rs (equal winding areas for primary and secondary) . You then have Z
> = 29.3 and Rtotal = 29.4!!! That does not mean that you have a negative
> leakage inductance: it means that it is so small that the measurements
> are not accurate enough to determine it.
> --
> Regards, John Woodgate, Phone +44 (0)1268 747839 Fax +44 (0)1268 777124.
> OOO - Own Opinions Only. The first glass for thirst, the second for
> nourishment, the third for pleasure and the fourth for madness.
--
John Woodgate
In article <34ABD604...@epix.net>, "Christopher R. Carlen"
<cr...@epix.net> writes
>John Woodgate wrote:
>
>> You don't know what leakage inductance is, but you know 29 mH is 'a bit
>> large'?
>
>I really didn't know what leakage inductance is before picking up some
>hints from other posts. The reason I knew 29 mH was too large, is
>because I have a mathematical model that I'm working on to simulate the
>capacitor current waveforms, and when I put in 29 mH, or something in
>that range, I got waveforms that were very different from what I saw on
>the scope. When I took the inductance down to about 1 mH, then my scope
>traces look almost exactly like the simulated waveforms. Hence, I
>concluded that there must be something about leakage inductance I didn't
>know yet. What that turns out to be is all this reflected stuff.
>Something I had heard about many times, but hadn't studied yet, well,
>because I'm a chemist not an EE.
Ah, well, that explains it. In isolation, your remarkably insightful
comment looked a bit suspicious.
>
>The following information really straightens out my puzzle. I will try
>to get the most accurate correlation possible between my simulated
>waveforms and the measured ones before I post the model. Thanks for the
>help and stay tuned...
>
Sure will!
Jerry Codner
Christopher R. Carlen wrote:
>
> I am interested in making a measurement of leakage inductance of a power
> supply transformer secondary. As with my post about capacitor ripple current,
> I don't know for sure what leakage inductance is either. But I suspect that it
> appears as a series inductance in the (simplified) equivalent circuit for the
> transformer secondary:
>
> (I have shown some component values measured below:)
>
> O------Rs-----Ls-----+
> |
> |
> |
> E(t) Rl
> |
> |
> |
> O--------------------+
>
> E(t) is the open circuit transformer secondary voltage = 14.2V (RMS) .
>
> Rs is the transformer secondary resistance = 2.1 ohm .
>
> Rl is the load resistance = 25.2 ohm .
>
> Ls is the leakage inductance, which appears as a series inductance (I'm
> really guessing, so correct me if I am wrong). This is unknown.
>
> circuit.
> The total circuit impedance that appears to E(t) is thus
>
> Z = 14.2 V / 0.484 A = 29.3 ohm
>
> and Ls should be
>
>
>
> Ls = 29 mH .
>
> approach, and
> some explanations of leakage inductance.
>
>
> _______________________
> Christopher R. Carlen
Christopher, you have received information from excellent sources such
as R. McCammon, J. Woodgate and W. Hill. Information from these people
should be well regarded. Here are my two cents.
You are correct that leakage inductance is a essentially a series
inductance. The inductance you are trying to measure is sometimes
called the short circuit secondary inductance because it is the secondary
inductance measured with the primary short-circuited (see W. Hill's
suggestion). If the secondary winding resistance is large, your method of
measuring the inductance may be prone to substantial error (as pointed out
by J. Woodgate). Although the transformer is designed for 60 Hz, it is
probably easier to measure the short circuit secondary inductance with an
impedance bridge using 1 kHz (usually) so that the inductive impedance is
appreciable compared to the resistive impedance and also so that the
reactance is extracted from the total impedance. You have to be careful
to avoid the series resonant frequency (leakage inductance and winding
capacitance), but 1 kHz should be sufficiently low, even for a 60 Hz
power transformer. If you don't have an impedance bridge, just use a
signal generator, an oscilloscope and a resistor for current sensing.
The oscilloscope allows you to measure phase so you can extract the real
and imaginary parts of the impedance.
Leakage inductance between two windings of a transformer is due to incomplete
magnetic coupling of the windings and is a function of the geometry of the
windings. Incomplete magnetic coupling means that there is magnetic flux in
one winding that is not present in the other winding.
I agree, 29 mH is rather large for any transformer. 1 mH sounds
more reasonable, but what is the construction? What is the primary to
secondary voltage isolation? What is the turns ratio? What are the current
ratings? All of these things affect the physical layout of the transformer
and therefore leakage inductance.
I read in a later post that you will use this information to analyze ripple
in a power supply. What method will you use for the analysis?
--
Jerry Codner
Unsolicited email sent to this address will
result in an administrative fee of $50.
On-topic replies are excluded, of course.
Christopher R. Carlen
Jerry Codner wrote:
> You are correct that leakage inductance is a essentially a series
> inductance. The inductance you are trying to measure is sometimes
> called the short circuit secondary inductance because it is the secondary
> inductance measured with the primary short-circuited (see W. Hill's
> suggestion). If the secondary winding resistance is large, your method of
> measuring the inductance may be prone to substantial error (as pointed out
> by J. Woodgate). Although the transformer is designed for 60 Hz, it is
> probably easier to measure the short circuit secondary inductance with an
> impedance bridge using 1 kHz (usually) so that the inductive impedance is
> appreciable compared to the resistive impedance and also so that the
> reactance is extracted from the total impedance. You have to be careful
> to avoid the series resonant frequency (leakage inductance and winding
> capacitance), but 1 kHz should be sufficiently low, even for a 60 Hz
> power transformer. If you don't have an impedance bridge, just use a
> signal generator, an oscilloscope and a resistor for current sensing.
> The oscilloscope allows you to measure phase so you can extract the real
> and imaginary parts of the impedance.
It would be nice if I had a real electronics lab, but I don't. I have
to jerry-rig everything. The method of using the scope and a signal
generator is available to me. (Actually, I don't have a signal
generator of my own, buy I could probably borrow one. Or use my ICL8038
chips and my stereo amplifier.) At some time, I may tinker with that.
At this point, I have a way to get a good enough estimate of leakage
inductance to make my model happy.
> I read in a later post that you will use this information to analyze ripple
> in a power supply. What method will you use for the analysis?
> Jerry Codner
I am using MAthematica 3.0 to numerically solve the system of
differential equations that results from Kirchhoff's analysis of a full
wave bridge rectifier circuit. I account for the transformer secondary
equivalent circuit, the voltage vs. current behavior of the diodes, and
the filter cap. and load resistance. At this point, the model is
working fine, after yesterday's measurements of leakage inductance in
the correct manner. (See thread "Leakage inductance measurements"). My
model is giving peak cap. current, RMS cap. current and peak and average
load currents to within about 5% of my measurements.
The tricky part of the model is of course, the diodes. I don't know if
it is typical or not, but I wrote a piecewise linear function to give
diode voltage drop as a function of current through the diode. With a
Kirchhoff's analysis, it didn't seem like there was any other way but
this. (I mean the form could be different, ie. a smooth function
instead of piece-linear, but the functional relation of V(I) must be
used for Kirchhoff's.) I suppose there are fundamentally different
analysis methods, such as those network theorems I learned 12 years ago
in high school, but I have forgotten them and I don't have a book.
(I'll be buying a good book soon though.)
Nonetheless, the diode model I have chosen does a good job, because the
results are sufficiently accurate in quantitative terms, and
qualitatively, the waveforms are almost identical.
The only problem that remains is I must figure out how to get my
Mathematica graphs onto my web page so I can reference them in the next
USENET posts on the subject. I am not experienced in this task, so I
must figure it out over the weekend and hopefully post the whole thing
in a few days.
Roy McCammon
John Woodgate wrote:
>
> In article <34ABD604...@epix.net>, "Christopher R. Carlen"
> <cr...@epix.net> writes
> >John Woodgate wrote:
> >
> >> You don't know what leakage inductance is, but you know 29 mH is 'a bit
> >> large'?
In the old days, you would see power transformers with the secondary
wound directly on the primary with just a little bit of tape separating
them. Then along came some safety regulations (IEC?) that encouraged
designers to use a split bobbin to segregate the primary and secondary,
which sure appears a lot safer and has much lower capacitive coupling
between the windings. A wholly worthy development.
I wonder though if this did cause an overall increase in leakage
inductences for a given VA and size etc.
John Woodgate
In article <34AD06...@mmm.com>, Roy McCammon <rbmcc...@mmm.com>
writes
results: I think I have two transformers with the same spec but
dofferent constructions.
John Woodgate
In article <34ACA33E...@lightlink.com>, Jerry Codner
<gco...@lightlink.com> writes
>inductance. The inductance you are trying to measure is sometimes
>called the short circuit secondary inductance because it is the secondary
>inductance measured with the primary short-circuited (see W. Hill's
>suggestion). If the secondary winding resistance is large, your method of
>measuring the inductance may be prone to substantial error (as pointed out
>by J. Woodgate). Although the transformer is designed for 60 Hz, it is
>probably easier to measure the short circuit secondary inductance with an
>impedance bridge using 1 kHz (usually) so that the inductive impedance is
>appreciable compared to the resistive impedance and also so that the
>reactance is extracted from the total impedance. You have to be careful
>to avoid the series resonant frequency (leakage inductance and winding
>capacitance), but 1 kHz should be sufficiently low, even for a 60 Hz
>power transformer. If you don't have an impedance bridge, just use a
>signal generator, an oscilloscope and a resistor for current sensing.
>The oscilloscope allows you to measure phase so you can extract the real
>and imaginary parts of the impedance.
H'mmm. Although most of the flux path of leakage inductance is air, some
is iron, and using a different frequency and voltage may well not give
the same answer. 'Bottom bend', for example, in a parallel thread here.
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
>
> Roy McCammon <rbmcc...@mmm.com> writes
>>
>> In the old days, you would see power transformers with the secondary
>> wound directly on the primary with just a little bit of tape separating
>> them. Then along came some safety regulations (IEC?) that encouraged
>> designers to use a split bobbin to segregate the primary and secondary,
>> which sure appears a lot safer and has much lower capacitive coupling
>> between the windings. A wholly worthy development.
>>
>> I wonder though if this did cause an overall increase in leakage
>> inductences for a given VA and size etc.
>
> results: I think I have two transformers with the same spec but
John, I'd expect the difference to be considerable, and I eagerly look
forward to your results. Anyone else?
-- Win
Perry L. Vale
hi...@rowland.org (Winfield Hill) wrote:
>-- Win
I design distribution and power transformers for a living, and I can
say from direct experience, that a transformer that has the secondary
wound over the primary directly will have less leakage flux(impedance)
than one with the windings separtated.
Perry Vale <pvale at midusa dot net>
Winfield Hill
Perry L. Vale, <pvale...@midusa.net> said...
>
> hi...@rowland.org (Winfield Hill) wrote:
>
>> John Woodgate, <j...@jmwa.demon.co.uk> said...
>>>
>>> Roy McCammon <rbmcc...@mmm.com> writes
>>>>
>>>> In the old days, you would see power transformers with the secondary
>>>> wound directly on the primary with just a little bit of tape separating
>>>> them. Then along came some safety regulations (IEC?) that encouraged
>>>> designers to use a split bobbin to segregate the primary and secondary,
>>>> which sure appears a lot safer and has much lower capacitive coupling
>>>> between the windings. A wholly worthy development.
>>>>
>>>> I wonder though if this did cause an overall increase in leakage
>>>> inductences for a given VA and size etc.
>>>
>>> Yes, some, but not much. I may get round to providing some comparative
>>> results: I think I have two transformers with the same spec but
>>> different constructions.
>
>> John, I'd expect the difference to be considerable, and I eagerly look
>> forward to your results. Anyone else?
>
> say from direct experience, that a transformer that has the secondary
> wound over the primary directly will have less leakage flux(impedance)
> than one with the windings separtated.
Thanks, Perry that's been my (more limited) experience as well. Also,
looking through the leakage-inductance-calculation sections of a few
reference books this morning, an inability to interleave, let alone
make primary and secondaries overlapping carries a substantial penalty.
But I can't help wondering if the designers have found a way out, and
am curious to see some comparative measurements.
John Woodgate
In article <68lb2a$6...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
>John Woodgate, <j...@jmwa.demon.co.uk> said...
>>
>> Roy McCammon <rbmcc...@mmm.com> writes
>>>
>>> In the old days, you would see power transformers with the secondary
>>> wound directly on the primary with just a little bit of tape separating
>>> them. Then along came some safety regulations (IEC?) that encouraged
>>> designers to use a split bobbin to segregate the primary and secondary,
>>> which sure appears a lot safer and has much lower capacitive coupling
>>> between the windings. A wholly worthy development.
>>>
>>> I wonder though if this did cause an overall increase in leakage
>>> inductences for a given VA and size etc.
>>
>> Yes, some, but not much. I may get round to providing some comparative
>> results: I think I have two transformers with the same spec but
>> different constructions.
>
> John, I'd expect the difference to be considerable, and I eagerly look
> forward to your results. Anyone else?
>
>
Noted. Watch this space! I can envisage that, in microhenrys, the
difference could be considerable, but how much difference it makes to
overall performance as a power transformer is another matter. It would
be different if it were an audio transformer, such as for matching a
loudspeaker to 100 V line: then the impact on overall performance would
likely be very significant. I MAY include something on that as well.
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
>
> Winfield Hill <hi...@rowland.org> writes
>>>> I wonder though if this did cause an overall increase in leakage
>>>> inductences for a given VA and size etc.
>>>
>>> Yes, some, but not much. I may get round to providing some comparative
>>> results: I think I have two transformers with the same spec but
>>> different constructions.
>>
>> John, I'd expect the difference to be considerable, and I eagerly look
>> forward to your results. Anyone else?
>>
>>-- Win
>
> Noted. Watch this space! I can envisage that, in microhenrys, the
> difference could be considerable, but how much difference it makes to
> overall performance as a power transformer is another matter. It would
> be different if it were an audio transformer, such as for matching a
> loudspeaker to 100 V line: then the impact on overall performance would
> likely be very significant. I MAY include something on that as well.
Yes, John, I agree completely. Note my leakage-inductance posting just
now, which includes calculations showing very little impact (HAH! - for
resistive loads - hah!).
Winfield Hill
Jerry Codner, <gco...@lightlink.com> said...
>
>Christopher R. Carlen wrote:
>>
>> I am interested in making a measurement of leakage inductance of a power
>
> Christopher, you have received information from excellent sources such
> as R. McCammon, J. Woodgate and W. Hill. Information from these people
> should be well regarded. Here are my two cents.
>
> You are correct that leakage inductance is a essentially a series
> inductance. The inductance you are trying to measure is sometimes
> called the short circuit secondary inductance because it is the secondary
> inductance measured with the primary short-circuited (see W. Hill's
> suggestion). If the secondary winding resistance is large, your method of
> measuring the inductance may be prone to substantial error (as pointed out
> by J. Woodgate). Although the transformer is designed for 60 Hz, it is
> probably easier to measure the short circuit secondary inductance with an
> impedance bridge using 1 kHz (usually) so that the inductive impedance is
> appreciable compared to the resistive impedance and also so that the
> reactance is extracted from the total impedance. You have to be careful
> to avoid the series resonant frequency (leakage inductance and winding
> capacitance), but 1 kHz should be sufficiently low, even for a 60 Hz
>
> Leakage inductance between two windings of a transformer is due to incomplete
> magnetic coupling of the windings and is a function of the geometry of the
> windings. Incomplete magnetic coupling means that there is magnetic flux in
> one winding that is not present in the other winding.
>
> I agree, 29 mH is rather large for any transformer. 1 mH sounds more
> reasonable, but what is the construction? What is the primary to secondary
> voltage isolation? What is the turns ratio? What are the current ratings?
> All of these things affect the physical layout of the transformer and
> therefore leakage inductance.
With all the discussion of AC line transformers and leakage inductance here
lately, I felt some actual data might be useful. The Signal Transformer 241
series are popular here at the Institute, and here're some AC measurements
from my files. The 5 transformers detailed below are all 24-volt CT, which
is available in 6 sizes (numbered 3 to 8) with rated power levels from about
2.5 watts to 100 watts, and currents from 0.1 to 4.0 amps rms.
The first table below shows measurements relating to the primary. In the
last 5 columns are the DC resistance, magnetizing and leakage inductances,
and the primary magnetizing current at 60Hz and 120.0 Vac, measured in rms
(meter) and peak (scope).
magnetizing
24-volt --- primary --- current
model no rms xfmr DC L_m L_ell rms peak
241-x-24 amps watts ohms mH mH mA mA
--- --- ----- ---- ---- ---- --- ---
3 0.1 2.4 372 4600 377 15 28
4 0.2 5 5 138 3780 178 22 45
5 0.5 12 - - - 27 55 <- #5 no meas avail.
6 1.25 30 21.5 947 58.8 45 90
7 2.4 58 9.64 1748 39.6 79 155
8 4.0 96 4.21 1102 25.4 110 225
Secondary measurements for the same transformers (2.5 watts to 100 watts);
Secondary voltage and impedance measurements in the middle 4 columns below.
------ secondary ------
model no rms xfmr ac V DC L_m L_ell
241-x-24 amps watts rms ohms mH mH
--- --- ----- ---- ---- ---- ----
3 0.1 2.4 30.5 23.8 412 25.1
4 0.2 5 5 30.0 8.33 215 11.3
5 0.5 12 - - - -
6 1.25 30 29.3 1.31 66 3.42
7 2.4 58 28.3 0.55 108 2.18
8 4.0 96 27.8 0.235 66 1.34
It's useful to reflect the primary DC resistance to the secondary (by 1/N^2),
leaving just one number for calculations. In a textbook transformer, the
primary and secondary loss contributions are identical, this is essentially
true for these real transformers. The last column below shows the leakage-
inductance reactance at 60Hz.
sec ----- DC ohms ----- L_leakage
model rms xfmr ac V xfmr'd effective X_L 60Hz
241-x-24 amps watts rms N pri sec total mH ohms
--- --- ----- ---- ---- ---- ---- ---- ---- ----
3 0.1 2.4 30.5 3.93 24.0 23.8 47.8 25.1 9.46
4 0.2 5 5 30.0 4.00 8.63 8.33 17.0 11.3 4.25
5 0.5 12 - - - - - - -
6 1.25 30 29.3 4.10 1.28 1.31 2.59 3.42 1.29
7 2.4 58 28.3 4.24 0.54 0.55 1.09 2.18 1.82
8 4.0 96 27.8 4.31 0.227 0.235 0.462 1.34 0.505
Leakage inductance effects are usually ignored when analyzing transformer
performance, and in fact generally contributes little to the analysis. For
these 24V transformers, the true ac output impedance ranges from 2% above
the effective DC output resistance for the smallest 2.5-watt part (which can
safely be ignored), to a noticable 48% for the 100-watt transformer.
rated Vsec net effective sec. @ rated I
model rms xfmr o/c DC R 60Hz impedance angle Vsec X_L
241-x-24 amps watts Vrms ohms X_L ohms degrees rms attn
--- --- ----- ---- ---- ----- ------ ---- ---- -----
3 0.1 2.4 30.5 47.8 j9.46 48.8 11 25.7 -0.05%
4 0.2 5 5 30.0 17.0 j4.25 17.5 14 26.6 -0.04%
5 0.5 12 - - - - - - -
6 1.25 30 29.3 2.59 j1.29 2.89 26 26.1 -0.15%
7 2.4 58 28.3 1.09 j1.82 1.36 37 25.7 -0.24%
8 4.0 96 27.8 0.462 j0.505 0.684 48 26.0 -0.26%
However, even the high leakage-inductance reactance of the size 8 transformer
does not translate into a high reduction in output voltage. This is because
the reactance should not be compared to Rsource, but to Rsource + Rload, viz,
Vo Rload Rload / ( Rload + Rsource )
-- = ---------------------------- = ---------------------------------
Vs Rload + Rsource + j 2 pi f L sqrt [ 1 + X_L^2 / (Rs + RL)^2 ]
Using a 6.5-ohm resistive load (4 amps), the output voltage is only 0.26%
lower with inductance than without, and the phase shifted only -4 degrees.
(The inductor's effect is larger with a rectifier & capacitor-input-filter.)
It should be noted that if the transformer's center-tapped secondary is used,
the output impedance is much less than the value for entire secondary shown
in the table. For example, a conventional dc rectifier with a center tap and
two diodes (i.e. no bridge rectifier), only one secondary side is used at a
time so I think the effective output resistance is about 3/4, and the leakage
inductance is about 1/4, of the values in the table.
As an exercise for the reader, show that the above 3/4 factor is true, why
this parameter could never approach 1/2, and therefore why a full-bridge
rectifier should be more efficient.
Tony Williams
In article <68mbev$b...@fridge.shore.net>, Winfield Hill
> > I design distribution and power transformers for a living, and I can
> > say from direct experience, that a transformer that has the secondary
> > wound over the primary directly will have less leakage flux(impedance)
> > than one with the windings separtated.
>
> Thanks, Perry that's been my (more limited) experience as well. Also,
> looking through the leakage-inductance-calculation sections of a few
> reference books this morning, an inability to interleave, let alone
> make primary and secondaries overlapping carries a substantial penalty.
> But I can't help wondering if the designers have found a way out, and
> am curious to see some comparative measurements.
For some strange reason I happen to have the following book open here:-
Small Transformers and Inductors. K.A. MacFadyen. Chapman and Hall. 1952.
Chapter 6, Calculation and Control of Inductances, carries a detailed
section on winding methods and the resultant comparative leakage
inductance factors. 4 winding schemes are detailed, superimposed, side
by side, and the sectional versions of those two. For each scheme there
is a diagram of height/width ratios and a detailed formula that produces
what he calls 'K1', the comparative leakage inductance factor.
If you cannot get access to this book I suppose a .gif could be done.
In the meantime, he gives some approximate formulae that you might like
to look (drool?) over......
'h' is the height, or build, of a winding.
'x' is the width of a winding.
1. Simple superimposed windings, Pri first, Sec on top.
Leakage Inductance= 0.97*10^-8 * h/x.
2. Side by Side.
Leakage Inductance= 1.07*10^-8 * x/h.
3. Sandwiched superimposed, Sec-Pri-Sec.
Leakage Inductance= 0.24*10^-8 * h/x.
4. Sandwiched side by side, Pri between two Secs.
Leakage Inductance= 0.27*10^-8 * x/h
Note the change from x/h to h/x, they are not typos (not by me anyway),
so be careful when comparing (1) with (2).
The more complicated formulae consider the relative dimensions of
windings and have an extra term, the gap between windings.
Mac-F seems also to be a great fan of more complicated sectionalising,
not only for leakage inductance but also to balance winding R's and even
to balance capacitances, winding-winding or winding-earth.
--
Tony Williams, Ledbury, UK.
James P. Meyer
On 3 Jan 1998, Winfield Hill wrote:
> Thanks, Perry that's been my (more limited) experience as well. Also,
> looking through the leakage-inductance-calculation sections of a few
> reference books this morning, an inability to interleave, let alone
> make primary and secondaries overlapping carries a substantial penalty.
> But I can't help wondering if the designers have found a way out, and
> am curious to see some comparative measurements.
"Substantial" is one of those "weasle-words" that can mean
anything you want it to mean. 8-)
Interleaving is necessary if one wants to improve coupling in air
cored transformers, but as the premeability of an iron core goes up, the
necessity for tedious winding methods goes down.
Jim
James P. Meyer
On Sat, 3 Jan 1998, John Woodgate wrote:
> Noted. Watch this space! I can envisage that, in microhenrys, the
> difference could be considerable, but how much difference it makes to
> overall performance as a power transformer is another matter. It would
> be different if it were an audio transformer, such as for matching a
> loudspeaker to 100 V line: then the impact on overall performance would
> likely be very significant. I MAY include something on that as well.
Did you see my post on the use of 1:1, 600 ohm, split-bobbin,
audio coupling transformers as common mode chokes?
The measurements I made led me to conclude that split bobbins
have negligible impact on even audio transformers.
Jim
Kevin McMurtrie
In article <ant03235...@ledelec.demon.co.uk>, Tony Williams <to...@ledelec.demon.co.uk> wrote:
- - -
> For some strange reason I happen to have the following book open here:-
>
> Small Transformers and Inductors. K.A. MacFadyen. Chapman and Hall. 1952.
>
> Chapter 6, Calculation and Control of Inductances, carries a detailed
> section on winding methods and the resultant comparative leakage
> inductance factors. 4 winding schemes are detailed, superimposed, side
> by side, and the sectional versions of those two. For each scheme there
> is a diagram of height/width ratios and a detailed formula that produces
> what he calls 'K1', the comparative leakage inductance factor.
- - -
How much difference does wrapping the transformer in a thick copper band have? I see this done in audio amps but it doesn't seem like it would work very well the way they usually have a little spot of solder connecting the two ends.
Tony Williams
In article <ant03235...@ledelec.demon.co.uk>, Tony Williams
>
> In the meantime, he gives some approximate formulae that you might like
> to look (drool?) over......
>
> 'h' is the height, or build, of a winding.
> 'x' is the width of a winding.
>
> 1. Simple superimposed windings, Pri first, Sec on top.
>
> Leakage Inductance= 0.97*10^-8 * h/x.
>
>
> 2. Side by Side.
>
> Leakage Inductance= 1.07*10^-8 * x/h.
>
>
> 3. Sandwiched superimposed, Sec-Pri-Sec.
>
> Leakage Inductance= 0.24*10^-8 * h/x.
>
>
> 4. Sandwiched side by side, Pri between two Secs.
>
> Leakage Inductance= 0.27*10^-8 * x/h
Correction and apologies... It was late at night and I was too keen to
finish, close down and climb the wooden hill. Have some more information.
Those figures above are what he calls 'K1' leakage inductance factor,
(approximate versions). They are to be used in the equation below.
Leakage Inductance referred to Pri, L-l(pri) = K1 * Np^2 * q'.
K1 is one of the shape-factors above, (0.27*10^-8 * x/h), etc.
Np is primary turns-count.
q' is mean turn-length, in inches.
Chuck Parsons
Winfield Hill wrote:
> It's useful to reflect the primary DC resistance to the secondary (by 1/N^2),
> leaving just one number for calculations. In a textbook transformer, the
> primary and secondary loss contributions are identical, this is essentially
> true for these real transformers. The last column below shows the leakage-
> inductance reactance at 60Hz.
>
> sec ----- DC ohms ----- L_leakage
> model rms xfmr ac V xfmr'd effective X_L 60Hz
> 241-x-24 amps watts rms N pri sec total mH ohms
> --- --- ----- ---- ---- ---- ---- ---- ---- ----
> 3 0.1 2.4 30.5 3.93 24.0 23.8 47.8 25.1 9.46
> 4 0.2 5 5 30.0 4.00 8.63 8.33 17.0 11.3 4.25
> 5 0.5 12 - - - - - - -
> 6 1.25 30 29.3 4.10 1.28 1.31 2.59 3.42 1.29
> 7 2.4 58 28.3 4.24 0.54 0.55 1.09 2.18 1.82
> 8 4.0 96 27.8 4.31 0.227 0.235 0.462 1.34 0.505
>
>
Win, its great the way you are not above going back to the bench. O.K. .5 ohms
from both the leakage inductance and resistance of the windings. This should lead
to very noticeable phase shift from the 'expected' 90 between current and voltage
for the the no load condition. I know, wouldn't want to reference to the secondary
in that case but it is still very significant. (Just noting another factoid to
file in neuron pool. )
> It should be noted that if the transformer's center-tapped secondary is used,
> the output impedance is much less than the value for entire secondary shown
> in the table. For example, a conventional dc rectifier with a center tap and
> two diodes (i.e. no bridge rectifier), only one secondary side is used at a
> time so I think the effective output resistance is about 3/4, and the leakage
> inductance is about 1/4, of the values in the table.
>
> As an exercise for the reader, show that the above 3/4 factor is true, why
> this parameter could never approach 1/2, and therefore why a full-bridge
> rectifier should be more efficient.
>
I get the same answer, but it seems more relevant to compare using the centertap
as you describe versus putting the windings in parallel. In that case the
inductance is the same, but the resistance is now 4/3 as much, not 3/4.
Bringing this out differently,
"this parameter could never approach 1/2, and therefore why a full-bridge
rectifier should be more efficient."
I don't think the factor of 1/2 is what you should be comparing to. If we take
both halves of the center tap windings, and put full-wave bridges on them in
parallel, then sure for the secondary windings Rsec -> Rsec/2. But for the
secondary referred primary windings, the loads are in phase and that means Rprimary
-> 2*Rprimary/2 = Rprimary.
Chuck Parsons
Tony Williams wrote:
> In article <68mbev$b...@fridge.shore.net>, Winfield Hill
> <URL:mailto:hi...@rowland.org> wrote:
> >
> > Perry L. Vale, <pvale...@midusa.net> said...
>
> > > I design distribution and power transformers for a living, and I can
> > > say from direct experience, that a transformer that has the secondary
> > > wound over the primary directly will have less leakage flux(impedance)
> > > than one with the windings separtated.
> For some strange reason I happen to have the following book open here:-
>
> Small Transformers and Inductors. K.A. MacFadyen. Chapman and Hall. 1952.
>
> Chapter 6, Calculation and Control of Inductances, carries a detailed
> section on winding methods and the resultant comparative leakage
> inductance factors. 4 winding schemes are detailed, superimposed, side
> by side, and the sectional versions of those two. For each scheme there
> is a diagram of height/width ratios and a detailed formula that produces
> what he calls 'K1', the comparative leakage inductance factor.
>
> If you cannot get access to this book I suppose a .gif could be done.
>
> In the meantime, he gives some approximate formulae that you might like
> to look (drool?) over......
>
> 'h' is the height, or build, of a winding.
> 'x' is the width of a winding.
>
> 1. Simple superimposed windings, Pri first, Sec on top.
>
> Leakage Inductance= 0.97*10^-8 * h/x.
>
> 2. Side by Side.
>
> Leakage Inductance= 1.07*10^-8 * x/h.
>
> 3. Sandwiched superimposed, Sec-Pri-Sec.
>
> Leakage Inductance= 0.24*10^-8 * h/x.
>
> 4. Sandwiched side by side, Pri between two Secs.
>
> Leakage Inductance= 0.27*10^-8 * x/h
>
> Note the change from x/h to h/x, they are not typos (not by me anyway),
> so be careful when comparing (1) with (2).
>
> The more complicated formulae consider the relative dimensions of
> windings and have an extra term, the gap between windings.
>
> Mac-F seems also to be a great fan of more complicated sectionalising,
> not only for leakage inductance but also to balance winding R's and even
> to balance capacitances, winding-winding or winding-earth.
Pretty good stuff.
I assume that x and h are always for all the windings taken as a unit, not
each layer individually. If not comparing the 2 layer and 3 layer configs
changes a bit.
In the same light I assume that x*h is held constant. That is the total core
cross-sectional area stays the same. In particular, assume we have a two
windings one on top of the other (case 1) that are only a single layer width
thick. If when we wrap the core we use a loose helix instead of a tight one
(big gaps between successive turns ) we can increase x, while holding the
number of turns and h constant. I would not expect that to lower the leakage
inductance by a factor of two, the way that x -> sqrt(2)*x, h -> h/sqrt(2)
would in a multilayer single winding.
I would not of guessed that a factor of four improvement would result from a
splitting a single winding into two layers. I hypothesize that once again flux
cancellation is important. Let me explain my reasoning. Consider a 1:2
windings, superimposed, with a double layer of turns on the bottom for the
"2" winding, followed by another winding of a single layer of turns.
For two windings superimposed, and a core of finite permeability, some of
the flux from the bottom winding will return through the air. If the flux lines
travel outside the second winding, all is well we just have a slightly lower
inductance than if we could of kept all the flux in the core. Some of the flux
however, will return beneath the second winding, and reduces the flux enclosed
by the second winding. For example, assume 1% of the flux returns below the
second winding, then the second winding sees a net flux, integrating B dot dA
of 100%-1% = 99%.
Now split the lower winding into two layers and consider each separately,
moving one of the layers to the top to create configuration #3. The lower half
winding is quite similar, except the secondary winding is now closer to the
core allowing less area for flux to return. If we assume a linear loss then the
1% return becomes .5%. For the outside winding, it is not the return flux that
we are concerned with but the flux in the same direction of the core flux, but
which does not go through the core. Assuming linearity with distance between
the winding versus core diameter one would expect this to be 0.5% also.
(I am assuming the winding thickness to be small compared to the core
diameter, I know not usually true, but lets try and keep it simple ).
Since each of the spilt windings has half of the original AmpTurns, the net
loss is 0.5*0.5% + 0.5*0.5% = 0.5%. Or is it???
This is where the flux cancellation comes in. We have considered the air flux
from the bottom most winding just below the middle winding. But it also has air
return flux just above the middle winding, but below its' outer twin. This
flux neatly cancels with the air flux from the outer winding, leading to
0.5*0.5% + 0.5*(0.5% -0.5%) = 0.25%
Anybody buy this? A similar thing happens if you consider the reverse
problem considering the flux from the middle winding on the inner and outer
windings.
It might help to consider the original configuration as two windings just as
I did the split configuration. The math ( if you can call it that ) is then
0.5*1.5% ( for the innermost layer ) + 0.5*0.5%( for the
outer layer) = 1%
You might suggest that there is cancelation in this configuration as well,
and there is, but there is also addition, it still works out the same.
I have not considered what happens in the core for good reasons. The air flux
will make a difference in the core flux for the two configurations, leading to
slightly different magnetizing inductances. But the differance between 99mH and
99.75mH is small for that term. While the difference, in leakage inductance
which comes soley from the the difference of the flux inclosed in the two
windings ( well leads count too, but that is a different problem ) changes from
1mH to 0.25mH a huge factor.
Cheers, Chuck ( thinking out loud again )
John Woodgate
>In article <68mbev$b...@fridge.shore.net>, Winfield Hill
><URL:mailto:hi...@rowland.org> wrote:
>>
>> Perry L. Vale, <pvale...@midusa.net> said...
>
>> > I design distribution and power transformers for a living, and I can
>> > say from direct experience, that a transformer that has the secondary
>> > wound over the primary directly will have less leakage flux(impedance)
>> > than one with the windings separtated.
>>
>> looking through the leakage-inductance-calculation sections of a few
>> reference books this morning, an inability to interleave, let alone
>> make primary and secondaries overlapping carries a substantial penalty.
>> But I can't help wondering if the designers have found a way out, and
>> am curious to see some comparative measurements.
>
Yes, well, he's presumably dealing with _signal_ transformers, for which
the considerations are more extensive than for power transformers.
Sectionalizing is very valuable in many types of signal transformers,
and is also treated in Radio(tron) Designer's Handbook, which shows that
some relatively simple sectionalizing can be as effective as some more
complex arrangements. This can save much money.
John Woodgate
In article <68mgbf$i...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
>John Woodgate, <j...@jmwa.demon.co.uk> said...
>>
>> Winfield Hill <hi...@rowland.org> writes
>
>>>>> I wonder though if this did cause an overall increase in leakage
>>>>> inductences for a given VA and size etc.
>>>>
>>>> Yes, some, but not much. I may get round to providing some comparative
>>>> results: I think I have two transformers with the same spec but
>>>> different constructions.
>>>
>>> John, I'd expect the difference to be considerable, and I eagerly look
>>> forward to your results. Anyone else?
>>>
>>>-- Win
>>
>> difference could be considerable, but how much difference it makes to
>> overall performance as a power transformer is another matter. It would
>> be different if it were an audio transformer, such as for matching a
>> loudspeaker to 100 V line: then the impact on overall performance would
>> likely be very significant. I MAY include something on that as well.
>
> now, which includes calculations showing very little impact (HAH! - for
> resistive loads - hah!).
>
if large enough, is a positive help in extending the conduction angle
losslessly (as opposed to adding series resistance, anyway) so as to
reduce mains harmonic current emissions to meet future EMC requirements,
especially in Europe but in Japan as well. There are several fairly
recent IEEE papers on this subject.
John Woodgate
In article <68mg2f$i...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
> With all the discussion of AC line transformers and leakage inductance here
> lately, I felt some actual data might be useful. The Signal Transformer 241
> series are popular here at the Institute, and here're some AC measurements
> from my files. The 5 transformers detailed below are all 24-volt CT, which
> is available in 6 sizes (numbered 3 to 8) with rated power levels from about
> 2.5 watts to 100 watts, and currents from 0.1 to 4.0 amps rms.
>
> The first table below shows measurements relating to the primary. In the
> last 5 columns are the DC resistance, magnetizing and leakage inductances,
> and the primary magnetizing current at 60Hz and 120.0 Vac, measured in rms
> (meter) and peak (scope).
> magnetizing
> 24-volt --- primary --- current
> model no rms xfmr DC L_m L_ell rms peak
> 241-x-24 amps watts ohms mH mH mA mA
> --- --- ----- ---- ---- ---- --- ---
> 3 0.1 2.4 372 4600 377 15 28
> 4 0.2 5 5 138 3780 178 22 45
> 5 0.5 12 - - - 27 55 <- #5 no meas avail.
> 6 1.25 30 21.5 947 58.8 45 90
> 7 2.4 58 9.64 1748 39.6 79 155
> 8 4.0 96 4.21 1102 25.4 110 225
>
How did you measure the inductances? For example, for transformer 6, the
fact that the mag current is 45 mA rms at 120 V 60 Hz, indicates that
the primary inductance under operating conditions is 7 H, not 947 mH. Of
course, the fact that the peak current is 90 mA, not 63.6 mA indicates
the non-linearity of the BH curve.
I think you used a method that runs into the 'bottom bend' low-
permeability region of the BH curve.
John Woodgate
In article <Pine.SOL.3.91.98010...@godzilla2.acpub.duke
.edu>, "James P. Meyer" <jim...@acpub.duke.edu> writes
>On Sat, 3 Jan 1998, John Woodgate wrote:
>
>> difference could be considerable, but how much difference it makes to
>> overall performance as a power transformer is another matter. It would
>> be different if it were an audio transformer, such as for matching a
>> loudspeaker to 100 V line: then the impact on overall performance would
>> likely be very significant. I MAY include something on that as well.
>
>audio coupling transformers as common mode chokes?
>
> The measurements I made led me to conclude that split bobbins
>have negligible impact on even audio transformers.
>
> Jim
>
>
generality, although it is clearly true under some conditions.
John Woodgate
In article <EYEdooAi...@jmwa.demon.co.uk>, John Woodgate
<j...@jmwa.demon.co.uk> writes, following up his own post:
>In article <34AD06...@mmm.com>, Roy McCammon <rbmcc...@mmm.com>
>writes
>>John Woodgate wrote:
>>>
>>> In article <34ABD604...@epix.net>, "Christopher R. Carlen"
>>> <cr...@epix.net> writes
>>> >John Woodgate wrote:
>>> >
>>> >> You don't know what leakage inductance is, but you know 29 mH is 'a bit
>>> >> large'?
>>
>>In the old days, you would see power transformers with the secondary
>>wound directly on the primary with just a little bit of tape separating
>>them. Then along came some safety regulations (IEC?) that encouraged
>>designers to use a split bobbin to segregate the primary and secondary,
>>which sure appears a lot safer and has much lower capacitive coupling
>>between the windings. A wholly worthy development.
>>
>>I wonder though if this did cause an overall increase in leakage
>>inductences for a given VA and size etc.
>>
>>
>>
>>Opinions expressed herein are my own and may not represent those of my
>employer.
>>
>Yes, some, but not much. I may get round to providing some comparative
>results: I think I have two transformers with the same spec but
>dofferent constructions.
Unfotunately, now I have found them, I remember that the difference is
not that one is double-bobbin and the other is single, but that, athough
they are the same size, one is rated at 20 VA (and presumably has grain-
oriented laminations) while the other is rated at 12 VA (and presumably
has the usual ex-sardine-tin laminations) (;-). Both are double-bob-bob-
bobbin.
I will have a further search for a more suitable pair.
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
>
>In article <EYEdooAi...@jmwa.demon.co.uk>, John Woodgate
><j...@jmwa.demon.co.uk> writes, following up his own post:
>> results: I think I have two transformers with the same spec but
>
> Unfortunately, now I have found them, I remember that the difference is
> not that one is double-bobbin and the other is single, but ...
>I will have a further search for a more suitable pair.
Yes, and since the theoretical difference is only 10%, according to most
authorities, they'll have to be otherwise exactly similar, so the effect
isn't buried in other factors... sigh.
Tony Williams
In article <Oak$oUAgc0...@jmwa.demon.co.uk>, John Woodgate
<URL:mailto:j...@jmwa.demon.co.uk> wrote:
> >
> > Mac-F seems also to be a great fan of more complicated sectionalising,
> > not only for leakage inductance but also to balance winding R's and even
> > to balance capacitances, winding-winding or winding-earth.
>
> Yes, well, he's presumably dealing with _signal_ transformers, for which
> the considerations are more extensive than for power transformers.
> Sectionalizing is very valuable in many types of signal transformers,
> and is also treated in Radio(tron) Designer's Handbook, which shows that
> some relatively simple sectionalizing can be as effective as some more
> complex arrangements. This can save much money.
Know when/where/how to sectionalise windings is part of the art
of transformer design. It is not easy to believe the improvement
possible until actually shown it for the first time.
John Woodgate
In article <mcmurtri-ya0240800...@news.wco.com>, Kevin
McMurtrie <mcmu...@wco.com> writes><to...@ledelec.demon.co.uk> wrote:
>
> - - -
>>
>> Small Transformers and Inductors. K.A. MacFadyen. Chapman and Hall. 1952.
>>
>> Chapter 6, Calculation and Control of Inductances, carries a detailed
>> section on winding methods and the resultant comparative leakage
>> inductance factors. 4 winding schemes are detailed, superimposed, side
>> by side, and the sectional versions of those two. For each scheme there
>> is a diagram of height/width ratios and a detailed formula that produces
>> what he calls 'K1', the comparative leakage inductance factor.
>
>How much difference does wrapping the transformer in a thick copper band have?
>I see this done in audio amps but it doesn't seem like it would work very well
>the way they usually have a little spot of solder connecting the two ends.
Well, it should be a BIG spot of solder. The copper band is to confine
th external field by putting a VERY low impedance shorted-turn around
it. Some of that external field is true leakage field and soem still
coubples to both windings. While the band probably does reduce the
leakage inductance, its primary purpose is to reduce hum induction into
neighbouring circuits and, in particular, tape heads. It can often make
the difference between 'good' and 'not quite good enough', which is
important.
Winfield Hill
>> 241 series are popular here at the Institute, here're some AC measurements
>> from my files. The 5 transformers detailed below are all 24-volt CT,
>> which is available in 6 sizes (numbered 3 to 8) with rated power levels
>> from about 2.5 watts to 100 watts, and currents from 0.1 to 4.0 amps rms.
>>
>> The first table below shows measurements relating to the primary. In the
>> last 5 columns are the DC resistance, magnetizing and leakage inductances,
>> and the primary magnetizing current at 60Hz and 120.0 Vac, measured in
>> rms (meter) and peak (scope).
>> magnetizing
>> 24-volt --- primary --- current
>> model no rms xfmr DC L_m L_ell rms peak
>> 241-x-24 amps watts ohms mH mH mA mA
>> --- --- ----- ---- ---- ---- --- ---
>> 3 0.1 2.4 372 4600 377 15 28
>> 4 0.2 5 5 138 3780 178 22 45
>> 6 1.25 30 21.5 947 58.8 45 90
>> 7 2.4 58 9.64 1748 39.6 79 155
>> 8 4.0 96 4.21 1102 25.4 110 225
> How did you measure the inductances? For example, for transformer 6, the
> fact that the mag current is 45 mA rms at 120 V 60 Hz, indicates that
> the primary inductance under operating conditions is 7 H, not 947 mH. Of
> course, the fact that the peak current is 90 mA, not 63.6 mA indicates
> the non-linearity of the BH curve.
I used a Philips inductance meter, working at 1kHz and 1V rms. This should
yield reliable leakage-inductance measurements, although the magnetizing
inductance values have to be taken with a grain of salt. During the 120V
ac excitation current measurements, I examined the current waveform with a
scope to determine the Ipeak values. In all instances the apparent middle
of the B-H curve currents were say 1/5 of the peak; Most of the measured
rms current was due to the rms meter's response to the peak as the core was
approaching saturation.
I was also disturbed by the low Lm values. If they're incorrectly too low,
this could also explain the seemingly too low inductance ratios (see added
column). Yes you're right, using the measured rms ac current in the formula
L = V / I * 377 results in much higher inductance values, (see next-to-last
column) despite the fact that they should be lower due to saturation! Ah,
that yields much better L_ell ratios!
- 60Hz Lm calculated -
model rms xfmr L_m L_ell rms peak calc L_m Lm / Lell
241-x-24 A watts mH mH ratio mA mA from Irms ratio
----- --- ---- ---- ---- ---- ---- --- ------- ----
3 0.1 2.4 4600 377 12.2 15 28 21 H 56
4 0.2 5 3780 178 21.2 22 45 15 84
5 0.5 12 - - - 27 55 12
6 1.25 30 947 58.8 16.1 45 90 7 119
7 2.4 58 1748 39.6 44.1 79 155 4.0 101
8 4.0 96 1102 25.4 43.4 110 225 2.9 114
> I think you used a method that runs into the 'bottom bend'
> low-permeability region of the BH curve.
Thanks for the suggestion, I'll investigate further. Hmm, time to examine
the full B-H curves for these babies. Isn't it also possible that 1kHz
eddy-current losses in the moderately-laminated iron core are reducing the
permeability and thus my measured Lm inductance? Hah! I see that Terman
called this the "magnetic skin effect."
Winfield Hill
>>>
>>> ...... Watch this space! I can envisage that, in microhenrys, the
>>> difference could be considerable, but how much difference it makes
>>
>> Yes, John, I agree completely. Note my leakage-inductance posting just
>> now, which includes calculations showing very little impact (HAH! - for
>> resistive loads - hah!).
>
> Well, now. When we come to look at rectifiers, the leakage inductance,
> if large enough, is a positive help in extending the conduction angle
> losslessly (as opposed to adding series resistance, anyway) so as to
> reduce mains harmonic current emissions to meet future EMC requirements,
> especially in Europe but in Japan as well. There are several fairly
> recent IEEE papers on this subject.
This is an interesting point, but doesn't it also increase the energy
in the ringing, at the snap when the rectifier diode's reverse-storage
time finishes? Sounds like a tradeoff.
Winfield Hill
Chuck Parsons, <Ch...@CatenaryScientific.com> said...
>
> Winfield Hill wrote:
>
>> It's useful to reflect the primary DC resistance to the secondary (by
>> 1/N^2), leaving just one number for calculations. In a textbook
>> transformer, the primary and secondary loss contributions are identical,
>> this is essentially true for these real transformers. The last column
>>
>> sec ----- DC ohms ----- L_leakage
>> model rms xfmr ac V xfmr'd effective X_L 60Hz
>> 241-x-24 amps watts rms N pri sec total mH ohms
>> --- --- ----- ---- ---- ---- ---- ---- ---- ----
>> 3 0.1 2.4 30.5 3.93 24.0 23.8 47.8 25.1 9.46
>> 4 0.2 5 5 30.0 4.00 8.63 8.33 17.0 11.3 4.25
>> 5 0.5 12 - - - - - - -
>> 6 1.25 30 29.3 4.10 1.28 1.31 2.59 3.42 1.29
>> 7 2.4 58 28.3 4.24 0.54 0.55 1.09 2.18 1.82
>> 8 4.0 96 27.8 4.31 0.227 0.235 0.462 1.34 0.505
> Win, its great the way you are not above going back to the bench.
> O.K. .5 ohms from both the leakage inductance and resistance of the
> windings. This should lead to very noticeable phase shift from the
> 'expected' 90 between current and voltage for the no load condition.
Noooo. You mention current and 90 degrees - are you talking about the
core magnetizing current (btw, another pair of columns in another of the
tables in the post) - this isn't affected significantly by the leakage
inductance, especially with no load. The only significant phase angle
that can result from this is the reactive nature of the transformer's
source impedance. And even that has very little effect on the voltage
phase to the load, in the end.
>> It should be noted that if the transformer's center-tapped secondary
>> is used, the output impedance is much less than the value for entire
>> secondary shown in the table. For example, a conventional dc rectifier
>> with a center tap and two diodes (i.e. no bridge rectifier), only one
>> secondary side is used at a time so I think the effective output
>> resistance is about 3/4, and the leakage inductance is about 1/4, of
>> the values in the table.
>
> I get the same answer, but it seems more relevant to compare using the
> centertap as you describe versus putting the windings in parallel. In
> that case the inductance is the same, but the resistance is now 4/3 as
> much, not 3/4.
Perhaps, except these transformers don't give you two secondaries, but
just one with a center tap. Actually, I think I miscalculated, and should
have gotten 0.5 sqrt 2 + 0.5 = 0.6035 times, for the series resistance for
half-voltage one-side usage. Of course, you'd expect 2x current for half
voltage, so you end up with 1.207 more voltage drop, for a 20.7% penalty.
> Bringing this out differently,
>
> "this parameter could never approach 1/2, and therefore why a full-bridge
> rectifier should be more efficient."
>
> I don't think the factor of 1/2 is what you should be comparing to. If we
> take both halves of the center tap windings, and put full-wave bridges on
> them in parallel, then sure for the secondary windings Rsec -> Rsec/2. But
> for the secondary referred primary windings, the loads are in phase and
> that means Rprimary -> 2*Rprimary/2 = Rprimary.
I'm not sure what you're driving at, but the 0.5 factor I was thinking of
should be more clear from the simple argument above, where the effective
60% dcr of each half causes a 20% penalty at the same power level. Hence
a 50% dcr would have been "perfect."
Note, all these arguments assume the transformer copper has been equally
divided between the primary and secondary. Hmmm. What if the secondary
has more copper?
Winfield Hill
Tony Williams, <to...@ledelec.demon.co.uk> said...
>
> Tony Williams
>
>> In the meantime, he gives some approximate formulae that you might
>> like to look (drool?) over......
>>
>> 'h' is the height, or build, of a winding.
>> 'x' is the width of a winding.
>>
>> 1. Simple superimposed windings, Pri first, Sec on top.
>> Leakage Inductance= 0.97*10^-8 * h/x.
>>
>> 2. Side by Side.
>> Leakage Inductance= 1.07*10^-8 * x/h.
>>
>> 3. Sandwiched superimposed, Sec-Pri-Sec.
>> Leakage Inductance= 0.24*10^-8 * h/x.
>>
>> 4. Sandwiched side by side, Pri between two Secs.
>>
>> Leakage Inductance= 0.27*10^-8 * x/h
>
> (approximate versions). They are to be used in the equation below.
>
> Leakage Inductance referred to Pri, L-l(pri) = K1 * Np^2 * q'.
>
> K1 is one of the shape-factors above, (0.27*10^-8 * x/h), etc.
> Np is primary turns-count.
> q' is mean turn-length, in inches.
It's interesting to note that all the side-by-side winding leakage
inductances are only about 10% higher than the superimposed layered
windings. These are the same differences in slightly different-looking
equations in texts I have. Other formulas use the entire winding area
for both primary and secondary, rather than just one winding (or is that
what you meant? As you stated, they also have a term (magnified 3x) for
insulation or other winding-winding spacing.
It's also very nice to see the dramatic reduction in leakage inductance,
about 4x (others say 3.75), for even the simplest interleaving, achieved
by splitting one winding in two and sandwiching the other. I've seen
this called coaxial or pancake windings. Some carry this much further,
for example the Radiotron Designer's Handbook (thanks Jim Meyer) with 6
secondary and 7 primary, or n = 12 interleaved pancake layers, achieving
n^2 = 144 reduction in Lell. They say the n-sectionalization can be
continued until the ratio n = sqrt (x/3c) approaches unity, where c is
the total insulation width. The RDH ref. is N.H. Crowhurst, Electronic
Engineering, 21.254 (April 1949) p 129, and 21.261 (Nov 1949) p 417.
It's interesting to review textbooks to see the various values for the
multiplicative constant, which range from 11.4 to 13.2 nH/cm. Consider
the table below for the usual layered pri-sec windings, with everybody
adjusted to a single equation,
L_ell = K q' n^2 (c + x/3) / b
where q' is the mean turn-length (for all the copper), n is pri or sec
turns, x and b are the total copper wire-wind widths and heights, resp.,
and c is the total insulation width. The theoretical value for K is the
permeability of free space, K = 4 pi nH/cm, 12.57 nH/cm = 31.9 nH/in.
values for K
source nH/cm nH/in comments, ref. cited
------ ----- ----- ---------------------
T.McLyman 12 (30.5) R. Landee
Terman (11.4) 29 Crowhurst
L. Smith 16 ** Crowhurst (trouble with the RDH chart?)
Pender McIlwain 32 R.R. Lawrence
MacFadyen 29 (see notes)
Grossner 4 pi (31.9) MIT staff, L. Blume
Flamagan 4 pi " --
E.C. Snelling 4 pi " MIT staff, Grossner
Colonel Wm T McLyman (he's not a real Colonel, that's his name) includes
some measurements of a 3-phase transformer, in which the calculated
leakage inductances come out about 8.5 to 10% below his measurements.
We could therefore empirically adjust his constant +10% to 13.2 nH/cm
(33.5 nH/in).
Terman says he uses 2.9x10^-8, rather than 3.2, "found in the textbooks"
(note, 4 pi 2.54 * 10 = 3.19), to adjust by -10% to agree with empirical
results as recommended by J.G. Story, Wireless Eng. 15, p 69 Feb 1938.
Sounds like this is a +/- 10% business!!
Ah, I adjusted MacFadyen's value, as you reported it, by 3 to account for
the 1/3 in most other formulas (and assuming h is the sum of both pri and
sec winding heights).
Chuck Parsons
Winfield Hill wrote:
> Chuck Parsons, <Ch...@CatenaryScientific.com> said...
> > O.K. .5 ohms from both the leakage inductance and resistance of the
> > windings. This should lead to very noticeable phase shift from the
> > 'expected' 90 between current and voltage for the no load condition.
>
> Noooo. You mention current and 90 degrees - are you talking about the
> core magnetizing current (btw, another pair of columns in another of the
> tables in the post)
You are quite right, brain (table) overload on my part. Forgot where I was.
> >> It should be noted that if the transformer's center-tapped secondary
> >> is used, the output impedance is much less than the value for entire
> >> secondary shown in the table. For example, a conventional dc rectifier
> >> with a center tap and two diodes (i.e. no bridge rectifier), only one
> >> secondary side is used at a time so I think the effective output
> >> resistance is about 3/4, and the leakage inductance is about 1/4, of
> >> the values in the table.
> >
> > I get the same answer, but it seems more relevant to compare using the
> > centertap as you describe versus putting the windings in parallel. In
> > that case the inductance is the same, but the resistance is now 4/3 as
> > much, not 3/4.
>
> Perhaps, except these transformers don't give you two secondaries, but
> just one with a center tap. Actually, I think I miscalculated, and should
> have gotten 0.5 sqrt 2 + 0.5 = 0.6035 times, for the series resistance for
> half-voltage one-side usage. Of course, you'd expect 2x current for half
> voltage, so you end up with 1.207 more voltage drop, for a 20.7% penalty.
>
>
Boy, Even when we agree I'm wrong! No I don't follow the above, no doubt you
are considering three constraints simultaneously, that haven't occurred to me
yet. but how does
0.5 sqrt 2 + 0.5 = 0.6305 ?
Try as I might to add pluses minuses or slashes I can't get 0.6305
> > Bringing this out differently,
> >
> > "this parameter could never approach 1/2, and therefore why a full-bridge
> > rectifier should be more efficient."
> >
> > I don't think the factor of 1/2 is what you should be comparing to. If we
> > take both halves of the center tap windings, and put full-wave bridges on
> > them in parallel, then sure for the secondary windings Rsec -> Rsec/2. But
> > for the secondary referred primary windings, the loads are in phase and
> > that means Rprimary -> 2*Rprimary/2 = Rprimary.
>
> I'm not sure what you're driving at, but the 0.5 factor I was thinking of
> should be more clear from the simple argument above, where the effective
> 60% dcr of each half causes a 20% penalty at the same power level. Hence
> a 50% dcr would have been "perfect."
>
Well, would it, in that energy losses are 1/2*RI^2? I mean if the voltage
penalty is 20% shoudn't the energy penalty be 40% since we expect twice the
current? I was starting from the assumption that R needs to be 1/4. The
secondary refered primary part is (I think ), but the secondary winding
resistance is only 1/2 because it's cross-sectional area is half of what it
could be due to the other half taking up space. 0.5*1/2 + 0.5/(1/4) = 3/8. For
an energy penalty of 1/2 * 3/8 * 2^2 / [1/2 * 1 * 1*2] = 1.5. I was trying to
understand your need for 1/2 and the only way I could do it was to assume that
you were saying each winding had to be 1/2 so that when they were put in
parallel they would equal 1/4. As you well know, ( just thought you had slipped
a factor ) that is true for the secondary resistance but not the secondary
referred primary resistance. My argument was to say that in this case 1/2 || 1/2
is 3/8, not 1/4 in this case due to the single primary.
So, it all stems from me not understanding your post, care to elucidate a bit
more? I will go back and re-read your equations, so don't bother if it is all
there already.
The long and the short of it is that in the center tap system half your
copper isn't doing anything at a given time.
Certainly, it would pay to reduce the copper in the primary and increase it in
the secondary to lower the losses. I'll bet that the optimum is once again to
equalize the resistance. Don't you have to say something about winding length vs
cross-sectional area though? Before Ampturns and cross-sectional area were both
constant for both windings, now AmpTurns must stay equal but area is divided
unequally. Without a specified winding form factor I hesitate to say that R is
just proportional to 1/Area in that case.
> Note, all these arguments assume the transformer copper has been equally
> divided between the primary and secondary. Hmmm. What if the secondary
> has more copper?
>
John Woodgate
In article <68p0uv$a...@fridge.shore.net>, Winfield Hill
>> if large enough, is a positive help in extending the conduction angle
>> losslessly (as opposed to adding series resistance, anyway) so as to
>> reduce mains harmonic current emissions to meet future EMC requirements,
>> especially in Europe but in Japan as well. There are several fairly
>> recent IEEE papers on this subject.
>
> This is an interesting point, but doesn't it also increase the energy
> in the ringing, at the snap when the rectifier diode's reverse-storage
> time finishes? Sounds like a tradeoff.
>
energy is involved? I don't recall seeing any ringing on a 60 MHz scope
when looking at waveforms for a 30 V d.c. 5 A supply (J002 bridge
rectifier, with no caps across diodes or windings), although I am not
denying that 'snap' occurs or can occur. In any case, the increase in
conducted emissions above 150 kHz is likely to be much less with the
inductor solution than with active power-factor correction applied to a
transformer/rectifier supply, which is normally very quiet above 150
kHz. That is not necessarily true for a direct-on-line supply as part of
a switch-mode unit, of course.
John Woodgate
In article <68p0u8$a...@fridge.shore.net>, Winfield Hill
>> low-permeability region of the BH curve.
>
> Thanks for the suggestion, I'll investigate further. Hmm, time to examine
> the full B-H curves for these babies. Isn't it also possible that 1kHz
> eddy-current losses in the moderately-laminated iron core are reducing the
> permeability and thus my measured Lm inductance? Hah! I see that Terman
> called this the "magnetic skin effect."
>
significant in 0.5 mm Si-Fe lams at 1 kHz, to add an 'imaginary' part to
the permeability, like cap. losses add one to the permittivity. This
then appears as a shunt resistor across the primary self-inductance in
the conventional equivalent circuit, just as hysteresis loss does.
Winfield Hill
Chuck Parsons, <Ch...@CatenaryScientific.com> said...
>
> Winfield Hill wrote:
>
>> Chuck Parsons, <Ch...@CatenaryScientific.com> said...
>>>> It should be noted that if the transformer's center-tapped secondary
>>>> is used, the output impedance is much less than the value for entire
>>>> secondary shown in the table. For example, a conventional dc rectifier
>>>> with a center tap and two diodes (i.e. no bridge rectifier), only one
>>>> secondary side is used at a time so I think the effective output
>>>> resistance is about 3/4, and the leakage inductance is about 1/4, of
>>>> the values in the table.
>>>
>>> centertap as you describe versus putting the windings in parallel. In
>>> that case the inductance is the same, but the resistance is now 4/3 as
>>> much, not 3/4.
>>
>> Perhaps, except these transformers don't give you two secondaries, but
>> just one with a center tap. Actually, I think I miscalculated, and
>> should have gotten 0.5 sqrt 2 + 0.5 = 0.6035 times, for the series
>> resistance for half-voltage one-side usage. Of course, you'd expect
>> 2x current for half voltage, so you end up with 1.207 more voltage drop,
>> for a 20.7% penalty.
>
> Boy, Even when we agree I'm wrong! No I don't follow the above, no
> doubt you are considering three constraints simultaneously, that haven't
> occurred to me yet. but how does
>
> 0.5 sqrt 2 + 0.5 = 0.6305 ?
Excuse me for the typo, it was, 1/2 ( 0.5 sqrt 2 + 0.5 ) = 0.6305
> Try as I might to add pluses minuses or slashes I can't get 0.6305
This isn't rocket science (but if it were, my rocket engine would have
no doubted have exploded during 2nd-stage separation!!), it's just my
simple-minded analysis of Signal's well-designed transformer, which has
equal resistive-loss contributions from both the primary and secondary
resistances, so Rsec' = Rp/N^2 + Rs, and these are equal terms.
>> I'm not sure what you're driving at, but the 0.5 factor I was thinking
>> of should be more clear from the simple argument above, where the
>> effective 60% dcr of each half causes a 20% penalty at the same power
>> level. Hence a 50% dcr would have been "perfect."
>
> Well, would it, in that energy losses are 1/2*RI^2? I mean if the voltage
> penalty is 20% shouldn't the energy penalty be 40% since we expect twice
> the current? I was starting from the assumption that R needs to be 1/4.
OK, yes that's better, think about power, rather than voltage loss. And
that's exactly right, R/4 is the relevant goal. But of course, with two
windings, only one in use at a time, we can't reach it. Thinking with
pencil and paper now, rather than my typing fingers, hmmm, I've convinced
myself the resistive-loss power penalty for abusing the transformer isn't
40%, but actually 50%. Anyway, whatever, it's way too high!!
>> Note, all these arguments assume the transformer copper has been equally
>> divided between the primary and secondary. Hmmm. What if the secondary
>> has more copper?
>
> The long and the short of it is that in the center tap system half your
> copper isn't doing anything at a given time. Certainly, it would pay to
> reduce the copper in the primary and increase it in the secondary to lower
> the losses. I'll bet that the optimum is once again to equalize the
> resistance.
Yes that's my (uneducated) intuition as well (oops! the fingers again!),
and it likely means 1/3 copper area each to the primary and the two
secondary halves. Resulting in a transformer no longer well suited for
full-bridge rectification.
> Don't you have to say something about winding length vs cross-sectional
> area though? Before Ampturns and cross-sectional area were both constant
> for both windings, now AmpTurns must stay equal but area is divided
> unequally. Without a specified winding form factor I hesitate to say
> that R is just proportional to 1/Area in that case.
OK, now you lost me. Anyway, it's late, and I have to walk our puppy!
Jerry Codner
Jerry Codner, <gco...@lightlink.com> said (regretting part of it)...
> >
> > Leakage inductance between two windings of a transformer is due to incomplete
> > magnetic coupling of the windings and is a function of the geometry of the
> > windings. Incomplete magnetic coupling means that there is magnetic flux in
> > one winding that is not present in the other winding.
> >
> > I agree, 29 mH is rather large for any transformer. 1 mH sounds more
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> > reasonable, but what is the construction? What is the primary to secondary
> > voltage isolation? What is the turns ratio? What are the current ratings?
> > All of these things affect the physical layout of the transformer and
> > therefore leakage inductance.
Ouch, did I really say for *any* transformer. That was a silly generalization
as anyone can see from Winfield's data. [more comments below also].
Winfield Hill replied ...
I certainly made a grossly wrong generalization about the range of leakage
inductances for all transformers, and thanks, Winfield for providing some
hard numbers to work with.
While leakage inductance contributes little to the analysis for your linear
case, it can contribute quite a bit to the case of a transformer driving a
rectifier/capacitive filter because of the effect on conduction angle which
may be of interest to the designer. It has a small effect on ripple and load
regulation for this case. For the inductor input filter, I believe the affect
on load regulation and ripple is more pronounced. This is because the leakage
inductance delays commutation of the rectifiers and the delay produces a notch
in the rectifier output voltage.
If anybody expresses interested, I will provide a SPICE analysis for one of the
above transformers driving a full-wave rectifier with both a capacitive input
and an inductive input filter comparing the negligible leakage inductance case
to the real leakage inductance case. I would examine the following parameters:
line current waveform (conduction angle and rms value), load regulation
(at 5%, 50% and 100% load) and full load output ripple.
--
Jerry Codner
Unsolicited email sent to this address will
result in an administrative fee of $50.
On-topic replies are excluded, of course.
Tony Williams
In article <68p45i$d...@fridge.shore.net>, Winfield Hill
<URL:mailto:hi...@rowland.org> wrote:
>
> It's interesting to note that all the side-by-side winding leakage
> inductances are only about 10% higher than the superimposed layered
> windings. These are the same differences in slightly different-looking
> equations in texts I have. Other formulas use the entire winding area
> for both primary and secondary, rather than just one winding (or is that
> what you meant? As you stated, they also have a term (magnified 3x) for
> insulation or other winding-winding spacing.
I think you can squeeze a bit more out of the sums by looking at the
ratio of winding height(build) v width. Clearly, the optimum leakage
inductance is obtained by having superimposed windings of minimal build.
However, bobbins are bought-out and the bobbin is really the decider.
A wide, flat bobbin would be best wound as superimposed whereas a tall,
narrow bobbin would be best wound as side-by-side pancakes.
Note: It is noticeable that some of the Siemen power ferrites are designed
for a very wide, flat bobbin. They've probably already been here.
Gapped cores introduce another winding-pattern decision, dependant on
where the gap is in relation to the windings... E-I or E-E.
[snip]
> It's interesting to review textbooks to see the various values for the
> multiplicative constant, which range from 11.4 to 13.2 nH/cm. Consider
> the table below for the usual layered pri-sec windings, with everybody
> adjusted to a single equation,
>
> L_ell = K q' n^2 (c + x/3) / b
>
> where q' is the mean turn-length (for all the copper), n is pri or sec
> turns, x and b are the total copper wire-wind widths and heights, resp.,
> and c is the total insulation width. The theoretical value for K is the
> permeability of free space, K = 4 pi nH/cm, 12.57 nH/cm = 31.9 nH/in.
>
> values for K
> source nH/cm nH/in comments, ref. cited
> ------ ----- ----- ---------------------
> T.McLyman 12 (30.5) R. Landee
> Terman (11.4) 29 Crowhurst
> L. Smith 16 ** Crowhurst (trouble with the RDH chart?)
> Pender McIlwain 32 R.R. Lawrence
> MacFadyen 29 (see notes)
> Grossner 4 pi (31.9) MIT staff, L. Blume
> Flamagan 4 pi " --
> E.C. Snelling 4 pi " MIT staff, Grossner
>
[snip]
Fascinating table. Umm...In my little backwoods I forgot that this
subject is so important in tr-design that it was bound to be well
researched and documented in great detail. You've ruined my gloat
over having MacF's detailed formulae.
> Sounds like this is a +/- 10% business!!
Mac-F, page 91, line.2.... "The foregoing formulae are considered to have
an accuracy of about +/- 10%".
> Ah, I adjusted MacFadyen's value, as you reported it, by 3 to account
> for the 1/3 in most other formulas (and assuming h is the sum of both
> pri and sec winding heights).
Just to make it clear, below is the full formula and his approximation
for a simple superimposed winding pattern.
-8 -8
K1 = 2.9*10 ( d + h1+h2 ) Reduces to:- K1 = 0.97*10 *(h/x)
------- ( ----- )
x ( 3 )
d = Interwinding gap.
h1,h2 = Height(build) of windings W1 and W2, and h = h1+h2.
x = Width of the windings.
The difference between the two formulae is merely setting of d=0.
Does this put MacF at odds with other people?
(The cited ref is J.G. Story, 1938 and the phrase "semi-empirical
equations based on a simplified picture of the leakage field" is
used in relation to these formulae.)
BTW:
Isn't it interesting (and reasonable) that because the leakage is assumed
to be in air then the permeability of the ferromagnetic material employed
takes no part in any equations.
Kendall Castor-Perry
[snipped all this good stuff, to save bw]
Keeping these threads builds a valuable reference database - thanks,
Win!
I feel the same about this thread as the one about "Accuracy of..." -
you know which one! 'Surely' someone out there must have the time, the
knowledge and sufficient access to compute resources to do some
numerical simulations to test these various forms? (Someone in a
scientific research institute, for instance...). Calculator-capable
approximations are necessary for the day-to-day but a definitive
calculation of a set of configurations (different sectionalisings,
perhaps) would settle the question as a final analysis.
ISTM that there must be some kind of lower limit to the achievable
leakage inductance *for any particular desired value of interwinding
capacitance* with the inductance going down as the iwc is allowed to
rise. This seems obvious because as you move away from truly coincident
windings (of which multifilar windings are the closest practical
approximation) the current paths are concentrated in zones (sections)
embedded in a matrix of low permeability (copper) and therefore are
likely to share less than 100% of their flux.
If you treat a set of windings as a single current-carrying solenoid
formed as a solid of revolution from the winding cross-section, and
embed it in its magnetic circuit, you can plot the flux pattern to be
expected for excitation at a low frequency (OK, at higher frequencies
there are going to be crowding effects). ISTM that the contours of the
flux pattern represent the most efficient boundaries for the
sectionalisation of windings because by definition there can be no flux
lost through a surface which is simultaneously an equal flux contour.
The trouble is that this requires a wierd form of concentric windings
which would appear to be rather difficult to manufacture!
Anyway, that's about the limit of my ramblings for the moment. Happy
New Year to all you reluctant designers out there (dah!)
Kendall Castor-Perry
If there's a NOSPAM in the reply-to, remove it...
my employer pays for my net connection but I speak for myself.
Chuck Parsons
Jerry Codner wrote:
> If anybody expresses interested, I will provide a SPICE analysis for one of the
> above transformers driving a full-wave rectifier with both a capacitive input
> and an inductive input filter comparing the negligible leakage inductance case
> to the real leakage inductance case. I would examine the following parameters:
> line current waveform (conduction angle and rms value), load regulation
> (at 5%, 50% and 100% load) and full load output ripple.
>
I would certainly like to see it. Jerry and Win will probably yawn, but not all of
us have looked into this subject in such detail. Speaking for myself, I would
appreciate a brief description of the circuit involved, just to ensure that my
guesses about the topology from the lingo (e.g. inductive input filter) are not wrong
in some way. I mean, "it must be "there", right. Whoops noooo..."
I ordered Snelling's "Soft Ferrites" (Ouch! $$ and out of print, but they think
they can get it) but right now my reference shelf is pretty much just AoE, and about
two tons of data books. Well, lots of physics and math books, but certainly nothing
dedicated to power supply design.
Winfield Hill
Kendall Castor-Perry, <ken...@rather.not> said...
>
> I feel the same about this thread as the one about "Accuracy of..." -
> you know which one! 'Surely' someone out there must have the time, the
> knowledge and sufficient access to compute resources to do some
>
> ISTM that there must be some kind of lower limit to the achievable
> leakage inductance *for any particular desired value of interwinding
> capacitance* with the inductance going down as the iwc is allowed to
Just in case it's not obvious to some, the leakage inductance and self
capacitance forms an LC filter setting the high-frequency limit to most
transformer designs. Sectionalization reduces winding capacitance, but
interleaving raises it. It's a kind of Murphy's law here, that the very
scheme most helpful for reducing leakage inductance raises inter-winding
capacitance. However, doesn't it reduce L_ell faster than it raises Cw?
> The trouble is that this requires a wierd form of concentric windings
> which would appear to be rather difficult to manufacture!
I think here we are getting into the area of specialized knowledge and
experience which is the stock-in-trade, tricks-of-the-trade, or perhaps
I should say, trade secrets of wideband audio transformer design, etc.
Clearly those who have learned how to make 5, 6, or even 7-decade (wow!)
transformers aren't going around telling everyone how it's done!
For example, Plitron gets $260 $US for their model 1070-UC 6.7-decade
toroidal trasnformer, which has a magnetizing inductance of 1574H and
leakage inductance of 0.67mH (an amazing 2.3 x 10^6 ratio). It has a
winding capacitance of 388pF (for an L_ell C resonance of 312kHz), so
it can sport a -3dB response of 0.09Hz to 451kHz (a 5 x 10^6 ratio)!!!
And, that's without any feedback!
Hmm, I wonder how much can be achieved by simply using bifilar windings?
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
>
>In article <68p0u8$a...@fridge.shore.net>, Winfield Hill
><hi...@rowland.org> writes
>>John Woodgate, <j...@jmwa.demon.co.uk> said...
>>> I think you used a method that runs into the 'bottom bend'
>>> low-permeability region of the BH curve.
>>
>> Thanks for the suggestion, I'll investigate further. Hmm, time to examine
>> the full B-H curves for these babies. Isn't it also possible that 1kHz
>> eddy-current losses in the moderately-laminated iron core are reducing the
>> permeability and thus my measured Lm inductance? Hah! I see that Terman
>> called this the "magnetic skin effect."
>
> Well, I would expect the eddy-current losses, which can certainly be
> significant in 0.5 mm Si-Fe lams at 1 kHz, to add an 'imaginary' part to
> the permeability, like cap. losses add one to the permittivity. This
> then appears as a shunt resistor across the primary self-inductance in
> the conventional equivalent circuit, just as hysteresis loss does.
Well, yes, but the simple reduction of permeability is a very serious issue
as well, analagous to the reduction in a wire's conductivity in a transformer
at high frequencies due to proximity effect, etc.
Winfield Hill
Jerry Codner, <gco...@lightlink.com> said...
>
>Jerry Codner, <gco...@lightlink.com> said (regretting part of it)...
>
>>> Leakage inductance between two windings of a transformer is due to incomplete
>>> magnetic coupling of the windings and is a function of the geometry of the
>>> windings. Incomplete magnetic coupling means that there is magnetic flux in
>>> one winding that is not present in the other winding.
>>>
>>> I agree, 29 mH is rather large for any transformer. 1 mH sounds more
>
> Ouch, did I really say for *any* transformer. That was a silly generalization
> as anyone can see from Winfield's data. [more comments below also].
>
> Winfield Hill replied ...
>
>> With all the discussion of AC line transformers and leakage inductance here
>
> While leakage inductance contributes little to the analysis for your linear
> case, it can contribute quite a bit to the case of a transformer driving a
> rectifier/capacitive filter because of the effect on conduction angle which
> may be of interest to the designer. It has a small effect on ripple and load
> regulation for this case. For the inductor input filter, I believe the affect
> on load regulation and ripple is more pronounced. This is because the leakage
> inductance delays commutation of the rectifiers and the delay produces a notch
> in the rectifier output voltage.
>
> If anybody expresses interested, I will provide a SPICE analysis for one of the
> above transformers driving a full-wave rectifier with both a capacitive input
> and an inductive input filter comparing the negligible leakage inductance case
> to the real leakage inductance case. I would examine the following parameters:
> line current waveform (conduction angle and rms value), load regulation
> (at 5%, 50% and 100% load) and full load output ripple.
Yes, Jerry, we're more than interested. Sounds like it's time for a few small
binary attachments!
John Woodgate
In article <68p46n$d...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
>
> Note, all these arguments assume the transformer copper has been equally
> divided between the primary and secondary. Hmmm. What if the secondary
> has more copper?
>
bridge is _always_ overall the better solution than a FWCT. If the
secondary has more copper, that *inevitably* means a bigger core,
becuase the primary winding area is fixed for a given core size by the
maximum permissible induction (-> no. of turns) and the full load
current (-> wire gauge). (Note that the supply voltage actually cancels
out in those two terms.)
John Woodgate
In article <68p45i$d...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
> It's interesting to note that all the side-by-side winding leakage
> inductances are only about 10% higher than the superimposed layered
> windings. These are the same differences in slightly different-looking
> equations in texts I have. Other formulas use the entire winding area
> for both primary and secondary, rather than just one winding (or is that
> what you meant? As you stated, they also have a term (magnified 3x) for
> insulation or other winding-winding spacing.
>
> It's also very nice to see the dramatic reduction in leakage inductance,
> about 4x (others say 3.75), for even the simplest interleaving, achieved
> by splitting one winding in two and sandwiching the other. I've seen
> this called coaxial or pancake windings.
I thought pancake winding was the splitting of a wave-wound r.f. choke
into sections with much greater radial dpeth than width, so as to reduce
self-capacitance. Maybe one term used with two different meanings?
>Some carry this much further,
> for example the Radiotron Designer's Handbook (thanks Jim Meyer) with 6
> secondary and 7 primary, or n = 12 interleaved pancake layers, achieving
> n^2 = 144 reduction in Lell. They say the n-sectionalization can be
> continued until the ratio n = sqrt (x/3c) approaches unity, where c is
> the total insulation width. The RDH ref. is N.H. Crowhurst, Electronic
> Engineering, 21.254 (April 1949) p 129, and 21.261 (Nov 1949) p 417.
>
Yes, and such elaborate sectionalization has been used in practice.
However, I think the most interesting cases are N^2 = 9 or 36 in Fig
5.13E of RDH, where the use of half-depth sections allows a considerable
decrease in teh number of sections for a given N^2.
> It's interesting to review textbooks to see the various values for the
> multiplicative constant, which range from 11.4 to 13.2 nH/cm. Consider
> the table below for the usual layered pri-sec windings, with everybody
> adjusted to a single equation,
>
> L_ell = K q' n^2 (c + x/3) / b
>
> where q' is the mean turn-length (for all the copper), n is pri or sec
> turns, x and b are the total copper wire-wind widths and heights, resp.,
> and c is the total insulation width.
Can you please define c more clearly? The insulation width of a paper-
interleaved winding is a little greater than the winding width, but I
don't see what that has to do with L_ell. In RDH, c is the _thickness_
of insulation between layered sections. The chart for L_ell in Fig.
5.13H is a real intellectual teaser!
> The theoretical value for K is the
> permeability of free space, K = 4 pi nH/cm, 12.57 nH/cm = 31.9 nH/in.
[snip]
>
> Sounds like this is a +/- 10% business!!
>
It seems to me that what may be missing is a complete accounting for the
effect of core geometry. For most of the leakage flux, part of the path
is through the core - it must link the winding that generates it. (It
can be proved by experiment, anyway: in spite of the large 'air-gap,
L_ell is non-linear with respect to primary voltage at constant
frequency.)
What I have in mind are two effects:
a) some old lamination patterns have side limbs that are much thinner
than half the centre limb width, so they have a higher reluctance,
tending to allow centre-limb flux to take the air route.
b) the lamination stack may be far from square in cross-section in some
old designs (and even in special cases in some new ones, e.g. equipment,
not using a toroidal transformer for some reason, but still squeezed
into a 1u or 2u rack case). 'Thin' cores give more exposed winding ends
in proportion than square or 'thick' cores.
Kendall Castor-Perry
In article <68qrda$4...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
[snip snip]
>
> Just in case it's not obvious to some, the leakage inductance and self
> capacitance forms an LC filter setting the high-frequency limit to most
> transformer designs. Sectionalization reduces winding capacitance, but
> interleaving raises it. It's a kind of Murphy's law here, that the very
> scheme most helpful for reducing leakage inductance raises inter-winding
> capacitance. However, doesn't it reduce L_ell faster than it raises Cw?
>
*intra*winding (=self-) capacitance. As with your inductor threads,
different approaches to sectionalising or bunching result in differing
Cs. However, ISTM that the *inter*winding capacitance is determined by
the boundary conditions between sections of an inductor which is in fact
wired up to be used as a transformer, whereas *intra* winding
capacitance can be affected by the degree of scramble amongst windings
in a given section. For some applications involving isolation,
capacitance between windings can be more important that the self-
capacitance and its effect on the transformer bandwidth.
I have been trying to get my head around the variation of self-resonant
frequency of a simplified system (say two parallel conductors) because
it's clear that the capacitance falls *very roughly* like 1/r while the
inductance [of the element of loop formed] goes *very roughly* with
area, i.e. as r - so the product is constant. Don't know how to drive
my library well enough to pull the standard formulae for C and L per
unit length of some standard lines, to test not Z but w instead.
Bookworms? I'm going to browse Feynmann (that's an evening gone!) -
anyone got a spare copy of Jackson?
>
> For example, Plitron gets $260 $US for their model 1070-UC 6.7-decade
> toroidal trasnformer, which has a magnetizing inductance of 1574H and
> leakage inductance of 0.67mH (an amazing 2.3 x 10^6 ratio). It has a
> winding capacitance of 388pF (for an L_ell C resonance of 312kHz), so
> it can sport a -3dB response of 0.09Hz to 451kHz (a 5 x 10^6 ratio)!!!
> And, that's without any feedback!
Mein goodness! However, a toroid [presumably ferrite or iron dust in
the case of the Plitron one] with (as you say) bifilar windings has to
be about the best shot you will get at driving the leakage inductance
down simply because very little flux can get out of any given turn
without linking another one. If you could put some permeable material
outside as well you might improve matters further for the windings on
the 'outside' of the core (I was thinking about potting it in epoxy
loaded with ferrite or iron dust?).
Of course if you could find a wire material with more permeability than
copper you could improve the flux linkage still further, by getting some
permeability into the space round which the current is flowing.
At high frequencies, do you get a proximity effect which pushes
current away from the arid centre of the winding pack? I'm wondering
about using a ductile nickel/iron wire plated with copper, or some other
arrangement which gets a little bit of mu in there (trading off on
winding resistance at lower frequencies).
Or is this just counterproductive because increasing mu in the winding
space will intrinsically increase the leakage inductance which, as
MacFadyen observes, is largely independent of core permeability because
most of the leakage occurs elsewhere?
Goodness, don't I ramble. Hope something interesting for someone!
Chuck Parsons
Kendall Castor-Perry wrote:
> In article <68qrda$4...@fridge.shore.net>, Winfield Hill
> <hi...@rowland.org> writes
> >
> > For example, Plitron gets $260 $US for their model 1070-UC 6.7-decade
> > toroidal trasnformer, which has a magnetizing inductance of 1574H and
> > leakage inductance of 0.67mH (an amazing 2.3 x 10^6 ratio). It has a
> > winding capacitance of 388pF (for an L_ell C resonance of 312kHz), so
> > it can sport a -3dB response of 0.09Hz to 451kHz (a 5 x 10^6 ratio)!!!
> > And, that's without any feedback!
>
> Mein goodness! However, a toroid [presumably ferrite or iron dust in
> the case of the Plitron one] with (as you say) bifilar windings has to
> be about the best shot you will get at driving the leakage inductance
> down simply because very little flux can get out of any given turn
> without linking another one.
With you so far
> If you could put some permeable material
> outside as well you might improve matters further for the windings on
> the 'outside' of the core (I was thinking about potting it in epoxy
> loaded with ferrite or iron dust?).
>
No! it would drag flux out of the core by giving it alternate high
permeability return path. For a pot core sure, the outside is the return flux
path, but a torroid? Maybe Win could be persuaded to to a quick test. Take a
torrid transformer, lay a unwound core of same diameter on top of it, and
below it if you have two. My guess is the leakage current goes way up. If I
had a transformer and a core I would do it myself.
>
Torroid side view, normal case.
ooooooooooooooooooooooooooo Windings
---------------------------------------->
-------------> Flux lines normal case
------------------------------------------>
ooooooooooooooooooooooooooo
>
leakage paths caused by high permeability over
layer --------<-------
oooooooo | oooooooooo |oooooo
--------->------
------------------------------------>
-------------------------------------->
oooooooooooooooooooooooooo
> Of course if you could find a wire material with more permeability than
> copper you could improve the flux linkage still further, by getting some
> permeability into the space round which the current is flowing.
> At high frequencies, do you get a proximity effect which pushes
> current away from the arid centre of the winding pack? I'm wondering
> about using a ductile nickel/iron wire plated with copper, or some other
> arrangement which gets a little bit of mu in there (trading off on
> winding resistance at lower frequencies).
>
Huh?! Seems this would be a disaster, all kinds of flux lines that don't
enclose both windings, but just a single strand. Inductance proportional to N
not N^2 in the worst case horrors! I always thought you wanted the windings
and the cores to be like links on a chain. Thick chain for sure, but never
mingle the flux carrier with the current carrier. Vacuum already does that.
> Or is this just counterproductive because increasing mu in the winding
> space will intrinsically increase the leakage inductance which, as
> MacFadyen observes, is largely independent of core permeability because
> most of the leakage occurs elsewhere?
>
Well noowwww you take it all back! ;-) ( maybe ) I don't see how having a
high permeability in the winding space wouldn't cause the flux from outer
layers to not go through the inner layers, thus causing poor coupling and
consequent high leakage inductance, as well as raising the number of turns
needed for a specific leakage inductance.
> Goodness, don't I ramble. Hope something interesting for someone!
>
Interesting to me.
Trivial fact of the day,
Loooooooooooooooooooooooooooooong lines of o's make the Netscape spell
checker really think hard.
Winfield Hill
John Woodgate at j...@jmwa.demon.co.uk says...
>
> Winfield Hill <hi...@rowland.org> writes
>
>> It's also very nice to see the dramatic reduction in leakage inductance,
>> about 4x (others say 3.75), for even the simplest interleaving, achieved
>> by splitting one winding in two and sandwiching the other. I've seen
>> this called coaxial or pancake windings.
>
> I thought pancake winding was the splitting of a wave-wound r.f. choke
> into sections with much greater radial dpeth than width, so as to reduce
> self-capacitance. Maybe one term used with two different meanings?
Perhaps, but several texts use the term.
>> Some carry this much further, for example the Radiotron Designer's
>> Handbook (thanks Jim Meyer) with 6 secondary and 7 primary, or n = 12
>> interleaved pancake layers, achieving n^2 = 144 reduction in Lell.
>> They say the n-sectionalization can be continued until the ratio
>> n = sqrt (x/3c) approaches unity, where c is the total insulation
>> width. The RDH ref. is N.H. Crowhurst, Electronic Engineering,
>> 21.254 (April 1949) p 129, and 21.261 (Nov 1949) p 417.
>
> Yes, and such elaborate sectionalization has been used in practice.
> However, I think the most interesting cases are N^2 = 9 or 36 in Fig
> 5.13E of RDH, where the use of half-depth sections allows a considerable
> decrease in the number of sections for a given N^2.
>
>> It's interesting to review textbooks to see the various values for the
>> multiplicative constant, which range from 11.4 to 13.2 nH/cm. Consider
>> the table below for the usual layered pri-sec windings, with everybody
>> adjusted to a single equation,
>>
>> L_ell = K q' n^2 (c + x/3) / b
>>
>> where q' is the mean turn-length (for all the copper), n is pri or sec
>> turns, x and b are the total copper wire-wind widths and heights, resp.,
>> and c is the total insulation width.
>
> Can you please define c more clearly? The insulation width of a paper-
> interleaved winding is a little greater than the winding width, but I
> don't see what that has to do with L_ell. In RDH, c is the _thickness_
> of insulation between layered sections.
Yes, but it's the sum for all the sections.
> The chart for L_ell in Fig. 5.13H is a real intellectual teaser!
I find it hard to use and can't seem to get the right answer from it.
Too bad they didn't simply print the formula like everyone else!
--
Winfield Hill hi...@rowland.org
Rowland Institute for Science
Cambridge, MA 02142
Winfield Hill
>>>
>>> Well, now. When we come to look at rectifiers, the leakage inductance,
>>> if large enough, is a positive help in extending the conduction angle
>>> losslessly (as opposed to adding series resistance, anyway) so as to
>>> reduce mains harmonic current emissions to meet future EMC requirements,
>>> especially in Europe but in Japan as well.
>>
>> in the ringing, at the snap when the rectifier diode's reverse-storage
>> time finishes? Sounds like a tradeoff.
>>> There are several fairly recent IEEE papers on this subject.
> Well, maybe: the papers I've read don't cover that aspect.
Can you recommend a few?
> How much energy is involved? I don't recall seeing any ringing on a
> 60 MHz scope when looking at waveforms for a 30 V d.c. 5 A supply
> (J002 bridge rectifier, with no caps across diodes or windings),
> although I am not denying that 'snap' occurs or can occur.
Normally we're looking at a 50 to 60Hz wave with 1ms/div, not the
right way to see ringing. But a good question, let's see, maybe a
quick crude calculation will be useful ...
For a 120Hz full-wave rectifier, we could estimate +15A peak pulses,
and a dI/dt of say 20A/ms for the trailing part of the diode current
waveform as it goes through zero. Estimating a 2.5us reverse recovery
time for the diode, that's - 0.05A left flowing when the charge cycle
is over. For the 4A transformer for which I posted data, the leakage
inductance was a modest 1.34mH, so the energy for 50mA would be 1.7uJ.
If the capacitance was say 200pF, the ringing frequency would be about
300kHz. If these numbers were right, we should expect a V = I sqrt (L/C)
= -130V pulse at the transformer secondary-to-diode node, ringing for
one half cycle, afterwhich it would lose its energy through the diode,
back into the storage capacitor. This shouldn't be too hard to see if
the scope was properly triggered.
If only one side is used (center-tap grounded) the leakage inductance
would be 1/4 as large, so we'd get a -32V spike ringing at 600kHz.
OK, not one to leave such an obvious stone unturned, I hooked up the
241-8-24 transformer in question, CT with one side only, with a 1N5404
diode (slower than I'd usually use...), and ran it at 3.5A average load
(I know, way past the ratings!). The current probe showed a 19A peak
2.5ms-long charging waveform, dropping from 15A to zero in the final
1ms, with a 20A/ms finish. Aha! A nice -35V spike was right there,
as expected at xero current, lasting 1.0us (one half-cycle of 500kHz).
Returning to our scenario, if the transformer's reasonably-low leakage
inductance was increased, say 10x, we'd have 1.7uJ of energy 120 times
a second, or about 0.2 mW with a 100kHz ringing frequency.
Hmmm, a fast-recovery diode would greatly reduce the spike power.
altavoz
Win
Just in case it's not obvious to some, the leakage inductance and self
capacitance forms an LC filter setting the high-frequency limit to most
transformer designs. Sectionalization reduces winding capacitance, but
interleaving raises it. It's a kind of Murphy's law here, that the
very
scheme most helpful for reducing leakage inductance raises
inter-winding
capacitance. However, doesn't it reduce L_ell faster than it raises
Cw?
altavoz
Isn't this where your copper foil wound , "100 times lower leak induct'
"
transformer comes in ? In other words , what was it's inter winding
C penalty vs leak induct' ?
John Woodgate
> Clearly those who have learned how to make 5, 6, or even 7-decade (wow!)
> transformers aren't going around telling everyone how it's done!
Well, no, but it's much a matter of applying the (fairly) well-known
rules with insight, as is so much design work. With transformers (and I
suspect SMPS are the same) it helps to have designed a few hundred good
ones previously.
>
> For example, Plitron gets $260 $US for their model 1070-UC 6.7-decade
> toroidal trasnformer, which has a magnetizing inductance of 1574H and
> leakage inductance of 0.67mH (an amazing 2.3 x 10^6 ratio). It has a
> winding capacitance of 388pF (for an L_ell C resonance of 312kHz), so
> it can sport a -3dB response of 0.09Hz to 451kHz (a 5 x 10^6 ratio)!!!
> And, that's without any feedback!
>
> Hmm, I wonder how much can be achieved by simply using bifilar windings?
>
Yes, that's an exceptional spec, but nickel-iron toroids do allow very
high performance to be achieved rather readily. Incidentally, the spec
as you've reported it raises the 'spectre' (sorry) of the largely
illusory '-3 dB' low-frequency limit of a signal transformer. But if we
go into that now, the thread will become dreadfully confused, so I'll
say no more at present (unless provoked!).
John Woodgate
In article <ant05100...@ledelec.demon.co.uk>, Tony Williams
<to...@ledelec.demon.co.uk> writes
[petasnip]
> BTW:
>
> Isn't it interesting (and reasonable) that because the leakage is assumed
> to be in air then the permeability of the ferromagnetic material employed
> takes no part in any equations.
Yes, particularly as experiment shows that the leakage inductance varies
with the applied voltage, so the core material does have some effect.
Leakage flux does not only escape from the exposed ends of the windings:
some goes through the wound part of the core but returns through the
winding that generated it, for example.
John Woodgate
In article <68qreu$4...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
>John Woodgate, <j...@jmwa.demon.co.uk> said...
>>
>>In article <68p0u8$a...@fridge.shore.net>, Winfield Hill
>><hi...@rowland.org> writes
>>>John Woodgate, <j...@jmwa.demon.co.uk> said...
>>>> I think you used a method that runs into the 'bottom bend'
>>>> low-permeability region of the BH curve.
>>>
>>> Thanks for the suggestion, I'll investigate further. Hmm, time to examine
>>> the full B-H curves for these babies. Isn't it also possible that 1kHz
>>> eddy-current losses in the moderately-laminated iron core are reducing the
>>> permeability and thus my measured Lm inductance? Hah! I see that Terman
>>> called this the "magnetic skin effect."
>>
>> Well, I would expect the eddy-current losses, which can certainly be
>> significant in 0.5 mm Si-Fe lams at 1 kHz, to add an 'imaginary' part to
>> the permeability, like cap. losses add one to the permittivity. This
>> then appears as a shunt resistor across the primary self-inductance in
>> the conventional equivalent circuit, just as hysteresis loss does.
>
> Well, yes, but the simple reduction of permeability is a very serious issue
> as well, analagous to the reduction in a wire's conductivity in a transformer
> at high frequencies due to proximity effect, etc.
>
then at 1 kHz with 20 times the voltage so as to keep the same
induction. At constant voltage, yes, the permeability drops off as the
BH loops get smaller and the slopes reduce.
Winfield Hill
Winfield Hill, <hi...@rowland.org> said...
> [ snip ]
> For a 120Hz full-wave rectifier, we could estimate +15A peak pulses,
> and a dI/dt of say 20A/ms for the trailing part of the diode current
> waveform as it goes through zero. Estimating a 2.5us reverse recovery
> time for the diode, that's - 0.05A left flowing when the charge cycle
> is over. For the 4A transformer for which I posted data, the leakage
> inductance was a modest 1.34mH, so the energy for 50mA would be 1.7uJ.
> reasonably-low leakage inductance was increased, say 10x, we'd have
^^^^^ ^^^^^^
17uJ 2 mW
-- Win
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
>
> Winfield Hill <hi...@rowland.org> writes
>>
>> Note, all these arguments assume the transformer copper has been
>> equally divided between the primary and secondary. Hmmm. What if
>> the secondary has more copper?
>
> We've already been through that in the 'private fight' over whether a
> bridge is _always_ overall the better solution than a FWCT. If the
> secondary has more copper, that *inevitably* means a bigger core,
> becuase the primary winding area is fixed for a given core size by
> the maximum permissible induction (-> no. of turns) and the full load
> current (-> wire gauge).
I'm surprised at the apparent rigidity of that conclusion, because the
copper losses are generally equal in the pri and sec, say for full-wave
bridge use, and ...
> (Note that the supply voltage actually cancels out in those two terms.)
Showing that the balance can be skewed somewhat without serious penalty.
On the other hand, when the same transformer is used with a full-wave
diode (not bridge) rectifier, the secondary copper loss is excessive, so
a skewing to the secondary would reduce it's heating at the expense of
some additional in the primary. Note, we're not talking fewer turns, but
just a slightly smaller wire gauge. Naturally this can't at all remedy
the disadvantage incurred from not using a bridge, but surely it can make
a slight improvement.
If this has been sucessfully hashed out, with a conclusion, I'm curious
to see the arguments. What's the relevant info, thread and key dates?
-- Win
Jerry Codner
Jerry Codner said...
> > If anybody expresses interested, I will provide a SPICE analysis for one of the
> > above transformers driving a full-wave rectifier with both a capacitive input
> > and an inductive input filter comparing the negligible leakage inductance case
> > to the real leakage inductance case. I would examine the following parameters:
> > line current waveform (conduction angle and rms value), load regulation
> > (at 5%, 50% and 100% load) and full load output ripple.
>
> Yes, Jerry, we're more than interested. Sounds like it's time for a few small
> binary attachments!
>
> --
> Winfield Hill hi...@rowland.org
OK, but I will use a web page rather than a news post for the 8 x 10 color
glossies with the pictures and arrows rather than binary posting, which is
frowned upon. I'll post a summary in text form also.
I'll use the #8 transformer rated at approximately 96 watts with a filter
capacitor giving <5% ripple. I prefer to use a model that avoids the turns ratio,
since I don't want to get into arguments over my choice of ideal transformer
model, i.e. a transformer that works at frequencies from dc to synchrotron
light. Therefore, I will use the elements referred to the secondary:
rms rms
Xfmr sec I watts sec V pri DCR sec DCR Lleak Lm
8 4.0 96 27.8 0.227 0.235 1.34 mH 66 mH
I will ignore magnetizing inductance, Lm, unless someone strongly objects.
Tony Williams
>
> For example, Plitron gets $260 $US for their model 1070-UC 6.7-decade
> toroidal trasnformer, which has a magnetizing inductance of 1574H and
> leakage inductance of 0.67mH (an amazing 2.3 x 10^6 ratio). It has a
> winding capacitance of 388pF (for an L_ell C resonance of 312kHz), so
> it can sport a -3dB response of 0.09Hz to 451kHz (a 5 x 10^6 ratio)!!!
> And, that's without any feedback!
>
> Hmm, I wonder how much can be achieved by simply using bifilar windings?
Umm... think 'octofilar', or better (worse?), separate windings connected
as req'd to get the turns ratio. Think also autotransformer if possible,
not all apps need galvanic isolation.
Winfield Hill
Jerry Codner, <gco...@lightlink.com> said...
>
>Jerry Codner said...
>
>>> If anybody expresses interested, I will provide a SPICE analysis for
>>> one of the above transformers driving a full-wave rectifier with both
>>> a capacitive input and an inductive input filter comparing the
>>> negligible leakage inductance case to the real leakage inductance
>
>> Yes, Jerry, we're more than interested.
>
> color glossies with the pictures and arrows rather than binary posting,
> which is frowned upon. I'll post a summary in text form also.
I'd say if you can gif the illustaration dwn to 10k (about 2.5 secs of
download time) it should be OK as an attachment. But that's often hard
to do, and the permanacy of a web site is attractive as well.
>
> I'll use the #8 transformer rated at approximately 96 watts with a
> filter capacitor giving <5% ripple. I prefer to use a model that avoids
> the turns ratio, since I don't want to get into arguments over my choice
> of ideal transformer model, i.e. a transformer that works at frequencies
> from dc to synchrotron light. Therefore, I will use the elements
> referred to the secondary:
>
> rms rms
> Xfmr sec I watts sec V pri DCR sec DCR Lleak Lm
> 8 4.0 96 27.8 0.227 0.235 1.34 mH 66 mH
OK, Jerry we're ready ...
-- Win
Winfield Hill
Tony Williams, <to...@ledelec.demon.co.uk> said...
>
> Winfield Hill <hi...@rowland.org> wrote:
>
>> For example, Plitron gets $260 $US for their model 1070-UC 6.7-decade
>> toroidal trasnformer, which has a magnetizing inductance of 1574H and
>> leakage inductance of 0.67mH (an amazing 2.3 x 10^6 ratio). It has a
>> winding capacitance of 388pF (for an L_ell C resonance of 312kHz), so
>> it can sport a -3dB response of 0.09Hz to 451kHz (a 5 x 10^6 ratio)!!!
>> And, that's without any feedback!
>>
>
> Umm... think 'octofilar', or better (worse?), separate windings
> connected as req'd to get the turns ratio.
Yes, multifilar would be the right word no doubt!
> Think also autotransformer if possible, not all apps need galvanic
> isolation.
Of course, tube amplifier audio output transformers certainly do!
BTW, I noticed a remark in Flanagan's book "Handbook of Transformer
Applications" (section 10.6) stating that the autoformer configuration
reduces leakage inductance by [(n-1)/n]^2 over that of a normal ratio-n
step-up transformer. That would mean 1/4 for n=2.
Winfield Hill
altavoz, <zalt...@mail.idt.net> said...
>
>Win
> Just in case it's not obvious to some, the leakage inductance and self
> capacitance forms an LC filter setting the high-frequency limit to most
> transformer designs. Sectionalization reduces winding capacitance, but
> interleaving raises it. It's a kind of Murphy's law here, that the
> very scheme most helpful for reducing leakage inductance raises inter-
> raises Cw?
>----------------------------------------------------------------------
>altavoz
> Isn't this where your copper foil wound , "100 times lower leak induct"
> transformer comes in ? In other words , what was it's inter winding
> C penalty vs leak induct' ?
That design is an intersting example, because it achieved minimum leakage
inductance with only one turn and modest sectionalizing. However, the
whole idea of sectionalizing is to thoroughly intermix the primary and
secondary so they share exactly the same flux lines. The full-width
copper strip was a big start, but there were also several of these strips
connected in parallel for the one-turn primary, with the secondary
winding sectioned between each strip and connected in series.
I never thought to measure the capacitances for those transformers, but
I'm sure there was a substantial penalty for the best version. Maybe
I'll get a chance to revisit that someday.
John Woodgate
In article <68roie$l...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
>John Woodgate at j...@jmwa.demon.co.uk says...
>>>> if large enough, is a positive help in extending the conduction angle
>>>> losslessly (as opposed to adding series resistance, anyway) so as to
>>>> reduce mains harmonic current emissions to meet future EMC requirements,
>>>> especially in Europe but in Japan as well.
>>>
>>> This is an interesting point, but doesn't it also increase the energy
>>> in the ringing, at the snap when the rectifier diode's reverse-storage
>>> time finishes? Sounds like a tradeoff.
>
>>>> There are several fairly recent IEEE papers on this subject.
>> Well, maybe: the papers I've read don't cover that aspect.
>
> Can you recommend a few?
>
Several of these were courtesy of Kendall C-P. I don't fully understand
the IEEE reference codes myself, so I can't help if they can't be
understood:
'Merits and limitations of full-bridge rectifier with LC filter in
meeting IEC1000-3-2 harmonic limits specifications', Milan M Jovanovic
and David E Crow, 0-7803-7/96. [Includes a little theory, practical
results and winding data for inductors!]
'Optimum inputs and output filters for a single-phase rectifier power
supply', Shashi B Dewan, 0093-0094/81/0500-0282 [Highly mathematical:
sketchy indeed on experiment.]
'Improved passive solutions to meet IEC1000-3-2 [now IEC61000-3-2]
regulation in low-cost power supplies', Enrique Masset,Esteban Sanchez,
Javier Sebastian and E de la Cruz, 0-7803-3507-4/96 [This one caused a
bit of consternation in the committee responsible for IEC61000-3-2,
becuase the resulting current waveforms may be consistent with the
limits in the standard but they are dreadful!]
'An approach to the powerfactor compensated and efficiency improved
rectification', Yasanobu Suziki and Toru Teshima, 0-7803-2750-0/95
{Some ingenious rectifier circuits: the information density is high and
the paper is somewhat opaque.]
'Costs and benefits of harmonic current reduction for switch-mode power
supplies in a commercial office building', Thomas S Key, 0094-9994/96
[Rationale for reducing mains harmonic currents. Has been challenged on
its assumptions.]
'Single phase active power filters for multiple non-linear loads', David
A Torrey and Adele M A M Al-Zamel, 0885-8993/95 [About centralized
'filtering'.]
'Poprawa ksztaltu pradu pobieranego z sieci przez prostownik z filtrem
pojemnosciowym', Jozef Borecki, Leslaw Ladniak and Nabel Sobh, Przeglad
Elektrotechniczny R. LXIX Z 11/93 [I don't read Polish, and I apologize
for any spelling mistakes/incorrect rendition of Polish characters in
ASCII. Inthis case, also, the current waveforms do not look good. A
Polish person told me that he thought the paper was a valuable one, but
I don't have a translation yet.]
Most of these are about solutions to the mains harmonic current
emissions problem, but they explain about conduction angle and its
implications as a step along the way.
John Woodgate
In article <68sa3g$d...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
Why do you say 'can't at all'? What other disadvantage of FWCT do you
have in mind apart from the inefficient use of winding area in the
transformer (becuae only one half of the secondary is conducting at any
instant)?
>
> If this has been sucessfully hashed out, with a conclusion, I'm curious
> to see the arguments. What's the relevant info, thread and key dates?
>
>-- Win
>
The thread is 'FWCT vs. bridge ps', and the 'private fight' was between
me and Tony Williams - stemming from the fact that I made a BIG
assumption and didn't state it at the beginning. (I'm just about to add
a new idea to the thread, as well.) But I thought you knew that anyway,
because you've been making comments about bridges being 'more efficient'
than full-wave in this thread.
I have expired most of the articles, but Tony quoted MacFayden(? I have
poor memory for names) saying that, to obtain equal performance (I can't
remember precisely the criteria for 'equal', if they were specified at
all) the FW whole secondary needs, IIRC, 157/110 = 1.43 times the
winding area of the bridge secondary (or the primary). I objected to
that as being a generalization which could be misleading, but (as you
anticipated above) it requires only a bit more than one AWG gauge
downsize (which means two gauges in practice unless you can 'squeeze'
the winding) of the primary to allow the secondary its extra winding
area (SWG and standard metric gauges aren't based on a constant ratio of
areas). Of course, the primary resistance will have gone up,
counteracting the effect of the reduced secondary resistance - may be
less than completely, maybe more than completely, and that applies in
spades to SWG or metric wire, becuase of the non-constant area ratios.
MacFayden, IIRC, takes the view that the only way to get the extra
winding area is to go to a bigger core.
Using the 'rules of thumb' for rms currents, 1.0*Idc for the FWCT (but
two windings) and 1.6*Idc, for the bridge gives a ratio of winding areas
for equal I^2R losses of (2/1.6)^2 = 1.56, which is just under two AWG
gauges.
Now, you see, when I challenge MacFayden for being too 'broad brush',
Tony tells me off. When I quote MacFayden (correctly, I hope), you tell
me off. I think I'll go and eat worms!
And I'm certainly changing my .sig back to the original!
--
Regards, John Woodgate, Phone +44 (0)1268 747839 Fax +44 (0)1268 777124.
OOO - Own Opinions Only. You can fool all of the people some of the time, but
you can't please some of the people any of the time.
Tony Williams
In article <m$rlRDADx...@jmwa.demon.co.uk>, John Woodgate
<URL:mailto:j...@jmwa.demon.co.uk> wrote:
[snip]
> Now, you see, when I challenge MacFayden for being too 'broad brush',
> Tony tells me off. When I quote MacFayden (correctly, I hope), you tell
> me off. I think I'll go and eat worms!
Don't worry John, if you are prepared to quote MacF I'll join in, against
allcomers. Anyway, he may be trying to wind you up.
If I've interpreted the tables correctly, MacF says that for 100 watts
of dc output from a (typical) capacitive-input filter, a bridge rectifier
would need a 140 VA-sized transformer and a bi-phase would need 170 VA.
Putting it more usefully, for the same-size transformer you can develop
more dc-power with a bridge than a bi-phase, 170:140, or about 20% more.
(That might be a useful expt to do, see what relative dc-powers can
be obtained for the same nominal 'full-power' primary rms-current.
Probably best done at reduced V-pri to avoid the expt being affected
by saturation effects.)
I am only just re-acquainting myself with Flanagan and have not found any
refs about tr-size there yet. He does however make a general reference
to the "substantial increase in the VA of power transformers serving
rectifiers with capacitive-input filters, etc", "unpleasant surprise".
The following are suggested refs, not seen by me:-
Schade O.H. "Analysis of Rectifier Operation" Proc IRE 1943 Vol.31 No.7.
Weidelich D.L. "Analysis of Full-Wave Rectifier Circuits and Capacitive-
-Input Circuits". Electronics, 1947, Vol.20, No.9.
cc'd and posted.
Tony Williams
Just out of interest, I've done a .gif of the various winding schemes
and associated leakage inductance factors for them, as detailed by
MacFadyen. They are slightly old (compared to Flanagan say) but do
give a useful insight into improvements possible.
Although not done on a PC, they have been kindly tested as receivable
and readable by a PC.
The .gif is 24k long, about 35k uuencoded. If anyone wants a copy
please drop an Email with the title MACF? and no need for anything
in the body. Please unmangle your return addr if you normally do so.
Cheers,
altavoz
Can i suggest a direction . Can we discuss the actual flux
path thru the windings to the core .
For a guess , does the core attract the flux so forcefully that
as soon it appears at the surface of the pri' wire , it is pulled
DIRECTLY into the core over the shortest path . I think this is true.
Which means the sec' had better be directly in the "way" , in order
to "cut" flux lines . If sec was a sheet of copper , completely
obscuring the core , all the pri flux would need to cut the sec .
Nix Z for email
Winfield Hill
Tony Williams, <to...@ledelec.demon.co.uk> said...
>
> John Woodgate <j...@jmwa.demon.co.uk> wrote:
>
> [snip]
>> Now, you see, when I challenge MacFayden for being too 'broad brush',
>> Tony tells me off. When I quote MacFayden (correctly, I hope), you tell
>> me off. I think I'll go and eat worms!
>
> Don't worry John, if you are prepared to quote MacF I'll join in, against
> allcomers. Anyway, he may be trying to wind you up.
Gentleman! Please gentleman! Cool your jets!
> If I've interpreted the tables correctly, MacF says that for 100 watts
> of dc output from a (typical) capacitive-input filter, a bridge rectifier
> would need a 140 VA-sized transformer and a bi-phase would need 170 VA.
>
> Putting it more usefully, for the same-size transformer you can develop
> more dc-power with a bridge than a bi-phase, 170:140, or about 20% more.
I have no argument with these conclusions, although I would like to know
the criteria. Some of my own calculations arrive at a different value.
I'll go back and read your discussion first, so you won't have to
rewrite it.
Tony, did you ever give us the full name and details for the MacFayden
book, ISBN etc.?
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
No, no, I simply re-emphasized that optimizing a transfomer for FW-CT
in no way makes up for its inferior performance in that mode.
>>> secondary has more copper, that *inevitably* means a bigger core,
This is the statement that botehrs me. To begin with, it means the
primary has less copper, taken entirely out of the wire size. So the
primary will have more copper loss, but the secondary will have less,
with a total loss below that possible with equal areas. This can be
easily shown mathematically, although I haven't seen it in any of my
books (OK, haven't really looked either!)
>> If this has been sucessfully hashed out, with a conclusion, I'm curious
>> to see the arguments. What's the relevant info, thread and key dates?
> The thread is 'FWCT vs. bridge ps', and the 'private fight' was between
> me and Tony Williams ...
OK, I know the one you mean... my eyes glazed over early on and I simply
didn't absorb the meat of your discussion (weren't there lots of you said,
I said exchanges?) so I'll try absorbing something from deja-news.
> ... Tony quoted MacFayden ...
Did we ever get a full reference for this book, title, ISBN number?
[ snip nice synopsis of some MacFayden points... ]
> Now, you see, when I challenge MacFayden for being too 'broad brush',
> Tony tells me off. When I quote MacFayden (correctly, I hope), you tell
> me off. I think I'll go and eat worms!
Well, I'm truly sorry for pushing the buttons, I'll study what you all
said and then re-ask my question.
> And I'm certainly changing my .sig back to the original!
> --
> Regards, John Woodgate, Phone +44 (0)1268 747839 Fax +44 (0)1268 777124.
> OOO - Own Opinions Only. You can fool all of the people some of the time,
> but you can't please some of the people any of the time.
Now John, let me state here, your writings please me most of the time!
Hey, Tony's as well, for that matter.
James Meyer
On 6 Jan 1998 15:23:04 GMT, hi...@rowland.org (Winfield Hill) wrote:
> BTW, I noticed a remark in Flanagan's book "Handbook of Transformer
> Applications" (section 10.6) stating that the autoformer configuration
> reduces leakage inductance by [(n-1)/n]^2 over that of a normal ratio-n
> step-up transformer. That would mean 1/4 for n=2.
Speaking of remarks in books... Norman Gilbert, in his 1941
book "Electricity and Magnetism", has this little "throw-away" line...
"When the secondary current leads the emf the effect of
magnetic leakage is to increase the secondary voltage. This effect
may be sufficient to more than balance the drop due to the resistance
of the coils and so give a net increase in the voltage of the
secondary with increase of load."
As soon as you guys figure out conventional transformers, you
can start in on things like Sola constant voltage transformers where
leakage inductance is purposely introduced. 8-)
Jim
JoseSainz
One good book dealing in leakage inductance and transformer design is
Transformer and Inductor Design Handbook
Col. Wm. T. McLyman
ISBN: 0-8247-6801-9 1978
See page 76
José
Roy McCammon
John Woodgate wrote:
> I think I'll go and eat worms!
anecdotal evidence suggests that it won't be fatal.
Opinions expressed herein are my own and may not represent those of my employer.
Roy McCammon
If I could single handedly change the terms people use,
I'd have every body start saying "flux circles" instead
of "flux lines". Of course, the curves are neither lines
nor circles (usually) but they are closed, so I think
"circles of flux captures their essence batter than
lines of flux.
Now, knowing that it is a circle (closed curve) and that
the primary winding is inside the circle, if the entire
circle of flux is pulled into the core, it must
cut the secondary in the process of entering the core.
So, the circles that don't cut the secondary are either
entirely in the air of partly in the air and partly
in the core. Don't know, but I think there are not
so many of this second type.
John Woodgate
In article <ant06235...@ledelec.demon.co.uk>, Tony Williams
<to...@ledelec.demon.co.uk> writes
>In article <m$rlRDADx...@jmwa.demon.co.uk>, John Woodgate
><URL:mailto:j...@jmwa.demon.co.uk> wrote:
>
>[snip]
>> Tony tells me off. When I quote MacFayden (correctly, I hope), you tell
>
> Don't worry John, if you are prepared to quote MacF I'll join in, against
> allcomers. Anyway, he may be trying to wind you up.
Is that a punny?
>
> If I've interpreted the tables correctly, MacF says that for 100 watts
> of dc output from a (typical) capacitive-input filter, a bridge rectifier
> would need a 140 VA-sized transformer and a bi-phase would need 170 VA.
>
> Putting it more usefully, for the same-size transformer you can develop
> more dc-power with a bridge than a bi-phase, 170:140, or about 20% more.
>
> (That might be a useful expt to do, see what relative dc-powers can
> be obtained for the same nominal 'full-power' primary rms-current.
> Probably best done at reduced V-pri to avoid the expt being affected
> by saturation effects.)
No, one should do the experiment under realistic practical conditions.
If not, I'll use my 100 mV, 200 A supply to prove that FWCT is better!
>
> I am only just re-acquainting myself with Flanagan and have not found any
> refs about tr-size there yet. He does however make a general reference
> to the "substantial increase in the VA of power transformers serving
> rectifiers with capacitive-input filters, etc", "unpleasant surprise".
>
> The following are suggested refs, not seen by me:-
>
> Schade O.H. "Analysis of Rectifier Operation" Proc IRE 1943 Vol.31 No.7.
>
> Weidelich D.L. "Analysis of Full-Wave Rectifier Circuits and Capacitive-
> -Input Circuits". Electronics, 1947, Vol.20, No.9.
>
Both of these deal with tube rectifiers which have, by modern standards,
huge resistive losses. To cope with semiconductor rectifiers, you
practically have to rewrite the papers. What they say is valid, but it
rarely applies to current designs.
Kendall Castor-Perry
In article <69047v$5...@fridge.shore.net>, Winfield Hill
<hi...@rowland.org> writes
[snip]
>
> >>> secondary has more copper, that *inevitably* means a bigger core,
>
> This is the statement that botehrs me. To begin with, it means the
> primary has less copper, taken entirely out of the wire size. So the
> primary will have more copper loss, but the secondary will have less,
> with a total loss below that possible with equal areas. This can be
> easily shown mathematically, although I haven't seen it in any of my
> books (OK, haven't really looked either!)
>
Is this comment specifically relevant to the case where the secondary
windings are underused by the FWCT arrangement? As a general statement,
'surely' it is incorrect? It looks like a classic " x + 1/x " situation
with the turning point at x=1. Let's see if this works out.
Think of a general purpose transformer wound with a single turn of
multifilar winding wire, forming a large number N of closely linked
turns which can be connected up as primaries or secondaries. Make up a
1:1 transformer by using half the windings in parallel as a primary,
other half as a secondary. Load up the transformer, each winding
dissipates a bit due to its copper loss. The current in both windings
is equal to I.
Now transfer some windings from one winding to the other, so that there
are n {little n} turns on one winding and N-n turns on the other. By
inspection, with a resistance per turn of r, the total power dissipated
is
P = rI^2 * ( 1 1 )
(____ + ____)
( N-n n )
[boy, this ASCII equation stuff is harder than I thought!]
Differentiate P wrt n to get how dissipation varies - I won't write it
down! - and you find it's positive for any n apart from N/2. This shows
that the optimum configuration in this case is equal area for each
winding, and with this analogy I reckon it extends to any usage of
individual turns, i.e. to any ratios.
What will *always* shift the equilibrium point is the presence of an
external term. Suppose you wanted to minimise the total dissipation in
the transformer *and the rectifiers* (a resonable target since there is
nothing which can be done about the load dissipation and only so much to
be done about regulator dissipation. The capacitors may also get
involved - check their ESR). With a few basic assumptions, you could
extend the above argument to get optimum pri:sec area ratios for any
given configuration. In fact, this would be quite easy to incorporate
into f.i. a spreadsheet method for calculating the currents and powers
in the supply design itself.
Kendall Castor-Perry
If there's a NOSPAM in the reply-to, remove it...
my employer pays for my net connection but I speak for myself.
Tony Williams
In article <690492$5...@fridge.shore.net>, Winfield Hill
<URL:mailto:hi...@rowland.org> wrote:
>
> Tony Williams, <to...@ledelec.demon.co.uk> said...
> > If I've interpreted the tables correctly, MacF says that for 100 watts
> > of dc output from a (typical) capacitive-input filter, a bridge rectifier
> > would need a 140 VA-sized transformer and a bi-phase would need 170 VA.
> >
> > Putting it more usefully, for the same-size transformer you can develop
> > more dc-power with a bridge than a bi-phase, 170:140, or about 20% more.
>
> I have no argument with these conclusions, although I would like to know
> the criteria. Some of my own calculations arrive at a different value.
I have not yet found any better advice on relating the output dc Watts
to the required VA of the transformer. Would be interested to find some.
> Tony, did you ever give us the full name and details for the MacFayden
> book, ISBN etc.?
Yes, and of Schade and Weiderlich. Think you now have all these refs.
Our old friend Samuel Seely also has a slight look into this question
of rectifiers and capacitor-filters... "Electron Tube Circuits". He
gives some very useful approximations based on triangular ripple and
they do give reasonably accurate results.
I still have to re-read Flanagan, but my Terman has been 'borrowed'.
Tony Williams
In article <8NJD9TAJ...@jmwa.demon.co.uk>, John Woodgate
<URL:mailto:j...@jmwa.demon.co.uk> wrote:
>
> In article <ant06235...@ledelec.demon.co.uk>, Tony Williams
> <to...@ledelec.demon.co.uk> writes
> > Don't worry John, if you are prepared to quote MacF I'll join in, against
> > allcomers. Anyway, he may be trying to wind you up.
>
> Is that a punny?
Already explained by email.
> > (That might be a useful expt to do, see what relative dc-powers can
> > be obtained for the same nominal 'full-power' primary rms-current.
> > Probably best done at reduced V-pri to avoid the expt being affected
> > by saturation effects.)
>
> No, one should do the experiment under realistic practical conditions.
This thread is all about doing measurements on transformers, doing sums
to find certain parameters, and then using the whole to see if it is
possible to predict the performance of, say, rectifier+filter circuits.
We are heading towards seeing if theory and practical results tie up.
Consequently it would seem prudent to pick measurement points that are
consistent with the final working points, or vice-versa.
Today's commodity transformers run so near the knuckle of Bsat that is
is dangerous to casually do, say apparent Shunt-Inductance or Turns
Ratio at full working voltage and then use those results in downstream
sums. It would be more sensible avoid the worst of the distortions
caused by saturation and to back down a little, to 80/90% of rated supply.
Therefore it is then perfectly consistent to suggest running the final
power supply at 80/90% stimulus in order to verify upstream predictions.
My suggestions for consistent results are:-
1. Test at the nominal frequency (may not be practical for L-ell).
2. Do R-windings at dc currents equal to the full load rms currents, or
some other method intended to account for the ppm/C of copper.
3. Shunt-Inductance at 80% of nominal supply. (I-mag + Watts-Loss.)
Don't trust bridge measurements, ie wimpy little voltages.
4. Leakage Inductance at full-load rms currents, if possible.
5. Voltage Ratio (to get Turns Ratio) at 80% of nominal supply, or less.
6. Test the dc supply at those same points, ie 80% of nominal supply and
referenced to full rms primary current. (Taking I-mag into account.)
> If not, I'll use my 100 mV, 200 A supply to prove that FWCT is better!
[snip]
> Both of these deal with tube rectifiers which have, by modern standards,
> huge resistive losses. To cope with semiconductor rectifiers, you
> practically have to rewrite the papers. What they say is valid, but it
> rarely applies to current designs.
Whatever.
Winfield Hill
John Woodgate, <j...@jmwa.demon.co.uk> said...
>
>In article <68sa3g$d...@fridge.shore.net>, Winfield Hill
><hi...@rowland.org> writes
>>John Woodgate, <j...@jmwa.demon.co.uk> said...
>>>
>>> Winfield Hill <hi...@rowland.org> writes
>>>>
>>>> Note, all these arguments assume the transformer copper has been
>>>> equally divided between the primary and secondary. Hmmm. What if
>>>> the secondary has more copper?
>>>
>>> We've already been through that in the 'private fight' over whether a
>>> bridge is _always_ overall the better solution than a FWCT. If the
>>> the maximum permissible induction (-> no. of turns) and the full load
>>> current (-> wire gauge).
>>
>> I'm surprised at the apparent rigidity of that conclusion, because the
>> copper losses are generally equal in the pri and sec, say for full-wave
>> bridge use, and ...
>>
>>> (Note that the supply voltage actually cancels out in those two terms.)
>>
>> Showing that the balance can be skewed somewhat without serious penalty.
>> On the other hand, when the same transformer is used with a full-wave
>> diode (not bridge) rectifier, the secondary copper loss is excessive, so
>> a skewing to the secondary would reduce it's heating at the expense of
>> some additional in the primary. Note, we're not talking fewer turns,
but
>> just a slightly smaller wire gauge. Naturally this can't at all remedy
>> the disadvantage incurred from not using a bridge, but surely it can
make
>> If this has been sucessfully hashed out, with a conclusion, I'm curious
>> to see the arguments. What's the relevant info, thread and key dates?
>
> The thread is 'FWCT vs. bridge ps', and the 'private fight' was between
> assumption and didn't state it at the beginning. (I'm just about to add
> a new idea to the thread, as well.) But I thought you knew that anyway,
> because you've been making comments about bridges being 'more efficient'
> than full-wave in this thread.
>
> that, to obtain equal performance (I can't remember precisely the
> criteria for 'equal', if they were specified at all) the FW whole
> secondary needs, IIRC, 157/110 = 1.43 times the winding area of the
Convinced that an unbalanced primary/secondary copper ratio was a good
idea, I approached the issue of optimizing the copper fraction a bit
differently, and came up with 41.5% more secondary than primary copper
as an optimum for a FW-CT transformer, only 45.7% higher loss resistance
than a FW-bridge of the same size, vs. 50% for an ordinary transformer
with equal copper cross-section areas. Strictly for a simplified
transformer, natch.
The bottom line is a 3% savings with an optimized transformer. Is this
what MacFayden was saying?
See the details in the Re: FWCT vs Bridge PS thread where I dredged up
an entry spot.
John Woodgate
In article <34B31B...@mail.idt.net>, altavoz
<zalt...@mail.idt.net> writes
>Can i suggest a direction . Can we discuss the actual flux
>path thru the windings to the core .
> For a guess , does the core attract the flux so forcefully that
>as soon it appears at the surface of the pri' wire , it is pulled
>DIRECTLY into the core over the shortest path . I think this is true.
> Which means the sec' had better be directly in the "way" , in order
>to "cut" flux lines . If sec was a sheet of copper , completely
>obscuring the core , all the pri flux would need to cut the sec .
>
Oh, my goodness me! Oh, my fur and whiskers! We've now got on to flux
cutting versus flux linking, which is a BIG bugaboo.
The Faraday induction equation is difficult to reproduce in ASCII:
Line integral([epsilon]dl = -d/dt{area integral(B dA)}
[epsilon] is the induced voltage in an element of conducting path dl
(and I'm not arguing over the formal problem of the infinitesimal nature
of dl).
B is the induction (used to be 'flux density') and A is the area
enclosed by the path.
This equation shows that [epsilon] is non-zero if _either_ B or A varies
with time.
For the transformer, the vast majority of the flux through the core
links both primary and secondary, whether the secondary is 'inside' or
'outside' the primary. If this were not so, a transformer would only
work one way round - i.e. with the energized winding on the outside.
Some leakage flux escapes from the exposed end faces of the primary
winding and does not travel through the core at all, while some makes a
complete loop around the primary winding only, and thus effectively does
not link with the secondary..
Winfield Hill
Tony Williams at to...@ledelec.demon.co.uk says...
Very nice set of shapes and formulas, Tony.
--
Winfield Hill hi...@rowland.org
Rowland Institute for Science
Cambridge, MA 02142
John Woodgate
In article <34b4b9b9...@news.duke.edu>, James Meyer
<jim...@acpub.duke.edu> writes
>On 6 Jan 1998 15:23:04 GMT, hi...@rowland.org (Winfield Hill) wrote:
>
>
>> BTW, I noticed a remark in Flanagan's book "Handbook of Transformer
>> Applications" (section 10.6) stating that the autoformer configuration
>> reduces leakage inductance by [(n-1)/n]^2 over that of a normal ratio-n
>> step-up transformer. That would mean 1/4 for n=2.
>
> Speaking of remarks in books... Norman Gilbert, in his 1941
>book "Electricity and Magnetism", has this little "throw-away" line...
>
> "When the secondary current leads the emf the effect of
>magnetic leakage is to increase the secondary voltage. This effect
>may be sufficient to more than balance the drop due to the resistance
>of the coils and so give a net increase in the voltage of the
>secondary with increase of load."
H'mmm. Yes, if the leakage inductance and the capacitive load approach
resonance (and the approach ned not be very close, no doubt).
>
> As soon as you guys figure out conventional transformers, you
>can start in on things like Sola constant voltage transformers where
>leakage inductance is purposely introduced. 8-)
>
> Jim
>
>
Well, I have just done some measurements on a high-leakage transformer
which is intended as a *constant-current* source. How do you get
constant voltage - by tuning the leakage inductance with an added
capacitor?
John Woodgate
In article <ant08133...@ledelec.demon.co.uk>, Tony Williams
<to...@ledelec.demon.co.uk> writes
> This thread is all about doing measurements on transformers, doing sums
> to find certain parameters, and then using the whole to see if it is
> possible to predict the performance of, say, rectifier+filter circuits.
> We are heading towards seeing if theory and practical results tie up.
>
> Consequently it would seem prudent to pick measurement points that are
> consistent with the final working points, or vice-versa.
>
> Today's commodity transformers run so near the knuckle of Bsat that is
> is dangerous to casually do, say apparent Shunt-Inductance or Turns
> Ratio at full working voltage and then use those results in downstream
> sums. It would be more sensible avoid the worst of the distortions
> caused by saturation and to back down a little, to 80/90% of rated supply.
>
> Therefore it is then perfectly consistent to suggest running the final
> power supply at 80/90% stimulus in order to verify upstream predictions.
>
> My suggestions for consistent results are:-
>
> 1. Test at the nominal frequency (may not be practical for L-ell).
Essential: Lell is non-linear and if you use a different frequency you
get the wrong answer.
>
> 2. Do R-windings at dc currents equal to the full load rms currents, or
> some other method intended to account for the ppm/C of copper.
H'mmm, yes, ideally, but it's awfully difficult to control winding
temeprature in some of the other tests.
>
> 3. Shunt-Inductance at 80% of nominal supply. (I-mag + Watts-Loss.)
> Don't trust bridge measurements, ie wimpy little voltages.
No, why at 80%? It gives the wrong value: what matters is the value
under working conditions, i.e. 100%.
>
> 4. Leakage Inductance at full-load rms currents, if possible.
Yes, it is normally possible, but not realistic, because you short-
circuit the secondary and apply reduced primary volts to get full-load
primary current. So the core induction is lower than normal, which means
that the reluctance is lower and the leakage inductance is thus (likely
to be) lower. I don't see an easy solution to this: trying to measure it
with the secondary loaded with its appropriate full-load resistance does
not give good accuracy, simply because the phase-angle, which is how you
get to the leakage inductance, is so small.
>
> 5. Voltage Ratio (to get Turns Ratio) at 80% of nominal supply, or less.
Well, this does not vary significantly with applied voltage (except at
very low voltages).
>
> 6. Test the dc supply at those same points, ie 80% of nominal supply and
> referenced to full rms primary current. (Taking I-mag into account.)
Oh, I haven't progressed beyond the transformer itself, yet. I haven't
dared to contemplate adding the rectifier and reservoir cap (;-).
John Woodgate
In article <69047v$5...@fridge.shore.net>, Winfield Hill
>> have in mind apart from the inefficient use of winding area in the
>> transformer ....
>
> No, no, I simply re-emphasized that optimizing a transfomer for FW-CT
> in no way makes up for its inferior performance in that mode.
>
>
> primary has less copper, taken entirely out of the wire size. So the
> primary will have more copper loss, but the secondary will have less,
> with a total loss below that possible with equal areas. This can be
> easily shown mathematically, although I haven't seen it in any of my
> books (OK, haven't really looked either!)
>
>>> to see the arguments. What's the relevant info, thread and key dates?
>
>> The thread is 'FWCT vs. bridge ps', and the 'private fight' was between
>
> OK, I know the one you mean... my eyes glazed over early on and I simply
> didn't absorb the meat of your discussion (weren't there lots of you said,
> I said exchanges?) so I'll try absorbing something from deja-news.
>
>> ... Tony quoted MacFayden ...
>
> Did we ever get a full reference for this book, title, ISBN number?
>
> [ snip nice synopsis of some MacFayden points... ]
>
>> Tony tells me off. When I quote MacFayden (correctly, I hope), you tell
>> me off. I think I'll go and eat worms!
>
> said and then re-ask my question.
>
>> And I'm certainly changing my .sig back to the original!
>
> Hey, Tony's as well, for that matter.
>
John Woodgate
In article <GtQHTEAx...@kemo1.demon.co.uk>, Kendall Castor-Perry
<ken...@rather.not> writes>[snip]
>>
>> >>> secondary has more copper, that *inevitably* means a bigger core,
>>
>> This is the statement that botehrs me. To begin with, it means the
>> primary has less copper, taken entirely out of the wire size. So the
>> primary will have more copper loss, but the secondary will have less,
>> with a total loss below that possible with equal areas. This can be
>> easily shown mathematically, although I haven't seen it in any of my
>> books (OK, haven't really looked either!)
>>
>windings are underused by the FWCT arrangement? As a general statement,
>'surely' it is incorrect? It looks like a classic " x + 1/x " situation
>with the turning point at x=1. Let's see if this works out.
>
>Think of a general purpose transformer wound with a single turn of
>multifilar winding wire, forming a large number N of closely linked
>turns which can be connected up as primaries or secondaries. Make up a
>1:1 transformer by using half the windings in parallel as a primary,
>other half as a secondary. Load up the transformer, each winding
>dissipates a bit due to its copper loss. The current in both windings
>is equal to I.
>
>Now transfer some windings from one winding to the other, so that there
>are n {little n} turns on one winding and N-n turns on the other. By
>inspection, with a resistance per turn of r, the total power dissipated
>is
>
>P = rI^2 * ( 1 1 )
> (____ + ____)
> ( N-n n )
>
>[boy, this ASCII equation stuff is harder than I thought!]
Try: P = rI^2*{1/(N-n) +1/n}?
>
>Differentiate P wrt n to get how dissipation varies - I won't write it
>down! - and you find it's positive for any n apart from N/2. This shows
>that the optimum configuration in this case is equal area for each
>winding, and with this analogy I reckon it extends to any usage of
>individual turns, i.e. to any ratios.
>
Yes, but does it apply if there are two secondaries, which do not carry
current simultaneously? That is the FWCT case, of course.
Jerry Codner
> Jerry Codner said...
>
> > > If anybody expresses interested, I will provide a SPICE analysis for one of the
> > > above transformers driving a full-wave rectifier with both a capacitive input
> > > and an inductive input filter comparing the negligible leakage inductance case
> > > line current waveform (conduction angle and rms value), load regulation
> > > (at 5%, 50% and 100% load) and full load output ripple.
> >
> rms rms
> Xfmr sec I watts sec V pri DCR sec DCR Lleak Lm
> 8 4.0 96 27.8 0.227 0.235 1.34 mH 66 mH
>
>
This analysis was very educational. If this is a typical transformer then
I am now calibrated to the world of 100 volt-amperes and you will have to
forgive me for past discussions in which I was looking at things from the
point-of-view of 1000 VA and above. Sorry, I am not going to analyze the
case of the inductive input filter. I think that would be pointless for this
transformer. See the results below or:
http://www.lightlink.com/gcodner/rectifier/
for current waveforms but before doing so, why not predict the average dc
output voltage yourselves? I think this transformer will challenge a lot
of people's assumptions, even if they usually live below 100 VA.
Circuit:
__________
Li Rp+Rs | FW |
Vin-----()()()--/\/\/\----| bridge |---|-------
(27.8 V RMS, |________| \ |
39.3 V pk) | / RL --- CL
gnd \ ---
| |
gnd gnd
The full load condition for the rectifier circuit was determined
using the secondary current rating of 4 amperes RMS. A 10,000 uF
filter capacitance (CL) was used. With this capacitor, a load
resistor of 13.9 ohms is full load.
Results Summary
(see http://www.lightlink.com/gcodner/rectifier/ for waveforms, a SPICE listing
and a much better looking schematic)
This transformer is lossy! At a secondary current of 4 amperes RMS and total
winding resistance of almost 0.5 ohm, the copper losses are 4*4*0.5 = 8 watts
with dc output power of approximately 75 watts. That's less than 90% efficient.
It's no wonder that the leakage inductance could be ignored for this heater.
The time constant of the winding resistances and the filter capacitance
is 5 ms!
For the low leakage inductance simulation at full load and for equal dc power
out, both RMS line current and dc output voltage were approximately 10% higher
(11.7% and 8% respectively) and voltage ripple was 14.7% higher than for the
simulation using the actual transformer leakage inductance.
The dc load regulation is very poor with nearly 20% difference in dc output
from 5% to 100% load and about 30% from no load to full load (the no load
voltage is 39.3 volts).
Run # RL Rp + Rs Li DC Vout RMS line I pk line I p-p ripple
[ohms] [ohms] [uH] [volts] [amperes] [amperes] [volts]
5% Load
1a 278 0.462 1340 36.18 0.31 0.91 0.087
1b 236 0.462 1340 36.07 0.34 1.00 0.098
1c 278 0.462 1.34 36.90 0.41 1.57 0.097
Full Load
2a 13.9 0.462 1340 30.70 3.54 7.33 1.04
2b 11.8 0.462 1340 30.12 4.01 8.15 1.17
2c 13.9 0.462 1.34 32.50 4.48 10.76 1.34
Conclusions
Although the line current waveforms are significantly different, leakage
inductance has only a 10% effect on RMS line/rectifier current for this
transformer when delivering equal dc output power. However, it has rather
high winding resistances that, on a per-unit basis, would be unacceptable
at higher power levels. It's load regulation is also very poor.
--
Jerry Codner
Unsolicited email sent to this address will
result in an administrative fee of $50.
On-topic replies are excluded, of course.
--
Jerry Codner
Unsolicited email sent to this address will
result in an administrative fee of $50.
On-topic replies are excluded, of course.
Jerry Codner
I don't know who wrote (but let's not worry about it) ...
> > > (That might be a useful expt to do, see what relative dc-powers can
> > > be obtained for the same nominal 'full-power' primary rms-current.
> > > Probably best done at reduced V-pri to avoid the expt being affected
> > > by saturation effects.)
and someone else wrote (something I partially agree with) ...
> >
> > No, one should do the experiment under realistic practical conditions.
>
But Tony Williams definitely wrote ...
> This thread is all about doing measurements on transformers, doing sums
> to find certain parameters, and then using the whole to see if it is
> possible to predict the performance of, say, rectifier+filter circuits.
> We are heading towards seeing if theory and practical results tie up.
>
> Consequently it would seem prudent to pick measurement points that are
> consistent with the final working points, or vice-versa.
>
> Today's commodity transformers run so near the knuckle of Bsat that is
> is dangerous to casually do, say apparent Shunt-Inductance or Turns
> Ratio at full working voltage and then use those results in downstream
> sums. It would be more sensible avoid the worst of the distortions
> caused by saturation and to back down a little, to 80/90% of rated supply.
>
> Therefore it is then perfectly consistent to suggest running the final
> power supply at 80/90% stimulus in order to verify upstream predictions.
>
> My suggestions for consistent results are:-
>
> 1. Test at the nominal frequency (may not be practical for L-ell).
>
> 2. Do R-windings at dc currents equal to the full load rms currents, or
> some other method intended to account for the ppm/C of copper.
>
> 3. Shunt-Inductance at 80% of nominal supply. (I-mag + Watts-Loss.)
> Don't trust bridge measurements, ie wimpy little voltages.
>
> 4. Leakage Inductance at full-load rms currents, if possible.
>
> 5. Voltage Ratio (to get Turns Ratio) at 80% of nominal supply, or less.
>
> 6. Test the dc supply at those same points, ie 80% of nominal supply and
> referenced to full rms primary current. (Taking I-mag into account.)
>
>
> Tony Williams, Ledbury, UK.
Tony, I agree with most of it, but why measure leakage inductance at full
load currents? I can't see why. As for measuring it at the nominal
frequency, I disagree. True, one must ensure that the inductance does
not appear abnormally high because of resonance with the winding capacitance,
especially when measuring "blind" using a bridge. A low frequency network
analyzer is really nice for this (ah, those were the days in the corporate
world when we had a Venable frequency response analysis system).
I agree with the other points and strongly with the winding resistance
measurement. Thermal runaway could be a real problem, especially with
a constant power load, such as a switching regulator. And the temperature
rise has as much to do with the physical design as the electrical load.
Jerry Codner
Jerry Codner tried to draw in ASCII:
>
> Circuit:
> __________
> Li Rp+Rs | FW |
> Vin-----()()()--/\/\/\----| bridge |---|-------
> (27.8 V RMS, |________| \ |
> 39.3 V pk) | / RL --- CL
> gnd \ ---
> | |
I hope it was simple enough to figure out and if it still doesn't work,
I give up.
Look here: http://www.lightlink.com/gcodner/rectifier/
altavoz
> altavoz
>Can i suggest a direction . Can we discuss the actual flux
>path thru the windings to the core .
> For a guess , does the core attract the flux so forcefully that
>as soon it appears at the surface of the pri' wire , it is pulled
>DIRECTLY into the core over the shortest path . I think this is true.
> Which means the sec' had better be directly in the "way" , in order
>to "cut" flux lines . If sec was a sheet of copper , completely
>obscuring the core , all the pri flux would need to cut the sec .
>Nix Z for email
John Woodgate,
Oh, my goodness me! Oh, my fur and whiskers! We've now got on to flux
cutting versus flux linking, which is a BIG bugaboo.
The Faraday induction equation is difficult to reproduce in ASCII:
Line integral([epsilon]dl = -d/dt{area integral(B dA)}
altavoz
STOP , ALTO ! We will be talking concepts of direction and force
of flux attraction to the core , BUT W/O THE NEED TO EXACTLY QUANTISIZE
that flux . Store your math til we get to a place where we can not
proceed
w/o it .
----------------------------------------------------------
John
For the transformer, the vast majority of the flux through the core
links both primary and secondary, whether the secondary is 'inside' or
'outside' the primary. If this were not so, a transformer would only
work one way round - i.e. with the energized winding on the outside.
Some leakage flux escapes from the exposed end faces of the primary
winding and does not travel through the core at all, while some makes a
complete loop around the primary winding only, and thus effectively does
not link with the secondary..
altavoz
"flux through the core links both primary and secondary,"
to reduce leak induct' w/o increasing other factors such that the theory
works but the transformer does not .
Given a single turn sec wind , located close to one of the 4 winding
window
faces ( say 1mm away), and the primary single turn placed close to same
face
but 2 mm away .
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| s p |
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I think the core attracts the flux so strongly that the sec will be
tightly
coupled because of proximity of pri turn to nearest face . But poor
coupling
if :
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| p s |
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Nix Z for email
altavoz
John
Yes, it is normally possible, but not realistic, because you short-
circuit the secondary and apply reduced primary volts to get full-load
primary current. So the core induction is lower than normal, which means
that the reluctance is lower and the leakage inductance is thus (likely
to be) lower .
altavoz
It won't be lower . This is the std text to measure leak induct .
It reduces eddy currents ( core loss currents) and the magnetizing
currents to better measure leak induct.
Tony Williams
In article <34B55658...@lightlink.com>, Jerry Codner
<URL:mailto:gco...@lightlink.com> wrote:
> Tony, I agree with most of it, but why measure leakage inductance at full
> load currents? I can't see why. As for measuring it at the nominal
> frequency, I disagree. True, one must ensure that the inductance does
> not appear abnormally high because of resonance with the winding capacitance,
> especially when measuring "blind" using a bridge. A low frequency network
> analyzer is really nice for this (ah, those were the days in the corporate
> world when we had a Venable frequency response analysis system).
I was really, really, hoping that no one would spot that, and query why
L-ll should be measured at full load currents.
The confession here is that I have never been able to dispense with a
mental imagining that L-ell is affected by secondary load-current. The
books just describe it as a fixed parameter, but this nagging feeling
remains and causes me to play-safe.
Authorities also say that since most of L-ell is in air (with only a
small part going through the core) then L-ell is relatively unaffected
by the permeability of the ferromagnetic material. This permits the
measurement of L-ll by bridges, at up to 1000Hz. That also does not
sit well.
> I agree with the other points and strongly with the winding resistance
> measurement. Thermal runaway could be a real problem, especially with
> a constant power load, such as a switching regulator. And the temperature
> rise has as much to do with the physical design as the electrical load.
If the winding-resistance is just measured with an ohmeter, something
has to be done to factor-in the increase in winding-resistance at full
load I. Copper is roughly +5000 ppm/C, so that's about +15% increase in
winding-R for a 30C temperature rise.
If you are trying to measure L-ell with ammeter/voltmeter the errors in
winding-R swamp the calcs completely.
Kendall Castor-Perry
In article <+qpz0DBN...@jmwa.demon.co.uk>, John Woodgate
<j...@jmwa.demon.co.uk> writes[snip]
>>>
>>Is this comment specifically relevant to the case where the secondary
>>windings are underused by the FWCT arrangement? As a general statement,
>>'surely' it is incorrect? It looks like a classic " x + 1/x " situation
>>with the turning point at x=1. Let's see if this works out.
>>
elsewhere in the thread]
>>
>Yes, but does it apply if there are two secondaries, which do not carry
>current simultaneously? That is the FWCT case, of course.
No it doesn't, as I thought I pointed out at the top of the post. If
there is a reduced utilisation in the sense of only half-wave in each
half, that will have to be allowed for.
This was only a highly simplified analysis for an unrealistic
transformer and I think Chuck's post (which I haven't had time to read
fully) captures much more of the next level of 'real world'.
I'm not sure why I'm contributing to the noise level on this thread
anyway since most of my work is done with split supplies, and given the
various options available IMO you'd need your head examined to plump for
a rectifier configuration which only used the secondary windings for
half the time.
Glen Walpert
In article <8KEzs0A7...@jmwa.demon.co.uk>, John Woodgate <j...@jmwa.demon.co.uk> wrote:
>In article <34b4b9b9...@news.duke.edu>, James Meyer
>> As soon as you guys figure out conventional transformers, you
>>can start in on things like Sola constant voltage transformers where
>>leakage inductance is purposely introduced. 8-)
>>
>> Jim
>>
>Well, I have just done some measurements on a high-leakage transformer
>which is intended as a *constant-current* source. How do you get
>constant voltage - by tuning the leakage inductance with an added
>capacitor?
AFIK, the Sola constant voltage transformers use core saturation for voltage
regulation, not leakage inductance; they are a special case of magnetic
amplifier. Unfortunately, my only transformer design book has no information
on the Sola design.
(from Amazon.com:)
Magnetic Components : Design and Applications
by Steve Smith
Availability: This title is out of print, but if you place an order we may be
able to find you a used copy within 1-3 months.
Published by Van Nostrand Reinhold
Publication date: October 1984
ISBN: 0442203977
This book does have a good discusion of magnetic amplifier design (as well as
extensive information about optimizing transformer design for efficiency). I
can't compare to other books on the subject since this is the only one I have
read, but will say that I think it provides enough information and design
examples that it could be used to design a viariey of real transformers for
special applications where a standard is not optimal.
Glen Walpert
Winfield Hill
altavoz at zalt...@mail.idt.net says...
Yes, altavoz is right, the shorted winding is the textbook method,
but John are you saying that the paralleled (and much higher - say
100x so therefore usually ignored) magnetizing inductance will be
a bit lower and thus distorting the leakage-inductance measurement?
By what, 1%. Hmmmmm, can you even get a 1% L_ell measurement under
actual operating conditions?