PSU's: Exceeding transformer rated current
Richard
In a half-wave rectifier circuit, a purely resistive load, no filtering, could you
double the rated secondary current? The transformer will not get any hotter than
being loaded at rated current output, but would the secondary windings be prone to
fail for overcurrent? TIA. Rich.
Richard
"Richard" <nearlyne...@ntlworld.com> wrote in message
news:avegas$ek3hp$1...@ID-150848.news.dfncis.de...
Actually, I'm not so sure that you double the current to have transfomer supply it's
maximum power. Whatever the current is I'm seeking to figure it out so that the
transformer supplies it's rated maximum. For a transformer secondary rating of 10v
(rms) and 1A (rms) the maximum load would be 10 watts, or I think that should be
10VA.
ro...@mauve.demon.co.uk
> Don't know whether this forum is appropriate, but here goes.
>
> In a half-wave rectifier circuit, a purely resistive load, no filtering,
> could you double the rated secondary current? The transformer will not
In general, no, though not all transformers will die when you do this.
> get any hotter than being loaded at rated current output,
> but would the secondary windings be prone to fail for overcurrent?
Neglecting the core for a moment, you'll double the copper losses in both
the primary and secondary windings.
However, in practice you won't ever have this problem, because the core
will bite you first.
Any DC current in either the secondary or primary windings will drive
the core towards saturation, reducing the maximum primary AC voltage
that can be sustained without saturation causing massive heating and
rapid failure. (minutes/seconds)
I don't off-hand know the derating factor if you were to half-wave
rectify the output of the transformer, I'd guess for most transformers,
10-20% of the AC rating would cause the same heating.
--
http://inquisitor.i.am/ | mailto:inqui...@i.am | Ian Stirling.
---------------------------+-------------------------+--------------------------
Windows 2000, software for next millenia. <latin pun alert> - Ian Stirling.
Richard
<ro...@mauve.demon.co.uk> wrote in message
news:avehv7$3qr$1$8302...@news.demon.co.uk...
> Richard <nearlyne...@ntlworld.com> wrote:
> > Don't know whether this forum is appropriate, but here goes.
> >
> > In a half-wave rectifier circuit, a purely resistive load, no filtering,
> > could you double the rated secondary current? The transformer will not
>
> In general, no, though not all transformers will die when you do this.
>
> > get any hotter than being loaded at rated current output,
> > but would the secondary windings be prone to fail for overcurrent?
>
> Neglecting the core for a moment, you'll double the copper losses in both
> the primary and secondary windings.
>
> However, in practice you won't ever have this problem, because the core
> will bite you first.
>
> Any DC current in either the secondary or primary windings will drive
> the core towards saturation, reducing the maximum primary AC voltage
> that can be sustained without saturation causing massive heating and
> rapid failure. (minutes/seconds)
>
> I don't off-hand know the derating factor if you were to half-wave
> rectify the output of the transformer, I'd guess for most transformers,
> 10-20% of the AC rating would cause the same heating.
Thanks. What looks fairly simple at first turns out to be a tricky problem. I was
going to work out what the secondary current would be, maintaining the rated
secondary voltage (and I'm sure current would not be x2 rated) and then multiplying
that by the voltage to get the maximum load supplied, but I was tackling that without
taking into consideration the things you bring to light, like core saturation, copper
losses etc.
Current would be DC, but varying, almost like ac as far as the transformer is
concerned I think.
Richard
No, I think you do double current.
Say secondary is 10.5v open circuit, 0.5R winding resistance. When 1A rms (ac) flows
the voltage drop in winding is 0.5v, thus we have 10v to supply. The load will be
10R and 10 watts/VA is dissipated.
If now the current flows for only half the cycle we must half our values. Watts will
be 5W. Wattage is a function of time the current flows; given the supply is 10v rms,
a peak current of 1.414A produces the heating effect of a dc current of 1A. That is
true for every half a cycle. So, if we half the number of cycles the average power
dissipated is half a full rms current.
Now I think this average of a half applies also to the voltage drop across the
winding resistance. A 2A rms ac current flowing thru 0.5R produces a voltage drop of
1v rms. So output of winding is 9.5v rms. If you half the cycles then you have to
average the voltage drop, and the average is a half. So, on average the voltage drop
is 0.5v rms. Because really, 1A rms *ac* is equal 2A rms "dc* in these terms.
As to copper losses, these stay the same.
So, on the face of things, to have the transformer supply it's maximum rated power in
a half wave rectifier circuit as mentioned above you can double the current.
So, I think the only reason you could not do so is something to do with saturating
the core. I don't think the windings would melt for eccess current. Copper losses
stay the same on the face of things.
Joop
>However, in practice you won't ever have this problem, because the core
>will bite you first.
>
>Any DC current in either the secondary or primary windings will drive
>the core towards saturation, reducing the maximum primary AC voltage
>that can be sustained without saturation causing massive heating and
>rapid failure. (minutes/seconds)
>
>I don't off-hand know the derating factor if you were to half-wave
>rectify the output of the transformer, I'd guess for most transformers,
>10-20% of the AC rating would cause the same heating.
Correct. I did this once with only a 100mA half-wave rectified load
and the 1A rated transformer became very hot. When also adding the
rest of the planned circuit that used the full-wave rectified 800mA or
so, the transformer actually cooled down again...
In the end I believe I decided to draw 100mA current only when needed
for short periods (temp use of 24V/100mA, where the 8V/800mA was used
to always power the main part of the circuit)
Joop
Phil Allison
"Richard" <nearlyne...@ntlworld.com> wrote in message
news:avegas$ek3hp$1...@ID-150848.news.dfncis.de...
> Don't know whether this forum is appropriate, but here goes.
** It is bad practice to half wave load any mains transformer - and
never do it to a toroidal type. The core will be pushed into saturation
very easily and the unit will protest by getting hot under the collar and
growling at you !!
............... Phil
John Woodgate
.com> wrote (in <avegas$ek3hp$1...@ID-150848.news.dfncis.de>) about 'PSU's:
Exceeding transformer rated current', on Tue, 7 Jan 2003:
You have a direct current in the secondary winding, which is the average
load current. This greatly increases the losses in the core, resulting
in the transformer heating up much more that you would expect from the
fact that the secondary current appears to be flowing for only half the
time.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
Richard
>
> "Richard" <nearlyne...@ntlworld.com> wrote in message
> news:avehi2$eo0st$1...@ID-150848.news.dfncis.de...
> No, I think you do double current.
>
> Say secondary is 10.5v open circuit, 0.5R winding resistance. When 1A rms (ac)
flows
> the voltage drop in winding is 0.5v, thus we have 10v to supply. The load will be
> 10R and 10 watts/VA is dissipated.
>
> If now the current flows for only half the cycle we must half our values. Watts
will
> be 5W. Wattage is a function of time the current flows; given the supply is 10v
rms,
> a peak current of 1.414A produces the heating effect of a dc current of 1A. That
is
> true for every half a cycle. So, if we half the number of cycles the average power
> dissipated is half a full rms current.
>
> Now I think this average of a half applies also to the voltage drop across the
> winding resistance. A 2A rms ac current flowing thru 0.5R produces a voltage drop
of
> 1v rms. So output of winding is 9.5v rms. If you half the cycles then you have to
> average the voltage drop, and the average is a half. So, on average the voltage
drop
> is 0.5v rms. Because really, 1A rms *ac* is equal 2A rms "dc* in these terms.
Actually the matter is more complicated, a doubling the current is not the only value
that will produce 10VA in a load.
If the diode is perfect, if you only see the primary as a perfect source with
resistance in series, if the load is pure resistance, then it seems the current
flowing in the half wave circuit described to get 10VA dissipated in a load, is
dependent upon the internal secondary resistance.
For an internal resistance of 0.5R with an EMF of 5.25v rms, I've worked out you need
a 1.6R load carrying 2.5A rms
Working:
Since voltage is only available for half the cycle, the open circuit voltage is 10.5v
rms /2 = 5.25v rms.
R int = 0.5R
R load = 1.6R
R total = 2.1R
I rms = 2.5A
V drop R int = 1.25v
V load = 4v
Watts = 10VA
My goal was to see what current you would need to draw to get a dissipation of 10VA.
I see that it all depends on internal resistance of the primary. Quite clearly if a
transformer had no inductance, core saturation etc, and it was rated at 10v 1A you
could never get above 5.25v rms on the load given the circuit as described. Depending
on internal resistance you can get more that 10VA dissipated.
But, things are far removed from reality, because we have inductance, core saturation
etc. Next step to add inductance parameter at least and some realistic rectifier
resistance and see what happens.
Richard
"John Woodgate" <j...@jmwa.demon.contraspam.yuk> wrote in message
news:nwAc6rBN...@jmwa.demon.co.uk...
> You have a direct current in the secondary winding, which is the average
> load current. This greatly increases the losses in the core, resulting
> in the transformer heating up much more that you would expect from the
> fact that the secondary current appears to be flowing for only half the
> time.
I don't doubt it, but I cannot see why. Rich.
Richard
Actually I think I see a part of the problem.
In my example there are two situations. The internal resistance of the primary is
0.5R.
Where the transformer supplies an ac current, ie say without any rectification, the
transformer rated at 10v rms @ 1A supplying it's max rating has 0.5 watts dissipated
in the internal resistance (I^2 x R; 1A x 0.5R) . Connect as half wave rectifier and
the available voltage for R load is 4v when dissipating 10 watts in R load and the
current is 2.4A. Dissipation in R internal is 2.4A ^2 x 0.5R = 2.88 watts. So, we
are supplying 10watts in both instances to a load, but greater copper losses where we
have used half-wave rectification. And that's assuming the 2.4A rms current, which
would be 6.7892504A peak would not melt the wire.
Rich.
Richard
>
> Actually I think I see a part of the problem.
>
> In my example there are two situations. The internal resistance of the primary is
> 0.5R.
>
> Where the transformer supplies an ac current, ie say without any rectification, the
> transformer rated at 10v rms @ 1A supplying it's max rating has 0.5 watts
dissipated
> in the internal resistance (I^2 x R; 1A x 0.5R) . Connect as half wave rectifier
and
> the available voltage for R load is 4v when dissipating 10 watts in R load and the
> current is 2.4A. Dissipation in R internal is 2.4A ^2 x 0.5R = 2.88 watts. So, we
> are supplying 10watts in both instances to a load, but greater copper losses where
we
> have used half-wave rectification. And that's assuming the 2.4A rms current, which
> would be 6.7892504A peak would not melt the wire.
>
> Rich.
Oh drat the current is supposed to be 2.5A rms in the half-wave circuit, not 2.4A
rms. That makes copper losses 3.125 watts as against 0.5 watts.
Richard
"John Woodgate" <j...@jmwa.demon.contraspam.yuk> wrote in message
news:nwAc6rBN...@jmwa.demon.co.uk...
> load current. This greatly increases the losses in the core, resulting
> in the transformer heating up much more that you would expect from the
> fact that the secondary current appears to be flowing for only half the
> time.
You are saying that a transformer rated at 1A rms current when supplying power during
both halves of the cycle cannot supply 1A rms if supplying power for only half the
cycle?
In both cases I^2 losses (I think) would at least be the same. But the peak
currents would be different, In first case, the peak current is 1.414A, I'm not
really sure what the peak current would have to be for 1A rms in the second case.
Mind gone blank. Bedtime for me. Rich.
John Popelish
The limitation is not caused by the heating of the copper windings
during the half cycle that current flows, but during the other half
cycle. With a half wave output, the primary winding has IR drop only
during one half cycle, so there is a bit more voltage applied during
the unused half cycle. The core integrates the total volts over time,
so this total, instead of adding up to zero, accumulates flux in the
direction that the stronger half cycles drive it. In a few cycles,
the core saturates before the end of the unused half cycle and the
winding inductance essentially disappears, leaving the primary winding
as a low value resistor across the line for the remainder of that
half cycle and for the same fraction of all those future half cycles.
This big blast of primary current once a cycle is what accounts for
the rapid heating of a transformer loaded with a half wave rectified
load. In effect, the transformer is a magnetic amp that switches from
an inductor to a short circuit every cycle, with a switching delay
based on the DC passing through the secondary.
--
John Popelish
John Woodgate
Exceeding transformer rated current', on Tue, 7 Jan 2003:
>the
>transformer rated at 10v rms @ 1A supplying it's max rating has 0.5 watts
>dissipated
>in the internal resistance (I^2 x R; 1A x 0.5R) . Connect as half wave
>rectifier and
>the available voltage for R load is 4v when dissipating 10 watts in R load and
>the
>current is 2.4A. Dissipation in R internal is 2.4A ^2 x 0.5R = 2.88 watts. So,
>we
>are supplying 10watts in both instances to a load, but greater copper losses
>where we
>have used half-wave rectification. And that's assuming the 2.4A rms current,
>which
>would be 6.7892504A peak would not melt the wire.
Ah, well, in that case, your original question meant 'can I get 20 W
*into the load* from a 10 VA transformer, using a half-wave rectifier?'.
The answer is that you can't even get 10 W, *because the rectifier
circuit is nowhere near 100% efficient*.
The question of whether you could get 20 W *out of the transformer* is
more reasonable, but the answer is still 'no', and again you can't even
get 10 W out.
John Woodgate
Exceeding transformer rated current', on Tue, 7 Jan 2003:
>I don't doubt it, but I cannot see why. Rich.
Why what? If you mean 'why does the core loss go up?', the answer is
difficult to quantify, because it depends on the non-linear properties
of the iron used in the core of your particular transformer, and, in
particular, non-linear properties that aren't given on the data sheet -
the hysteresis loss due to magnetization by the half-wave rectified
current. This opens up a distorted incremental hysteresis loop (B-H
graph), whose area represents the power loss. All very empirical,
because there is no mathematical simulation of the behaviour of real
iron.
John Woodgate
wrote (in <3E1CD631...@rica.net>) about 'PSU's: Exceeding
transformer rated current', on Wed, 8 Jan 2003:
>The limitation is not caused by the heating of the copper windings
>during the half cycle that current flows, but during the other half
>cycle. With a half wave output, the primary winding has IR drop only
>during one half cycle, so there is a bit more voltage applied during
>the unused half cycle. The core integrates the total volts over time,
>so this total, instead of adding up to zero, accumulates flux in the
>direction that the stronger half cycles drive it. In a few cycles,
>the core saturates before the end of the unused half cycle and the
>winding inductance essentially disappears, leaving the primary winding
>as a low value resistor across the line for the remainder of that
>half cycle and for the same fraction of all those future half cycles.
>This big blast of primary current once a cycle is what accounts for
>the rapid heating of a transformer loaded with a half wave rectified
>load. In effect, the transformer is a magnetic amp that switches from
>an inductor to a short circuit every cycle, with a switching delay
>based on the DC passing through the secondary.
Well, it's an interesting idea that the core 'walks' towards saturation
by that means, but it is also necessary to take into account that the
*secondary* current includes a far more substantial d.c. component,
equal to the d.c. load current of course. I think that swamps any core-
walking effect.
John Popelish
>
> John Popelish wrote:
>
> >The limitation is not caused by the heating of the copper windings
> >during the half cycle that current flows, but during the other half
> >cycle. With a half wave output, the primary winding has IR drop only
> >during one half cycle, so there is a bit more voltage applied during
> >the unused half cycle. The core integrates the total volts over time,
> >so this total, instead of adding up to zero, accumulates flux in the
> >direction that the stronger half cycles drive it. In a few cycles,
> >the core saturates before the end of the unused half cycle and the
> >winding inductance essentially disappears, leaving the primary winding
> >as a low value resistor across the line for the remainder of that
> >half cycle and for the same fraction of all those future half cycles.
> >This big blast of primary current once a cycle is what accounts for
> >the rapid heating of a transformer loaded with a half wave rectified
> >load. In effect, the transformer is a magnetic amp that switches from
> >an inductor to a short circuit every cycle, with a switching delay
> >based on the DC passing through the secondary.
>
> Well, it's an interesting idea that the core 'walks' towards saturation
> by that means, but it is also necessary to take into account that the
> *secondary* current includes a far more substantial d.c. component,
> equal to the d.c. load current of course. I think that swamps any core-
> walking effect.
That secondary DC has no higher magnitude than than the secondary
would normally have during that half cycle if full wave operation were
taking place, so the peak value of that current does not account for
saturated magnetics. The core flux is not instantaneously related to
the secondary current, since secondary current is not supplied from an
external source to the secondary, but is driven by current in the
primary. The flux is just the integral of coil voltage, and half wave
operation with a non zero impedance source and non zero impedance
coils produces an imbalance between half cycle integrations. If the
voltage waveform is suitably rebalanced to compensate for the DC load,
you get forward converter magnetics with no saturation and half wave
operation. Same DC secondary current. The key is voltage. What am I
missing?
--
John Popelish
Phil Allison
"John Woodgate" <j...@jmwa.demon.contraspam.yuk> wrote in message
>
> Well, it's an interesting idea that the core 'walks' towards saturation
> by that means, but it is also necessary to take into account that the
> *secondary* current includes a far more substantial d.c. component,
> equal to the d.c. load current of course. I think that swamps any core-
> walking effect.
** I took a 15 VA E-Core and added a diode and 22 ohms resistor in
series across the 15 volt secondary. Running it at 200 volts AC (to
minimise the magnetising current) I monitored the primary waveform while
switching the half wave load on and off.
The current in the primary built up to its final value in just under
1 second, it took a very similar time to return to the oroginal value when
the load was removed.
I guess that means it took about 50 "steps" to walk the core into and
out of saturation.
.................... Phil
John Popelish
> ** I took a 15 VA E-Core and added a diode and 22 ohms resistor in
> series across the 15 volt secondary. Running it at 200 volts AC (to
> minimise the magnetising current) I monitored the primary waveform while
> switching the half wave load on and off.
>
> The current in the primary built up to its final value in just under
> 1 second, it took a very similar time to return to the oroginal value when
> the load was removed.
>
> I guess that means it took about 50 "steps" to walk the core into and
> out of saturation.
Phil, if your experimental setup is still handy, could you try a
modification for me?
I am interested in the result when you add a half cycle voltage
corrector to the primary, consisting of a series diode that carries
the 'on' half cycle current, paralleled with two or more diodes in
inverse parallel that carry the 'off' half cycle current, but lower
the voltage a bit extra, to compensate for the absence of IR drop in
the primary for that half cycle. If the secondary current is not
involved in the saturation, except for the primary IR drop effect on
integrated voltage, some small number of diodes should prevent
saturation when the secondary load is connected, but produce the same
effect when it is open circuited.
--
John Popelish
Phil Allison
"John Popelish" <jpop...@rica.net> wrote in message
news:3E1E6684...@rica.net...
** I placed a 6.8 volt, 3 watt zener in series with the primary. This
creates a DC component in the supply and did appear to compensate for the
effect of the half wave secondary load.
I tried the same set up with a 160 VA 36 volt toroidal and got
text book results in terms of waveforms.
................. Phil
John Woodgate
transformer rated current', on Thu, 9 Jan 2003:
>What am I
>missing?
I don't follow all of your argument, but I think my argument applies
only if there is a filter capacitor.
John Popelish
With reference to the toroidal transformer: Was the text book result
with or without the zener? An for those of us without the test book,
please describe the waveforms. Thanks.
--
John Popelish
Richard
"John Woodgate" <j...@jmwa.demon.contraspam.yuk> wrote in message
> Ah, well, in that case, your original question meant 'can I get 20 W
> *into the load* from a 10 VA transformer, using a half-wave rectifier?'.
> The answer is that you can't even get 10 W, *because the rectifier
> circuit is nowhere near 100% efficient*.
>
> The question of whether you could get 20 W *out of the transformer* is
> more reasonable, but the answer is still 'no', and again you can't even
> get 10 W out.
"Don't know whether this forum is appropriate, but here goes.
In a half-wave rectifier circuit, a purely resistive load, no filtering, could you
double the rated secondary current? The transformer will not get any hotter than
being loaded at rated current output, but would the secondary windings be prone to
fail for overcurrent? TIA. Rich."
I probably did not pose my question very well. What the query is about is current
flowing into the primary for half the cycle rather than the full cycle. Before I had
thought it through I was thinking about the possibility, in a case of half-wave
rectification, of drawing double the current from the secondary, assuming the
secondary voltage being the same, with a view of supplying the rated transformer
wattage (as per FW rectification) to a load. Immediate thought was why not, since if
current is only flowing in the secondary for half the time, drawing twice the current
should not overload the tfmr, one rthing balances another. I think it turns out in
my not realistic example of perfect diode etc, that you could *as a simple matter of
ohms law* supply the same wattage to a load be there half-wave rectification or
full-wave rectification. However, there are reasons why this cannot be for a real
transformer, because in a real transformer you cannot just rely on ohms law because
you have core saturation issues etc. One thing also is that you cannot possibly
maintain, in the example I gave, the FW rated voltage. The rms voltage to the load
has to reduce with HW rectication.
Phil Allison
> Phil Allison wrote:
>
> > ** I placed a 6.8 volt, 3 watt zener in series with the primary.
This
> > creates a DC component in the supply and did appear to compensate for
the
> > effect of the half wave secondary load.
> >
> > I tried the same set up with a 160 VA 36 volt toroidal and got
> > text book results in terms of waveforms.
>
> With reference to the toroidal transformer: Was the text book result
> with or without the zener?
** With of course.
An for those of us without the test book,
> please describe the waveforms. Thanks.
** The toroid's primary current was a perfect half sine - there being no
magetising current to confuse the matter. I use a LEM LTA 50P hall effect
transducer to monitor and measure such things.
................ Phil
John Popelish
> "Don't know whether this forum is appropriate, but here goes.
>
> In a half-wave rectifier circuit, a purely resistive load, no filtering, could you
> double the rated secondary current? The transformer will not get any hotter than
> being loaded at rated current output, but would the secondary windings be prone to
> fail for overcurrent? TIA. Rich."
Double the current produces 4 times the resistive heating. If that
current occurs only half the time (half wave operation) then the
effective heat would be half of that 4 times. If the half wave
current was 1.4 times the rated current, this would double the IR
heating, but the 50% duty cycle would reduce the average heat back to
rated. All this assumes that you somehow compensate for the DC
component (caused by IR drop reducing only one half of the applied
voltage) added to the primary voltage that will soon saturate the
core.
> I probably did not pose my question very well. What the query is about is current
> flowing into the primary for half the cycle rather than the full cycle. Before I had
> thought it through I was thinking about the possibility, in a case of half-wave
> rectification, of drawing double the current from the secondary, assuming the
> secondary voltage being the same, with a view of supplying the rated transformer
> wattage (as per FW rectification) to a load. Immediate thought was why not, since if
> current is only flowing in the secondary for half the time, drawing twice the current
> should not overload the tfmr, one rthing balances another. I think it turns out in
> my not realistic example of perfect diode etc, that you could *as a simple matter of
> ohms law* supply the same wattage to a load be there half-wave rectification or
> full-wave rectification. However, there are reasons why this cannot be for a real
> transformer, because in a real transformer you cannot just rely on ohms law because
> you have core saturation issues etc. One thing also is that you cannot possibly
> maintain, in the example I gave, the FW rated voltage. The rms voltage to the load
> has to reduce with HW rectication.
--
John Popelish
Richard
> Richard wrote:
>
> > "Don't know whether this forum is appropriate, but here goes.
> >
> > In a half-wave rectifier circuit, a purely resistive load, no filtering, could
you
> > double the rated secondary current? The transformer will not get any hotter than
> > being loaded at rated current output, but would the secondary windings be prone
to
> > fail for overcurrent? TIA. Rich."
>
> Double the current produces 4 times the resistive heating. If that
> current occurs only half the time (half wave operation) then the
> effective heat would be half of that 4 times. If the half wave
> current was 1.4 times the rated current, this would double the IR
> heating, but the 50% duty cycle would reduce the average heat back to
> rated. All this assumes that you somehow compensate for the DC
> component (caused by IR drop reducing only one half of the applied
> voltage) added to the primary voltage that will soon saturate the
> core.
As you know, doubling the current produces 4 times the resistive heating, if the
resistance stays the same. Because then the voltage has to double to produce the
double in current.
With HW operation, the effective (rms) voltage reduces by half, 10v rms becomes 5 v
rms to the load. So, if rms current were to be the same, changing from FW to HW, we
would need to half the load resistance. Power dissipated then would be half the FW
heating position. So, you then have to half the load resistance again to bring it up
to the original resistive heating that we had with FW.
All this as you say is ignoring the complications introduced from parameters
associated with the supplying source, such as source internal resistance, core
saturation etc.