Center Tapped and Regular Transformer

[Anand P. Paralkar]

Hi,

Could you please explain the following points regarding a center tapped
transformer (some questions apply to a non-center tapped "regular"
transformer as well):

a. Is the center tapped transformer wound differently than a
non-center tapped transformer? Or is it just a regular transformer for
which the center point of the secondary winding is "brought outside".

b. Considering the secondary voltage of a transformer is Vs, the two
terminals of the secondary are at +Vs/2 and -Vs/2. This implies a
voltage gradient across the secondary. The gradient passes through a
zero point which we "tap". What causes this voltage gradient?

c. Can we say that all the turns in the secondary winding of a
transformer have the same amount of flux passing through them at a given
instant or do they have a different amount of flux (with the flux
depending on the position of the turn)?

Thanks,
Anand

[Daniel Pitts]

I don't know the answer to these, and I'm interested to know, but they
are sound very much like homework questions and I wouldn't be surprised
if the answers you get are along the lines of "do your own homework".

Now, I would guess that Wikipedia and Google are going to help you in
your quest for information. Good luck.

[John Larkin]

On Mon, 27 Aug 2012 23:50:24 +0530, "Anand P. Paralkar"
<anand.p...@gnospammale.com> wrote:

>Hi,
>
>Could you please explain the following points regarding a center tapped
>transformer (some questions apply to a non-center tapped "regular"
>transformer as well):
>
> a. Is the center tapped transformer wound differently than a
>non-center tapped transformer? Or is it just a regular transformer for
>which the center point of the secondary winding is "brought outside".

For power and low frequuency transformers, they just bring out the
midpoint of a consecutive, layered winding. For high frequency stuff,
the winding tends to be bifalar, with the wires of the two windings
side-by-side. You can tell which is which by measuring the DC
resistances. For bifalar, the resistance from CT to the two ends will
be very close. For a tapped layered consecutive winding, the outer
winding has a bigger radius so the wire is longer and has more
resistance than the inner one.

>
> b. Considering the secondary voltage of a transformer is Vs, the two
>terminals of the secondary are at +Vs/2 and -Vs/2. This implies a
>voltage gradient across the secondary. The gradient passes through a
>zero point which we "tap". What causes this voltage gradient?

Each turn of the secondary has a time-varying magnetic field pass
through it, so each turn has a voltage induced into it. The magnitude
is typically on the order of 1 volt RMS per turn, for power
transformers at least.

>
> c. Can we say that all the turns in the secondary winding of a
>transformer have the same amount of flux passing through them at a given
>instant or do they have a different amount of flux (with the flux
>depending on the position of the turn)?

Pretty much the same. The flux is concentrated in the iron core, and
all turns loop that core. There is a small amount of magnetic flux
that cheats and flows through space, not iron, that can make small
variations in the flux on different turns. This causes "leakage
inductance", usually under 1% difference between different turns.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation

[Phil Allison]

"John Larkin"

>> The gradient passes through a
>>zero point which we "tap". What causes this voltage gradient?
>
> Each turn of the secondary has a time-varying magnetic field pass
> through it, so each turn has a voltage induced into it. The magnitude
> is typically on the order of 1 volt RMS per turn, for power
> transformers at least.

** That would need to be a rather large ( multi kVA) transformer.

The "rule of thumb" for a laminated iron E-core is 5 to 6 turns per volt,
for 1 square inch core cross section.

( 5 turns if it is for 60 Hz only and 6 turns if it is for 50Hz )

If the core is a strip wound toroidal, then the numbers is reduce to 4 and 5
turns.

So, for 1 turn per volt, the cross section needs to be around 5 square
inches.


... Phil

[Tim Wescott]

On Mon, 27 Aug 2012 23:50:24 +0530, Anand P. Paralkar wrote:

> Hi,
>
> Could you please explain the following points regarding a center tapped
> transformer (some questions apply to a non-center tapped "regular"
> transformer as well):
>
> a. Is the center tapped transformer wound differently than a
> non-center tapped transformer? Or is it just a regular transformer for
> which the center point of the secondary winding is "brought outside".

Is a non-center tapped transformer wound the same as the non-center
tapped transformer sitting next to it?

Maybe.

Like Larkin said, line power and audio transformers are wound in a fairly
straightforward way. But as the frequency goes up the inter-winding
capacitances start to get important, and all sorts of winding schemes (of
which bifilar is but one) come into play.

> b. Considering the secondary voltage of a transformer is Vs, the two
> terminals of the secondary are at +Vs/2 and -Vs/2. This implies a
> voltage gradient across the secondary. The gradient passes through a
> zero point which we "tap". What causes this voltage gradient?

Considering that the secondary voltage of a non-center tapped transformer
is non-zero, this implies a voltage gradient across the secondary. What
causes this voltage gradient?

(translation: do your homework. Google is your friend. Etc.)

> c. Can we say that all the turns in the secondary winding of a
> transformer have the same amount of flux passing through them at a given
> instant or do they have a different amount of flux (with the flux
> depending on the position of the turn)?

Yes and no. It depends on the transformer, and on what your threshold is
for considering an amount of flux to be "different".

For nearly all line voltage or audio, iron core transformers, the amount
of flux per turn is practically the same. But there are exceptions.

Power transformers for microwave ovens, OTOH, have intentionally high
leakage inductance to limit the current to the magnetron. They keep the
flux/turn constant (or nearly so) for the HV secondary, though. Given
that power transformers have been with us for over a century, I wouldn't
be at all surprised at applications (particularly 1950's-era mil-aero)
that intentionally split the magnetic flux between two or more
secondaries.

More prosaically, any transformer that is air-wound or has a large gap is
going to have different flux densities in different windings. I doubt
that there are many such in current production electronics, but again, I
wouldn't be surprised at any weird thing found in consumer kit.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

[Anand P. Paralkar]

Daniel, John, Phil and Tim, thanks to each one of you for your reply.

Let me assure you, this isn't a homework question that I am trying to
push on to the newsgroup. (If it was homework, I wouldn't do it. :) )

I stumbled on these doubts while trying to develop a negative DC voltage
source. Amongst the other features in the requirement spec., one
requirement is that we would like to make this negative voltage source
as far as possible without using a center tapped transformer. As it so
"turns" out - its difficult.

What was irritating is that I could not find (printed or online) any
standard text which discusses these points:

a. A transformer with its primary terminals connected to Vp and
ground (zero volts) will have its secondary terminals as +Vs/2 and
-Vs/2. (For a small bit of time, I continued to think that the
secondary terminals would be: +Vs and 0V, mimicking the single ended
primary.)

b. If the flux crossing all the coils is the same (as in same
density), what causes the gradient in the voltage across the coil? You
see, the gradient on the coil is a change in the voltage magnitude and
polarity from one end to the other! I have concluded that this happens
because at one end, the flux is *entering* the coil and *leaving* at the
other. This *entering* and *leaving* causes the opposite polarity (some
"thumb rule"), but what about the drop in magnitude? Then again I
"imagine" that the voltage generated all across the coil is such that
the positive voltage is completely worn out by the time it reaches the
other end and vice versa for the negative voltage. The positive and
negative voltages meet at the mid-point, cancel each other and give us a
beautiful zero! But why?

I will continue to be worried that my understanding is a figment of
imagination until I see some text/formal/mathematical treatment which
confirms/denies all this.

I will be embarrassed to see any URLs which discuss these points
straight away. But then they are welcome anyway.

Thanks once again,
Anand

[John Robertson]

And why is producing a negative voltage with a non-center tapped
transformer a problem? Use a couple of bridge rectifiers (fuse AC side
please). Tie one of the bridges negative terminal to common, the second
bridge with the positive to common. Now you have an equal positive and
negative supply that you can filter and regulate as required.

If you only need negative then just use one bridge and ties its'
positive terminal to common...

John :-#)#

--
(Please post followups or tech enquiries to the newsgroup)
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
Call (604)872-5757 or Fax 872-2010 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."

[Peter Bennett]

On Wed, 29 Aug 2012 19:55:10 +0530, "Anand P. Paralkar"
<anand.p...@gnospammale.com> wrote:


>
>Daniel, John, Phil and Tim, thanks to each one of you for your reply.
>
>Let me assure you, this isn't a homework question that I am trying to
>push on to the newsgroup. (If it was homework, I wouldn't do it. :) )
>
>I stumbled on these doubts while trying to develop a negative DC voltage
>source. Amongst the other features in the requirement spec., one
>requirement is that we would like to make this negative voltage source
>as far as possible without using a center tapped transformer. As it so
>"turns" out - its difficult.
>
>What was irritating is that I could not find (printed or online) any
>standard text which discusses these points:
>
> a. A transformer with its primary terminals connected to Vp and
>ground (zero volts) will have its secondary terminals as +Vs/2 and
>-Vs/2. (For a small bit of time, I continued to think that the
>secondary terminals would be: +Vs and 0V, mimicking the single ended
>primary.)

In a normal transformer, the primary and secondary windings are
electrically isolated from each other. If you connect the primary
winding to Vp and ground, then try measuring the voltage between
ground and either end of the secondary, you will not measure any
voltage, as there is no current path between the secondary and
primary. You will, however, measure a voltage between the ends of the
secondary winding.

>
> b. If the flux crossing all the coils is the same (as in same
>density), what causes the gradient in the voltage across the coil? You
>see, the gradient on the coil is a change in the voltage magnitude and
>polarity from one end to the other! I have concluded that this happens
>because at one end, the flux is *entering* the coil and *leaving* at the
>other. This *entering* and *leaving* causes the opposite polarity (some
>"thumb rule"), but what about the drop in magnitude? Then again I
>"imagine" that the voltage generated all across the coil is such that
>the positive voltage is completely worn out by the time it reaches the
>other end and vice versa for the negative voltage. The positive and
>negative voltages meet at the mid-point, cancel each other and give us a
>beautiful zero! But why?
>
>I will continue to be worried that my understanding is a figment of
>imagination until I see some text/formal/mathematical treatment which
>confirms/denies all this.
>
>I will be embarrassed to see any URLs which discuss these points
>straight away. But then they are welcome anyway.

You are over-analyzing this, and are being misled by your assumption
stated above that the voltage on the secondary is galvanically related
to the voltage on the primary.

To make your negative voltage source, you can simply make a positive
supply, and call the "+" terminal "ground", and use the terminal you
would normally connect to ground as the negative terminal.
Alternatively, make a normal half-wave rectifier circuit, but reverse
the polarity of the diode and capacitor - then the junction of diode
anode and capacitor "-" will be your negative terminal, and the
capacitor "+" terminal will be ground.

>
>Thanks once again,
>Anand

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

[Anand P. Paralkar]

>
> And why is producing a negative voltage with a non-center tapped
> transformer a problem? Use a couple of bridge rectifiers (fuse AC side
> please). Tie one of the bridges negative terminal to common, the second
> bridge with the positive to common. Now you have an equal positive and
> negative supply that you can filter and regulate as required.
>
> If you only need negative then just use one bridge and ties its'
> positive terminal to common...
>
> John :-#)#
>

To "use a couple of bridge rectifiers.....", you are assuming that the
bridge rectifiers are galvanically isolated from each other. That is,
the source for positive voltage regulator and the source for the
negative voltage regulator should not come from the secondary of the
same transformer.

If you do "tie one of the bridges negative terminal...." when the source
of the positive and negative regulators are the same there is a problem.
For a full wave bridge rectifier, the capacitor charges to 2Vpeak (no
load condition) as in: +Vpeak on its positive terminal and -Vpeak on its
negative terminal. So connecting the positive of one bridge to the
negative of one bridge is not a solution.

You could simulate this or build this. Just measure the voltage between
the two points you want to tie as common before connecting them. You
will see a 2Vpeak voltage between them. (Obviously, you don't go ahead
and connect them.)

If using two separate windings or separate transformers is your
solution, then I agree. It is easy.

Thanks,
Anand

[Anand P. Paralkar]

>
> You are over-analyzing this, and are being misled by your assumption
> stated above that the voltage on the secondary is galvanically related
> to the voltage on the primary.

I don't see where I have assumed or stated that the secondary is
galvanically related to the primary. All that I have said is valid with
or because of the galvanic isolation.

>
> To make your negative voltage source, you can simply make a positive
> supply, and call the "+" terminal "ground", and use the terminal you
> would normally connect to ground as the negative terminal.
> Alternatively, make a normal half-wave rectifier circuit, but reverse
> the polarity of the diode and capacitor - then the junction of diode
> anode and capacitor "-" will be your negative terminal, and the
> capacitor "+" terminal will be ground.

Sorry I wasn't clear about this, the idea is to make a positive *and*
negative supply from the same source (as in from the same secondary
winding) without using a center tap. (In which case "calling the "+"
terminal ground doesn't help. And flipping the diode's orientation just
moves which side of the bridge the positive and negative voltage appears.)

Thanks,
Anand

[Anand P. Paralkar]

On 27-08-2012 23:50, Anand P. Paralkar wrote:

I am sorry, I haven't been clear. We need to generate a positive *and*
negative voltage using the same transformer secondary without a center tap.

(I kind of started writing where I was irritated thinking about the
negative voltage.....)

[Dave Platt]

In article <aa733f...@mid.individual.net>,

Anand P. Paralkar <anand.p...@gnospammale.com> wrote:

>I am sorry, I haven't been clear. We need to generate a positive *and*
>negative voltage using the same transformer secondary without a center tap.
>
>(I kind of started writing where I was irritated thinking about the
>negative voltage.....)

http://www.bristolwatch.com/ele/power_supplies.htm shows how (see the
first example).

Basically: using a single-winding (non-centertapped) secondary, you
just ground one end of the secondary, and use two half-wave rectifying
circuits (one diode and one capacitor each) to generate the positive
and negative DC supplies.

I believe this approach will have somewhat higher amounts of ripple on
the unregulated DC supplies, because each capacitor will be recharged
during only half of the powerline cycle.

--
Dave Platt <dpl...@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

[Michael Black]

But you have to examine the whole cost, and it may be simpler to go with
the centre tapped transformer.

You could use a second transformer. This works especially well if the
second voltage is relativley low current, you can use a much smaller
transformer.

Got to a switching supply, more complicated but with multiple voltages it
may work out cheaper.

Especially if the second voltag is low current, use a dc to dc converter,
ie connect to the positive supply an oscillator and rectify the output for
the lower current negative voltage.

Use a higher voltage supply and split the DC voltage in half. This works
well if the two voltages are the same but of different polarity, and works
best if the current is relatively low. You in effect are setting an
artificial ground at the output of the rectifier.

Michael

[George Herold]

On Aug 29, 1:42 pm, "Anand P. Paralkar"

OK can I add more components to the none ceter tapped secondary?

Then make yourself (say) a 30V supply. Add a power opamp rail
splitter*, and tie the center point to ground. Viola... +/-15 Volt
supply.

George H.

*in it's simplist opamp is a buffer with non inverting input driven
from two resistors one tied to each rail of the power supply.

[Phil Allison]

"Anand P. Paralkar"

>
> Sorry I wasn't clear about this, the idea is to make a positive *and*
> negative supply from the same source (as in from the same secondary
> winding) without using a center tap. (In which case "calling the "+"
> terminal ground doesn't help. And flipping the diode's orientation just
> moves which side of the bridge the positive and negative voltage appears.)

** It is ridiculously simple to make equal "+" and "-" voltages from a
single winding.

The circuit is called a "full wave voltage doubler". Two diodes and two
filter caps, each pair wired as a half wave rectifier gives two DC rails of
opposite polarity from one winding.

http://electrapk.com/wp-content/uploads/2011/08/full-wave-voltage-doubler-1.jpg

The centre of the two caps is the zero volts point.


.... Phil

[P E Schoen]

"Phil Allison" wrote in message news:aa7d18...@mid.individual.net...

> ** It is ridiculously simple to make equal "+" and "-" voltages
> from a single winding.

> The circuit is called a "full wave voltage doubler". Two diodes and
> two filter caps, each pair wired as a half wave rectifier gives two
> DC rails of opposite polarity from one winding.

> http://electrapk.com/wp-content/uploads/2011/08/full-wave-voltage-doubler-1.jpg

> The centre of the two caps is the zero volts point.

Agreed. Here is a simulation of a circuit that I actually built:
http://www.enginuitysystems.com/pix/120Sine-320DC_Doubler.png

Here's the ASCII file if you want to play with it:
http://www.enginuitysystems.com/pix/12V-300V_CT_Doubler.asc

Paul
www.muttleydog.com

[John Larkin]

Ground one end of the secondary. Note that "ground" might be a
chassis, might be a water pipe, or might be anything else that we care
to use as a reference level. It's where we're going to clip the black
wire of our voltmeter, not much more.

Connect the other end of the secondary through a diode, to a
capacitor, and call that junction "DC+". Ground the other side of the
capacitor. This diode has anode on the transformer, cathode on the
cap.

Now the diode rectifies the AC from the transformer, dumping the AC
peaks into the capacitor. You have DC voltage across the cap, with a
bit of ripple.

If you reverse the diode, you'd get negative DC at the cap.

With two diodes and two capacitors, you can do both, namely get DC+
and DC- simultaneously, both relative to ground.

http://en.wikipedia.org/wiki/Rectifier#Half-wave_rectification

http://metroamp.com/wiki/index.php/Half_Wave_Dual_Polarity_Rectifier

(In that last link, he has the current arrows backwards. The Internet
is a mess.)

[Jamie]

sure that's easy..

use one leg of the secondary as the common and the other leg will have
2 diodes branching from it. Each diode will be such that you'll get a
+ and a - source...

THere is one big problem with this how ever, you need to use a cap
double or more than what you'd normally would to filter it on the DC
side. This is because you'll only get 30 hz per diode. Basically, you'll
have a gap on each side when the opposite side is conducting.

Of course, if your requirement for the - source is a low current type
that could be done with a added circuit that operates from a single
power source.

Jamie

[Mark Zenier]

In article <ns22h9-...@radagast.org>,

Dave Platt <dpl...@radagast.org> wrote:
>In article <aa733f...@mid.individual.net>,
>Anand P. Paralkar <anand.p...@gnospammale.com> wrote:
>
>>I am sorry, I haven't been clear. We need to generate a positive *and*
>>negative voltage using the same transformer secondary without a center tap.
>>
>>(I kind of started writing where I was irritated thinking about the
>>negative voltage.....)
>
>http://www.bristolwatch.com/ele/power_supplies.htm shows how (see the
>first example).
>
>Basically: using a single-winding (non-centertapped) secondary, you
>just ground one end of the secondary, and use two half-wave rectifying
>circuits (one diode and one capacitor each) to generate the positive
>and negative DC supplies.
>
>I believe this approach will have somewhat higher amounts of ripple on
>the unregulated DC supplies, because each capacitor will be recharged
>during only half of the powerline cycle.

Another way, that I saw in an Elektor project, is to use a full wave
rectifier for the main supply and use two half wave doublers hooked
to each end of the secondary. (Or you could call it a capacitivly
coupled full wave rectifier).


------|(-----+----|<---------+----- V-?
| |
+-->|--gnd +--|(--gnd
|
+-->|--gnd |
| |
------|(-----+----|<---------+


Mark Zenier mze...@eskimo.com
Googleproofaddress(account:mzenier provider:eskimo domain:com)

[Anand P. Paralkar]

On 27-08-2012 23:50, Anand P. Paralkar wrote:

On 27-08-2012 23:50, Anand P. Paralkar wrote:

Hi Everybody,

Firstly, thanks a lot for your detailed replies. Although this
started-off as a thread on my doubts on the transformer, this thread got
(unintentionally) drawn to another topic. One that I was planning to
post here anyway.

The reason I said that getting both - the positive and negative source
from a transformer without a center tap is difficult is that I tried
what a lot of people have recommended here. I built a circuit that had
the two diodes connected to the transformer secondary. One diode had
its anode while the other had its cathode connected to the transformer
secondary. The other end of these diodes were connected to a capacitor
each. These capacitors had a common point which we could call the ground.

I was surprised to find that this ground actually drifted! When I
measured the voltage across the ends of the capacitors (the end
connected to the diodes), the voltage measured was constant. However,
when I measured the voltage across the ground and the other end of the
capacitors, I saw that this voltage changed. So the V+ source and V-
source with respect to the ground was not constant!

I don't know the exact reason what causes this drift. But as a remedy,
I put a resistor in parallel to each of the capacitors. (P. E. Schoen
has posted this).

This stopped the ground from drifting but I don't think one could use
this solution in a practical circuit. A resistor in parallel at the
output of a voltage source will not hold up in case of a heavy load (low
load resistance).

That's why it seemed difficult.

There are ofcourse many other circuits that have been suggested here.
Thanks a ton gentlemen.

Thanks,
Anand

[Jamie]

For some reason that circuit just does not look appealing..

Not knowing the application it is kind of hard to come up with the
proper solution however....


Transformer Bridge
+-+-++---+--------+
-. ,+--------+ A A + | |
)|( +--++ | --- | |
)|( +--+(-+ --- | |
-' '+--------+ A A + | |
+-+-++ + +------------------+
| .-. | | |
| | | | + |
R1, R2 100K | | | | |\| Ilimit |
| '-' +-+|-\ ___ |
| | | >--|___|-+---+ Commom
| +-----+|+/ GND
| .-. |/+
| | | |
| | | |
| '+' |
+---+--------+

Jamie

[Phil Hobbs]

You did connect the midpoint of the capacitor string to the other end of
the transformer secondary, right? Otherwise the whole thing won't do
much at all

Putting a bleed resistor on the caps reduces the supply impedance by
making the diodes conduct harder on each peak.


Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

[John Fields]

---
As Phil Hobbs has posted, it sounds like your caps aren't connected to
the other end of the transformer.

Or you've made a wiring error...

Try this: (view using a fixed-pitch font)


AC>----+ +----+--[DIODE>]-------+---->DC+
| | | |
P||S +--[<DIODE]-+-----|---->DC-
R||E |- |+
I||C [BFC] [BFC]
| | | |
AC>----+ +----------------+-----+---->GND

--
JF

[John Robertson]

Adding a resistor across each the caps is just loading down the circuit
enough so that stray fluctuations disappear.

Back in the Tube Amplifier days a heavy ballast resistor was often
placed across the B+ to ground (eg. 5K @ 25W - 50W) to stabilize the
voltage when there wasn't a loud signal to process - kept the B+ within
the capacitors' and other components tolerances...

[Michael A. Terrell]

John Robertson wrote:
>
> Adding a resistor across each the caps is just loading down the circuit
> enough so that stray fluctuations disappear.
>
> Back in the Tube Amplifier days a heavy ballast resistor was often
> placed across the B+ to ground (eg. 5K @ 25W - 50W) to stabilize the
> voltage when there wasn't a loud signal to process - kept the B+ within
> the capacitors' and other components tolerances...

It also discharged the capacitor, to protect the 'Darwin types'.

[John Robertson]

Yeouch....charged caps can be 'fun'.

And from rec.humour.funny quoting - From *Orbit*, the Journal of the
Rutherford High Energy Laboratory,Didcot, England (31 January 1965) p.12


Ten Commandments of Electrical Safety

(1) Beware of the lightning that lurks in an undischarged
capacitor lest it cause thee to be bounced upon thy backside in a
most ungainly manner.

[Michael A. Terrell]

John Robertson wrote:
>
> Michael A. Terrell wrote:
> > John Robertson wrote:
> >> Adding a resistor across each the caps is just loading down the circuit
> >> enough so that stray fluctuations disappear.
> >>
> >> Back in the Tube Amplifier days a heavy ballast resistor was often
> >> placed across the B+ to ground (eg. 5K @ 25W - 50W) to stabilize the
> >> voltage when there wasn't a loud signal to process - kept the B+ within
> >> the capacitors' and other components tolerances...
> >
> >
> > It also discharged the capacitor, to protect the 'Darwin types'.
>
> Yeouch....charged caps can be 'fun'.
>
> And from rec.humour.funny quoting - From *Orbit*, the Journal of the
> Rutherford High Energy Laboratory,Didcot, England (31 January 1965) p.12
>
> Ten Commandments of Electrical Safety
>
> (1) Beware of the lightning that lurks in an undischarged
> capacitor lest it cause thee to be bounced upon thy backside in a
> most ungainly manner.

That's why no broadcast engineer worth his pay ever trusted bleeder
resistors. They had a nasty habit of opening up, with no physical signs
of the failure. A shorting stick inside every cabinet with HV.

[Jamie]

John Robertson wrote:
> Michael A. Terrell wrote:
>
>> John Robertson wrote:
>>
>>> Adding a resistor across each the caps is just loading down the circuit
>>> enough so that stray fluctuations disappear.
>>>
>>> Back in the Tube Amplifier days a heavy ballast resistor was often
>>> placed across the B+ to ground (eg. 5K @ 25W - 50W) to stabilize the
>>> voltage when there wasn't a loud signal to process - kept the B+ within
>>> the capacitors' and other components tolerances...
>>
>>
>>
>> It also discharged the capacitor, to protect the 'Darwin types'.
>
>
> Yeouch....charged caps can be 'fun'.
>
> And from rec.humour.funny quoting - From *Orbit*, the Journal of the
> Rutherford High Energy Laboratory,Didcot, England (31 January 1965) p.12
>
>
> Ten Commandments of Electrical Safety
>
> (1) Beware of the lightning that lurks in an undischarged
> capacitor lest it cause thee to be bounced upon thy backside in a
> most ungainly manner.
>
>
> John ;-#)#
>

Is that how "Fly Backs" got their name ? :)

Jamie

[John Fields]

---
No.

The voltage appearing across a charged capacitor, once the charging
source is removed, is static and is:

Q
V = ---,
C

While the "flyback" voltage appearing across a charged inductor, once
the charging source is removed, is dynamic and is:

L dI
E = ------
dt

--
JF

[Jamie]

unbelievable.

Jamie

[John Fields]

On Mon, 03 Sep 2012 18:28:46 -0400, Jamie

---
Tua culpa, non mea.

--
JF

[Jamie]

I think you got that backwards, bud!

Jamie

[John Fields]

On Tue, 04 Sep 2012 08:04:43 -0400, Jamie

---
Shouldn't that be "Lamie"???

--
JF

[Michael A. Terrell]

John Fields wrote:
>
> Shouldn't that be "Lamie"???

'Lame-nard'.


Maynard A Philbrook JR KA1LPA
Willimantic, CT 06226
We are the original "Brand Rex" company

[John Fields]

On Tue, 04 Sep 2012 10:12:08 -0700, Fred Abse
<excret...@invalid.invalid> wrote:

>On Mon, 03 Sep 2012 16:08:21 -0500, John Fields wrote:
>
>> While the "flyback" voltage appearing across a charged inductor, once the
>> charging source is removed, is dynamic and is:
>>
>> L dI
>> E = ------
>> dt
>

>Please:
>
>E = -L di/dt
>
>Opposing the change that produces it.
>
>This is .basics, so we need a little pedantry.

---
Right you are! :-)

--
JF

[John Larkin]

On Tue, 04 Sep 2012 10:12:08 -0700, Fred Abse
<excret...@invalid.invalid> wrote:

>On Mon, 03 Sep 2012 16:08:21 -0500, John Fields wrote:
>

>> While the "flyback" voltage appearing across a charged inductor, once the
>> charging source is removed, is dynamic and is:
>>
>> L dI
>> E = ------
>> dt
>

>Please:
>
>E = -L di/dt
>
>Opposing the change that produces it.
>
>This is .basics, so we need a little pedantry.

Ground one end of an inductor. Apply a ramp of increasingly-positive
current into the free end. Measure the voltage at that inductor
terminal. It will be positive. So, with that sign convention,

>> L dI
>> E = ------
>> dt

That's the way most people do it.

http://en.wikipedia.org/wiki/Inductor#In_electric_circuits



--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

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