Building a transformer

[Jim Thompson]

Jim Thompson wrote:
[snip]

I said it was silicon. The Ge device I used in HS was a Delco
doorknob power device. I did 10W class-A ;-)

...Jim Thompson

OK. You have used a "doorknob" power device. What the heck did it look
like? Where can i find outline drawings? You are about the third person i
have heard admit of their existence.

I probably have an appropriate data book around here somewhere. I'll
look for it.

...Jim Thompson

See....

http://analog-innovations.com/SED/2N1358.jpg

...Jim Thompson

 

[John Woodgate]

In message <[email protected]>, dated Mon, 14

See....

http://analog-innovations.com/SED/2N1358.jpg

I discovered, to my surprise, that you can still get ADZ12s, although
they are EUR 32 each. The fbeta of these is measured in inverse minutes.
(;-)

 

[Michael A. Terrell]

OK. You have used a "doorknob" power device. What the heck did it look
like? Where can i find outline drawings? You are about the third person i
have heard admit of their existence.

The DS501/2N441 was a "Doorknob", AKA "JEDEC TO-36" package. I can
put some pictures on alt.binaries.schematics.electronic if you want to
see a real part.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[John Fields]

To get +-50v dc you need the same transformer, no reason to have 2
windings at all. So you just need one 50v winding.

---
I don't think that's true. (View in Courier)


Consider a conventional full wave bridge:

+----+---------+
| | |
MAINS>----+ | [CR1] [CR3]
| | |A |K
P||S +--[C1+]--+--->V+
R||E | |
I||C +---------|--->V-
| | |A |K
MAINS>----+ | [CR2] [CR4]
| | |
+----+---------+

Note that (neglecting the secondary I²R losses) for a 50VRMS
secondary winding the voltage from V+ to V- will be:


VDC = (VRMS * sqrt(2)) - (2Vf) = (50 * 1.414) - 1.5V ~ 69.2V


Since either V+ or V- could be used as a [ground] reference, then
the output of the supply could be either +62.5V or -62.5V, but not +
and - at the same time. An artificial reference could be
constructed so that the output of the supply would be symmetrical
and "simultaneously bipolar", but in that case the output would be
+/- 34.6VDC, not +/- 50VDC.

In order to get simultaneous + and - outputs what's usually done (if
separate transformers aren't used), is this:

+--[CR2>]--+--->V+
| |
+--[CR1>]--+
| |+
MAINS>----+ | [C1]
P||S |
R||E----------+--->GND
I||C |+
| | [C2]
MAINS>----+ | |
+--[<CR3]--+
| |
+--[<CR4]--+--->V-

Now, if we have a single 50V winding it'll have to be center-tapped,
with the result that there'll be 25V RMS on either side of the
secondary, and the DC output will be:


VDC = (VRMS * sqrt(2)) - (Vf) = (25 * 1.414) - .75 ~ 34.6V


For the negative side the result will be the same, but the sign will
be different.

So, we still didn't get +/- 50VDC from a transformer with a single
50V winding.

 

[[email protected]]

Youre not the only person to say this, and I find it very puzzling, as
the means to get +/- 50v is elementary:

+-----|<|------- V-
MAINS>----+ |
| +-----|>|------- V+
P||S
R||E
I||C
| |
MAINS>----+ |
|
+--------------- 0V


NT

 

[John Fields]

To get +-50v dc you need the same transformer, no reason to have 2
windings at all. So you just need one 50v winding.

---
I don't think that's true. (View in Courier)


Consider a conventional full wave bridge:

+----+---------+
| | |
MAINS>----+ | [CR1] [CR3]
| | |A |K
P||S +--[C1+]--+--->V+
R||E | |
I||C +---------|--->V-
| | |A |K
MAINS>----+ | [CR2] [CR4]
| | |
+----+---------+

Note that (neglecting the secondary I²R losses) for a 50VRMS
secondary winding the voltage from V+ to V- will be:


VDC = (VRMS * sqrt(2)) - (2Vf) = (50 * 1.414) - 1.5V ~ 69.2V


Since either V+ or V- could be used as a [ground] reference, then
the output of the supply could be either +62.5V or -62.5V, but not +
and - at the same time. An artificial reference could be
constructed so that the output of the supply would be symmetrical
and "simultaneously bipolar", but in that case the output would be
+/- 34.6VDC, not +/- 50VDC.

In order to get simultaneous + and - outputs what's usually done (if
separate transformers aren't used), is this:

+--[CR2>]--+--->V+
| |
+--[CR1>]--+
| |+
MAINS>----+ | [C1]
P||S |
R||E----------+--->GND
I||C |+
| | [C2]
MAINS>----+ | |
+--[<CR3]--+
| |
+--[<CR4]--+--->V-

---
Aarghhhh!!! Big OOPS...


+-----[CR1>]--+--->V+
| |
| +--[CR2>]--+
| | |
MAINS>----+ +------+ | |+
| | | | [C1]
P||S | | |
R||E------|--|----------+--->GND
I||C | | |+
| | | | [C2]
MAINS>----+ +------|--+ |
| | |
| +--[<CR3]--+
| |
+-----[<CR4]--+--->V-

 

[John Fields]

Youre not the only person to say this, and I find it very puzzling, as
the means to get +/- 50v is elementary:

+-----|<|------- V-
MAINS>----+ |
| +-----|>|------- V+
P||S
R||E
I||C
| |
MAINS>----+ |
|
+--------------- 0V

---
Well, I can't argue with that, LOL, but by doing it that way you've
just made the smoothing caps twice as big as they have to be since
you've halved the frequency of the rectified AC. Also, you've
increased the current the diodes _and_ the transformer need to
handle.

 

[[email protected]]

Well, I can't argue with that, LOL, but by doing it that way you've
just made the smoothing caps twice as big as they have to be since
you've halved the frequency of the rectified AC. Also, you've
increased the current the diodes _and_ the transformer need to
handle.

Secondary is twice the current, half the V. Higher diode i, half the
number of diodes.

Smoothing caps would need to be a little over twice the capacitance
irl, as i is flowing out of the caps for more than twice the length of
time between topups. When youre building from salvage the choice comes
down to what you happen to find.


NT

 

[John Woodgate]

---
Well, I can't argue with that, LOL, but by doing it that way you've
just made the smoothing caps twice as big as they have to be since
you've halved the frequency of the rectified AC. Also, you've
increased the current the diodes _and_ the transformer need to
handle.

It's a very convenient way of getting + and - rails from an AC wall-wart
for low-current op-amp applications, where the larger filter caps aren't
TOO large.

 

[John Popelish]

Youre not the only person to say this, and I find it very puzzling, as
the means to get +/- 50v is elementary:

+-----|<|------- V-
MAINS>----+ |
| +-----|>|------- V+
P||S
R||E
I||C
| |
MAINS>----+ |
|
+--------------- 0V

In addition to the filter and secondary current comments:
If you don't load the two outputs equally, there will be Dc in the
transformer core, causing it to saturate at the end of alternating
half cycles. It will hum and get hot.

 

[jasen]

What do you mean by 'common-mode ripple'?

both go negatve or positive at the same time.

to an op-amp connected accross the rails it looks like the earth is moving,

with a full-wave rectifier and centre tap the ripple ripple is symmetric
and at twice the frequency, which makes it easier to filter out.

Bye.
Jasen

 

[John Woodgate]

both go negatve or positive at the same time.

Thanks; that's what I thought.

If you draw out what happens (and look with a scope, of course), you
find that the two ripple waveforms are unsymmetrical sawtooth waveforms,
displaced in time. So there is some common-mode and some differential
mode ripple.

I would use this technique only for low-current applications, where low
ripple can be achieved with reasonable filter capacitor values, or for
feeding regulators.

to an op-amp connected accross the rails it looks like the earth is
moving,

It seems to me that the op-amp doesn't 'see' the rails until its input
or output voltage gets close to them. If it did, PSRR wouldn't be so
high.

 

[J. B. Wood]

In addition to the filter and secondary current comments:
If you don't load the two outputs equally, there will be Dc in the
transformer core, causing it to saturate at the end of alternating
half cycles. It will hum and get hot.

Hello, and direct current in the core as a result of current flow in the
secondary windings? Just how would that be possible? Is the core part of
some external circuit? If you mean eddy currents they would be AC and a
power transformer core is laminated to minimize this effect. Either I
misconstrued your comment or I must have been sleeping in my EE101
electric machinery course years ago. Sincerely,

John Wood (Code 5550) e-mail: [email protected]
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

 

[Phil Allison]

"J. B. Wood"

John Popelish



Hello, and direct current in the core as a result of current flow in the
secondary windings?

** Of course not.

Always exclude the impossible when interpreting someone's language.

In this case "DC" is an abbreviation of "DC offset " = magnetic field
offset.

If you mean eddy currents they would be AC and a
power transformer core is laminated to minimize this effect.

** WRONG.

The heat is generated by simple I squared R loss in the copper wire due to
excess primary current as a result of core saturation.

The audible hum is the iron core protesting about being driven into
saturation 50 or 60 times per second.

Either I misconstrued your comment

** You did.

or I must have been sleeping in my EE101
electric machinery course years ago.

** Maybe you were - but the behaviour of a transformer subjected to an
asymmetrical current load is rarely mentioned in elementary texts.

John Wood (Code 5550) e-mail: [email protected]
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

** Hmmmm - sounds kinda serious like ......



........ Phil

 

[John Woodgate]

Hello, and direct current in the core as a result of current flow in
the secondary windings?

Not direct current but DC magnetic flux.

 

[John Popelish]

Hello, and direct current in the core as a result of current flow in the
secondary windings? Just how would that be possible? Is the core part of
some external circuit? If you mean eddy currents they would be AC and a
power transformer core is laminated to minimize this effect. Either I
misconstrued your comment or I must have been sleeping in my EE101
electric machinery course years ago. Sincerely,

I have been waiting for this.

I misspoke. There is no current in the core (except for the normal AC
eddy current). I meant to say that there will be a net DC in the
windings, which will cause the core to magnetically saturate at the
end of alternating half cycles.

 

[joseph2k]

Jim Thompson wrote:
[snip]

I said it was silicon. The Ge device I used in HS was a Delco
doorknob power device. I did 10W class-A ;-)

...Jim Thompson

OK. You have used a "doorknob" power device. What the heck did it look
like? Where can i find outline drawings? You are about the third person
i have heard admit of their existence.

I probably have an appropriate data book around here somewhere. I'll
look for it.

...Jim Thompson

See....

http://analog-innovations.com/SED/2N1358.jpg

...Jim Thompson

Thanks a bunch. Never knew that is what doorknob is. I have used a few
also, a long time ago (30 years +).

 

[joseph2k]

There are also a lot of 'mistakes' in it. Schade's curves, for example,
were derived in 1943 for valve/tube rectifier diodes and aren't correct
for semiconductor diodes, which have a much lower 'on-state' resistance.
Some of the waveform diagrams are positively weird, and don't aid
understanding at all.

Thanks for the heads up.

 

[John Woodgate]

Thanks for the heads up.

The source I use is 'Single-phase rectifier circuits with CR filters', A
Lieders, Electronic Components and Applications, Part 1 Vol.1, no.3 p
153-163 (May 1979), Part 2 Vol.1 no.4 p 216-230 (August 1979) (Philips,
Netherlands).

A simplified treatment that works well is given in 'DC supplies from AC
sources - 3', K L Smith, Electronics and Wireless World, February 1985.

 

[werty]

You want 600 va ? But that means 90% effeciency !
That means you must use switch mode .

You must choose an ungapped ferrite core with 2000 or more MU ,

it will need litz wire on output ( simply use smaller wire and mo
strands ) .
Tiny cores will easy do 600 watts .
MJE13005's are the most popular but the Cxxxx ( 2SCxxxx)
equivalents are seen more cause all are built in China .

You will see all switch mode pow supplies changing to single
drive , for they are just as eff as push pull , and cheaper .
Thery self oscillate ! Control circuit simply pulls down the Base
of bipolar Xisistor to adj output voltage .
It is a myth that more power can be sent to output using push pull !
All low end switch mode PS are using single transistor and if you
measure
the Xformer and other parts , SURPRISE , it is same size and cheaper
to build and its also an easier circuit to control ! Simpler parts .
You still need a volt ref and Opto and a LM393 to sense current .

The myth started when they said the dead time would waste time
but as we EE's know the dead time dont exist ! And we dont have to
control it as LM494 crowd must do !

Its a simplification made in heaven , no compromises al all .
You simply dbl up the Primary winding transistors !
Remember ! Transistors in class D dont worry about balancing !
If standard PCPS has push pull using 2 each C2335/or MJE13005
i simply put them in parallel ! Same power output !
Of course in dumpster diving , you will need to be clever on the
2ndaries to get same voltages

Dont be put off , this is so simple it should be used in school
to teach . You will be amased at how many parts are tossed out !


BTW Free OpSys for PDA . I'll buy a ARM Eval board (
www.littlechips.com)
and create a Forth like integrated OpSys , call it NewForth for now .
Wanna learn to do top System software ? Follow me .....