Building a transformer

See....

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...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson
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In message , dated Mon, 14 Aug 2006, Jim Thompson writes

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. (;-)

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
Reply to
John Woodgate

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
Reply to
Michael A. Terrell

--- 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>----+ | | +--[

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

NT

Reply to
meow2222

--
Aarghhhh!!! Big OOPS...


                    +-----[CR1>]--+--->V+
                    |             |
                    |  +--[CR2>]--+ 
                    |  |          |
MAINS>----+  +------+  |          |+ 
          |  |      |  |         [C1] 
          P||S      |  |          |
          R||E------|--|----------+--->GND      
          I||C      |  |          |+
          |  |      |  |         [C2]       
MAINS>----+  +------|--+          |
                    |  |          |
                    |  +--[
Reply to
John Fields

--
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.
Reply to
John Fields

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

Reply to
meow2222

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.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
Reply to
John Woodgate

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.

Reply to
John Popelish

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

Reply to
jasen

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.

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.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
Reply to
John Woodgate

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: snipped-for-privacy@itd.nrl.navy.mil Naval Research Laboratory

4555 Overlook Avenue, SW Washington, DC 20375-5337
Reply to
J. B. Wood

"J. B. Wood"

** 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.

** 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.

** You did.

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

....... Phil

Reply to
Phil Allison

In message , dated Tue,

15 Aug 2006, J. B. Wood writes

Not direct current but DC magnetic flux.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
Reply to
John Woodgate

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.

Reply to
John Popelish

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

--
 JosephKK
 Gegen dummheit kampfen die Gotter Selbst, vergebens.  
  --Schiller
Reply to
joseph2k

Thanks for the heads up.

--
 JosephKK
 Gegen dummheit kampfen die Gotter Selbst, vergebens.  
  --Schiller
Reply to
joseph2k

In message , dated Thu,

17 Aug 2006, joseph2k writes

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.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
Reply to
John Woodgate

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 (

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and create a Forth like integrated OpSys , call it NewForth for now . Wanna learn to do top System software ? Follow me .....

Mark Fortune wrote:

Reply to
werty

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