Center Tapped and Regular Transformer

Anand P. Paralkar · 2012-08-27T18:20:24+00:00

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

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Jamie 14 years ago

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

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Phil Hobbs 14 years ago

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

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John Fields 14 years ago

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 +--[DC- R||E |- |+ I||C [BFC] [BFC] | | | | AC>----+ +----------------+-----+---->GND

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John Robertson 14 years ago

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...

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."

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Michael A. Terrell 14 years ago

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

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John Robertson 14 years ago

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 ;-#)#

(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."

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Michael A. Terrell 14 years ago

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.

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Jamie 14 years ago

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

Jamie

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John Fields 14 years ago

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

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Jamie 14 years ago

unbelievable.

Jamie

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John Fields 14 years ago

Tua culpa, non mea.

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Jamie 14 years ago

I think you got that backwards, bud!

Jamie

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John Fields 14 years ago

Shouldn't that be "Lamie"???

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Fred Abse 14 years ago

Please:

E = -L di/dt

Opposing the change that produces it.

This is .basics, so we need a little pedantry.

"For a successful technology, reality must take precedence over public relations, for nature cannot be fooled." (Richard Feynman)

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Michael A. Terrell 14 years ago

'Lame-nard'.

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

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John Fields 14 years ago

Right you are! :-)

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John Larkin 14 years ago

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,

That's the way most people do it.

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John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators

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