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Old 21st October 2009, 04:33 PM   #1
rman is offline rman  Canada
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Default Some general opt questions.

Hi.

I just built a guitar amp using an Edcor gxpp output transformer. the primary winding did not have equal dc resistance on each half. One side read IIRC about 93 Ohms while the other side was around 84 Ohms. This was my first push pull amp as well. I was wondering if this would be of concern for the sound quality if building a hi-fi amplifier.

Also could anyone give a quick explanation for why the impedance ratio of a transformer is the square of the turns ratio?

Thanks.
Rolf.
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Old 21st October 2009, 05:34 PM   #2
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Quote:
Originally Posted by rman View Post
Hi.

I just built a guitar amp using an Edcor gxpp output transformer. the primary winding did not have equal dc resistance on each half. One side read IIRC about 93 Ohms while the other side was around 84 Ohms. This was my first push pull amp as well. I was wondering if this would be of concern for the sound quality if building a hi-fi amplifier.

Also could anyone give a quick explanation for why the impedance ratio of a transformer is the square of the turns ratio?

Thanks.
Rolf.
It's not of concern to me. Some really nice OPTs like my vintage UTCs have unequal resistance. It just means one side is larger diameter turns. Maybe there are some reasons it's better to have equal resistance and equal diameter turns, but it's a small compromise if any.

The impedance changing as the square of the turns ratio can be explained mathematically. Faraday's law gives the basic relations of induced voltage and current. The voltage ratio is equal to the turns ratio (unloaded, etc) vsec = vpri * (tsec/tpri) ; the current ratio is equal to the inverse of the turns ratio isec = ipri * (tpri/tsec) ; the impedance (v/i) is thus (tsec/tpri) * (tsec/tpr) or simply the turns ratio squared.
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Old 21st October 2009, 05:57 PM   #3
Yvesm is offline Yvesm  France
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Quote:
Originally Posted by rman View Post
. . .
Also could anyone give a quick explanation for why the impedance ratio of a transformer is the square of the turns ratio?
Or you may simply apply Ohm law because the turn ratio IS the voltage (U) ratio and the inverse of the current (I) ratio.

Since Z = U / I, if voltage (turn) ratio is 4, then Z = 4/(1/4).
Rearanging gives 4 * 4

Yves.

Ooops ... bis repetitae ... sorry.
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Old 21st October 2009, 11:26 PM   #4
rman is offline rman  Canada
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Thanks guys.

Both answers are valuable because they are stated differently. I will ponder them both for awhile and the light will come on slowly like a tube filament. My brain doesn't work fast like turning on an led. The different ways of saying the same thing will help. Thank's again.

Rolf.
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Old 22nd October 2009, 12:10 AM   #5
rman is offline rman  Canada
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I get it!

suppose a transformer has a ratio of 10:1 and you input 100 volt AC, the output is 10 volt AC. Now put an 8 Ohm resistor across the output, 1.25 amps will flow through it. Now since the transformer is stepping up the current by 10, 0.125 amps flow trough the primary. R=V/I so R= 100/0.125 = 800. So as far as the 100 volt AC source is concerned, it has a 800Ohm resistor across it!

That is so cool
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Old 22nd October 2009, 12:14 AM   #6
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wow rman! yours is the best explanation I have ever read! Seriously! You should teach!
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Old 23rd October 2009, 01:56 AM   #7
rman is offline rman  Canada
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Thank's!

I hope my head doesn't swell up too big Once I figured it out I figured I would throw in a practical example so any one else with the same question would have no trouble understanding. Then again, maybe working it out for themselves would be more educational. Anyway it's out there and might help someone.

Peace.
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