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Old 30th May 2013, 08:16 PM   #1
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Default Configuring an Audio Transformer for 8 Ohms

I have a question regarding an audio output transformer I want to incorporate into a tube amplifier I am building.

The output transformers are taken from an ‘old’ Northcourt ‘Stereo Ninety-Five’ amplifier someone once gave me (an obscure UK amp?). Please refer to the attached marked up schematic I found online of a tube amp output stage, which shows the exact same color code as the transformer I want to use.

The primary (HV) leads are pretty straightforward and I understand the color code and how to connect them (colors are identical to the one I have). However, there are 4 output (speaker) wires at the secondary output. Based on the explanation in the schematic, I understand that a ‘15 ohm’ output impedance is available between the green and black wires, and that a ‘3 ohm’ impedance is available between the brown and orange wires. The original amp also had outputs for either 3 or 15 ohms, which I consider to be somewhat oddball impedances.

My question is if I want to configure the output for 8 ohms (or close-by), can this be obtained between, say, the black and brown wires (see red mark-up in attachment) or other combination? Is anyone familiar with this transformer or has some idea whether it is more ‘flexible’ than shown on the schematic.

Any info or advice anyone can kindly provide on this matter would be greatly appreciated.
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File Type: jpg Tube Amp Output schematic.jpg (167.3 KB, 83 views)

Last edited by metalman96; 31st May 2013 at 06:59 AM.
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Old 30th May 2013, 09:43 PM   #2
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You can measure transformer ratio by applying a voltage to the primary (I use a variac set to 100v) and measuring voltages at the secondary. Impedance ratio is the square of voltage ratio. Most audio outputs will be wired for 4, 8, 16 Ohms or a subset of those. Sometimes those ratios are not exact - for instance, four equal winding segments would give 1,4, 9, and 16 Ohms if connected in the possible series and parallel combinations.

I would guess that the schematic is wrong and that the taps are 4, 8, 15 Ohms with respect to the common black wire. And that Brown to Orange is about 1 Ohm.
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Old 30th May 2013, 10:30 PM   #3
FoMoCo is offline FoMoCo  United States
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By my calculations I figured that knowing the impedance for two combinations does not give enough information to know the answer. We have two windings that we can figure the total percentage of turns for the sum of the two, but not each winding.

I then assumed this was the standard 16-8-4 winding used back before I was born. That doesn't work either. If we assume that and connect our load between the 4 and 8 taps, like the 3 ohm wiring is here, we see an expected load of less than an ohm.

That being the case... The only sure solution is the one Tom gave: Measure the turns ratio and remember that impedance is the square.
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Old 31st May 2013, 07:32 AM   #4
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Thanks, I really appreciate the advice. With regard to measuring the impedance ratio, a silly question - between which primary leads would I apply the input (i.e. 100V) voltage? I'm guessing between the outer (green and blue) ones, I guess it shouldn't matter if I'm measuring a ratio.
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Old 31st May 2013, 11:37 AM   #5
FoMoCo is offline FoMoCo  United States
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Quote:
Originally Posted by metalman96 View Post
Thanks, I really appreciate the advice. With regard to measuring the impedance ratio, a silly question - between which primary leads would I apply the input (i.e. 100V) voltage? I'm guessing between the outer (green and blue) ones, I guess it shouldn't matter if I'm measuring a ratio.
It doesn't matter. I'd use the outer two (CT left open.) You also don't have to use 100 volts. But, it does make the math easier than, say, 59.534, or so.
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