# Verifying turns ratios on an OT by measuring AC signals thru the amp

Discussion in 'Amp Tech Center' started by peteb, Dec 7, 2017 at 11:27 AM.

1. ### petebFriend of Leo's

Apr 25, 2003
I have looked into the impedance ratio of an OT before and it's function, and probably forgotten what I learned because I had no practical application of it.

Now more recently I have been looking at the AC signal size as it progresses through the amp, it is interesting and I've learned a lot and now want to complete my measurements in the last place, around the OT, and at the same time see if I can verify the impedance ratio, turns ratio, voltage ratio in the theory of audio transformers.

I'm going to do the impedance ratios calculations, and measure the signal voltage on both sides of the OT.

Where to measure these voltages?

On the output side the place to measure the signal voltage is the speaker terminals. On the input side the two choices are to measure the signal voltage at the plate, or accross the OT primary. I will try both ways and see which provides the better results.

I just had the best spell check fix: I thought I typed measure and it turned into misjudge.

2. ### petebFriend of Leo's

Apr 25, 2003
The impedance ratio calculation for a push pull 6V6 amp driving an 8 ohm speaker:

Plate to plate load resistance from the tube spec sheets for a pair of 6V6 output tubes is 8,000 ohms

This, I take it is measured across both OT primaries in series, from plate to plate.

I want to point out that the tube Spec Sheets call this a resistance even though it is an impedance to the AC signal.

The impedance ratio is the impedance of OT primary/impedance of the speaker

Impedance ratio = 8,000 / 8 = 1,000 / 1

The square root of this ratio gives the needed turns ratio for the OT:

# of Primary windings / # of secondary windings = 31.6 / 1

This gives the voltage ratio on the windings

Voltage drop across the primary windings / voltage drop across the secondary windings = 31.6 / 1

I already did some testing at a certain signal, loud guitar, at a certain middle volume, 6 and found the AC signal at the speaker maxes out around 3 VAC.

I should see 31.6 X 3 VAC = 100 VAC or close, at some place around the OT.

I already tried measuring the signal at the plate and did get around 100 VAC under these certain playing conditions.

This surprised me. I thought the place to measure is across the OT windings. I will try both and see what I get.

To me, the voltage at the plate does not seem like the right place to measure the signal for this, but it worked.

3. ### petebFriend of Leo's

Apr 25, 2003

The OT primary has relatively low resistance to DC, and is like wise known to have small DC voltage drops, less than 10v.

Now I've learned that when the low resistance winding is connected thru an OT to a speaker, the small resistance becomes a larger impedance, 8,000 ohms.

It reasons that the impedance being many times larger than the resistance, the AC voltage drop across the primary will be many times larger than the DC voltage drop across the same primary.

THE AC in the amp uses itself up working against the speaker, and the DC uses itself up working against the power tube. The tubes have large resistance to DC, like 10k, they might also have a large impedance to AC. We may see that the drops in AC may be divided some what equally between the tube and the OT, but DC goes right thru the OT like its nothin.

4. ### petebFriend of Leo's

Apr 25, 2003
Has any one ever done this, with the OT in the amp?

Of course a signal generator would give better results but the same principles apply.

It's pretty interesting how transformers work.

I've always wanted to do this. Wrap some coils around a bar of iron. Wrap some more coils around the same bar with a different piece of wire. Count the turns ratio, apply a voltage, and see if a transformer really does what it is supposed to do.

It's cool stuff.

5. ### petebFriend of Leo's

Apr 25, 2003
And

The whole thing about tube amp sound is the interaction between the power tube and the OT and the speaker

6. ### dsutton24Poster ExtraordinaireGold Supporter

Dec 29, 2010
Illinois
You're creating your own theory, and it's a little sideways...

The low d.c resistance is because the only thing that contributes to the d.c. resistance of a transformer is the resistance per foot of the wire. That d.c. resistance is meaningless it the PA.

The transformer has a relatively large impedance on the primary side. Impedance is the aggregate of resistance, capacitive reactance and inductive reactance. Resistance is a very small part of the impedance of an output transformer.

It's possible, as you see in the output transformer. for a device to have a low resistance, but a high impedance.

The output transformer isn't really there to raise or lower voltage. The tubes in the PA are voltage devices, they amplify voltage. Speakers are power devices, you can throw a lot of voltage at a speaker, but if that voltage source can't supply a fair amount of current at the same time, the speaker won't work very well. That's what the output transformer, more properly called a matching transformer, does.

Apply a signal to your amp and scope the primary of your output transformer, you'll see a relatively high a.c voltage. Now scope the secondary side, you'll see a much lower a.c. voltage. Now, bear in mind that a lower turn count in the secondary means a lower output voltage, but it also means a proportionately higher output current.

Power equals current times voltage...

That, Charlie Brown, is the meaning of output transformers.

7. ### zookFriend of Leo's

Aug 6, 2003
Cochise, AZ
Output transformer impedance

You determine the turns ratio by using the outside windings of a push-pull transformer, ignoring the center tap. You can determine the turns ratio by using 117VAC (or a known low voltage AC power source) on the primary. Example: 117 VAC on the primary = 3.5 VAC on secondary. Turns ratio = 117/3.5 = 33:1. Impedance ratio = (33 x 33) = 1089:1 which would equal an impedance of 8712 ohms working into an 8 ohm load (1089 x 8).

8. ### petebFriend of Leo's

Apr 25, 2003
Dsutton,

Thank you for the response.

I don't think it's my theory, I'm just trying to confirm existing theories.

This is what we are finding, the OT impedance is much larger than its DC rsistance.

Because the AC powers the speaker, and the DC doesn't.

I read thru what you said, twice, but it does not disagree with what I'm saying.

The main point:

Low resistance = low DC voltage drop

High impedance = high AC voltage drop - EDITED

Last edited: Dec 8, 2017 at 11:10 AM

9. ### petebFriend of Leo's

Apr 25, 2003
Zook,

Thank you

I'm not sure what new information you've added. Do you have a source?

What I did read is that it mentions the center tap. I'm finding that the center tap confuses the interpretation of the turns ratio. I'm think I'm going to look at single ended to get my bearings straight.

10. ### petebFriend of Leo's

Apr 25, 2003
I think the answer to the thread is that the turns ratio is supposed to be proportionate to the voltage ratios across the windings.

This means that the Ac voltage across the OT primary will be proportionate with the Ac voltage across the Speaker in the same proportion as the number of turns in the primary and secondary windings of the OT.

The other part, how much AC drop in the tube versus how much AC drop ion the OT, may be new material but there is no surprise that there would be a significant drop on both.

11. ### robrobPoster ExtraordinaireAd Free Member

Dec 29, 2012
United States
No, high impedance = high AC voltage drop.

It really depends on what the impedance is made up of. If the impedance is made up of mostly resistance then it will cause a DC voltage drop. If the impedance is made up of mostly reactance (inductance and capacitance) then it will not cause a DC voltage drop.

12. ### petebFriend of Leo's

Apr 25, 2003
Rob, I think that's exactly what I meant

High impedance will cause a high AC voltage drop - I will edit the above

Side side side question: how is a voltage drop any different than gain in reverse?

13. ### petebFriend of Leo's

Apr 25, 2003
The new question for today

Where exactly in the amp does the signal voltage reach its maximum?

The DC is pretty easy to read from the schematic.

Who really knows amps? Where exactly is the signal voltage at its very highest?

I don't know yet but I'm going to find out.

14. ### petebFriend of Leo's

Apr 25, 2003
Per the above I measured 3 VAC at the speaker under some certain, repeatable, middle of the range playing conditions.

I used the impedance ratio to find the turns ratio and the voltage ratio is 31.7

3 X 32 = 100 VAC

I went looking for 100 VAC signal in the area of the primary of the push pull OT, under these certain playing conditions.

Where will it be?

The plate voltage?

Across one primary winding?

Across both primary windings in series?

First I found the 100 VAC at the plate and this surprised me.

Then across a single primary I saw voltages, the same, I couldn't tell a difference from the plate voltage, it seemed to hit 100 also. I kind of thought the correct one would take the voltage drop across both primaries, because the combined impedance of both is used in the impedance ratio calculation.

Should the volatage be from both primaries? Or one? Or one and double?

I did not read AC from plate to plate because I thought that it would give a weird result. Isn't only one plate conducting at a time, or at least they are opposite phase. It almost seems like the AC voltage from plate to plate would be the same as the AC from plate to center tap. This is an important detail, but not one that has to be resolved right now.

I think this question Needs to be answered first:

If the plate has 100 AC on it, and one half of the OT primary also has 100 VAC across it, what is the AC voltage at the center tap? 100 + 100 = 200? Or is it 100 - 100 = 0 VAC

And

Why is the plate voltage the same as across the OT? Are they actually the same thing?

When a drop in AC voltage is measured, unlike DC, the measurement alone does not tell you which side is at the highest VAC potential. The way to determine which side is at the higher potential is to compare each VAC potential to ground and compare.

Which gets back to the question of where in the amp has the highest AC?

The AC increases from the grid to the plate. Is this the high point and then the AC loses itself against the OT?

Or does the AC grow, gain, in the OT and the highest point of VAC is the center tap, or the connection to the high voltage DC power supply?

These are some good amp questions and I'm going to find the answers.

15. ### petebFriend of Leo's

Apr 25, 2003
Predictions?

I predict that the DC in the tube boosts the AC, that's why the AC increases sevenfold from the grid to the plate, and then it's all down hill from there, the AC can only lower itself as it works out against the combined speaker OT load.

My prediction is the AC at the center tap will have to be low to non existent.

16. ### petebFriend of Leo's

Apr 25, 2003
And if this prediction is correct, then that means the VAC on the plate is also the VAC drop across the OT primary.

17. ### robrobPoster ExtraordinaireAd Free Member

Dec 29, 2012
United States
The DC voltage doesn't boost the AC, the change in DC is the AC signal voltage.

The B+1 power supply node supplies the high voltage DC and the tube sends current to ground which causes the DC plate and output transformer voltage to drop--that drop in DC voltage is the AC signal. The AC signal is said to "ride on top" of the DC voltage.

When the power tube reduces the current to ground the plate and transformer voltage rise toward the B+1 voltage--that rise in DC voltage is the AC signal.

Last edited: Dec 9, 2017 at 7:58 AM

18. ### petebFriend of Leo's

Apr 25, 2003
I wasn't being real technical with the 'boost' term but the tubes affect in the AC reminds me of those hot wheel sets where every time the hot wheel car goes by the booster station it gets a boost in speed.

19. ### petebFriend of Leo's

Apr 25, 2003
The test was interesting and conclusive.

The plate of the power tubes is the highest point for the signal in the amp. It's like going over a mountain pass.

The entire signal expends itself against the OT, there is no AC left at the other end of the primary. The center tap is the end of the road for the AC, until it hops over the OT to the speaker thru the electromagnetic transformation of energy.

So the answer is that the AC signal measured at the plate relative to ground, is equivalent to the AC voltage drop across the primary. Either should have the correct voltage and voltage ratio to the speaker voltage, calculated from the impedance ratio of the OT.

It's a good example of how the AC and DC progress thru the amp in opposite directions. In the direction that the AC drops across the OT, the DC is increasing, but much less.

20. ### robrobPoster ExtraordinaireAd Free Member

Dec 29, 2012
United States
Highest voltage yes but highest current occurs in the output transformer secondary and the speaker coil.

D'tar likes this.

IMPORTANT: Treat everyone here with respect, no matter how difficult!
No sex, drug, political, religion or hate discussion permitted here.