How do you determine primary winding OHMS on an unknown Audio Transformer?

[audioguru]

47 years ago when all amplifiers used vacuum tubes (including mine) would you buy an unknown transformer without knowing its manufacturer's name, which model it was used in, its impedances and its power rating?
Would you buy it today?

I used my vacuum tube amplifier for two years. It played very well for two or three months then its output tubes needed replacement over and over to keep the distortion low. I still have the solid state receiver (HH Scott) I replaced it with about 47 years ago and it still works perfectly.

In 1970 I bought my first TV. It was a hybrid with some vacuum tubes and some solid state. I needed to replace its vacuum tubes many times.

In the "good old days" every corner store had a vacuum tube tester and sold replacements.

 

[MrAl]

Remember...reflected impedance figures are CALCULATED...not "measured". Turns ratio never changes, therefore impedance ratio never changes either. Yet in your chart the turns ratio keeps changing. Why is this?

You have a voltage transfer ratio of 24.35, which constitutes a turns ratio of 24.35. Your impedance ratio would be -

24.35 ^ 2 = 593

Now...impedance ratio is the turns ratio squared.Therefore, your chart should have looked more like this -

Secondary Load        Reflected Primary Impedance

1R            593R
2R            1186R
4R            2372R
8R            4744R
16R            9488R
32R            18976R
64R            37952R
128R            75904R

Unless you know something about valve amplifiers, it would seriously be in your best interests to refrain from posting on this thread. Seriously.

Hello again Jon,


Ideal devices have characteristics that most often dont peter out in real life applications because the ideal devices become real devices with secondary characteristics that have to be modeled appropriately in order to more accurately depict the true life operation. Using the ideal device parameters can lead to quite erroneous results in the real world. Part of this non ideal behavior comes in the form of limits placed on the device operation.

As almost everyone knows the wattage of a device (not just of transformers) plays a part in the limits on how that device can be effectively used in real life. Without knowing the wattage of a device we dont have critical information that tells us just how much we can stress the device before it no longer operates correctly or even breaks down completely.
The transformer is no exception here. The wattage, coupled with knowing the characteristic output impedance, tells us (through the indirect use of the Transformer Equation) how high the primary voltage can be.
For example, if we have an 8 watt transformer with a specified output impedance of 8 ohms, that means we can have up to 80 volts on the primary. Without knowing that additional specification we might assume we could put 160v on the primary, and that would either burn up the transformer or at best just cause tremendous signal distortion.

Also, you seem to be saying both that the transformer impedance ratio is constant, but also that it is not constant. So which is it? Since the transformer has reactance that means it has a variable impedance which varies with frequency. This also has an impact on the allowable source impedance which restricts our use to certain output impedances. Note that if in the above example if we increase the output load to 16 ohms we can only use the transformer for 4 watts not its rated 8 watts, so we can no longer use the transformer as effectively. Also, if we try to use it at the full 8 watts we'd have to increase the input voltage which would either burn up the transformer or at best create very bad odd harmonic distortion.

 

[Jon Wilder]

Also, you seem to be saying both that the transformer impedance ratio is constant, but also that it is not constant. So which is it?

Not sure where I appeared to state that impedance ratio was not constant. Zratio = Tratio ^ 2. Since turns ratio is constant, so is Zratio.

Now...this "ideal vs real world" stuff really needs to go away. While there is SOME variance between the two, it's not really enough to cause huge concern and create the confusion that it does.

Note that if in the above example if we increase the output load to 16 ohms we can only use the transformer for 4 watts not its rated 8 watts, so we can no longer use the transformer as effectively. Also, if we try to use it at the full 8 watts we'd have to increase the input voltage which would either burn up the transformer or at best create very bad odd harmonic distortion.

If we double the load impedance on the secondary, the load impedance also doubles on the primary so theoretically this will halve the output power. However...the load line and where it crosses the plate/grid characteristic curves on a given valve type also comes into play on this. If the power transfer ratio is not kept the same, it won't be exactly halved.

In regards to how "effectively" we use a given transformer, this all depends on what kind of power transfer ratio you're after.

A common trick we do on parallel push-pull valve amps is to double the load impedance, then remove either the outer or the inner pair of valves from the circuit. This accurately halves the output power as it keeps the load line crossing the plate curves at the same place it crossed it with both valves installed and the correct load impedance so the power transfer ratio remains the same.

 

[The Electrician]

did you notice that at 8 ohms, the results of the impedance ratio calculation closely matched the measured voltage ratio? the reflected primary impedance with a secondary load of 8 ohms was also very close to 5k.

I selected a transformer from my stash that had a rated 5k primary (and 8Ω secondary) because that's what the OP mentioned; so, yes, I noticed it.

In post #4, you said:

"Impedance Ratio = (Vpri / Vsec) ^ 2

Once the impedance ratio has been determined, you can multiply the impedance ratio by the speaker load that you plan to drive with it in order to determine what the reflected primary impedance with that load will be."

Then in response, in post #6, MrAL said:

"Also, what Jon mentioned about the impedance of a transformer is interesting, but with audio transformers they are often designed with a given load in mind. This means there is some optimization that goes into the design of many transformers regarding a given load impedance (and consequently a given source impedance). The difference between the theory and practice here is that the audio transformer does best when used as the manufacturer recommends, so if they say "16 ohms" then they mean 16 ohms and not 4 ohms."

Then you said in post #7:

"Y'all are seriously making this out to be harder than it is...

If you double the load impedance on the secondary, the reflected primary impedance ALSO doubles. If you halve the load impedance on the secondary, the reflected primary impedance also halves.

It really is that simple."

My purpose in making the measurements I posted is to point out that MrAL isn't "making it out to be harder than it is". It is more complicated than just connecting any value of load impedance and expecting the primary impedance to be the load impedance times the voltage transfer ratio squared.

Now, it's possible you were only referring to the behavior of an ideal transformer, but in posts #4 and #7, you described how you make real measurements of a real transformer, so it was reasonable to infer that you were describing the behavior of real transformers, and the simple rule you gave is substantially in error for real transformers, which I pointed out.

The measurements I posted (for my particular transformer) show that simply doubling the load from 8 to 16 ohms leads to a primary impedance that is about 17% lower than theoretically ideal. A 4 times larger load leads to a primary impedance that is about 37% too low compared to a theoretical ideal. These are substantial errors.

MrAL is right; one should use the manufacturer's rated impedances.

 

[MrAl]

Not sure where I appeared to state that impedance ratio was not constant. Zratio = Tratio ^ 2. Since turns ratio is constant, so is Zratio.

Now...this "ideal vs real world" stuff really needs to go away. While there is SOME variance between the two, it's not really enough to cause huge concern and create the confusion that it does.

Ok then i guess i can toss real vs ideal out the window and use one of those audio transformers on the input to my scope which has a 1 megohm input impedance, so i can boost the input impedance to 600 megohms for testing audio circuits, right? That's nice i think i'll hook up my 1000watt audio transformer with 8 ohm characteristic output and 600:1 impedance ratio right now so i can start testing my audio pre amp stuff with a nice near 1 gigaohm probe

If we double the load impedance on the secondary, the load impedance also doubles on the primary so theoretically this will halve the output power. However...the load line and where it crosses the plate/grid characteristic curves on a given valve type also comes into play on this. If the power transfer ratio is not kept the same, it won't be exactly halved.

In regards to how "effectively" we use a given transformer, this all depends on what kind of power transfer ratio you're after.

That's not the point. The point is that you said that the wattage wasnt important and that you can change the impedance to whatever you wanted it to be. I showed you that there are negative consequences to this and that we need to know the wattage of the transformer.

A common trick we do on parallel push-pull valve amps is to double the load impedance, then remove either the outer or the inner pair of valves from the circuit. This accurately halves the output power as it keeps the load line crossing the plate curves at the same place it crossed it with both valves installed and the correct load impedance so the power transfer ratio remains the same.

The point is not to "accurately halve the output power". The point is to use the transformer core metal mass and winding copper content as effectively as possible. By doubling the output load impedance on a transformer designed for non doubling (ie 8 ohms vs 16 ohms) we can not use the transformer as effectively. We can not ignore the Transformer Equation.

We dont care about valve amplifiers here, we only care about transformers

 

[The Electrician]

Remember...reflected impedance figures are CALCULATED...not "measured". Turns ratio never changes, therefore impedance ratio never changes either. Yet in your chart the turns ratio keeps changing. Why is this?

The turns ratio doesn't change; the impedance ratio changes. When you say "Turns ratio never changes, therefore impedance ratio never changes either.", this is only true for an ideal transformer; this is the distinction unclejed613 failed to make. The whole purpose of my measurements is to show just how much the impedance ratio does change with different load impedances for a real transformer.

You have a voltage transfer ratio of 24.35, which constitutes a turns ratio of 24.35. Your impedance ratio would be -

24.35 ^ 2 = 593

Now...impedance ratio is the turns ratio squared.Therefore, your chart should have looked more like this -

Secondary Load		Reflected Primary Impedance

1R			593R
2R			1186R
4R			2372R
8R			4744R
16R			9488R
32R			18976R
64R			37952R
128R			75904R

My chart didn't look like that because the values in my chart are the measured impedances of a real transformer with varying load impedances; the values in your chart are for an ideal transformer.

 

[Jon Wilder]

Ok then i guess i can toss real vs ideal out the window and use one of those audio transformers on the input to my scope which has a 1 megohm input impedance, so i can boost the input impedance to 600 megohms for testing audio circuits, right? That's nice i think i'll hook up by 1000watt audio transformer with 8 ohm characteristic output and 600:1 impedance ratio right now so i can start testing my audio pre amp stuff with a nice near 1 gigaohm probe

You're completely missing the point. Stop thinking impedance and start thinking VOLTAGE & CURRENT. It will make lots more sense that way.

That's not the point. The point is that you said that the wattage wasnt important and that you can change the impedance to whatever you wanted it to be. I showed you that there are negative consequences to this and that we need to know the wattage of the transformer.

In no post did I EVER state that wattage was not important. While load power consumption can be run BELOW what the transformer is rated at, you never want to exceed the transformer's power rating.

The point is not to "accurately halve the output power". The point is to use the transformer core metal mass and winding copper content as effectively as possible. By doubling the output load impedance on a transformer designed for non doubling (ie 8 ohms vs 16 ohms) we can not use the transformer as effectively. We can not ignore the Transformer Equation.

We dont care about valve amplifiers here, we only care about transformers

Depends on the application.

 

[MrAl]

Hi there Electrician,

I forgot to thank you for taking the time to perform those measurements. That was very nice of you so that we can all get a good picture of what is really happening with a real life device.

 

[unclejed613]

actually, there are tube amps that have very low output impedances. maybe not always AS low as can be achieved with SS amps, but low enough to have a damping factor of at least 10 or 20. Electro-Voice actually had a tube amp line that had current feedback as well as voltage feedback, and the output impedance was variable from +8R to -1R. of course, the output transformers for these amps were cathode driven, and not plate driven, so the primary impedance was a lot lower to begin with. the voltage swing required for this to work correctly was achieved by floating the plate supplies (separate floating supplies for each output tube were used). the common term for this type of output stage is "circlotron" in case anybody wants to google it. there are solid state versions of the same principle, used in QSC amps, and some Crown amps.

negative feedback in tube amps is generally tapped from the speaker winding, so it does compensate for variations in speaker and transformer impedance.

 

[MrAl]

You're completely missing the point. Stop thinking impedance and start thinking VOLTAGE & CURRENT. It will make lots more sense that way.



In no post did I EVER state that wattage was not important. While load power consumption can be run BELOW what the transformer is rated at, you never want to exceed the transformer's power rating.



Depends on the application.

Hi Jon,


I'm missing the point? So true or false, i can use the 1000 watt audio transformer on the input of my scope so i can perform tests on my audio pre amp stages now having a 600 megohm input probe?

Now you are saying that wattage is 'somewhat' important? The wattage is important not ONLY because of the transformers power rating. It's also about induction limits.

Depends on the application? What does? Transformer cost is or at least should be a concern in most but maybe the lowest power applications.

 

[Jon Wilder]

So true or false, i can use the 1000 watt audio transformer on the input of my scope so i can perform tests on my audio pre amp stages now having a 600 megohm input probe?

Did I ever say that? No I didn't. When you do scope tests on transformer coupled circuits you typically do them with a resistive dummy load across the secondary so that the proper load impedance is reflected to the primary.

But we're not talking about doing that here. We're talking about feeding a voltage to the primary, then measuring the UNLOADED voltage output at the secondary. This gives us the turns ratio of the transformer, which we use to calculate the impedance ratio, WHICH IS A FIXED VALUE BECAUSE THE TURNS RATIO IS ALSO FIXED.

Now you are saying that wattage is 'somewhat' important? The wattage is important not ONLY because of the transformers power rating. It's also about induction limits.

First off I never stated that wattage wasn't important in the first place. Stop attempting to make it appear as if I'm changing up my story here.

Depends on the application? What does? Transformer cost is or at least should be a concern in most but maybe the lowest power applications.

Yes...it depends on the application. More specifically...in a guitar amplifier where you're not trying to milk every last 1/10th of a watt out of the thing (which is the ONLY place where the variance between ideal vs real world even begins to matter). No valve amp requires that level of precision when selecting proper reflected load impedance.

 

[MrAl]

**broken link removed** Originally Posted by colin55
I am just saying you need to know the wattage of a transformer to see if it is suitable for an application.
It looks like he wants an output transformer.


Unless you know something about valve amplifiers, it would seriously be in your best interests to refrain from posting on this thread. Seriously.

Hi Jon,

Ok maybe i misunderstood you. Then why did you tell colin that he should refrain from posting in this thread after he just mentioned that the wattage is important in order to determine suitability?

 

[Nigel Goodwin]

I used my vacuum tube amplifier for two years. It played very well for two or three months then its output tubes needed replacement over and over to keep the distortion low. I still have the solid state receiver (HH Scott) I replaced it with about 47 years ago and it still works perfectly.

Then you had something SERIOUSLY wrong with it, valve amps weren't as reliable as modern transistor ones, but they certainly didn't require frequent valve replacement.

You 'might' change the output valves every 5-10 years?, and an occasional preamp valve VERY rarely, but that was about it.

I had a Leak valve based system for a while, it was fairly old when I got it, and still sounded great - never had a valve (or anything else) replaced.

Most common failures were anode load resistors, coupling capacitors and electrolytics.

I've been repairing valve amps for over 40 years, and still do odd ones now - but really only guitar gear these days.

 

[gary350]

I found this in a book. "The primary impedance of a transformer as it appears to the source of power is determined wholly by the load connected to the secondary and by the transformer turns ratio."

Np/Ns = Square root of Zp/Zs

Np = number of turns in the primary.
Ns = number of turns in the secondary.
Zp = impedance of the primary.
Zs = impedance of the secondary.

Example.
Zp = 5000 ohms and Zs = 8 ohms.
Ratio = Square root of 5000/8 = 25
Turns ratio is 25 to 1

This means the transformer can have 25 turns on the primary and 1 turn on the secondary OR 300 turns on the primary and 12 turns on the seconday OR many other choices.

The book says, the design of the transformer has to be determined by its needs and by testing to get frequency responce, distortion, etc. that you want. It says, if a transformer is designed to work best at mid range frequencys it will not work as well on low frequencies or high frequencies. The number of turns on the primary determines the frequency and frequency range of the transformer.

===============================================================================

Lets assume you want to find the impedance of an unknown audio transformer. Put 10 VAC on the 8 ohm secondary and you get 250 VAC on the primary.

Turns ratio = Np/Ns = 250/10 = 25 which means a ratio of 25 to 1

Primary Impedance = Ratio x ratio x secondary ohms = 25 x 25 x 8 = 5000 ohms.

 

[The Electrician]

I found this in a book. "The primary impedance of a transformer as it appears to the source of power is determined wholly by the load connected to the secondary and by the transformer turns ratio."

It should be understood that this statement, without further qualification, is only true for an ideal transformer.

Np/Ns = Square root of Zp/Zs

Np = number of turns in the primary.
Ns = number of turns in the secondary.
Zp = impedance of the primary.
Zs = impedance of the secondary.

Example.
Zp = 5000 ohms and Zs = 8 ohms.
Ratio = Square root of 5000/8 = 25
Turns ratio is 25 to 1

This means the transformer can have 25 turns on the primary and 1 turn on the secondary OR 300 turns on the primary and 12 turns on the seconday OR many other choices.

The last part in red is something that ideal transformers can do without any effect on the transformer's performance, and that real transformers cannot.

The book says, the design of the transformer has to be determined by its needs and by testing to get frequency responce, distortion, etc. that you want. It says, if a transformer is designed to work best at mid range frequencys it will not work as well on low frequencies or high frequencies. The number of turns on the primary determines the frequency and frequency range of the transformer.

This contradicts the earlier statements which I pointed out only apply to ideal transformers. If those statements applied without limitation to real transformers, then why would it be true that "if a transformer is designed to work best at mid range frequencys it will not work as well on low frequencies or high frequencies. The number of turns on the primary determines the frequency and frequency range of the transformer."?

===============================================================================

Lets assume you want to find the impedance of an unknown audio transformer. Put 10 VAC on the 8 ohm secondary and you get 250 VAC on the primary.

Turns ratio = Np/Ns = 250/10 = 25 which means a ratio of 25 to 1

Primary Impedance = Ratio x ratio x secondary ohms = 25 x 25 x 8 = 5000 ohms.

It is imprecise to use the phrase "the impedance of an unknown audio transformer" in engineering discussion. In the situation you have described, what you want to find is "the apparent impedance at the primary winding with a specified impedance connected to the secondary."

As I showed in the measurements of post #12, it is not true for all load impedances that "The primary impedance of a transformer as it appears to the source of power is determined wholly by the load connected to the secondary and by the transformer turns ratio." if the transformer is a real transformer.

There is a load impedance "range", where the turns ratio squared closely determines the impedance ratio, and when the load impedances are greater or less than this optimum range, the impedance ratio is no longer as close to the squared turns ratio. The further the load impedance is from the optimum, the further the actual impedance ratio deviates from the square of the turns ratio.

You have described a situation where the rated secondary load impedance is known. In that case, the voltage ratio can be measured and the apparent primary impedance will most likely be determined by the square of the voltage ratio when the rated load impedance is connected to the secondary. We assume that the manufacturer has designed the transformer so that the optimum impedance range is the one specified. The optimum range is the range where the impedance ratio is nearly the same as the square of the voltage ratio.

But, suppose you have a completely unknown transformer? How can we determine the optimum impedance range at primary and secondary for that transformer? This is a problem not well understood by non-specialists. Many people think the DC resistance of the windings is approximately equal to the "rated" impedance of the windings. This is a topic that is probably worthy of a separate thread.

 

[gary350]

It should be understood that this statement, without further qualification, is only true for an ideal transformer.



The last part in red is something that ideal transformers can do without any effect on the transformer's performance, and that real transformers cannot.



This contradicts the earlier statements which I pointed out only apply to ideal transformers. If those statements applied without limitation to real transformers, then why would it be true that "if a transformer is designed to work best at mid range frequencys it will not work as well on low frequencies or high frequencies. The number of turns on the primary determines the frequency and frequency range of the transformer."?

The book goes into great detail explaining this I did not want to type a novel so I made it short. I will try to explain this better but you need to read the book. The number of turns on the primary determines a lot of things, you can make the transformer work better at low frequency, better a mid range or better at high frequency but you can not have your cake and eat it too. The ideal transformer would work best at any frequency but the real transformer can not do that. If there are formulas to determines the number of turns on the primary to get a certain frequency the transformer operates best at I have not read that yet. The way I understand it is you wind a transformer with 300 turns on the primary then test it to see how it works. Then build 2 more transformers one with 275 turns and another with 325 turns and test them and so on. You may have to build several more transformers and test them to find the one that suits your needs for the application. I would assume if your needs involves a lot of low frequencies then you would want the transformer to operate best at low frequency. The ideal transformer would be 3 totally seperate transformers each one designed to operate best at low, medium and high frequency with 3 seperate sets of speakers for low, medium and high frequency.

It is imprecise to use the phrase "the impedance of an unknown audio transformer" in engineering discussion. In the situation you have described, what you want to find is "the apparent impedance at the primary winding with a specified impedance connected to the secondary."

An unknown transformer is a transformer that you removed from some old junk equipment and you know nothing about it other than it is an audio transformer, you will probably also know how it was used and the tubes it was used with but the RCA and GE tube manual is not going to tell you the impedance of the transformer. Now you need to do some tests to determine what this transformer turns ratio and impedence is.

As I showed in the measurements of post #12, it is not true for all load impedances that "The primary impedance of a transformer as it appears to the source of power is determined wholly by the load connected to the secondary and by the transformer turns ratio." if the transformer is a real transformer.

There is a load impedance "range", where the turns ratio squared closely determines the impedance ratio, and when the load impedances are greater or less than this optimum range, the impedance ratio is no longer as close to the squared turns ratio. The further the load impedance is from the optimum, the further the actual impedance ratio deviates from the square of the turns ratio.

You have described a situation where the rated secondary load impedance is known. In that case, the voltage ratio can be measured and the apparent primary impedance will most likely be determined by the square of the voltage ratio when the rated load impedance is connected to the secondary. We assume that the manufacturer has designed the transformer so that the optimum impedance range is the one specified. The optimum range is the range where the impedance ratio is nearly the same as the square of the voltage ratio.

But, suppose you have a completely unknown transformer? How can we determine the optimum impedance range at primary and secondary for that transformer? This is a problem not well understood by non-specialists. Many people think the DC resistance of the windings is approximately equal to the "rated" impedance of the windings. This is a topic that is probably worthy of a separate thread.

.......

 

[The Electrician]

I said: "It is imprecise to use the phrase "the impedance of an unknown audio transformer" in engineering discussion. In the situation you have described, what you want to find is "the apparent impedance at the primary winding with a specified impedance connected to the secondary."

You responded:

An unknown transformer is a transformer that you removed from some old junk equipment and you know nothing about it other than it is an audio transformer, you will probably also know how it was used and the tubes it was used with but the RCA and GE tube manual is not going to tell you the impedance of the transformer. Now you need to do some tests to determine what this transformer turns ratio and impedence is.

You've missed my point. A transformer doesn't have an "impedance" of its own; only the windings have impedances, and any one winding only has an impedance related to the load on other windings when the other winding(s) are terminated with some impedance(s). You can't use a value (5000 ohms; 8 ohms; et cetera) and say that that single number is the "impedance" of the transformer.

For a transformer with only two windings, you can speak of the "primary impedance" or "secondary impedance", but you can't leave out the words "primary" or "secondary" and just say impedance and somehow associate a single impedance with the transformer as a whole.

And for an ideal transformer, even the windings don't have any intrinsic impedance. They only have an apparent impedance when the other winding(s) are terminated with external impedance(s).

However, a real transformer does have an optimum impedance associated with each winding such that the transformer works "best" (in some sense) when each winding is loaded or driven with that "optimum" impedance.