Transformer Measurements 2

[The Electrician]

This will be a thread about impedances and audio transformers, particularly vacuum tube output transformers.
I'm going to treat the transformers as two-ports: https://en.wikipedia.org/wiki/Two-port_network
That page is full of high-powered math, but it's not necessary to get very deep into it.

Most modern methods of connecting audio components use bridging rather than impedance matching, but OPTs are one thing where impedance matching is necessary, and therefore interesting. In this case the goal is to achieve a simultaneous impedance match at both input and output. This leads to an interesting situation. I'm going to illustrate this with a very simple two-port consisting of two resistors; it's just a simple L pad with no particular purpose other than serving as an example:


This is shown using the usual convention of input on the left and output on the right. We can easily calculate the impedances of each port with nothing connected to the other port (the self impedances). The input self impedance is 900 ohms and the output self impedance is 500 ohms. The goal here is to have the two-port connected between a signal source on the left having some source impedance, and a load impedance on the right such that there is a simultaneous impedance match at both ports.

Let's start by connecting a 500 ohm load to the output port. Since the output impedance of the two-port is 500 ohms to start with, we have an impedance match.
Now on to the input port. We know that the input self impedance is 900 ohms we might think that a 900 ohm source impedance signal source would give us the desired match. But no! Connecting the 500 ohm load to the output has changed the input impedance of the two-port to 650 ohms (500 ohm load in parallel with the 500 ohm internal resistor giving 250 ohms, then in series with the 400 ohm internal resistor). OK, so connect a signal source of 650 ohm impedance to the input and we will have a match.

So, first we made a match at the output port and now we have a match at the input port; that's a simultaneous match at both ports, right? Wrong! The connection of a 650 ohm source impedance at the input port changed the output impedance of the two-port. The output impedance is no longer the 500 ohm self impedance; it's now 338.71 ohms and the 500 ohm load no longer provides a match. OK, then change the load to 338.71 ohms and that will restore the output match. But just like already happened at the input, changing the load impedance also changes the input impedance and we lose the match we just had at the input!

This is discouraging; does it never end? Is it not possible to achieve a simultaneous match? There is a light at the end of the tunnel. Continue this back and forth for many times and here are the various values of impedance that provide a temporary match at each step:

Output impedance 500
Input impedance 650
Output impedance 338.71
Input impedance 601.92
Output impedance 333.55
Input impedance 600.08
Output impedance 333.34
Input impedance 600.003
Output impedance 333.334
Input impedance 600
Output impedance 333.333

We're in luck! The process converges! So if we apply a signal source of 600 ohms impedance to the input and a load of 333.333... ohms to the output will we have a simultaneous match at both ports? Yes!

Let's check it out. The 333.333 output load in parallel with the internal 500 ohm resistor gives an equivalent resistance of 200 ohms, which in series with the internal 400 ohm resistor gives 600 ohms for the input impedance; that's equal to the signal source impedance of 600 ohms. Oh happy day so far!
The input source impedance of 600 ohms is in series with the internal 400 ohms, giving 1000 ohms. That combination is in parallel with the internal 500 ohm resistor, giving 333.333 ohms, which is equal to the load impedance of 333.333 ohms.

We have a simultaneous match to the two-port!

How does this apply to audio transformers? Stay tuned!

 

[The Electrician]

Certain types of two-ports have a property that is useful. A two-port that is passive and bilateral AND has both series losses and shunt losses will have an associated pair of finite impedances such that when the given two-port is operated between those impedances (one of the impedances being the impedance of a signal source applied to the input, the other being a load impedance), they will produce a simultaneous impedance match to the two-port. Those impedances are unique; there is only one such pair of impedances. In addition, they will allow power to be transferred to the load with the smallest loss of any pair of source and load impedances.

"Bilateral" means that the two-port has the property that an impedance connected to one of the ports will cause a change in the measured impedance at the other port. Bill Whitlock described it this way as it applies to transformers:

...transformers can simultaneously reflect two different impedances. One is the impedance of the driving source, as seen from the secondary, and the other is the impedance of the load, as seen from the primary.

Audio transfomers are perfect candidates for treatment as two-ports; they are passive and bilateral and they have both series (from the wire) and shunt (from the core) losses. Thus there will be a pair of impedances between which a transformer can operate with minimum loss.

There is a 20 year old thread What determines transformer impedance?

A lot of what I'm going to discuss is in that thread, but I'm going into more detail.

There can be many things that a user considers when deciding what quality of a transformer they value, and operating between different impedances can emphasize some such things. Some qualities are lowest distortion, maximum bandwidth, high signal level capability, etc. Some of these are subjective, some are not.

One objective quality is loss; for OPTs, we don't want to overheat the transformer, so minimum loss is a good thing.

A good design that emphasizes some quality other than minimum loss will start with near minimum loss impedances and then tweak the desired quality by adjusting the operating impedances. I've measured a lot of transformers and I often find that their impedance specs are not the impedances that give minimum loss. But usually the amount of tweaking to emphasize some quality of the best transformers only incurs extra loss of maybe 1 db or less, so it's a worthwhile tradeoff. If a transformer is top quality and does all the right stuff to achieve a low loss design, the operating impedance tweaking doesn't incur much more loss. In other words, the lowest loss a design is to start with, the further away it can be operated from its lowest loss impedances without much loss penalty.

Manufacturers build transformers to operate in ways their customers want, and they will have decided what impedance spec gets that result. The manufacturer impedance spec is often not the exact pair of minimum loss impedances.

Given an (possibly unknown) audio transformer, how can we determine the minimum loss impedances?

Stay tuned.

 

[The Electrician]

Consider the nature of an ideal audio transformer. It would have a core made of the finest unobtanium with infinite permeability and no losses. The windings would be wound with superconducting wire. The self inductance of each winding would be infinity. To get a turns ratio of, say, 10:1 we would only need 10 turns on the primary and one turn on the secondary.

Of course, a real transformer has losses in the copper wire, and hysteresis and eddy current losses.

In the second post of this 20 year old thread https://groupdiy.com/threads/transformer-z.1119/

The impedance ratio is never exactly the square of the turns ratio, because of losses. But losses at nominal impedance and frequency range are usually small, so the "error" is negligible.

This is quite true except for one exception. There is one particular load which when applied to the secondary of an OPT will result in a primary impedance whose ratio to the secondary load is exactly equal to the square of the turns ratio!

Here are some measurements I made of a small Bogen 70.7 volt transformer. The output winding is rated for 8 ohms and the 1 watt tap has a measured turns ratio of 24.40 @ 1kHz.

I connected various loads to the secondary ranging from 1 ohm to 64 ohms, and for each load I measured the impedance seen at the primary (1 watt tap). I then divided the measured primary impedance by the secondary load impedance and took the square root. That square root of the impedance ratio should equal the turns ratio.

Note how it is not at all constant. It is only equal to the turns ratio for one particular load impedance, 8.927 ohms.

Load Resistance          Primary Impedance               Square Root of
                            (measured)                    impedance ratio
      1 ohms                 1180 ohms                       34.35
      2 ohms                 1743 ohms                       29.52
      4 ohms                 2841 ohms                       26.65
      8 ohms                 4876 ohms                       24.69
   8.927 ohms                5315.8 ohms                     24.40
     16 ohms                 8268 ohms                       22.73
     32 ohms                 12770 ohms                      19.98
     64 ohms                 16913 ohms                      16.26

The result for 8.927 ohms is interpolated.
The OPT does transform the load impedance to some other value at the primary side, but it's not what one would calculate from the square of the turns ratio except for one particular load.

Notice how, starting at the 1 ohm load, the square root of the impedance ratio is much greater that the turns ratio of 24.40, and that square root is decreasing as the load increases. At the largest load the square root of the impedance ratio is much smaller than 24.40. Somewhere along the way the value of the square root must have been 24.40, and we see that with a load of 8.927 ohms the square root of the impedance ratio is exactly equal to the turns ratio.

These two impedances, 8.927 ohms and 5315.8 ohms have the property that the transformer transforms either one into the other. If we apply an impedance of 5315.8 to the primary and measure the impedance at the secondary, it will be 8.927 ohms. And applying an impedance of 8.927 ohms to the secondary will give a measured impedance at the primary of 5315.8 ohms.

Let's try this with an impedance of 8268 ohms connected to the primary. We get a measured impedance of 12.52 ohms, not the 16 ohms we expected. For all other load impedances than 8.927 ohms the square root of the impedance ratio is not 24.40.

The two impedances, 8.927 ohms and 5315.8 ohms are the "magic" impedances we're looking for. They give the lowest loss through the transformer. For this Bogen transformer the manufacturer was probably aiming for a spec of 5000:8 ohms for the 1 watt tap I measured.

This is a way to determine the "optimum" (lowest loss) impedances to operate the transformer between. Make a series of measurements and find the load which gives an impedance ratio equal to the square of the turns ratio. Of course there can be other criteria than lowest loss, and they may require operating between other impedances. Because of these other criteria the manufacturer's spec impedances may not be the lowest loss. Some transformers I've measured have specs that are some distance away from minimum loss, but the loss penalty is usually not much more than a db or so.

Finally, there is the question from that 20 year old thread, what determines transformer impedance? If by "transformer impedance" we mean impedances to operate betweeen for minimum loss, the answer is, the ratio of the transformer two-port model's series losses to the shunt losses. It's that simple.

There's a better way to find the lowest loss impedances. Stay tuned.

 

[The Electrician]

I have some older measurements I made on an older Bogen from 10 years ago still on my hard drive. I'm going to use these numbers in what follows; for these values I also measured the efficiency/loss.

Anybody who wants to duplicate my procedures will need an LCR meter that can measure impedance magnitude |Z| at 1 kHz. There are some fairly low cost LCR meters for $100 to $300 that can do this. I will make my measurements at 1 kHz unless specified differently. This transformer has a measured turns ratio of 24.6 which gives a theoretical impedance ratio of 605. I connected a 1 ohm resistor to the secondary and measured the reflected impedance at the primary, and I also measured the Effeciency/Loss. Next I doubled the secondary load to 2 ohms and repeated the measurements. I doubled the secondary load to 4 ohms, measured, and continued this until a last measurement was made with a secondary load of 128 ohms. Here are the results:

Load Resistance     Primary Impedance    Impedance ratio     Efficiency/Loss
                         (measured)
      1                     1190               1190          .498 (3.03 dB)
      2                     1642                821          .654 (1.84 dB)
      4                     2854                713          .768 (1.15 dB)
      8                     4876                609          .821 (.859 dB)
     16                     8068                504          .810 (.917 dB)
     32                     12215               382          .735 (1.34 dB)
     64                     15710               245          .604 (2.19 dB)
    128                     17896               140          .440 (3.56 dB)

There is much to say and learn about these measurements. If the transformer were ideal the measured impedance ratio would be 605 for every load resistance, but for this real transformer the impedance ratio is only ~605 for one of the eight load resistances.

Just from the theoretical impedance ratio of 605 with a load of 2 ohms we would expect the primary reflected impedance to be 1210 ohms, but its measured value is 1642 ohms. The measured impedance is quite a bit higher than the theoretical value, as we would expect if the resistance of the windings adds to the measured reflected impedance.

But apparently there's more to it than that. Consider the measurement with a secondary load of 32 ohms. The theoretical primary reflected impedance would be 19360 ohms (32*605), but in fact its measured value is 12215 ohms. This is considerably less than the theoretical value. Why didn't the resistance of the windings add to the theoretical value? This result is explained by the theory of two-ports.

Consider the 8 ohm load result. The value of the reflected primary impedance is almost exactly what it should be. The winding resistance has no effect on the reflected impedance in this case.

Look at the effeciency/loss measurements; the lowest loss occurs when the secondary load is ~8 ohms and the primary load is ~4876 ohms. There's something special about these impedances.

These are the impedances we are looking for. There is a 1925 book "Transmission Circuits for Telephonic Communication" wherein Bell Telephone engineers develop the theory of what I'll show. The Engineers at Bell Telephone Laboratories called these the "Image Impedances" or "Characteristic Impedances". They preferred the term Image Impedance.

In the audio world, if you want to match a source impedance to a load impedance you would like the impedance match to occur at both the source and load interface. You would like the source impedance which is connected to the primary to reflect to an impedance at the secondary whose value is equal to the load impedance. And the other way around; you would like the load impedance connected to the secondary to reflect to an impedance at the primary equal to the source impedance. In other words, you would like a simultaneous match at both windings.

If an impedance of ~4876 ohms is connected to the primary of this transformer, it reflects to an impedance of ~8 ohms at the secondary, and if an impedance of ~8 ohms is connected to the secondary, it reflects to an impedance of ~4876 ohms at the primary. These two impedances are the image impedances of this transformer.

If this transformer is used to connect a source impedance to a load impedance, the signal transferred from source to load will experience the lowest loss possible when this transformer is connected between source and load impedances that are equal to the transformer's image/characteristic impedances.

For any two winding audio transformer there will always be a pair of impedances (and only one such pair) between which the transformer will give lowest loss.

 

[The Electrician]

How do we find the image impedances? One way would be to take a guess of what the likely lowest loss impedances would be, and make a series of measurements bracketing the likely pair as I did above.. Look for the lowest loss pair.

Another way to do it is what I call the double reflection method. Consider the 2 ohm load example in the chart above. The 2 ohm load at the secondary reflects to 1642 ohms at the primary. Disconnect the 2 ohm from the secondary and connect 1642 ohms to the primary; it will reflect to 3.6 ohms at the secondary, which is not the 2 ohms we started with. But an 8 ohm load will reflect to 4876 ohms at the primary. Connect 4876 ohms to the primary and it will reflect to 8 ohms at the secondary, which is what we started with! Only two impedances will satisfy the double reflection test from a starting impedance connected to the secondary, to a reflected impedance at the primary, and back to the starting impedance at the secondary; those two impedances are the image impedances.

It's trial and error with this method because you have to guess a secondary impedance to start with, and keep trying different values until you get the one that satisfies the double reflection test.

Fortunately the Engineers at Bell Labs have given us a method that goes right to the image impedances.

Designate the windings of a two-winding transformer as winding 1 and winding 2. Let the impedance Z1O be the impedance measured at winding 1 when winding 2 is open circuited. Let Z1S be the impedance measured at winding 1when winding 2 is short circuited. Now calculate Z1C = SQRT(Z1O * Z1S); this is the image impedance of winding 1.

Let the impedance Z2O be the impedance measured at winding 2 when winding 1 is open circuited. Letet Z2S be the impedance measured at winding 2 when winding 1 is short circuited. Now calculate Z2C = SQRT(Z2O * Z2S); this is the image impedance of winding 2.

We now have the image impedances of the transformer. Even if a person wanted to optimize for some other property (like low distortion) by connecting the transformer between other mpedances, the image/characteristic impedances would be a good starting point.

The image/characteristic impedances, once found, exhibit an interesting property. Their ratio is equal to the square of the turns ratio, so finding them gives you the turns ratio without explicitly measuring it!

 

[The Electrician]

I should make it quite clear that measurements of the parameters of a device such as an audio transformer having a ferromagnetic core cannot be made to great accuracy; getting to within 10% for a particular measurement setup is good. Another setup could easily be 10% or more away.

The Bogen transformer whose measurements are earlier in the thread have these image impedances derived from these measurements:

The value of Z1O (primary impedance with secondary open) is 21956 ohms.
The value of Z1S (primary impedance with secondary shorted) is 620 ohms.
The primary image impedance is therefore Z1C = SQRT(21956 * 620) = 3689 ohms

The value of Z2O (secondary impedance with primary open) is 37 ohms
The value of Z2S (secondary impedance with primary shorted) is 1.04 ohms
The secondary image impedance is therefore Z2C = SQRT(37 * 1.04) = 6.2 ohms

You would think that this transformer should be specified as 3700:6 ohms. These should be the impedances that give lowest loss, but they aren't as I will explain.

The manufacturer's description specifies this transformer as a 5000:8 ohm 70.7 volt transformer. These impedances are very close to the lowest loss impedances which are 5315:8.9 ohms.

The explanation for these results lies in the very nature of impedance. The maximum power transfer theorem says that matching a source impedance to a load impedance gives maximum power transfer. (We don't always want to transfer maximum power when the source is a very low impedance and it would cause damage to match a load to that very low impedance).

When alternating currents and voltages are involved, such as audio, the full description of the impedances involved requires the use of complex numbers. A full specified impedance would be of the form Z = R + jX, where R is the resistive (real) part of the impedance, and X is the reactive (imaginary) part of the impedance. Often when just the word "impedance" is used, the magnitude of the impedance is meant: |z| = SQRT(R*R + X*X); I use this a lot of the time myself when it's appropriate.

When a source impedance is complex and has a non-zero reactive part, the maximum power transfer does not occur when the load has the same complex impedance. In that case, maximum power transfer occurs when the load impedance is the complex conjugate of the source; this is done in the RF world at a single frequency (or very narrow band). But it's not feasible in audio to provide a load that is the conjugate of the source impedance over the wide bandwidth that audio frequencies occupy.

In the case where source and load impedance are actually complex (have a non-zero reactive part) and a complex conjugate load is infeasible, the maximum power transfer occurs when the impedance magnitude |Z| of the load matches the |Z| of the source.

The reason that the 3689 : 6.2 ohm impedances don't provide maximum power transfer is that they are not the actual complex image impedances; they are the image impedance magnitudes. The actual complex image impedances are Z1C = (2560 + j2657) and Z2C = (4.295 + j4.466). The reactive parts of these impedances are nearly the same magnitude as the real parts. This complicates things.

For this transformer the two resistances 5315 and 8.927 are the two pure real (not having a complex value) practical image impedances that provide the lowest loss of all pure real impedance pairs. I'll say more about how to calculate these later.

I hope some people who have LCR meters will try this method to get the image impedances for a transformer and post their results.

 

[The Electrician]

As an example, here's the derivation of the image impedances for the low cost eBay OPT I've referenced in another thread.

Using only a hand held LCR meter, the DE5000 (which can't measure Z directly), measure the open circuit primary. At 1 kHz get the values of Rs(series mode) and Q. We use the fact that X = Rs*Q

Calculate Zpri with the secondary open circuited.
I get Rs = 7620 and Q = 9.76
Calculate Zpri(open) = Sqrt(Rs^2 and (Rs*Q)^2) This gives Zpri(open) = 74760

Now the Zpri with the secondary shorted.
I get Rs = 868.4 and Q = .305
Calculate Zpri(short) = Sqrt(Rs^2 and (Rs*Q)^2) This gives Zpri(short) = 907.9

Calculate Sqrt(74760*907.9) = 8238.6 This is the primary image impedance.

Calculate Zsec with the primary open circuited.
I get Rs = 36.7 and Q = 3.64
Calculate Zsec(open) = Sqrt(Rs^2 and (Rs*Q)^2) This gives Zsec(open) = 138.5

Now the Zsec with the primary shorted.
I get Rs = 1.108 and Q = .299
Calculate Zsec(short) = Sqrt(Rs^2 and (Rs*Q)^2) This gives Zsec(short) = 1.156

Calculate Sqrt(138.5 * 1.156) = 12.65 This is the secondary image impedance.

The primary and secondary image impedances are 8238.6 and 12.65 so the impedance spec for minimum loss could be 8000:12 ohms taking into account the impossibility of getting results better than 10% or maybe even 20% for a transformer with an iron core.

This OPT may have been intended as a compromise to give good results with either 8 or 16 ohm speakers.

 

[john12ax7]

Having a database of equivalent two port networks for various transformers would be very useful. Do any manufacturers publish this?

 

[The Electrician]

Having a database of equivalent two port networks for various transformers would be very useful. Do any manufacturers publish this?

I am unaware of any such data published by a manufacturer. I haven't even seen data for a simple transformer model. Has anyone else seen anything like this?

 

[Wavelength]

This is great info thanks!
I worked with Mike LaFevre who owned Peerless/Altec, MagneQuest had IP from Dynaco, WE, Triad and others. Mike and I worked for close to 30 years together I still have probably 300 of his transformers in stock since his passing last summer.
Mike use to tell me a simple method for determining the quality of a transformer for single ended outputs and chokes/inductors was R*Idc < 20V. Being a EE I listened to a lot of stuff and we made some pretty cool stuff over the years.
I have some 5K:8 single ended outputs (no DC) would with %99.999 pure silver, teflon insulation and pure cobalt cores. Crazy stuff but audio people like the best. Since these require a loading choke what we call parallel feed which requires a cap coupled output the choke for this circuit was 60ma/120H had a voltage drop of 9V. They sound good but really only like 6W max output. I use a VT52, 50 or other small directly heated triodes with these.

I have always wondered about the varying impedance of a speaker and how it effects the primary impedance. I use variovents on my sealed speakers to lower the Z at Fs because tube amps really do not like to see those huge impedance hump.

Looking forward to more info. If you want I have full lab setup and even access to some drawings in the peerless catalog.
Thanks,
Gordon