Imagine you've found a cool OPT at a yard sale or on eBay. It even looks like it might be an old Dynaco, but no markings are visible. The secondary appears to be a single winding with no taps. How does one determine the rated winding impedances?
This is a perennial question, and numerous descriptions of a suitable procedure can be found in this forum and on the web.
The procedure is to apply an AC voltage from a variac or a signal generator to one winding, measure that applied voltage, measure the voltage appearing at the other winding and calculate the ratio of the two voltages. The square of that turns ratio is the impedance ratio.
Usually the descriptions tell the reader that a transformer is an impedance changing device, and the windings don't have an intrinsic impedance of their own. It's only when a load is connected to the secondary that the primary then has a defined impedance. We are told that if we connect an 8 ohm load to the secondary, the primary will then exhibit an impedance equal to 8 times the impedance ratio. Sometimes it's said that we could equally well connect a 4 ohm load to the secondary and also see a primary impedance equal to 4 times the impedance ratio.
We seem to be left with the idea that the OPT will work with a load of 4 or 8 or 16 ohms. But will it work equally well in all these cases?
This thread:
http://www.diyaudio.com/forums/tube...6-ohm-output-transformer-what-difference.html
asks whether a 5k:8 ohm rated transformer would/could be physically identical to a 10k:16 ohm rated transformer? The impedance ratio is the same, so why not? After much discussion, it is concluded that an OPT works optimally between a single pair of impedances Zp:Zs. (For this purpose, "optimally" refers to the power transfer function of the transformer without considering whether the distortion may be less with some other impedances.)
The question I raise is this: the turns ratio finding procedure does tell us what the nominal impedance transformation ratio is, but it doesn't find the optimum pair of impedances; how can we find the optimum impedances?
This is a perennial question, and numerous descriptions of a suitable procedure can be found in this forum and on the web.
The procedure is to apply an AC voltage from a variac or a signal generator to one winding, measure that applied voltage, measure the voltage appearing at the other winding and calculate the ratio of the two voltages. The square of that turns ratio is the impedance ratio.
Usually the descriptions tell the reader that a transformer is an impedance changing device, and the windings don't have an intrinsic impedance of their own. It's only when a load is connected to the secondary that the primary then has a defined impedance. We are told that if we connect an 8 ohm load to the secondary, the primary will then exhibit an impedance equal to 8 times the impedance ratio. Sometimes it's said that we could equally well connect a 4 ohm load to the secondary and also see a primary impedance equal to 4 times the impedance ratio.
We seem to be left with the idea that the OPT will work with a load of 4 or 8 or 16 ohms. But will it work equally well in all these cases?
This thread:
http://www.diyaudio.com/forums/tube...6-ohm-output-transformer-what-difference.html
asks whether a 5k:8 ohm rated transformer would/could be physically identical to a 10k:16 ohm rated transformer? The impedance ratio is the same, so why not? After much discussion, it is concluded that an OPT works optimally between a single pair of impedances Zp:Zs. (For this purpose, "optimally" refers to the power transfer function of the transformer without considering whether the distortion may be less with some other impedances.)
The question I raise is this: the turns ratio finding procedure does tell us what the nominal impedance transformation ratio is, but it doesn't find the optimum pair of impedances; how can we find the optimum impedances?
Last edited:
Through test and measurement, probably with low frequencies to determine when the core goes into saturation.
What does "optimum pair" of impedances mean? Usually, you either optimize for output power or the lowest distortion, everything in between would be a compromise... So it depends on what your design goal is.
Measure the inductance of the primary under load? From that calculating the reactance at the desired frequency (probably 20Hz) and compare it to the reflected impedance. This reactance should preferably be much higher, but maybe twice the reflected impedance is ok (for a -3dB point at the desired frequency)?
just an idea.
BR, Erik
just an idea.
BR, Erik
As I said in the first post: "For this purpose, "optimally" refers to the power transfer function of the transformer without considering whether the distortion may be less with some other impedances."
The goal is to transfer power to the load with maximum efficiency.
Hi,
Maximum power throughput at any arbitrary level of distortion,
i.e. optimum source and load impedances for peak flux in the core.
rgds, sreten.
Maximum power throughput at any arbitrary level of distortion,
i.e. optimum source and load impedances for peak flux in the core.
rgds, sreten.
Here are some measurements I made of a small OPT. This OPT has a turns ratio of 24.35 @ 1kHz.
I connected various loads to the secondary ranging from 1 ohm to 128 ohms, and for each load I measured the impedance seen at the primary. I then divided the primary impedance by the secondary load impedance and took the square root. That value is the impedance ratio, and it should equal the turns ratio.
Note how it is not at all constant. It is only equal to the measured turns ratio for one particular load impedance.
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.
I connected various loads to the secondary ranging from 1 ohm to 128 ohms, and for each load I measured the impedance seen at the primary. I then divided the primary impedance by the secondary load impedance and took the square root. That value is the impedance ratio, and it should equal the turns ratio.
Note how it is not at all constant. It is only equal to the measured turns ratio for one particular load impedance.
Code:
Load Resistance Primary Impedance Square Root of impedance ratio
(measured)
1Ω 1190 Ω 34.50
2Ω 1642 Ω 28.65
4Ω 2854 Ω 26.71
8Ω 4842 Ω 24.60
16Ω 8068 Ω 22.46
32Ω 12215 Ω 19.54
64Ω 15710 Ω 15.67
128Ω 17896 Ω 11.82The 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.
How does one:
1. Measure the inductance of the primary under load?
2. For each (resistive?) load, measured the impedance seen at the primary?
I have an RLC meter. I once measured the primary inductance of a mains transformer with it and got around 12H. This is without anything connected to the primary nor secondary. 12H equals around 1K5 of Reactance at 20Hz. So if i am to use this as OPT and assuming a 2:1 ratio of reactance vs reflected impedance, then the maximum load i can connect at the secondary is one that reflects as 750R? Is this correct?
1. Measure the inductance of the primary under load?
2. For each (resistive?) load, measured the impedance seen at the primary?
I have an RLC meter. I once measured the primary inductance of a mains transformer with it and got around 12H. This is without anything connected to the primary nor secondary. 12H equals around 1K5 of Reactance at 20Hz. So if i am to use this as OPT and assuming a 2:1 ratio of reactance vs reflected impedance, then the maximum load i can connect at the secondary is one that reflects as 750R? Is this correct?
Last edited:
The ratio is only determined when measuring the voltage across unloaded windings.
This is not correct and is quite common mistake. Transformers do have a finite impedance when unloaded. The ideal transformer would have infinite and frequency independent inductance (and thus impedance) but real transformers don't.
In the attached picture you can see the primary impedance when the the secondary is open (blue), loaded with 6R (red) and shorted (green). The shorted secondary test is usually done to determine the self resonance which is that peak appearing at high frequency.
It will work but not equally well because losses will not be the same, inductance will be the same in all cases and parasitic will not change (but high frequency behaviour will change). Regarding copper losses, if you have a turn ratio of 25:1 the intended primary impedance is 5K for 8R load you will typically get some 120R primary resistance and 0.25R secondary resistance in quality transformers. Now you can connect 4R or 16 R to get 2.5K and 10K, respectively however those 120R and 0.25R will have a different weight when related to the impedance. It will be worse for 4R and better for 16R. However for 16R inductance will be very likely too low and so low frequency response will be worse. The power rating will be worse for 16R also because the saturation depends on the applied voltage and not on the impedance and so will get worse as the secondary load increases. Designing a good transformer is matter of balance among all requirements.
Attachments
You can measure the inductance of the primary under load with an LCR meter, but you won't get the same value as the unloaded inductance. The load changes the apparent inductance at the primary and you would have to do some math to extract the unloaded inductance from the value of the loaded impedance.
Most conveniently, you use a meter that directly displays the "impedance". When I gave values for the impedance in the table above, I could have more precisely called it the magnitude of the impedance.
Depending on the context, when someone refers to "impedance", they may mean the complex impedance, with a real and imaginary part; other times, as I did above, they may mean the magnitude of the impedance. You need the imaginary part of the impedance to calculate the apparent inductance, or you could start with the apparent inductance and get the imaginary part of the impedance from that.
Some handheld LCR meters don't display the magnitude of the impedance. In that case you have to use the measured inductance, multiply by 2*pi*f, where f is the measurement frequency, probably 1000 Hz; this gives you the inductive reactance, Xl. Then you use the meter's measurement of Rs (this is the ESR when measuring capacitors) and combine it vectorially with Xl (take the square root of the sum of the squares) to get the magnitude of the impedance.
You have to include the plate (or P to P) resistance in the calculation also. A bigger problem is that the inductance measured by your meter will depend on the voltage your meter applies to the primary when making the measurement. Most hand held LCR meters can only apply a single voltage when making a measurement.
Here are measurements made of a small OPT at 20 Hz with varying applied voltage:
Code:
Applied Inductance
Voltage
.5 19.3H
1 22.4H
2 26.9H
3 30.4H
4 33.4H
5 36.3H
9 48.1HIt would seem that the low frequency cutoff varies with signal amplitude. The phenomenon has been discussed in other threads.
I´d measure the primary inductance for low frequency cutoff (choose tube with appropiate rp). If single ended measure max dc current before saturation. Turns ratio (hence impedance ratio). And finally DCR to compute losses. Thats all you need to know how and where to use the transformer. (If its a critical application one should also measure leakage inductance to compute high freq rolloff)
As Bac increases -> incremental permeability increases -> hence inductance also increases until a point and then falls down to zero when the Bac is so big that saturation is reached
Sorry about that, I wasn't reading carefully and the words "optimum pair" jumped out at me... Anyway, the graph posted by 45, came from here. I believe the transformer characterization procedure used was similar to those used by Yves Monmagnon and Rod Elliott.
I communicated with Rod and Yves about this. Rod said "I haven't performed impedance plots on output transformers." Yves hasn't responded to my query. I would like very much to read whatever Yves may have done, but in the meantime I will go on with my post.
45 said:
This is quite true, but it is often said, even by experts, that transformers don't have any intrinsic impedance; for example, in Jensen Transformers app note AN002 by Bill Whitlock:
I'm sure that he knows, as you do and as I do, that real transformer windings do have a finite impedance. But the important thing he wants to convey is that it is the impedance transforming capability of transformers that matters. Whatever impedance a given winding has with no load on the other winding(s) is incidental most of the time, but not always. It's that "not always" part that I'm going to discuss.
Much can be learned from impedance plots of transformer windings.
Here is a plot of the primary impedance (magnitude) of an Edcor WSM600/600 transformer with the secondary open and then shorted. This is a line transformer with a 1 to 1 turns ratio, rated for 600 ohm impedances. The plot is logarithmic in both x and y directions, with 10 Hz at the left (the marker A at about 1 kHz), 1 MHz at the right, 10 ohms at the bottom, and 100k ohms at the top:
The upper curve is the impedance with the secondary open circuited, and the lower curve is with the secondary shorted.
The important thing to recognize right away is that no matter what load (excluding capacitors) is connected to the secondary, the reflected primary impedance curve must be below the upper curve in the above image, and above the lower curve.
It is the parasitics, core losses, copper losses, distributed capacitance, finite inductance, etc., that result in the boundaries represented by the upper and lower curves in the impedance plot above.
Attachments
Here's another impedance sweep, with the rated 600 ohm load connected to the secondary. The middle curve is the impedance reflected to the primary. We see that the reflected impedance at 1 kHz is about 720 ohms, which is reasonably close to the desired 600 ohms:
I'll have some more to say about this transformer.
I'll have some more to say about this transformer.
Attachments
- Status
- Not open for further replies.