Hi guys, I always wondered how the insertion loss of an output transformer is measured by the manufacturers.
Based on definition insertion loss of any network is the power which is lost when it is inserted in the signal chain; so based on that, we should measure OPT's insertion loss by applying a sine wave (lets say 1KHz - 100V) to the primary and measure the secondary voltage one time without load and one time with the nominal load, and then divide the unloaded voltage by the loaded voltage and calculate it in dBs, am I correct?
BTW I found in some books that the formula is such this:
Insertion Loss = 20 * log (V unloaded/ V loaded),
And in some other place the formula is like this:
Insertion Loss = 10 * log (V unloaded/ V loaded)
So what do you think about it? Which one of these two formulas can be correct?
My measurement setup is like this:
Assume the DUT is a 5K to 8 ohms OPT for UL 6L6 single ended amp; with the load on primary side disconnected I adjusted 4.04V on the secondary side and I get 100V on the primary; after that with the nominal Zp connected to primary I get 100V on the primary side again by applying 4.26V on the secondary side. So how I calculate insertion loss for this transformer?
Any help would be appreciated.
Based on definition insertion loss of any network is the power which is lost when it is inserted in the signal chain; so based on that, we should measure OPT's insertion loss by applying a sine wave (lets say 1KHz - 100V) to the primary and measure the secondary voltage one time without load and one time with the nominal load, and then divide the unloaded voltage by the loaded voltage and calculate it in dBs, am I correct?
BTW I found in some books that the formula is such this:
Insertion Loss = 20 * log (V unloaded/ V loaded),
And in some other place the formula is like this:
Insertion Loss = 10 * log (V unloaded/ V loaded)
So what do you think about it? Which one of these two formulas can be correct?
My measurement setup is like this:
Assume the DUT is a 5K to 8 ohms OPT for UL 6L6 single ended amp; with the load on primary side disconnected I adjusted 4.04V on the secondary side and I get 100V on the primary; after that with the nominal Zp connected to primary I get 100V on the primary side again by applying 4.26V on the secondary side. So how I calculate insertion loss for this transformer?
Any help would be appreciated.
Thank you, but which one used to report the insertion loss of OPT by manufacturers?
Second equation is not correct - should be: Insertion Loss = 10 * log (P unloaded/ P loaded) (P: Power) Then the two equations are equivalent.
And this measurement does not account for core loss (though it's minimal at mid frequencies). For radio frequency transformers, I measured two transformers back-to-back vs. a straight through connections (divided by two, of course).
And this measurement does not account for core loss (though it's minimal at mid frequencies). For radio frequency transformers, I measured two transformers back-to-back vs. a straight through connections (divided by two, of course).
Thank you so much, that's what I was looking for. Another question if I may; I've seen some people use percent for copper losses, how that can be converted to dBs?
Percent loss is power, so 10 log (1-loss). (1-loss) is efficiency, another way to look at it.
My rule-of-thumb Approximation for a medium insertion loss transformer, measured at 1kHz, is mostly due to the primary DCR, and the secondary DCR.
The following is not exact, but close enough . . .
A 5k primary, with 250 Ohms DCR, and an 8 Ohm secondary with 0.4 Ohms DCR:
There is 0.5 dB loss in the primary DCR, and 0.5 dB loss in the secondary DCR.
That amounts to 1 dB total insertion loss (please remember, I said @ 1kHz, Not @ 20Hz, and Not @ 20kHz).
For an approximation that works well at 1kHz, consider the winding impedance to be in series with the DCR;
That is a voltage divider.
(do not worry about the phase; it almost always is zero degrees at somewhere between 500Hz and 2kHz).
5000 Ohms of winding, in series with 250 Ohms of DCR: I call that a voltage divider
20 Log (5000/(5000 + 250)), 20 Log (5000/5250) = 0.42 dB loss of the primary . . Oh, my estimate was 0.5 dB, do you care about the 0.1 db error?
Now, do a repeat measurement and calculation for the secondary.
Add the primary and secondary dB losses . . . you are "In Like Flint".
If you experience insertion loss @ 1kHz, that is not Predominately due to the primary DCR, and the secondary DCR, then . . .
You need to get a transformer with better lamination material, or get a transformer with a better winding technique, Or Both.
YTMV (Your Transformer May Vary).
$0.03
Adjusted for inflation
The following is not exact, but close enough . . .
A 5k primary, with 250 Ohms DCR, and an 8 Ohm secondary with 0.4 Ohms DCR:
There is 0.5 dB loss in the primary DCR, and 0.5 dB loss in the secondary DCR.
That amounts to 1 dB total insertion loss (please remember, I said @ 1kHz, Not @ 20Hz, and Not @ 20kHz).
For an approximation that works well at 1kHz, consider the winding impedance to be in series with the DCR;
That is a voltage divider.
(do not worry about the phase; it almost always is zero degrees at somewhere between 500Hz and 2kHz).
5000 Ohms of winding, in series with 250 Ohms of DCR: I call that a voltage divider
20 Log (5000/(5000 + 250)), 20 Log (5000/5250) = 0.42 dB loss of the primary . . Oh, my estimate was 0.5 dB, do you care about the 0.1 db error?
Now, do a repeat measurement and calculation for the secondary.
Add the primary and secondary dB losses . . . you are "In Like Flint".
If you experience insertion loss @ 1kHz, that is not Predominately due to the primary DCR, and the secondary DCR, then . . .
You need to get a transformer with better lamination material, or get a transformer with a better winding technique, Or Both.
YTMV (Your Transformer May Vary).
$0.03
Adjusted for inflation
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The first part of measuring the insertion loss at 1kHz, is to measure the turns ratio.
Drive the primary with a 1kHz low impedance signal source (much lower than the expected primary impedance).
Leave the secondary Un-loaded.
Measure the ratio of the primary voltage / secondary voltage, Vp/Vs
Pick a secondary output impedance that the transformer was designed for, 8 Ohms for example.
Impedance is according to the square of the turns ratio.
Suppose the turns ration is 25:1
(Vp/Vs) Squared x 8 Ohms = primary impedance.
(25) squared x 8 = 5000 Ohms.
See how that works?
Because we did not load the secondary, and used the midband frequency of 1kHz, then . . .
almost all of the effect of the primary DCR and secondary DCR drops out of the measurement of turns ratio (the resulting turns ratio will be close enough).
Now, continue to my Post # 8.
Drive the primary with a 1kHz low impedance signal source (much lower than the expected primary impedance).
Leave the secondary Un-loaded.
Measure the ratio of the primary voltage / secondary voltage, Vp/Vs
Pick a secondary output impedance that the transformer was designed for, 8 Ohms for example.
Impedance is according to the square of the turns ratio.
Suppose the turns ration is 25:1
(Vp/Vs) Squared x 8 Ohms = primary impedance.
(25) squared x 8 = 5000 Ohms.
See how that works?
Because we did not load the secondary, and used the midband frequency of 1kHz, then . . .
almost all of the effect of the primary DCR and secondary DCR drops out of the measurement of turns ratio (the resulting turns ratio will be close enough).
Now, continue to my Post # 8.