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Hi kavkav,
Thanks for the measurements and of course it is not a problem you did not use a capacitor to tune out the inductive reactance of the secondary coil(s). I suggest it doing though when you have some more time because you have the chance for receiving the maximum output power possible and this goes together with Lenz manifesting at its worst effect (at least in a conventional setup and this was the main reason I suggested using a cap).
Trying to evaluate your data, first I assume the current in the second test was 216 to 218 uA (microAmper) into the primary coil and the milliAmper you wrote is but typo, right? (This is because I think it is unlikely you have 105 uA current at 1.1V RMS input to the primary in the first test and then you lower the input voltage to about 228 mV RMS and then the input current goes up to the two hundreds mA range.)
So your first test shows that loading one of the secondaries reflects on the primary input power in a negligible way and you found this same behaviour in your second test too.
Regarding the data in the second test, and trying to arrive at an output/input power ratio, the problem is that we do not know the phase angle at 385 Hz between the input current and voltage of the primary coil. Just multiplying current and voltage surely gives the worst case i.e. the highest input power mathematically because it would miss a multiplier in the formula, the cosine of phase angle. Thane Heins has always stressed to maintain a near 90 degree phase angle difference between the input current and voltage and if you use cos89° in the formula to multiply the product of the input current and voltage, mathematically you surely get a much lower input power draw from the formula, see this when using your 2nd test data: P=0.228V*0.000216A = 0.000049248W i.e. 49.25 uW when you disregard the phase angle and here is when you achieve say a 89° phase difference (cos89°=0.01745) between I and V: P=0.228V*0.000216A*0.01745 = 0.859 uW this latter is indeed a tiny input power in this calculation but it gives the effective power V*I*cos(theta) where theta is the phase angle.
However if you consider the so-called reactive power due to the fact that the primary coil of such transformer represents an inductive load, then the formula is V*I*sin(theta) i.e. the sine of 89° is 0.999 so reactive power Q=0.228*0.000216*0.999 = 49.19 uvar (var is volt-amper-reactive, I refer to this link on reactive power: Volt-ampere reactive - Wikipedia, the free encyclopedia )
Now let's compare any of these input powers to the output power dissipated in your 10 Ohm resistor. Output power is (0.011V*0.011V)/10 Ohm = 0.0000121 W = 12.1 uW here the voltage and current is in phase because of the resistive load, and I assume your power resistor is not wire-wound.
So it seems that in case we consider only the example of the 89° phase difference (that we are supposed to maintain) between the input current and voltage, we get (mathematically) a power gain of 12.1uW/0.859uW = 14 for the out/in ratio which sounds extremely good. When we do not consider the phase angle (i.e. we assume the input current and voltage into the primary are in phase), then there is a power loss (instead of gain): 12.1uW/49.25uW = 0.245
(In fact, if you study the "power triangle" in the Figure of this link: AC power - Wikipedia, the free encyclopedia you can calculate the complex power S which is consumed by the primary, using the Phitagoras formula for the right triangle where P and Q are already known when the phase angle is known that is.)
Question is whether we can neglect the reactive power input to the transformer primary or not? I do not think we can because the var power must also be supplied by our power source, in your case by the signal generator or the amplifier, in other cases by the AC mains.
I am puzzled by your third test: did you mean the secondary current was 0.835 mA ie you used an AC ampermeter for doing the actual short? You may wish to repeat this test because it would be good info how this transformer behaves at higher input AC voltages like a few volts or even at some 10 volts if possible.
I cannot answer whether you go on building another transformer (a prototype as you say), this must be decided by you, sorry. What I wrote above as an evaluation of your data comes from conventional science knowledge, and it is possible that in many countries the electric meter does not register var power...
Greetings, Gyula
Thanks for the measurements and of course it is not a problem you did not use a capacitor to tune out the inductive reactance of the secondary coil(s). I suggest it doing though when you have some more time because you have the chance for receiving the maximum output power possible and this goes together with Lenz manifesting at its worst effect (at least in a conventional setup and this was the main reason I suggested using a cap).
Trying to evaluate your data, first I assume the current in the second test was 216 to 218 uA (microAmper) into the primary coil and the milliAmper you wrote is but typo, right? (This is because I think it is unlikely you have 105 uA current at 1.1V RMS input to the primary in the first test and then you lower the input voltage to about 228 mV RMS and then the input current goes up to the two hundreds mA range.)
So your first test shows that loading one of the secondaries reflects on the primary input power in a negligible way and you found this same behaviour in your second test too.
Regarding the data in the second test, and trying to arrive at an output/input power ratio, the problem is that we do not know the phase angle at 385 Hz between the input current and voltage of the primary coil. Just multiplying current and voltage surely gives the worst case i.e. the highest input power mathematically because it would miss a multiplier in the formula, the cosine of phase angle. Thane Heins has always stressed to maintain a near 90 degree phase angle difference between the input current and voltage and if you use cos89° in the formula to multiply the product of the input current and voltage, mathematically you surely get a much lower input power draw from the formula, see this when using your 2nd test data: P=0.228V*0.000216A = 0.000049248W i.e. 49.25 uW when you disregard the phase angle and here is when you achieve say a 89° phase difference (cos89°=0.01745) between I and V: P=0.228V*0.000216A*0.01745 = 0.859 uW this latter is indeed a tiny input power in this calculation but it gives the effective power V*I*cos(theta) where theta is the phase angle.
However if you consider the so-called reactive power due to the fact that the primary coil of such transformer represents an inductive load, then the formula is V*I*sin(theta) i.e. the sine of 89° is 0.999 so reactive power Q=0.228*0.000216*0.999 = 49.19 uvar (var is volt-amper-reactive, I refer to this link on reactive power: Volt-ampere reactive - Wikipedia, the free encyclopedia )
Now let's compare any of these input powers to the output power dissipated in your 10 Ohm resistor. Output power is (0.011V*0.011V)/10 Ohm = 0.0000121 W = 12.1 uW here the voltage and current is in phase because of the resistive load, and I assume your power resistor is not wire-wound.
So it seems that in case we consider only the example of the 89° phase difference (that we are supposed to maintain) between the input current and voltage, we get (mathematically) a power gain of 12.1uW/0.859uW = 14 for the out/in ratio which sounds extremely good. When we do not consider the phase angle (i.e. we assume the input current and voltage into the primary are in phase), then there is a power loss (instead of gain): 12.1uW/49.25uW = 0.245
(In fact, if you study the "power triangle" in the Figure of this link: AC power - Wikipedia, the free encyclopedia you can calculate the complex power S which is consumed by the primary, using the Phitagoras formula for the right triangle where P and Q are already known when the phase angle is known that is.)
Question is whether we can neglect the reactive power input to the transformer primary or not? I do not think we can because the var power must also be supplied by our power source, in your case by the signal generator or the amplifier, in other cases by the AC mains.
I am puzzled by your third test: did you mean the secondary current was 0.835 mA ie you used an AC ampermeter for doing the actual short? You may wish to repeat this test because it would be good info how this transformer behaves at higher input AC voltages like a few volts or even at some 10 volts if possible.
I cannot answer whether you go on building another transformer (a prototype as you say), this must be decided by you, sorry. What I wrote above as an evaluation of your data comes from conventional science knowledge, and it is possible that in many countries the electric meter does not register var power...
Greetings, Gyula
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