Some tests on mono and bifilar coils

[conradelektro]

@MileHigh: good that you made me check the numbers, the self resonance frequencies were right (see the attached drawing, I checked with two methods) but I made an error when writing down the self capacitance numbers.

It is ~50 pF (not 5 pF) for the new monofilar coil.
And it is ~6 nF (was right) for the new bifilar coil.

The difference is still a factor 100.


With the pan cake coils the difference was a factor 5:

Self capacitance of monofilar pan cake coil 9 pF  -- calculated from 9 MHz and 34 µF
Self capacitance of bifilar pan cake coil 46 pF -- - calculated from 4 MHz and 34 µF


This can may be explained by the fact that the pan cake coils hat 32 turns and the new helical coils have 3000 turns. So, the difference in self capacitance should be more pronounced.

The new helical coils have paper between the layers which should in general lower the self capacitance (greater distance between the layers).


Greetings, Conrad

P-S.: I also attach a corrected coil parameter PDF-file.

[gyulasun]

....
The new helical coils have paper between the layers which should in general lower the self capacitance (greater distance between the layers).
....

Hi Conrad,

Regarding the effect of paper insulation on the self capacitance of your new coils, it is possible that its dielectric constant (which is > 1 as you looked it up on the web) actually help to counteract or even overcompensate the actual capacitance-reducing effect of the greater distance between the layers. I mean if the paper had exactly the same constant than that of the air, then the greater distance would surely reduce self capacitance versus the case when no paper (or no any other insulating material) were used between the winding layers. But this could only be tested by making another helical bifilar coil without any paper between the layers, using the same number of turns. Of course there is no real need to test that.

When you have a few minutes, please check the capacitance between the two winding end or start wires of your helical  bifilar coil, I would like to know.

You measured 25.6 V (rms) across the unloaded paralel tank circuit. Whatever resonant real impedance the tank has, the rotating magnet maintains that voltage across the tank and when you find a resistive load which just halves the unloaded 25.6 V then it is sure that the load has the same resistance what the unloaded LC tank has. So your 420 Ohm load just made that close: you measured 13.6V across it and 2*13.6=27.2 V, while the 300 Ohm load made 11.2 V from the 25.6 V, these two values indicate perhaps an interpolated 400 Ohm or so tank impedance and I guess when you would attach a 400 Ohm resistor, the voltage would probably be pretty close to 12.8 V across it (25.6/2). (This is why I mentioned earlier the 1 kOhm non wire wound potmeter to use as the load and adjust till half of the unloaded tank voltage can be measured across it.) This 400 Ohm or so tank impedance means that we have not considered or included something in the impedance calculation (probably still there is some loss which reduces the real Q of the tank to be less than the calculated 2.7).

Thanks for all your efforts.

Greetings, Gyula

[Farmhand]

Hi Conrad,

Regarding the effect of paper insulation on the self capacitance of your new coils, it is possible that its dielectric constant (which is > 1 as you looked it up on the web) actually help to counteract or even overcompensate the actual capacitance-reducing effect of the greater distance between the layers. I mean if the paper had exactly the same constant than that of the air, then the greater distance would surely reduce self capacitance versus the case when no paper (or no any other insulating material) were used between the winding layers. But this could only be tested by making another helical bifilar coil without any paper between the layers, using the same number of turns. Of course there is no real need to test that.

When you have a few minutes, please check the capacitance between the two winding end or start wires of your helical  bifilar coil, I would like to know.

You measured 25.6 V (rms) across the unloaded paralel tank circuit. Whatever resonant real impedance the tank has, the rotating magnet maintains that voltage across the tank and when you find a resistive load which just halves the unloaded 25.6 V then it is sure that the load has the same resistance what the unloaded LC tank has. So your 420 Ohm load just made that close: you measured 13.6V across it and 2*13.6=27.2 V, while the 300 Ohm load made 11.2 V from the 25.6 V, these two values indicate perhaps an interpolated 400 Ohm or so tank impedance and I guess when you would attach a 400 Ohm resistor, the voltage would probably be pretty close to 12.8 V across it (25.6/2). (This is why I mentioned earlier the 1 kOhm non wire wound potmeter to use as the load and adjust till half of the unloaded tank voltage can be measured across it.) This 400 Ohm or so tank impedance means that we have not considered or included something in the impedance calculation (probably still there is some loss which reduces the real Q of the tank to be less than the calculated 2.7).

Thanks for all your efforts.

Greetings, Gyula

Could it be a change in the rotor speed that causes the variance ?

Well the theorem seems to hold, it can tell us what resistance will show the most power for a given coil at a certain frequency. if the speed changes during tests and so forth (unlike a wall transformer) then some tuning would be needed, but the Theorem can guide us to max power output maybe. Interesting.

Cheers

..

[conradelektro]

When you have a few minutes, please check the capacitance between the two winding end or start wires of your helical  bifilar coil, I would like to know.

You measured 25.6 V (rms) across the unloaded paralel tank circuit. Whatever resonant real impedance the tank has, the rotating magnet maintains that voltage across the tank and when you find a resistive load which just halves the unloaded 25.6 V then it is sure that the load has the same resistance what the unloaded LC tank has. So your 420 Ohm load just made that close: you measured 13.6V across it and 2*13.6=27.2 V, while the 300 Ohm load made 11.2 V from the 25.6 V, these two values indicate perhaps an interpolated 400 Ohm or so tank impedance and I guess when you would attach a 400 Ohm resistor, the voltage would probably be pretty close to 12.8 V across it (25.6/2). (This is why I mentioned earlier the 1 kOhm non wire wound potmeter to use as the load and adjust till half of the unloaded tank voltage can be measured across it.) This 400 Ohm or so tank impedance means that we have not considered or included something in the impedance calculation (probably still there is some loss which reduces the real Q of the tank to be less than the calculated 2.7).

@Gyula:

The capacitance between the two winding end or start wires of the helical bifilar coil is 26 nF .

I did some measurements concerning the "halve of the unloaded tank voltage" with the monofilar coil. The load resistor (1 K Ohm pot) has to be about 345 Ohm in order to halve the Voltage:

no load, resonance at 90 Hz rotor speed, motor consumes 5.5 V and 1.20 A = 6.6 Watt
dissipation in LC circuit: 0.156 * 0.156 * 77 = 1.87 Watt 
Voltage over cap or coil 26.4 V

short circuit, 101 Hz rotor speed, motor consumes 5.5 V and 0.80 A = 4.4 Watt
dissipation in LC circuit: 0.064 * 0.064 * 77 = 0.31 Watt

345 Ohm load, 96 Hz rotor speed, motor consumes 5.5 V and 1.00 A = 5.5 Watt
dissipation in LC circuit: 0.088 * 0.088 * 77 = 0.59 Watt
dissipation in 345 Ohm load: (13.2 / 345) * 13.2 = 0.50 Watt
in sum 1.09 Watt
Voltage over cap, coil or "345 Ohm load" 13.2 V

The coil was newly mounted near the spinning magnet (because I took it off for other experiments) and it seems to be a millimetre closer to the spinning magnet which raises the power output slightly (in comparison to earlier measurements).

@Farmhand:

The problem with the calculations (they vary) is that everything depends on three critical values, first the 1 Ohm shut is not exactly 1 Ohm, second the DC resistance of the coil is not exactly 77 Ohm and the inductance of the coil is not exactly 357 mH. So, all calculations are slightly off. And of course, the frequency is critical for calculating Z and Q.

When calculating the impedance Z of the tank circuit with the Thevenin Theorem

26,4 V (Voltage open circuit) / 0.064 I (I short circuit) = 412 Ohm impedance of circuit

the critical value is 64 mA through the 1 Ohm shunt. It should be 76 mA to give Z = 345 Ohm. And if the 1 Ohm shunt is slightly less than 1 Ohm it could well be 76 mA.

When calculating Z via Q and L

96 Hz resonance frequency --> w = 2 * π * 96 = 603 radian per second
L of coil = 0.357 Henry, R of coil = 77 Ohm
Q = (w * L) / R = (603 * 0.357) / 77 = 2.8 Q factor
w * L = XL = 603 * 0.357 = 215
Z = XL * Q = 215 * 2.8 = 602 Ohm impedance of coil

the critical values are DC resistance of the coil and inductance of the coil. The inductance should be 201 mH to result in an impedance of 345 Ohm (instead of 602).

I measured Z and Q of the coil with my LCR meter at 100 Hz and the values are Z = 240 Ohm and Q = 2.9 (about the same for monofilar and bifilar coil).

There is also the problem of "impedance of the coil" and "impedance of the LC circuit", which should be different?

The experimental value of Z = 345 Ohm (halve of the unloaded tank voltage) seems to be the best. And it must be the Z of the tank circuit. The measured Z = 240 value of the coil (at 100 Hz) with the LCR meter is not far off if we consider that the LC circuit should have more losses than the coil alone. Also the vale Z = 412 Ohm for the tank circuit from the Thevenin Theorem is not too bad.

The value Z = 602 Ohm for the coil calculated via the inductance and the DC resistance of the coil is pretty bad and I can nor explain it?

Greetings, Conrad

[conradelektro]

I did a test with a "series LC circuit" at the magnet spinner (with the monofilar coil, but it is about the same with the bifilar coil). please see the attached drawing with the circuit diagram and the measured values.

When switching between "short circuit" and "1 K Ohm load" I had to regulate the power supply for the motor down in order to stay at 90 Hz.

When varying the 1 K Ohm pot from "1 K Ohm" to "short circuit" the measurements gradually change between the two extremes noted in the drawing.

Greetings, Conrad