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Author Topic: Investigating "anomalies" in Bifilar coils  (Read 133702 times)
Group: Tech Wizard
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The second pic shows the circuit,scope probe placement,and resultant VRMS value across the 10 ohm load resistor.
Our P/out is then 1.9vRMS across the 10 ohm load resistor.
P/out=361mW.
Note the current flowing through R2(the load resistor) is then 190mA,and the transformer turn ratio is 1:1  :o

Our COP in this test seems to be 331%  ???
...

Hi Brad,

You show the type of R2 resistor in the above pictures and most of such heat sinked power resistors are wire wound, embedded in cement.  If this is so, then its 10 Ohm nominal DC resistance represents a much higher inductive reactance as the load at the 12-13 MHz frequency you use here. 
I apologize if that resistor is really a non-inductive type!  The 2 Ohm CSR resistor is also a heat sinked type, suspicious for wire wound one.

Gyula
   
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Hi Brad,

Would like to suggest a simple test to check whether the 10 Ohm heat sinked power resistor is inductive or not.

If you have two 10 Ohm (say 5%) metal film or carbon or  even SMD chip resistors, then you could make a voltage divider circuit from them to check if they halve the output voltage from your FG at the 13 MHz frequency involved, see attached drawing.

So if you find that two such resistors divide the FG output by 2, then you could substitute the resistor at Z1 by your 10 Ohm (heat sinked 100W) type resistor and see how the divided voltage changes at the same 13 MHz or so frequency. I suspect the divided output voltage as seen on CH2 would be much lower than half of the input voltage seen on CH1, this would indicate a much higher than 10 Ohm (series R-L) impedance for that '10 Ohm'.

Notice that if the two metal film or carbon or chip resistors divide the FG's output voltage by 2 or near to that as they should, it may not mean that they are non inductive but they may have identical reactances.  And if you introduce the heat sinked 10 Ohm resistor in place of Z1 or then Z2, the output voltage may differ significantly from the half value.

Gyula
   

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Quote
Hi Brad,

You show the type of R2 resistor in the above pictures and most of such heat sinked power resistors are wire wound, embedded in cement.  If this is so, then its 10 Ohm nominal DC resistance represents a much higher inductive reactance as the load at the 12-13 MHz frequency you use here.
I apologize if that resistor is really a non-inductive type!  The 2 Ohm CSR resistor is also a heat sinked type, suspicious for wire wound one.

Gyula



My thoughts exacly, and it shows on the screenshots, both have significant phase shifts between voltage (yellow)
and current (blue), the 10 Ohm R2 more then the 1 Ohm R1.

By the way, in the diagram it says R1 1 Ohm, but the picture shows a 10 and a 2 Ohm resistor!  Is this being compensated for in the Math function and/or manual calculations?

Should the CH2 (blue) signal not be inverted?  (not sure if it will change much, but for the phase it does matter).


Itsu
   
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It's turtles all the way down
Hi Brad,

There can be another explanation besides what Ion proposes.

Your bifi coil developes a parallel resonance (by its two windings and their distributed capacitance and the scope probe self cap added).

So this parallel resonant 'LC tank' is able to produce 2.25 V across the 10 Ohm, this then involves 225 mA current in the tank.  So the input current to this resonant  'LC tank' should be Q time less than 225 mA  i.e. if the loaded Q is say 50, then the input current should be 225/50=4.5 mA,  this small current can cause 4.5 mA * 2 Ohm = 9mV drop across the CSR.  Can this small voltage difference be distinguished in the 2 V/DIV scope settings?  I think it is hard, so perhaps a direct differencial mode scope measurement is to be done to see the small voltage drop across the CSR.

Gyula

Dear Gyula
You have made an estimate of the loaded Q at around 50. Is this a reasonable value for a load of 10 ohms? How did you estimate this?

I agree that my wild guess (of reflected wave nulling) is probably erroneous and there is more likely a measurement error at play.

Regards


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Tech Wizard
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Dear Ion,

Well, my estimation for the loaded Q of 50 was surely high, my main aim was to indicate what difference can be between the current values flowing into an LC network and the current in the load across the so called output of such network, this was to help understand why Brad could not have measured a reasonable voltage drop across the 2 Ohm CSR placed directly in series with the LC network input.

You are right that a 10 Ohm load resistor is rather low for letting a loaded Q of 50 remain, it could surely attenuate an LC circuit or network too much, so that the in-out current difference (as shown measured by Brad) would become much less, certainly not a figure of 50 but less than 10 or so.  I gave Q value rather lavishly  :D

However, I think two things are to be considered when figuring on the loaded Q of Brad's circuit, one is the possibility that the 10 Ohm as a load is not 10 Ohm at around 13 MHz but has a higher (series RL) impedance and the other is that the load may get transformed back towards the input at a higher value than 10 Ohm (if it were 10 Ohm indeed but likely it is >10 Ohm). The higher_than_one impedance transformation in the backwards direction may sound unjustified at first because the turns ratio is 1:1 as Brad said.  But the input winding has a floating end so the two windings cannot work as a normal 1:1 transformer and input current can go through via the capacitive coupling between the bifilar windings, this very likely boils down to a resulting (parallel) LC circuit coupled capacitively to the function generator. 

Thanks for drawing attention to this.

Gyula


Dear Gyula
You have made an estimate of the loaded Q at around 50. Is this a reasonable value for a load of 10 ohms? How did you estimate this?

I agree that my wild guess (of reflected wave nulling) is probably erroneous and there is more likely a measurement error at play.

Regards
   
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It's turtles all the way down
Gyula brought up some interesting points to consider:

Quote
"However, I think two things are to be considered when figuring on the loaded Q of Brad's circuit, one is the possibility that the 10 Ohm as a load is not 10 Ohm at around 13 MHz but has a higher (series RL) impedance and the other is that the load may get transformed back towards the input at a higher value than 10 Ohm (if it were 10 Ohm indeed but likely it is >10 Ohm). The higher_than_one impedance transformation in the backwards direction may sound unjustified at first because the turns ratio is 1:1 as Brad said.  But the input winding has a floating end so the two windings cannot work as a normal 1:1 transformer and input current can go through via the capacitive coupling between the bifilar windings, this very likely boils down to a resulting (parallel) LC circuit coupled capacitively to the function generator". 

I had not considered that the load resistor might be inductive.  As I always use short leads and carbon composition or thin film or other non inductive types for these tests, therefore it was in my blind spot.

Regards


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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Buy me some coffee
Ok,taking your advice guys,i swapped out the large 10 ohm resistor for a small 1/2 watt carbon resistor,and my large COP disappeared

So it seems that these !so called! Large non inductive carbon resistors i bought from ebay are inductive wire wound types.
Never trust ebay sellers.

For clarification,i swapped out the 2 ohm CVR for a 1 ohm 1 watt carbon resistor in the last test.
When using the two ohm CVR-yes,i compensated with the scope--but pointless anyway,as that 2 ohm resistor is of the wire wound type as well-apparently.

One would think i would check these things by now,rather than take the suppliers word for it.


Brad


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Group: Mad Scientist
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A lot of non inductive resistors are for audio circuits. Possible that they show inductive qualities at higher than audio freq.

Mags
   

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Buy me some coffee
A lot of non inductive resistors are for audio circuits. Possible that they show inductive qualities at higher than audio freq.

Mags

Yes
I tested for inductance using a pulse from the sg,and i start seeing inductive spikes at around 2.3 MHz
Below that they seem to be ok,and show the same waveform as a carbon resistor.


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Never let your schooling get in the way of your education.
   
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