RESONANCE EFFECTS FOR EVERYONE TO SHARE

[DrSimon]

Hi Dr. Stiffler,

thank you for taking the time to share this very valid and valuable information. As I said above I was not happy with my attempt to measure the coil efficiency, so I made a new video and I would ask for you to look at it to see if you agree with this way of calculating the performance output of each coil. I do believe the use of the 10 ohm resistor as load on each 30,000uf cap is accurate enough since that large cap size should be more than enough to smooth it to DC. After doing the video and seeing your post I attached the probes to one of the coil output cap and I attached the scope shot below. Looks fairly flat to me and I am also happy to report my meters are accurately displaying the volts the scope is displaying.

The only thing I can see where there may be a problem is the 10 ohm input resistor. The frequency at this time is 300Khz, do you think if I pickup a accurate carbon resistor we could measure to a satisfactory result?

Link to new video: http://www.youtube.com/watch?v=ylnH1-UGITE

Thank you for sharing

Luc

@gotoluc

Please let me answer for Dr. Stiffler as he has left for a couple of days yet we have both been watching this.

Dr. Stiffler told me he had advised you to place a resistor of low value - I think < 10 ohms in the ground lead of your signal generator that is between the generator and all of your circuit. This resistor need to be non or minimally reactive and a preferred type would be a solid carbon. A film resistor may be fine but that would depend on how it was trimmed (i.e. spiral or end).

What we would like to see is what is going (returning) back to the generator and this resistor will do that provided it is minimally reactive (as stated above). The only problem that we can see is that your scope seems to have limited integration functionality but may be able to handle the complex waveform you will see across this resistor. By placing the resistor in the ground return circuit you will remove the problem of having to have a differential probe as you would need if this resistor were in the output lead of the generator as it would be above the reference ground on both sides.

Also you should try and terminate the generator in its characteristic impedance as you were advised in this thread. Normally unless this gen was modified it should be 50 ohms. If your coils were a low impedance to the output you could place a 50 ohm resistor in series with the output lead but it looks like your coils are above 50 ohms so you could place a 50 ohms across the output and get close. What you really would see at the gen would be the coils in parallel with the 50 ohm load. If the coils were high then minimal effect on overall loading.

You have a complex configuration because you have the coils close to each other and there is an interaction and this must be taken into account.

Simon

[gotoluc]

@everyone,

can someone please explain to me how to correctly measure and calculate my signal generator output ... please keep in mind that I have no electronic knowledge  so using words or letters that are related to this feild means nothing to me.

Please just explain it in plain English... like example: put scope prob A across SG and probe B across input resistor, then take RMS voltage reading of probe A and multiply by probe B RMS voltage, then divide by resistor value and this amount would be your watts in.

Can someone please do this favor for me.

Thank you for your time.

Luc

[watercellguy]

I would open the signal Generator, and would add connections so I could measure how many AMP and Volts is the signal Generator using.

I would set it so I could measure AMPS on the DC side between the rectifier and Capacitor.

Watercellguy

[turbo]

Help is on the way 

[Antimon]

You need to put the current in a voltage signal (with a shunt). Put it on channel 2. On channel 1 measure the voltage. If its a resistive load then multiply the values. If its a reactive load then you have to consider the cos phi, the power factor. Its then P=U*I*cos phi.

If you do that, you have to use the same ground point, then you have to invert channel 2 at your scope to measure the right signal.

If you want to know the power at a resistor, just measure the voltage across it and P=U²/R.

U ... Voltage
I ... Current
P ... Power

I hope i could help.

A.