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Registered Member #2680
Joined: Wed Feb 10 2010, 09:23PM
Location:
Posts: 45
I have been stumped by this for a while. I'm trying to determine experimentally, the coupling coefficient between a pancake coil and a hard drive disk (i like to launch them in the air). The main problem is that i cannot take measurements of the hard drive disk, so i took measurements of the pancake coil with and without the hard drive disk. However, I do not know how to apply the analysis to the data I collected.
The setup: Function generator in series with the pancake coil and a 5.00 ohm shunt resistor.
The data:
Inductance of the pancake coil: 18.59uH (measured with LCR meter). Resistance of the shunt: 5.00 ohms Voltage input: 1 volt peak to peak (sinusoid) Input frequency of 322 kHz
Without the hard drive disk 13.1 mA through the coil current leading the voltage by 80.2 degrees Power (avg): 0.56 mW
With hard drive disk 45.6 mA through coil current leading the voltage by 57.5 degrees Power (avg): 6.1 mW
I have tried creating a model to solve which involves a 1V pk-to-pk voltage source in series with an inductor (the pancake coil) and 5 ohm shunt. The pancake coil is coupled to an inductor (hard drive disk) with an unknown inductance in series with an unknown resistance. However, I am having trouble solving for even the mutual inductance :/
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Just treat it as a short-circuit test on a transformer. The pancake coil is the primary, the platter is the shorted secondary. That gives you the leakage inductance.
Removing the platter is an open-circuit test, the inductance of the coil without the platter is the "magnetising" inductance.
From those two inductance values you can calculate the coupling coefficient. The bad news is I forgot the formula.
Registered Member #2680
Joined: Wed Feb 10 2010, 09:23PM
Location:
Posts: 45
Using the equations WaveRider posted and the data I collected, k equals 0.868509+j*0.013645. Is the imaginary part important, or is k simply the magnitude of this answer?
Registered Member #29
Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
I did not include the effect of the shunt resistor in the calculations. Without the disk, you measure the inductance L=L1 with your RLC meter. With the disk in place you measure L=L1(1-k^2). Substitute your L1 in and solve for k. L should be a real number as well as k. Imaginary values for k and L have no meaning in this case...
Registered Member #2680
Joined: Wed Feb 10 2010, 09:23PM
Location:
Posts: 45
So the L1 in the equation: Vd = j*omega*(1-k^2)*L1*Id, the inductance of the coil whilst the hard drive disk is on it or without? I will try measuring it when i have access to the LCR meter again. I did factor in the shunt resistor into equations (but round off error might have introduced the imaginary component into my final result).
Registered Member #29
Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
It looks like it could be roundoff/measurement error.. In this case, just ignore the imaginary part of k.. 0.87 is a reasonably strong coupling coefficient..
Registered Member #2680
Joined: Wed Feb 10 2010, 09:23PM
Location:
Posts: 45
I took the inductance measurement of the coil today while the hard drive disk was on it. From that i calculated the coupling coefficient to be 0.86 using L=L1*(1-k^2). While a bit off topic but relevant, it there any way of determining the amount of current flowing through the hard drive disk? I don't know what the inductance of the hard drive disk is, nor do i have any means of measuring it.
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Experimentally determining coupling coefficient
