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#16
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Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
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I'm familiar with placing a reference resistance of known value in series with an inductor under test and finding the sinusoidal frequency at which the true RMS voltage drop across the coil equals the drop across the reference resistor, but that doesn't seem to be what you are describing. |
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#17
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Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
That's precisely what I'm doing, but I'm using the resistance of my source instead of a discrete resistor. Here's a link detailing the procedure.
Incidentally, we already had all the information necessary to compute inductance. Look at the current waveforms in the original post. During the on-time we apply ~13.8 volts for ~12.2 us and got a delta-current of ~2A comes out to about 84 uH. Of course, I'm not using cursors but rather eyeballing the PNGs I posted earlier, so it's completely expected that the number not come out the same. But it's right ballpark. -Sasha |
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#18
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Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
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#19
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Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
Well, not at the same time, but it's the same method. Basically I find a high-enough frequency where the voltage across at the output of the meter (post-50R source resistor) doesn't change when I attach the motor.
Then I lower the frequency until the voltage is exactly half that. At that point the voltage across the resistor and the voltage across the motor are the same, but I can't measure the voltage across the resistor directly. I have to trust that my instrument is consistent between the two measurements. It's probably a good assumption. |
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#20
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Re: Findings from a couple hours at the lab bench with a CIM and a Jaguar
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Consider: When the voltage across the motor is half the voltage across the resistor+motor series circuit, the frequency will be f = R/(2*pi*L*sqrt(3)). At that frequency, the magnitude of the motor impedance will be R/sqrt(3), so the voltage across the resistor will be sqrt(3) times the voltage across the motor. For the voltages across the resistor and motor to be equal, the frequency would have to be f = R/(2*pi*L). At that frequency, the magnitude of the motor's impedance would be R, and the voltage across the motor (and the resistor) would each be 0.707 times the voltage across the motor+resistor series circuit. Last edited by Ether : 02-02-2014 at 01:09. Reason: added calculation for equal voltages |
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