CJ
Well-known member
A lot of people seem to be doing eq's right now, so I thought I would start a thread on measuring the inductors.
I think it is best to run the inductor at it's desiginated frequency to get a good measurement. I usually use this simple resistor-inductor technique:
So lets say you wanted to measure the 108 millihenry inductor for the 1000 hz section of a MEQ-5. Well, at 1000 hz, the inductor should have a reactance of 2 pi F L , so we would have 6.28 x 1000 x 0.108 H.
This comes out to be about 678 ohms. So if we hook up a signal generator and scope to the above circuit, and use a 680 ohm series resistor (close enough), and apply a 10 volt peak to peak signal, we should see 5 volts across the resistor and inductor, right?
What do you think? Is this a legitimate method?
If you wanted to calculate the inductance when you did not get an equal voltage drop, say you only read 1 volt across the inductor, then you would just work backwards and do some algebra.
So 1 volt out of ten would indicate the resistor is responsible for 90 percent of the voltage drop. So if the resistor is taking 90 percent, that means the inductor is taking 10 percent. So the inductor would have to be 10 percent divided by 90 percent, or 1/9 the resistance of the resistor.
So in our above circuit, 1/9 th of 680 ohms is about 75.5 ohms. Plugging this back into the inductive reactance formula, we would have Xl=2 pi F L or 75.5 = 6.28 x 1000 x L. Solving for L gives 75.5/6280 or about 12 millihenries.
If this is too tough for some of you, we could write a simple excel program where you input all the parameters and it would spit out the answer.
This approach assumes a perfect inductor with zero ohms dc resistance, so it won't be perfect.
cj
I think it is best to run the inductor at it's desiginated frequency to get a good measurement. I usually use this simple resistor-inductor technique:

So lets say you wanted to measure the 108 millihenry inductor for the 1000 hz section of a MEQ-5. Well, at 1000 hz, the inductor should have a reactance of 2 pi F L , so we would have 6.28 x 1000 x 0.108 H.
This comes out to be about 678 ohms. So if we hook up a signal generator and scope to the above circuit, and use a 680 ohm series resistor (close enough), and apply a 10 volt peak to peak signal, we should see 5 volts across the resistor and inductor, right?
What do you think? Is this a legitimate method?
If you wanted to calculate the inductance when you did not get an equal voltage drop, say you only read 1 volt across the inductor, then you would just work backwards and do some algebra.
So 1 volt out of ten would indicate the resistor is responsible for 90 percent of the voltage drop. So if the resistor is taking 90 percent, that means the inductor is taking 10 percent. So the inductor would have to be 10 percent divided by 90 percent, or 1/9 the resistance of the resistor.
So in our above circuit, 1/9 th of 680 ohms is about 75.5 ohms. Plugging this back into the inductive reactance formula, we would have Xl=2 pi F L or 75.5 = 6.28 x 1000 x L. Solving for L gives 75.5/6280 or about 12 millihenries.
If this is too tough for some of you, we could write a simple excel program where you input all the parameters and it would spit out the answer.
This approach assumes a perfect inductor with zero ohms dc resistance, so it won't be perfect.
cj