Calculating q factor of an inductor

[hogshead]

I have some old ignition coils that I would like to compare. I have set the secondary to resonate with a .47 mf capacitor and these coils will resonate at 50 to 70 hz. The problem is, I can't really compare them as the center frequency is different on each. Is there a way to test their q factor at a certain frequency, say 60 hz, short of substituting in a bunch different caps?

 

[vk6kro]

Yes, you can.

Assuming you mean the large coil if the two in an ignition coil, you can calculate the inductance of the coil knowing the resonant frequency with a known capacitor.

You said 0.47 mF but this would probably be 0.47 μF and it resonates at 60 Hz approximately.

This would mean the coil had an inductance of 14.97 Henrys and a reactance (X_L) of 5643 ohms.
(Check these).

The Q of a coil is X_L / R so, you just need to measure the resistance of the coil and then Q = 5643 / R

 

[hogshead]

Yes, good idea. Here is my plan. I will set up a circuit with the coil and a variable resistor in series. I will drive the circuit with an ac source (at 60 hz) and measure the voltage at each leg on the source to the point where the resistor and coil join together. I'll adjust the resistor until the voltage is the same. I'll then measure the value of the resistor, which will equal the reactance of the coil at 60 hz. Easy! Thanx.

 

[Bob S]

Old automotive ignition coils usually resonated about 20 to 40 kHz. See simulation with points and condenser.

 

[hogshead]

Ok, I've tested a few with my method. They measure:

5.66K reactance 2.92K resistance Q 1.94
5.04K reactance 2.89K resistance Q 1.74
6.24K reactance 3.35K resistance Q 1.77

Does this seem correct? How can one test the Q of the primary? My method does not seem to work until I test at around 6000 hz. At which point I get 169 ohms reactance and .7 ohms resistance.

 

[vk6kro]

Yes, good idea. Here is my plan. I will set up a circuit with the coil and a variable resistor in series. I will drive the circuit with an ac source (at 60 hz) and measure the voltage at each leg on the source to the point where the resistor and coil join together. I'll adjust the resistor until the voltage is the same. I'll then measure the value of the resistor, which will equal the reactance of the coil at 60 hz. Easy! Thanx.

No, that will only give you an idea of the reactance. You can't measure resistance like that.

Measure the resistance with a multimeter and then put the result in the formula I gave you earlier.

You can use the same scheme to find the inducance of the primary. Use the same capacitor and find where it is resonant then calculate the inductance and then the reactance.

You can also vary the frequency to find the half voltage points of the resonant circuit. The difference between them is the bandwidth.
Then divide the resonant frequency by this bandwidth and this gives you the Q.
Because the coil has an iron core, the Q you get this way will be more accurate than the resistance method.

 

[hogshead]

No, that will only give you an idea of the reactance. You can't measure resistance like that.

Measure the resistance with a multimeter and then put the result in the formula I gave you earlier.

hmmm... I have tested several coils with this method and the results are textbook. I don't understand why I am not measuring the reactance.
https://www.physicsforums.com/attachment.php?attachmentid=46145&stc=1&d=1334214699


Sorry for the crappy schematic, but I don't know how to do it better. After I balance the (rms) voltage at the two test points, I remove the resistor from the circuit and note its value. It seems to work to get the reactance at 60 hz.

Now, I tried the bandwidth method, but that gives the q factor at resonance, which varies from coil to coil. I wanted the q factor at a fixed frequency.

 

[vk6kro]

That will give you the reactance, but you need to be measuring the resistance, since you already know the reactance.

It is 5643 ohms.

 

[Bob S]

The ignition coil primary series resistance is part of the Q factor, but the secondary series resistance is not. This is because there is no current in the secondary until the spark occurs, which quenches the Q factor.. The only other contributor to the Q factor are the eddy currents in the iron core.
The typical "condenser", about 0.02 uF, is in parallel with the coil primary (and in series with the "points"). See my attachment in post #4 for circuit diagram and simulation. The simulation shows that the spark occurs when the points open. When the points close, the inductor (coil primary) charges to about 1 or 2 amps, and when the points open, the coil primary discharges into the condenser.

 

[vk6kro]

I think he is just using the ignition coil secondary as a convenient inductor.

So, it has its own L, X_L, series R and Q as well as core losses.

 

[hogshead]

Ok, I was assuming that the q factor of the primary and secondary somehow corresponded. What I really want to know is the q factor of the primary. Later, I'd like to figure out how to calculate iron losses from hysteresis and eddy currents. These coils are not standard automotive coils, but are "buzz coils". They have points on them that are activated from the flux in the primary core.

Here is my new tack: I'm going to measure the voltage across a resistor in series with the primary. Then by dividing, I can calculate the current. That should allow me to calculate the reactance (voltage across coil/current through coil). From that I should be able to calculate inductance (reactance/2pi frequency).

Thanks for your help so far.

 

[vk6kro]

Yes, that will work, but be aware that inductance is not a constant if the coil has an iron core.

Inductance will vary with current through the coil, and also with frequency.

If you can try a few different currents, this effect may show up. You can do this with differing resistors or voltages.

Also, you still need to measure the resistance of the coil. This will affect any current reading you get with this method.