Testing Swinging Chokes

[k3msb]

I have a few chokes that I believe are swinging chokes and I would like to test them, which I really don’t know how to do.

I was planning on putting a current limiting resistor in series with the choke, as well as an AC Ammeter, and using my variac as a 60 Hz AC source.

Measuring the AC voltage across the choke, I can then calculate the reactance XL = V/I

Then the inductance can be computed from XL = 2 * pi * f * L

As I increase the voltage (and thus the current), I should see a change in L if it’s a swinging choke.  If L does not change then I have a filter choke.

Is this the best way to do it, or are them better / safer ways?

Thanks!

[K1JJ]

Hi Mark,

Frank/ GFZ describes a measuring technique that is similar to yours:


http://amfone.net/Amforum/index.php?topic=6909.0;wap2


Tom, K1JJ

[AB2EZ]

A swinging choke is typically intended to be used in a choke input power supply... where there will be DC flowing through it (to feed to load on the power supply), plus some residual AC ripple in the current flowing through it.

To test the choke, you should probably feed it with the sum of a constant (DC) voltage and a 60Hz sine wave voltage. A small sine wave voltage is okay.. e.g. 12V rms.

You can measure the current (both DC and AC components) passing through the choke by placing a small value resistor in series with the choke (e.g. 10 ohms). You can measure the AC voltage across the choke with an AC voltmeter

You should make the DC adjustable from 0mA up to whatever you think the maximum DC the choke is rated to handle. If you don't know the maximum DC rating, you can increase the DC voltage (and the corresponding current) slowly... while keeping a close watch on how hot the choke is getting.

By looking at the current passing through the choke (both a DC and the 60Hz components), and 60 Hz AC  voltage across the choke, you can obtain the impedance of the choke at 60Hz (for different values of DC passing through it).

You should also measure the resistance (R) of the choke when it is disconnected.

You can obtain the reactance (X) of the choke from the measured value of impedance (Z) by using the formula:

ZZ = RR + XX

Once you have X, you can calculate the inductance of the choke (for various values of DC).

Example:

Measured DC flowing through the choke = 200mA
Measured 60Hz current = 4mA
Measured 60Hz component of the voltage across the choke = 12V
Measured resistance of the choke = 100 ohms

First obtain Z= 12V/.004A = 3000 ohms

Then obtain X = the square root of [(3000 ohms x 3000 ohms) - (100 ohms x 100 ohms)]  = 2998 ohms

Then obtain L = 2998 ohms / [2pi x 60Hz] = 7.95H

So... with 200mA of DC flowing through the choke, its inductance is approximately 8H

Stu

[k3msb]

Thanks Guys --

So, is the the correct test setup?

I'm wondering if the AC Ammeter should be between the resistor and the choke, but I'm not sure if an AC ammeter will handle a large DC current through it.

Thanks!

[AB2EZ]

Hi

Your proposed setup will not produce (across the choke) AC voltage superimposed on the DC voltage. The DC voltage source will "short out" the AC voltage source.

One way to do what is needed is (see attachment) to use a filament transformer with a 12 volt output winding (as an example).  Connect one end of the 12V output winding to the small resistor, R, that leads to the choke. Connect the other end of the 12V output winding to one side of the DC supply. Connect the other side of the DC supply to ground. Connect the other side of the choke to ground. If you have a large filament transformer (capable of delivering several amps of filament current) use it... not because you will need anywhere near that much AC current, but because a large transformer will be able to handle the superimposed DC current passing through its secondary.

Knowing the DC resistance of the choke (measured with an ohmmeter), you can estimate how much DC voltage is needed to produce a target value of DC current through the choke.

Use the small resistor (e.g. 10 ohms), R, to measure both the DC current passing through the choke, and the AC current passing through the choke. I.e. measure the voltage across R with a DC voltmeter to determine the DC current passing through R. Measure the AC voltage across R with an AC voltmeter to determine the AC current passing through R.

Stu

[k3msb]

Hi Stu

So here's the new setup based upon your description:



I think I have one or two filament transformers lurking about.

As for the original schematic, I can't see how the DC supply will short out the AC supply as the DC is blocked by the capacitor.

Thanks again!

[k3msb]

Sorry Stu, I missed the attachment you provided and didn't see your schematic!

[AB2EZ]

The DC is blocked by the capacitor, but all of the AC current will flow from the top of the AC supply, through the capacitor, through the DC supply, and back to the bottom of the AC supply.

I.e. the DC supply (which is an almost ideal voltage source) looks like, essentially, a zero resistance path to ground with respect to the current exiting the capacitor. Therefore all of the AC current exiting the capacitor will flow to ground through the DC supply, and none of it will flow into the series combination of R and L.

Stu

Hi Stu

So here's the new setup based upon your description:



I think I have one or two filament transformers lurking about.

As for the original schematic, I can't see how the DC supply will short out the AC supply as the DC is blocked by the capacitor.

Thanks again!

[k3msb]

Got it.  For some reason I viewed the DC supply as an open, not a short.

Thanks for the lesson Stu!