# Measuring the capacity of a coil.

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#### MrAl

##### Well-Known Member
Hi Guys,
I've just joined. I'm located in the U.K. and am a retired electronics engineer. I was looking for a method of measuring the self-capacitance of a coil and found this site and this article. Am now reading it . . .very interesting . . . .
March 2015

Hi and welcome to the forum,

When i saw this thread i thought it was more recent, so i read a little more of it first. What really puzzled me was that so many people replied in a manner that would allude us into believing that there is no capacitance associated with a coil. That very surprising because i am pretty sure the people who replied would know that there is always some capacitance in very coil, and it is sometimes important to know, such as the reply from one member who talked about self resonance. Sometimes we can ignore, sometimes not, it all depends on the application.
I do have a feeling though that the OP was using the word "capacitance" in a more general way however, so more like asking about the 'capability' of a coil rather than the actual capacitance, but i just thought i would mention this in passing.

One way would be to see how it oscillates...at what frequency...then use the LC formula for frequency solving for the capacitance. So starting with:
w=1/sqrt(LC)
or in alternate form:
w=1/(sqrt(L)*sqrt(C))
then we solve for C:
w*sqrt(C)=1/sqrt(L)
and finally:
sqrt(C)=1/(w*sqrt(L))
and if you care to square that:
C=1/(w^2*L)

and there you have the capacitance.

You may wish to include an error capacitance term:
Cc+Cerr=1/(w^2*L)

where you might guess at the test jig capacitance Cerr as that would change the capacitance, or use a calculation.

#### BobW

##### Active Member
Mr. Al,
It appears that we were both posting at about the same time. See my post #20 regarding the problem using the natural self resonant frequency of the coil to calculate self-capacitance.

#### Alan Hall

##### New Member
Well for fun I'll carry on my hobby of radio and electronics - which I have been actively engaged in since I was a kiddie. A lifetime's employment in Electronics and Radio has covered a huge range of topics: radio transmitters and receivers (professional and domestic), personal computer maintenance, mainframe computer maintenance, field service engineer, technical trainer, salesman, printing industry, domestic electrical and electronics, power engineering, microwave engineering (incl. RADAR), photocopier design and production, some aircraft electronics, industrial heating and cooking systems, etc,. etc.

Just for the record: I really couldn't believe most of what I read by way of replies to that thread about coil self-capacitance. Anyway, it is an old thread: hopefully the questioner has solved his problem. I'll say no more now about that thread: let sleeping digs lie, eh?

Amongst my other interests are mathematics, physics, chemistry, human motivation and applied psychology, classical music, traditional art / paintings and D-I-Y around the house. My greatest fear is boredom - with lack of money coming in as a close second.

I'll pop in now and again. If I read anything to which I feel I can make a meaningful and worthwhile contribution, I'll do so. If I have an 'unresolved issue' (don't we all), I'll simply ask.

All said: ta ta.

Al. / March 30, '15 //

#### MrAl

##### Well-Known Member
Mr. Al,
It appears that we were both posting at about the same time. See my post #20 regarding the problem using the natural self resonant frequency of the coil to calculate self-capacitance.

Hello there Bob,

Ok good, i was going to suggest:
C=((f1^2-2*f2^2)*dC)/(2*f2^2-2*f1^2)

where dC is the parallel capacitance for test 1, and 2*dC is the parallel capacitance for test 2, so dC is the delta capacitance. #### Alan Hall

##### New Member
Thank you, Bob - that is what I usually do. But the following method has crossed my mind.

Choose a coil and a capacitor that will produce a resonant freq. close to that of interest. Using an RLC bridge, measure the L and the C. Now calculate the theoretical resonant freq.: call if Fc: (c for 'calculate')
Now build that circuit and measure the real resonant freq.: call it Fm (m for 'measure').
On account of the coil's self-capacitance, Fm < Fc.
By using the standard formula for the res. freq. for Fc and Fm and a bit of algebra, the self-capacitance of the coil can be calculated.
Obviously, in determining Fm, precautions must be taken to minimise stray capacitances. And the above assumes that the kit for measuring the inductance completely ignores the effect of the coil's self -capacity: not all inductance-measuring devices do that.
Alternative approach is to use a Q-meter. When I have repaired mine, I'll probably use that.

Cheers,
Al.

#### BobW

##### Active Member
It depends on how your RLC bridge measures inductance. One measurement at a single frequency isn't sufficient. An advanced test instrument may take multiple readings at different frequencies. Since it's not possible to separate the self-capacitance from the inductance during the measurement, most simple inductance meters will give you an inductance value that's slightly lower than the "true" (DC) inductance value because it can't compensate for self-capacitance. By measuring at two frequencies, you have enough data to determine both the self-capacitance and the "true" inductance.

To elaborate on the measurement technique I described earlier:
Suppose you connect various values of capacitance across the terminals of the inductor, and then measure the resulting resonant frequencies. The reciprocal of the square of the frequency will be directly proportional to the total capacitance. So, if you plot 1/F^2 vs. C on a graph, you will get a straight line. If the coil had no self-capacitance, then the line on the graph would pass through the origin, [0,0]. But because of self-capacitance the line will actually cross the C axis at –Cs, where Cs is the self-capacitance.

(The formula that was given by MrAl is a simplification of the general formula that results if you pick the external capacitance values such that one is twice the value of the other.)

Referring back to the straight line 1/F^2 vs. C relationship, the true inductance is proportional to the slope of the line, and is given by:
L=1/4π^2*(1/F1^2-1/F2^2)/(C1-C2)

So, two measurements are required in order to determine both the self-capacitance and the true inductance. However, it's better still to do more than two measurements using multiple external capacitance values and then do a linear regression. This gives much better accuracy, and in addition, it allows you to see just how straight the the line is, and consequently confirm that the theory is valid. I have a spreadsheet on my personal website that uses the linear regression method, if you're interested:

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#### MrAl

##### Well-Known Member
Hello again,

If you use an incremental capacitance value for changing the external capacitance you can also do this for N tests using this simple formula:
N*L*C=Ʃ[1 to N] ((1/wk^2)-k*dC*L)

Here, the capacitance on the left side is the average capacitance C, so divide both sides by N*L to get the average capacitance. Also, wk here is the angular frequency w sub k, so w1, w2, w3, etc.
dC is the increment in capacitance. It works out simpler with a constant set value for the incremental capacitance, and it's not too hard to compare capacitances of the exact same values.

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