Measuring the capacitance of an inductor or Inductors

[Magnut]

I do not remember where I found this but thought you might find it useful. PM me if you want the word/PDF version.

[bbsailor]

Belwar,

If you don't want to load a program such as MiscEL into your computer, go to this, one of many on-line electronic calculators for LC resonant frequency, and just enter the pickup inductance in Henries and the resonant frequency in KHz. This will tell you the pickup's self-capacitance. LC Resonance Frequency Calculator at WhatCircuits.com

Joseph Rogowski

[belwar]

Joseph,

Thanks for this method, I'm going to try this method as well as Joe Gwins method and compare the results. The online programs are great, but I dont learn anything by using them.. Do you happen to know the math used to determine the self capacitance from the Resonance Frequency and Inductance? I like to punish myself and do the math myself!

Also the Oscillator I have isnt very accurate on the sweep wheel so I use the DMM to measure the frequency now.

Matt

Belwar,

Try this using the equipment that you have.

1. Connect your pickup under test (PUT) to your old HP audio oscillator with a 1 Meg ohm resistor in series with that connection. The grounds are all a common connection point.
2. Connect your USB scope to the pickup (after the 1 meg ohm resistor and the common ground point).
3. Set your scope to read the range of 1KHz to 10KHz.
4. Sweep the HP audio oscillator from 1KHz to 10 KH and observe the scope pattern. When you hit the pickup resonance point you will see the voltage rise sharply.
5. Note the frequency on the oscillator where the pickup peak resonance occurs.
6. Use your Extech LCR meter to measure the capacitance of the probe. Note this measurement.
7. Use the free program MiscEL to enter the resonant frequency and the inductance in Henries of the PUT measured with the Extech LCR meter.
8. The result of this calculation will be the self capacitance of the pickup and the scope probe.
9. Subtract the capacitance of the scope probe and you will now have the self capacitance of the PUT.

This should do it for you with the equipment you have on hand.

Joseph Rogowski

P.S. If your HP oscillator frequency readings are not very accurate, attach your DMM, with frequency readout, to the oscillator output and use that DMM frequency reading as your frequency to enter into the MiscEL calaculator.

[bbsailor]

Joseph,

Thanks for this method, I'm going to try this method as well as Joe Gwins method and compare the results. The online programs are great, but I dont learn anything by using them.. Do you happen to know the math used to determine the self capacitance from the Resonance Frequency and Inductance? I like to punish myself and do the math myself!

Also the Oscillator I have isnt very accurate on the sweep wheel so I use the DMM to measure the frequency now.

Matt

Matt,

The pickup inductance is in parallel with the coil's self capacitance so we will use the concept of parallel resonance.

In a nutshell, here is the key concept. When the capacitative reactance (XC) of the capacitor (coil capacitance) equals the inductive reactance (XL) they cancel each other out and cause a resonance where the impedance is the highest.

XC = 1/2pi*f*C

XL = 2pi*f*L

The key concept to keep in mind is that capacitive reactance (XC) goes down as the frequency raises and the inductive reactance (XL) goes up as the frequency raises.

When you use the on line calculator, just put in the coil inductance and the resonant frequency to find C. Then use the formulas above to work out the math and see that the capacitive and inductive reactances will be very close, based on the accuracy of your measurements. This exercise should cement this resonance relationship in your mind.

Directly solving for C is a little more complicated. See this link for a formula to solve for C when you know L and the pickup self resonance. http://www.edn.com/contents/images/159688.pdf


I hope this helps.

Joseph Rogowski

[belwar]

So heres a question I didnt think of before.. So obviously I now know that I cant use an LCR meter to directly measure the self capacitance of a pickup, but what about the inductance if the coil has multiple coils? Are the inductance readings somehow off or incorrect? I would imagine that the reading im getting is the Mutal Inductance of the two coils in series.. Is there anything I should know about measuring the inductance of a humbucker that I am un aware of? I know that the inductance is different depending on whether the coil is in series or parallel.

[bbsailor]

So heres a question I didnt think of before.. So obviously I now know that I cant use an LCR meter to directly measure the self capacitance of a pickup, but what about the inductance if the coil has multiple coils? Are the inductance readings somehow off or incorrect? I would imagine that the reading im getting is the Mutal Inductance of the two coils in series.. Is there anything I should know about measuring the inductance of a humbucker that I am un aware of? I know that the inductance is different depending on whether the coil is in series or parallel.

Matt,

Inductance (Henries) changes by the square of the turns added or reduced.

Here is an example. If I double the amount of turns on the same coil then the inductance goes up by 2 squared or 4. So a coil wound with 2000 turns that is increased to 4000 turns (2x) will have an increased inductance by 2 squared or 4.

Now, since a humbucker coil is wound on independent bobbins, the independent inductors sum like resistors in series in that you add their individual values. So, a single humbucker coil of 2H will be 4H with two coils in series and be 1H in parallel. There may be some magnetic coupling between them but that coupling is very loose and can be ignored for this measurement. Measure the inductance of a single humbucker coil and then measure the full humbucker in series and see for yourself.

I had to edit this after I had my first cup of coffee!!!

Joseph Rogowski

[Possum]

...

Would you even need an oscilloscope for this method? Couldn't you just drive the coil with another coil, note the AC voltage peak coming out of the pickup on a DMM and measure that frequency on same? No need to measure probe capacitance. I'm not sure what value all this would be, if you're measuring a vintage pickup, every coil will be different and magnet wire and insulation varied and aren't something you can duplicate. It might be interesting to figure it out anyway just to see. The value of capacitance has alot to do with tension and insulation build, these are never identical on any two pickups. You might just find that like frequency response curves, that most humbuckers of the same DC resistance pretty much fall into the same capacitance range. I'll have to give it a try on the examples I have here and see if anything useful shows up....

[Possum]

.....

I see a potential flaw in this method, you're using henries in the equation. Now, if you have a humbucker with 1022 pole screws and change them to nearly pure electrical iron, the inductance reading will change, but this doesn't change the capacitance of the coil itself? How does what kind of core and what kind of metal the coil is sitting on figure into getting accurate information?

[Joe Gwinn]

So here's a question I didn't think of before.. So obviously I now know that I can't use an LCR meter to directly measure the self capacitance of a pickup, but what about the inductance if the coil has multiple coils? Are the inductance readings somehow off or incorrect? I would imagine that the reading I'm getting is the Mutual Inductance of the two coils in series.. Is there anything I should know about measuring the inductance of a humbucker that I am un aware of? I know that the inductance is different depending on whether the coil is in series or parallel.

There are lots of complexities when one measures something, and the trick is to arrange things such that the undesired effects are either too small to matter for the present purpose, or known well enough that one can compensate.

With multiple inductors in series or in parallel, what matters most is if any of the coils are close enough and/or arranged/designed such that significant mutual coupling exists.

What is mutual coupling? It is simply magnetic field generated by one coil that passes through the other coil. If the two coils are wound bifilar on the same coil former, essentially all flux from either coil passes through the other coil, so the coupling is essentially perfect. This is the desired condition in a transformer. If instead the coils are separated by many coil diameters, then essentially none of the flux from one coil passes through the other, and the mutual coupling is essentially zero.

If none of the inductors are coupled with any other inductor, the rule is very simple. It follows the resistor rule: Series and parallel inductors : INDUCTORS.

If there is significant magnetic coupling (the usual situation between coils of a humbucker), it gets a bit more complex, so we will focus on a pair of inductors.

Mutual Inductance | TutorVista.com

The total inductance of two inductors a and b in series is Lab=La + Lb +/- Mab. The +/- part is because it matters which way one connects the inductors. One can measure M in a pickup my measuring each coil alone and then measuring the total of the coils in series twice (reversing the connection for one), to yield four inductances. Then solve for M.

If the coils are in parallel, the formula becomes very messy, so we will limit the discussion to series for now.

[belwar]

I see a potential flaw in this method, you're using henries in the equation. Now, if you have a humbucker with 1022 pole screws and change them to nearly pure electrical iron, the inductance reading will change, but this doesn't change the capacitance of the coil itself? How does what kind of core and what kind of metal the coil is sitting on figure into getting accurate information?

I dont know if its going to change the self capacitance, but it will change the resonance peak.. You'd be changing the inductance, and the inductive reactance. If the inductive reactance changes, the point at which the inductive reactance and capacitive reactance are equal changes as well - changing your resonance peak.

Myself, I would think that changing the core material would alter the capacitance. I'll put that to the test when the new Capacitors get here - should be today

[Mike Sulzer]

Myself, I would think that changing the core material would alter the capacitance.

Funny you should mention that. Does it change the capacitance, or just change the measurement?

Over the weekend I measured a humbucker coil with and without the cores in place. I get 78 pdf with he cores in place and 88 pf with them out.

I measured the resonant frequency by measuring the zero of the phase using my I-V circuit and Mac with Electroacoustics toolbox. The measurement was done with no external capacitor and with 94.3 pf in parallel. Then I solved the equations for the square of the resonant frequency to get both L and C.

Joe might say the difference is due to eddy currents. I might say that it is due to the leakage flux from short open cores. But we really do not know. I doubt that it represents a real change in the C, but that is certainly not proven.

[belwar]

Funny you should mention that. Does it change the capacitance, or just change the measurement?

Ok, im a beginner at electronics so im probably totally wrong but here was my thinking:

The capacitance of an inductor is a result of how much energy the inductor stores as a magnetic field correct? That magnetic field is changed based on the core of the inductor.. An air core inductor, has a different field that a ferrite core inductor. The permeability of the core plays a role in the inductance, and should also play a role (although probably smaller) in capacitance.


Though the permeability of one set of screws to the next is probably minor. If it were a blade, that might be a different story.

[Mike Sulzer]

Ok, im a beginner at electronics so im probably totally wrong but here was my thinking:

The capacitance of an inductor is a result of how much energy the inductor stores as a magnetic field correct?

When a capacitor is charged, energy is stored in an electric field, not magnetic.

[David Schwab]

Joe might say the difference is due to eddy currents. I might say that it is due to the leakage flux from short open cores. But we really do not know. I doubt that it represents a real change in the C, but that is certainly not proven.

Would you even get any meaningful eddy currents with no magnetic in the circuit?

[Mike Sulzer]

Would you even get any meaningful eddy currents with no magnetic in the circuit?

AC current flowing in the coil causes a fluctuating magnetic field which induces a current in the core. This reduces the effective permeability, leading to a reduction in the inductance, more at higher frequencies. My claim is that this effect is very small when the cores are open (unlike, for example, a toroid). Joe is not convinced. We will eventually get the matter settled.