# Calculating the capacitance of solenoid

I am working on a project where I need to create a fairly strong magnetic field (15~30 mT) with a solenoid. I realized it is not so easy to create a strong electromagnet. The solenoid is 6.5 cm long, ~1000 windings, inner radius of 1.2 cm. The copper wire is 0.5 mm in diameter (a bit bigger with enamel layer probably around 0.55mm) there are 10 layers (so, the outer radius of the solenoid comes out to ~ 1.8 cm).

I am having difficulty understanding the behavior of the solenoid when connected to a function generator. Its resistance comes out as ~8.6 ohm, the inductance ~10.5 mH (rough calculation with the theoretical formula gave ~14 mH, so, I believe the result is correct). Due to multiple windings, I believe the solenoid has a rather large capacitance, and to be honest, I have no idea how to calculate or model it. My question is two folds.

1. I believe you can model solenoid as a resistor in series with an inductor and a capacitor in parallel with both the resistor and the inductor. Is this a correct picture (or at least a good approximation)?

2. Given a correct circuit model of a solenoid, how would you go about calculating its capacitance? Can it be done with an oscilloscope? I feel like an oscilloscope needs to be highly precise in this case.

Thank you in advance.

• I knew how to find the inductance of a solenoid, and I tried it successfully, using a scope. But I thought a solenoid's capacitance is too small to find. So I never tried it. Commented Jun 24, 2022 at 10:03
• I learnt how to measure inductance by watching Rose-Hulman's tutorial I guess the same method can also be used to measure capacitance. (1) How to measure inductance [using a scope] - Rose-Hulman youtube.com/watch?v=ajycdvKyCEw (2) forums.raspberrypi.com/… (3) forums.raspberrypi.com/… (4) forums.raspberrypi.com/… Commented Jun 24, 2022 at 12:44
• Commented Jun 24, 2022 at 16:48

I believe you can model solenoid as a resistor in series with an inductor and a capacitor in parallel with both the resistor and the inductor. Is this a correct picture (or at least a good approximation)?

Yes, you can do this reasonably accurately for an air-cored solenoid as I believe yours to be.

Given a correct circuit model of a solenoid, how would you go about calculating its capacitance? Can it be done with an oscilloscope? I feel like an oscilloscope needs to be highly precise in this case.

Use a signal generator driving the solenoid via a 1 kΩ resistor (there or thereabouts) and manually sweep the frequency up to about 10 MHz. Look at the solenoid waveform on the scope and where it peaks is where the coil's self-resonant frequency occurs. This is due to the parasitic capacitance.

It's about 300 pF$$\^{\text{ note 1}}\$$ (previously miscalculated at 3 pF due to thinking the answer was pF per metre) and, the self-resonant point is going to be about 1 MHz (now more like 100 kHz) reading your coil description but, testing is the best way. $$\f = \frac{1}{2\pi\sqrt{LC}}\$$

I am working on a project where I need to create a fairly strong magnetic field (15~30 mT) with a solenoid. I realized it is not so easy to create a strong electromagnet. The solenoid is 6.5 cm long

The shorter the better because the mean length in which the field travels is shorter and that improves magnetic field strength (H-field).

Of course, using a core will improve matters considerably too and, if you are trying to make an electromagnet then the south pole of the core should extend (through ferrous core material) around the coil to the north side leaving a small gap to attract iron.

$$\^{\text{ note 1}}\$$ - I used a coax cable capacitance calculator from Everything RF knowing that a reasonable approximation of capacitance could be made considering the outer winding layer as a solid tube and the inner winding layer also as a solid tube. So, outer diameter is 3.6 cm and inner diameter is 2.4 cm and got this: -

I then divided by 1000 (mm) and multiplied by 65 (mm) to get 2.714 pF (incorrectly). Only a hundred times out.

• Thanks again. but if you don't mind, how did you know that it should be about 3 pF? Is there a formula? Did you use formula for capacitor? (epsilon x Area / distance) Well, if the capacitance is that small, then I am missing something because it can be neglected at low frequency (<100 kHz). I will just have to study a bit more before posting another question. Again, thanks (again). Commented Jun 24, 2022 at 10:39
• I did calculate capacitance wrong. I used this coaxial calculator (knowing that the outer winding and inner winding are the relevant ones to consider) and got 42 pF and thought it was 42 pF per metre hence, 6.5 cm yields 2.73 pF. But it's 42 pF per cm hence, capacitance is actually 100 times higher. D'oh. OK resonant frequency will be about 100 kHz. Commented Jun 24, 2022 at 15:44
• @M.K.Saunders If you have access to a university technical library, "Soft Ferrites", by E. C. Snelling (ISBN 0592027902), pages 350-354, has a method to calculate self capacitance of a winding. Much easier to measure using the technique in Andy's answer.
– qrk
Commented Jun 24, 2022 at 21:48
• @Andyaka I looked up that link to your coaxial calculator, and further down the page, just under the 'Formulas ... ' section, it does say C = Capacitance in pF/Meter, εr = Relative Permeability of th .... So you were right the first time, it is 47 pF/m. O(100pF/m) is right for typical coaxial cable, 47 pF/cm would be a very strange cable indeed. Commented Jun 26, 2022 at 10:29

I believe the solenoid has a rather large capacitance, and to be honest, I have no idea how to calculate or model it.

A model of the self capacitance of an inductor will allow you to understand where the capacitance is coming from, and to wind alternative topologies to reduce it. It will also allow you to get a (very) ballpark figure for what the capacitance is, hopefully within an order of magnitude.

Let's assume that a full finite element or integral solution of the 3D mess of conductors, airgaps, insulation is not needed. We then make some sweeping approximations to get into the right order of magnitude.

An inductor not only stores energy in the magnetic field due to current flowing, it also stores energy in the electric field between turns. We can model this energy storage as a capacitor across its terminals. This capacitor will store energy given by CV2/2. Where is the energy stored? Any wire in the winding will have a numer of other wires next to it. Those in the same layer will only have a single turn of voltage between them. Those in different layers will have hundreds of turns of voltage between them. Obviously the energy storage of the former pales into insignificance compared to the latter. Therrefore ...

Approximation - no energy is stored between the turns of a layer, all the energy is stored between layers.

Now the solenoid has suddenly become more tractable, 10 layers instead of 1000 turns.

Approximation - we replace each layer of turns by a sheet of voltage

To find the energy stored between each pair of adjacent layers, we sum, or integrate (whichever you feel happier with) the contribution of each bit of area as we go across the layers, using the standard parallel plate capacitance formula. Note that the 'normal' method of winding is to go 'back and forth', which means the voltage between the layers will be zero at one end, and twice the layer voltage at the other. A lower capacitance method of winding is to go 'forth and forth', always winding from the same end, and retracing the wire quickly back to the start. Now the voltage between layers is constant across the layer at one times the layer voltage, with the retrace wire adding negligible extra capacitance. This gives a lower stored energy between any pair of layers.

Finally, sum up all the energy stored between all pairs of layers for a unit voltage across the coil, and substitute back into the normal formula for energy in the coil's self capacitance.

You can reduce the energy dramatically (it goes as the square of the voltage) by reducing the voltage between layers. Two independent solenoids each of half the length reduces the length of the layer, and so the inter-layer voltage. Taken to its extreme, this results in several 'pancakes' stacked along the solenoid, a method of winding often seen in high frequency filters. It's also seen in segmented high voltage high frequency transformers. While it's probably principally for the electrical insulation, the resulting higher SRF of the winding is a very useful by-product.

Perhaps the biggest guess to make here if we want a quantitative estimate is the effective electrical distance between the layers. If it's a high voltage winding with a significant thickness of extra insulation between layers, say more than the conductor diameter, then your guess will be reasonably convincing. If it's just the wire enamel, or a single thin layer of tape, then it's a bit trickier, but an upper and lower bound ought to be possible, or even do a quick 2D FEA of this bit of geometry. If it's a scramble wound solenoid, then you're lost!

• Please stick to the CoC, calling people out is not appropriate. Thanks Commented Jun 24, 2022 at 20:27