# Purely cubic capacitor

Discussion in 'General Electronics Chat' started by circuit_non, May 9, 2014.

1. ### circuit_non Thread Starter New Member

May 9, 2014
2
0
Hello,

I am wondering if it is possible to implement in circuit a capacitor which has the following voltage-charge relationship:
V(q)=C*q^3

Note, it is important that there is no linear term in there (i.e. I don't want a*q+C*q^3) etc.

Thank you.

2. ### wayneh Expert

Sep 9, 2010
12,388
3,245
It would be possible if there were such a thing.

Possible doesn't mean useful. I suppose there must be some niche application in which this behavior would be useful.

3. ### circuit_non Thread Starter New Member

May 9, 2014
2
0
Yes, I have application in mind. Do you know how to implement it or can you provide some sources ?

4. ### wayneh Expert

Sep 9, 2010
12,388
3,245
Oh, that's a different question. I know of no such thing. Depending what you're up to, you could get a microprocessor to simulate that behavior, i.e. to act like your hypothetical capacitor.

I think there is also such a thing as a voltage multiplier circuit in the analog world. You need (V/C)^3.

Sep 5, 2014
58
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Well, analog circuits were originally used to solve differential equations, so I'm sure there is a way to obtain that relationship. It seems like you know the circuit is going to be non-linear. I did some research and I don't think a capacitor will give you that relationship. However, you could use op-amps and diodes to do it. You may have to remove a linear term using a summing amplifier though. If your using a breadboard don't forget to attach capacitors across the voltage supply (I read a post that mentioned this).

You could chain two integral amplifiers together with a triangle wave input and you would obtain cubic sections if that's what your looking for.