Sorry, I forgot the [joke][/joke] tags

Ah! Got me!

 

With a varactor diode the capacitance is inversely proportional to the square-root of the applied voltage.

If the capacitance increases with the voltage does that make it a negative capacitance?

 

Hi,

Wouldnt a negative capacitance just be interpreted as -C instead of just C ?
So in other words, it would act to cancel out any existing 'regular' capacitance, or at least reduce the existing capacitance which would make some circuits switch a lot faster. If this could be implemented in a CPU we might see much higher CPU speeds, provided the effect is good enough to cancel a significant amount of the normal capacitance. The transistors would have an equivalent of less capacitance, so they would be more efficient.
It is most likely an effect that comes from some set of operating principles where the equivalent LOOKS like a negative capacitance (ie the normal capacitance decreases and there is no other explanation).
This is rather new to me too though so dont hold me to any of this

 

In general, the amount of charge stored on a capacitor is a function of voltage.

Q = Q(v)

For a linear capacitor this is a linear function of the form

Q = kV

and we call the proportionality constant, k, the capacitance, C.

Q = CV

For a non-linear capacitor we have some other functions form. It might be

Q = aV + bV² + c·sqrt(V)

In this case, the overall capacitance, or the large-signal capacitance is

C = Q/V = a + bV + c/sqrt(V)

The small-signal capacitance is the slope of the Q(v) line at a particular voltage

dC = dQ/dV = a + 2bV + c/(2sqrt(V))

If either b or c is negative, then it is possible for the dC to be negative over some region even though the large signal capacitance might always be positive.

So let's say that you have a device that has 10C of charge stored when there is 10V across it. You have a large-signal capacitance of 1F. If this is a linear capacitor, then if you increase the voltage by 10mV you will get an additional 10mC of charge stored and so you have a small-signal capacitance that is also 1F. But let's say that if you increase the voltage by 10mV you get 20mC of additional charge storage. Now you have a small signal capacitance of 2F even though the large-signal capacitance is still very close to 1F (10.020C/10.010V). If you increase the voltage by 10mC and you only get 1mC of additional charge storage, then the small signal capacitance is only 0.1F even though the large-signal capacitance is still very close to 1F (10.001C/10.010V). As long as the total charge stored increases with voltage, then the small-signal capacitance is positive. But if the increase in voltage causes a phase transformation in the dielectric (or some other comparable mechanism) such that an increase in voltage causes less charge to be stored, then that will be a negative small-signal capacitance. So if you increase the voltage by 10mV and the charge stored goes DOWN by 20mC, then you have a small-signal capacitance of -2F even though you still have a large-signal capacitance very close to 1F (9.980C/10.010V).