# Definition of capacitance in differential charge-voltage terms

Discussion in 'General Electronics Chat' started by aman92ullah, May 26, 2012.

1. ### aman92ullah Thread Starter New Member

Jan 19, 2011
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What is the difference between these two expressions:
C=Q/V
C=dQ/dV

I encountered the second one while studying MOSFET.

2. ### BillB3857 AAC Fanatic!

Feb 28, 2009
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I think the d in dQ/dT represents DELTA or change. Always willing to learn if I'm wrong! If they are the same value, they cancel each other, but if different, will have an effect on C

3. ### aman92ullah Thread Starter New Member

Jan 19, 2011
12
0
Yes, i realize 'd' means change. What I want to ask is in what situations do we use the 'differential' definition.

Can we use the analogy of average and instantaneous values, like we do for velocity?

In cases where the charge distribution is a spatially dependent, like in the case of the inversion layer under SiO2, the capacitance will vary from point to point; then,unlike parallel plate capacitor, average is different from instantaneous.
Is this reasoning correct?

4. ### crutschow Expert

Mar 14, 2008
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5,825
The second version is for a small change in charge for a small change in voltage (which still equals capacitance over that small range). They use that for semiconductor measurements where the incremental capacitance can change with DC bias. An extreme example of this is the varactor diode.

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5. ### #12 Expert

Nov 30, 2010
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It is also used to describe a defect in a capacitor. Oil filled motor capacitors have a large variation in capacitance with change in voltage, making it impossinble to get a good measurement with a small DC voltage. The only way to get a valid measurement is to apply the proper AC voltage and measure the resulting current.

6. ### steveb Senior Member

Jul 3, 2008
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The C=Q/V is the true definition of capacitance. The other relation is used when there is a nonlinear capacitance. In such cases, the effective capacitance for small AC signals (operating around a DC condition) can be related to dQ/dV.

For large signals, consider the capacitor circuit law, I=C dV/dt. This relation is not correct when capacitance is nonlinear, and the derivation of the correct law is as follows.

I=dQ/dt= d(CV)/dt=C dV/dt + V dC/dt=C dV/dt + V (dC/dV dV/dt)

Hence, I=C dV/dt (1+V/C dC/dV)

If we want to use the standard old (and incorrect) law, then we have to change it to

I=Ceff dV/dt where Ceff=C+V dC/dV

Here, the capacitance value Ceff is voltage dependent, and the value is clear if the function C versus V is known.

I don't think you should consider the spatially changing charge, or even time changing charge. Linear capacitance can have both of these. It is rather the way the spatial or dynamic changes in charge occur which determines if it is nonlinear. The simplest way to say it is that nonlinear capacitance varies in time (not space) and is dependent on charge (or equivalently dependent on voltage). In such cases the capacitance is nonlinear and the C=Q/V relation is not so simple any more.

It should be remembered that the concept of circuit capacitance is a way of wrapping all of the spatial dependence up into one number (for linear capacitance) or into a simple one dimensional function of voltage (for nonlinear capacitance).

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7. ### ErnieM AAC Fanatic!

Apr 24, 2011
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These two expressions are not interchangeable. The former is the fundamental definition of capacitance. Thus the second expression seems to be in error (in the general sense).

It would be interesting to see where this 2nd expression comes from.

8. ### crutschow Expert

Mar 14, 2008
20,546
5,825
The two are interchangeable and the second is not in error. Both equations define capacitance as the amount of charge moved (or stored) for a given change in voltage. In the first equation it is implied that the voltage change is from 0V to the measured voltage or V = dV and that the charge is the change in charge caused by going from 0V to V or Q = dQ.

The second is used to define the incremental capacitance at a particular operating point. For example, a varactor diode needs to be defined with the second equation so that the desired characteristic of a change in small signal capacitance with DC voltage can be quantified.

9. ### aman92ullah Thread Starter New Member

Jan 19, 2011
12
0
So, the thing I get is that capacitance can depend upon the voltage(that's the non-linear capacitor). Then it makes sense to calculate the 'small signal capacitance' around a bias point, where we are linearizing the capacitance around that point, like we do in case of BJT/MOS small signal model.
Thus if a small signal comes, it will feel the capacitance given by (dQ/dV) and not the absolute Q/V.

10. ### crutschow Expert

Mar 14, 2008
20,546
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I believe you've got it.

11. ### MrChips Moderator

Oct 2, 2009
17,600
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Imagine you charge a capacitor C with a voltage V.
The charge on the capacitor is Q.
Plot the graph of Q vs V.
Since a perfect capacitor is linear you would get a straight line going through the origin 0,0.
The capacitance is a constant value and is calculated as Q/V.
From the graph C is also the slope of the graph = (Q - 0)/(V - 0)
The slope of the graph is also the first derivative = dQ/dV which is the same at any point on the graph.
Thus C = Q/V = dQ/dV only if the capacitance is constant and the capacitor is linear with zero intercept.

12. ### Ron H AAC Fanatic!

Apr 14, 2005
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It can be enlightening to see the CV curve of a diode (or transistor junction). I have done this in simulations (see attached). Slew rates of 1v/usec or 1v/ns seem to work best. If you run the sim at both slew rates and get the same curve, you can be pretty sure it is accurate. When you check the plot against Cj0 (from the spice model), the results should also match.
The simulation subtracts leakage current, which can be significant in Schottky diodes. It ignores series resistance, which I believe is generally insignificant in this sim.

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