Capacitors, charge, and electrons

Hi guys,

I'm a CS major, but an EE major at heart

I know how to build circuits, I use SPICE, but I am now trying to
figure out the more physical properties of devices.

As for capacitors, I know how to charge them, discharge them, use them
for timing circuits, etc.

However, I'd like to know what it actually means at the physical level.

For example:

Let's take a Cap rated at 35V, 100uf.

Let's apply a voltage source to it, say 10V - in other words, charge it
up

Taking another Cap that is exactly the same, and charge it with 20V.

Now, let's take a Cap rated at 35V, but has 1000uf capacity.

We charge with 10V as before to fully charge.

What I would like to know now is what is the difference between all
three Caps at the physical property level?

For example, the two 100uf caps are charged with different voltage
levels. What does that mean exactly? Does the one charged with 20V
have more electrons stored than the one charged with 10V? Or are they
at different energy levels? What gives?

The one with 1000uf certainly has more 'charge', but again, does the
mean more electrons?

Please help shed some light on this.

Thanks,
Grug

 

[PeteS]

You are correct that the cap charged with the higher voltage has more
electrons on one plate (and a corresponding lack of them on the other
plate). The basic equations for capacitors (useful in both the
electrical and physics domains) are:

Q=CV, where Q = charge in coulombs, C the capacitance, and V the
voltage across the plates, which when integrated to get energy, gives

E = 1/2 [C (V^2)]

To know the current, or the rate of change of voltage, then take the
derivative of the first equation wrt time (Q/t = current, steady state)

I = C (dV/dt).

Giving the useful piece of information that charging a cap with a
current source gives a linear ramp of voltage (used extensively in
timing circuits).

A coulomb, incidentally, is roughly equal to 6.241506×10^18 electron
charges (One might say electrons for a generality).

plate
style capacitor is given by

C = (k(o) * k(r) * A) / d
where k(o) is the permittivity of free space (8.854 * 10^-12 F/m), k(r)
the relative permittivity of the dielectric, A the area of each plate
(one plate - remember they should be symmetric) and d the distance
between the plates.

That ignores leakage at the edges, but it's a starting point.

This comes in very handy when using power planes next to ground planes
in multilayer boards to figure out the effective capacitance (which I
have used as part of the local high speed filtering).

For an interesting experiment, charge two capacitors in parallel,
disconnect the supply and reconnect them in series. if the caps were
equal, you have a voltage doubler (because you now have half the
capacitance). Then do it with one cap 10 times the capacitance of the
other. Measure the new voltage. You'll be quite surprised. This trick
is used in Parametric amplifiers (common in Radars, for example).

Cheers

PeteS

 

[PeteS]

Let's do the actual mathematics

Assume we charge a 10uF and 1uF to 10V. The 10uF will have 100uC (Q=CV)
and the 1uF 10uC of charge. The total charge is 110uC.
Now lets stick them in series.
The first thing is to calculate the new effective capacitance =
C1*C2/(C1 + C2) = 0.909uF. A thing about series caps is the total
capacitance is always less than the smallest cap.

As Q = CV, then V = Q/C so 110uC/0.909uF = 121V (near enough).



So what you get is V(init) + [V(init] * C2/C1].
In Paramps, we use varactors (reverse biased diodes specifically
designed to be used as votage variable caps) to vary the effective
capacitance, and thence vary the plate voltage.

Cheers

PeteS