# Capacitors, charge, and electrons

Discussion in 'Electronic Basics' started by [email protected], Apr 4, 2005.

1. ### Guest

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

2. ### BobGuest

Grug,

The energy (E) stored in a capacitor is:

E=1/2 * C * V^2

where C is the capacitance (in Farads)
V is the voltage

You can do the math for yourself to see which cap has more or less energy
stored in it.

One thing to remember is that a capacitor always has zero net charge in it.
I know this seems strange, but it's true. The deletion of every electron
(from the more positive plate(s) of the capacitor) means an addition of
another electron on the other plate(s) (its more negative side). This
"separation" of charge is due to the transfer of energy from the external
source (in your case, your power supply). The more the charge is separated
then the more voltage there is across it (and vice versa).

Bob

3. ### PeteSGuest

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).
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

4. ### Michael RedmannGuest

Remarkable! If my computation is right, the factor is 11/sqrt(10). So
the total voltage will be 34.8 volts, which is more than the sum of
every single voltage. I'm ashamed but up to now I thought resonance is
the only effect to get voltages up.

Regards

5. ### PeteSGuest

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

6. ### Michael RedmannGuest

Hi Pete

I did it by calculating the total energy stored in both capacitors which
is 5.5 mJ. The corresponding voltage with both caps in series is V = V0
* (C1+C2)/sqrt(C1*C2) after charging each cap to V0.

So, what's correct? Conservation of charge or energy?

Let's test your calculation with two equal capacitors, say 10 uF at 10
volts. If you were right voltage would be 200 uC/5 uF = 40 V ! That's
obviously wrong. Of course the correct answer is 20 V. That's the same
voltage you get using my formula.

Regards

7. ### BobGuest

If the method is:

1) Place two paralled caps, of different values, across a voltage source (of
magnitude V)
2) Remove the voltage source
3) Separate the caps and reconnect them in series
4) Measure the voltage across the series combination

It doesn't make sense that the series voltage would be anything other than
2*V. How can the voltage across either of the caps suddenly increase when
you separate them?

Also, I tried it. I used a 220uF cap and a 2200uF cap. The supply was 5V.
After I separated them, and then place them in series, I measured 10V across
the series combination.

Was this the method described for the so-called parametric amplifier? I did
a little searching on this subject, but never found this method described
for parametric amplifiers.

???

Bob

8. ### Kevin AylwardGuest

Nope. Its because one cap has V and so does the other, the sum is 2V.
I would be, if it were anything other than V1 + V2. If this were not so
connecting different sized batteries in series would not produce V1 +
V2.

Charged caps look like voltage sources. Voltage sources in series add
voltages. I am at a complete loss as to why you think one can generate
high voltages in this manner. Maybe I have missed something in the last
25 years...
Oh?

Parametric amplifies typically use a non-linear capacitor, a pump
oscillator and a signal. The mixing produces gain at the signal
frequency due to the 3rd harmonic term non-linearity, much the same as
in http://www.anasoft.co.uk/EE/tapebias/tapebias.html.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

9. ### Kevin AylwardGuest

Amazing...Simply connecting two charged caps, of different capacitance
values produces, a higher voltage then either. Wow. Wonders will never
cease. Like, dude, you haven't ever tried this have you...You out to
lunch on this one.
What are you actually trying to say here?

A voltage varies a capacitor and hence another voltage varies means
what, exactly?

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

10. ### Kevin AylwardGuest

They cant.
Yep. No getting around emperical evidence.
Couldn't be. Parametric amplifiers use a different principle, more or
less
It works, essentially, the same as
http://www.anasoft.co.uk/EE/tapebias/tapebias.html. Change magnetic
non-linearity to capacitor non-linearity.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

11. ### Michael RedmannGuest

Bob and Kevin: you're absolutely right! After replying on Pete's post I
was in doubt if this strange effect could really be true. It can't.

But what's wrong with Pete's and my computation. Neither conservation of
charge nor of energy lead to the correct answer. (Okay, if nothing
really happens there's nothing to compute ;-)

regards

12. ### Kevin AylwardGuest

Of course they do. One just has to know what one is modelling!

The root cause of the confusion is that the series calculation of
capacitance is completely meaningless. I have no idea what Pete was
trying to do there.

Moving a cap and connecting one of its ends to a potential does nothing
to the charge on it or voltage across it. Hint: the voltage on a cap
don't want to change. It will conserve both charge and energy.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

13. ### John FieldsGuest

---
No analogy is perfect, but let's say that instead of capacitors we
have three pitchers which are right circular cylinders: two with
volumes (capacitances) of 100cm^3, one with a volume of 1000cm^3, and
all three with heights (voltage ratings) of 35cm. Let's also say
we've attached identical pressure gauges (voltmeters) which read gage
pressures from zero g/cm² to 35g/cm² to the bottoms of the pitchers,
and that the pitchers are all empty (discharged) and resting on their
bottoms with the gauges reading zero.

Now let's say that we pour a fluid (charge) with a specific gravity of
1g/cm^3 into the first pitcher until its pressure gauge reads 10g/cm²
(10V). Since the density of the fluid is 1g/cm^3, the height of the
fluid column will be 10cm, and the pitcher will be about 28.57% full.

If we now pour enough fluid into the second pitcher to cause its
pressure gauge to read 20g/cm², the height of the fluid column will be
20cm and the pitcher will be about 57.1% full, so the quantity of
fluid (charge) in the pitcher with the gauge reading twice the
pressure (charged to twice the voltage) will be double that of the
other pitcher.

Now let's pour enough fluid into the third pitcher to make its
pressure gauge read 10g/cm². As with the first pitcher, since the
density of the fluid is the same, the fluid column will be 10cm high,
but because the pitcher has a larger capacity it will contain more
fluid. How much more? Since the first pitcher had a capacity of
100cm^3 and the third pitcher a capacity ten times greater than that,
for the same height of column the third will contain ten times as much
fluid as the first.

So, since the quantity (Q)of fluid in the pitcher is going to depend
on the capacity (C) of the pitcher and the pressure (Voltage) exerted
by the fluid, we can write:

Q = CV

Where, for a capacitor,

Q = the quantity of charge in the capacitor in coulombs,
C = the capacitance of the capacitor in farads, and
V = the voltage across the capacitor in volts.

For the first capacitor, then:

Q = C * V = 100E-6F * 10V = 1E-3coul = 1 millicoulomb

for the second:

Q = C * V = 100E-6F * 20V = 2E-3coul = 2 millicoulomb

and for the third:

Q = C * V = 1000E-6F * 10V = 1E-2coul = 10 millicoulombs

Just like the fluid being poured into the pitchers was made out of
molecules, the fluid charging the capacitors will be made out of
electrons, with one coulomb of charge containing about 6.02E18
electrons.

14. ### Rich GriseGuest

But if you put a 220uF and 2200uF in series, and charge the series string
to, say, 11 volts, you'd get 10 and 1, right? Just an ordinary capacitive
voltage divider.

Thanks,
Rich

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