# Capacitor Discharge

1. Dec 25, 2008

### Pythagorean

Ok, so I've heard two conflicting things about real capacitors and I'm trying to clear up my understanding. It's possible they were in different contexts.

This pertains to a RC circuit. It's a real capacitor so it will have some internal resistance too. It's also a high-voltage capacitor (rated at about 40kV)

1) I first heard that the larger the resistor is, the less time it will take for the capacitor to discharge (the problem being it could discharge too fast with too large of a resistor). My intuition didn't agree with this, but I looked at the equation:

V = Vo*exp(-t/RC)

as R --> 0, V = 0 for all t
as R --> inf, V = Vo for all t

And I found I could interpret to agree with his claim, but I'm not sure about my interpretation (that V = 0 means no current flows at all and V = Vo means the applied voltage flows)

But if you look at the current:

I = (Vo/R)*e(-t/RC)

as R --> 0, I blows up
as R --> inf, I = 0

With this interpretation, the opposite is true:

2) that a smaller resistor will discharge the capacitor faster. This makes more sense with the time constant:

tau = RC

(bigger R means bigger tau)

It's possible that 1) was in the context of energy dissipation, considering the power rating of the components (namely the resistor).

Any thoughts on this?

2. Dec 25, 2008

### Defennder

Hmm, so doesn't this tell you that you want R to be as small as possible so that V would approach 0 sooner? EDIT: The comments for R-> infinity below also holds for this part. Assuming R->inf implicitly assumes open circuit.

I don't think so. The e^(-t/RC) term dominates since it's exponential. So taken together this means that I actually decreases quicker for small R. I don't think you can examine the case for R -> infinity simply by letting taking R to infinity and keeping t constant. If R = infinity, that means that you have an open circuit and unsurprisingly, I=0 since no currrent flows. So this means you must find another way to represent large R without implicitly assuming an open-circuit capacitor.

I do believe this is the case.

3. Dec 25, 2008

### Pythagorean

well, in the R = 0 case, it goes to zero for all t, so I thought that meant it was independent of time. Of course, this doesn't tell us any information about t<0, so it could mean that the instantaneous event has already happened.

ha, of course, thank you.

4. Dec 25, 2008

### Defennder

If R is zero that just means you are connecting a short-circuit wire across the two capactor plates. So ignoring the time taken for the electrons to discharge through the 0 resistance wire (as we usually do in quasi-static circuit theory) the discharge occurs simultaneously. But of course all real wires do have a small resistance.

5. Dec 25, 2008

### Pythagorean

So where does it discharge to? The charge leaves the capacitor and what? distributes itself evenly throughout the whole circuit? Leaves the circuit? Comes back to the other side of the capacitor and finds an equilibrium?

I'm imagining just a capacitor in a loop

6. Dec 25, 2008

### Staff: Mentor

The charge separation across the capacitor zeros itself out. The work that was done to separate the charge in the first place (charge up the cap) is lost in thermal and radiative losses. Interestingly, even if the resistance of the discharge path is vanishingly small, the EM radiated by the inductance of the connection causes a loss.

7. Dec 25, 2008

### Defennder

Sorry for the typo. I meant instantaneously instead.

8. Dec 25, 2008

### Pythagorean

The way I'm imagining it in my mind is that free electrons stack up on one plate in the capacitor, and that's where the charge is created. When we close the circuit, the electrons return to the other side of the capacitor, balancing the charge. Is this a correct view?

If there were no resistance in the wires and capacitor (which I imagine is the source of thermal loss) then would it discharge slower through radiative loss?

9. Dec 26, 2008

### Defennder

It depends on what you mean by "balancing the charge". A capacitor has equal amounts of positive and negative charge on each plate. So short-circuiting the capacitor would simply produce an uncharged wire in a loop.

10. Dec 26, 2008

### Pythagorean

you mean to say positive charges on one side and an equal amount of negative charges on the other side right? (as opposed to positive and negative charges on both sides in equal numbers)

by balancing the charge I meant that the electrons return to the side that lacked electrons when the capacitor had a charge separation.

11. Dec 26, 2008

### Defennder

Well then we mean the same thing.

12. Dec 26, 2008

### Staff: Mentor

Yes. The electrons flow until there is no net charge on either plate, and therefore no Electric field between the plates storing energy.

Interesting point. You will have an RLC (tank) oscillation in the general case, with the damping factor provided by the R. If the R is negligible, then yes, there will be ringing after you close the switch, and the energy will oscillate back and forth (E field stored each opposite way on the cap, and current flowing each opposite way through the switch/wire).

It would be fun to calculate what the maximum (probably with R=0 and the cap+wire very short compared to an EM wavelength for the oscillation frequency, so the radiating system is not very good at launching the EM wave) discharge time is, versus the minimum (probably with R chosen for overdamping, or maybe critical damping?).

13. Dec 27, 2008

### Pythagorean

There's no component inductor in the circuit, so I imagine you're talking about inductance inherent to the wire and capacitor. I also recall that you need an inductor to have oscillations.

I was already assuming a pure case where there was no inductance either, I realize.

So... Is the radiative loss significantly due to the inductance? I imagine that the acceleration of the charges when the switch is closed (on the charged capacitor) would cause some radiative loss until they reach a constant velocity in the circuit and then again as they slow down.

Also, when the capacitor balances the charge, to the electrons travel across the capacitor or go around through the wire to the other side of the capacitor? I imagine the conductor's easier to travel than the dielectric, but my intuition's been hazy before.

14. Dec 27, 2008

### Staff: Mentor

A good antenna radiates because of the motion of the current, so yes, I guess you could say it's mostly due to the inductance (althought it's not usually stated that way). Good antenna structures have inductance and capacitance that matches the characteristic impedance of the feed line, and they also have physical structure that is the correct size to be resonant at the frequency of the drive signal. There is a propagation velocity for the RF currents on the antenna conductors (generally close to c), and this prop time and the physical length is what gives the resonant frequency. As previously stated, many good antennas use 1/4 wave elements. Like a dipole antenna has two opposing rods, each 1/4 wavelength. Or a vertical monopole is a 1/4 wave element over a good ground plane.

The electrons travel throught the plates' conductive material and through the external conductive connection (wire).

15. Dec 27, 2008

### Pythagorean

thank you much for input, this has cleared up a lot of my understanding about the stuff they never talked about in Physics II (my only real exposure to circuits. I took the physics E&M too, but's it's so abstract and each concept is isolated so it's difficult to trust that you're putting them together right all the time. We never studied a single circuit in EM, mostly just electromagnetic properties in their pure cases.

As a tangent that your antennae discussion reminded me of, have you heard of these solar antennae for sustainable energies?

http://manufacture-engineering.suite101.com/article.cfm/solar_antennae_that_generate_power

Last edited by a moderator: Apr 24, 2017
16. Dec 27, 2008

### Staff: Mentor

I read an issued patent back in about 1985(?) about using micro antenna arrays to do solar energy conversion. The problem with them lies in the conversion of the light-frequency oscillations to DC or low-frequency power. The antenna terminal voltages are on the order of a volt or two, so if you just rectify the antenna terminal signals, you blow most of your power in the rectification. They seem to recognize this in the current work on these arrays at the Idaho National Lab (I didn't even know we had a NL in Idaho!):

Still, with some creative thinking, maybe there's a way to stack enough antenna voltages to make the light-frequency(!) rectifying diode drop negligible... Also, since all of the photons hitting the antennas are not coherent, you lose a lot of efficiency that way too. It would be nice to re-phase the photons on their way to the target antennas, and add up their antanna terminal voltages in phase somehow.... That would be a great breakthrough, if it could be a cheap adder to the basic antenna planar structure.

BTW, if you are interested in learning more about circuits on your own, to get a more practical knowledgebase that you can use to augment your Physics degree and work, I'd recommend the AofE by H&H. You may have seen us talking about it before here in the EE forum. Read it cover-to-cover on your own, and you will have a *much* more intuitive understanding about basic circuit concepts:

.

Last edited by a moderator: Apr 24, 2017
17. Dec 27, 2008

### Pythagorean

Yeah, I've actually decided on going for a master's in EE, so that book might be a good asset for my transition. I'll start taking the deficiency classes fall 2009. I'd like to bring a physics view there.

Not sure what I plan to do after the master's, but I'm inclined towards going back to physics for a PhD in Applied Optics, (or stay in engineering and go Optical Engineering.)

This makes the mention of an optical solution to the antennae problem alluring for me.

As another alternative, would it really be that difficult to harvest heat from the antennae? Heating is a major energy vampire in fair chunk of the globe.

I wonder how many direct applications like this are possible with high frequency energy. Sounds like you could make a cool weapon with it too :tongue2:

18. Dec 27, 2008

### Staff: Mentor

Check out the hits for "Maxwell's Demon" with respect to thermal-to-energy conversion.

And I think the optical aspects of the resonant visible light antenna problem are very interesting. I'd encourage you to pursue that area of interest. Good questions and thread, Pythag.

19. Dec 28, 2008

### Pythagorean

why does such a small voltage "blow most of your power"? My real-world circuit understanding fails here (or maybe I'm not considering some theory).

This sounds like there could be a materials science solution. I've always wanted to be able to keep up with materials science but so much of the terminology flies over my head. I still try to read the latest list whenever I can.

It's almost as if you'd need a "demon" in this case, too (to grab photons of different phase and make them all one phase) and that once again, hiring the demon wouldn't be worth what he charges.

20. Dec 29, 2008

### Staff: Mentor

If the antenna terminal voltage were 2.0Vpp and you used a full-wave light-speed diode bridge to rectify that, you'd end up with about 2.0-1.2V = 0.8Vdc. You're burning most of the insolation power in the rectification process.

Excellent observation!