Cpacitor discharge

I think they are.

Not if the current is constant. That form applies to a capacitor discharged by a resistor, so that as the voltage falls, so does the current.

Another handy form of the capacitor equation that relates voltage and current is I=C*(dv/dt). The current is proportional to the time rate of change of voltage. If the current is not a constant (where the C * V = I * T solution fits) or proportional to the total capacitor voltage (where the exponential solution fits) this completely general form may allow a solution. Note that there are non electrolytic capacitors for low voltages that have capacitances very much higher than 10,000 uF for low current supply hold up applications. e.g.:

(1 - e(t))

Yes. For a constant current charge or drain, the charge/discharge rates will vary with instantaneous voltage. See equations here: (Do I don't have to type them in....)

I actually just researched this exact same question two weeks ago... Keep in mind electrolytics can easily vary +/- 20% or more in value, and also age. This will very likely affect your timing calculations.

Also, I happened upon a neat little online simulator where you could (for free) "virtually" build your circuit online and run some simple simulations. This might help you verify your assumptions / math.

Visit:

-mpm

This program can do the math for your:

Thanks all who replied.

My assumption is the load will draw a constant 10mA.

My load is a microcontrolller circuit.... and I have made a big assumption - constant 10mA.

I will have to either build or model the microcontroller circuit to see f - over the range 3.3v to 2.4v the current is constant.

Regards Joe

On a sunny day (Sun, 11 Feb 2007 12:54:26 +1100) it happened "Joe G \\(Home\\)" wrote in :

Sounds correct.

No, because you discharge wit ha _constant_ current, not in a resistor.

The only way way you can get a current out of a capacitor is by using a resistor.It may have a small value but a resistor it is. All currents and voltages in an RC circuit have the form i= io*exp -t/Req*C v=voexp-t/Req*C

Andy

(snip)

You might want to rethink that generality a bit. The universe is more complicated than that.

On a sunny day (Tue, 13 Feb 2007 18:49:41 +0100) it happened Andrew Edge wrote in :

No way, 'current' equals electrons. There are many ways other then 'resistors'. Look up vacuum tube, transistor, semiconductor, and in this context especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear with the increase in resistance.

In something that allows control of the current, say a transistor, or tube, the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some minimum voltage, and then only depends on Ib. Same for a 'penthode' tube. Many integrated circuits use current sources and current mirrors, and as such Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

They all have rsistances.

Ouch ... That one has a damn large resistance ... Infinity! .

You said it.

Andy

It sure is. I am always willing to learn ... so give me an example.

Andy

Connect a super conducting inductor across it. There are lots of active circuit and nonlinear loads that would also pass a current that is not instantaneously proportional to the voltage. A constant power, switching regulator will draw an increasing current as the capacitor voltage falls.

In addition to what John wrote, consider LC filters and resonant tank circuits. There may be some resistance in there, but the reactance of the inductors generally dominates over the effects of the resistance.

Consider capacitors used for power factor correction across a low- frequency AC power line; the current is proportional to the rate of change of voltage, driven by the power line. Half the time, current is charging the capacitor and half the time, it's "coming out" and discharging the capacitor. It would be very difficult to relate the capacitor current to what happens in some resistor in that circuit, and resistance is not required to adequately understand what's going on in it.

You wrote in your previous posting about the exponential behaviour of current and voltage in RC circuits, but that's all derivable from the much more general

Q = C*V ----> i(t) = C*dv/dt + v*dC/dt and of course v=i*R.

Cheers, Tom

The Op is talking about discharging a capacitor. How do you normally do that? I use a resistance. If you use LC filters or resonant tank circuits well that is your choice. The case in consideration doesn't mention any of your choices if I remember right. Besides I would challenge that statement that the reactances dominate. If resistance was not a factor in an LC circuit the circuit would oscillate indefinitely AND that doesn't happen. Dissipation of energy in the resistances takes place. In the case the Op is considering, we talk about the reactance of the capacitor, and it DOES NOT dominate over the resistance.

It is quite a simple thing for an engineer. A Kirchoff loop equation with the Voltage and Current phasors would get you there in a second. Simulators use such equations. So why shouldn't we. But that is another case. There is no outside Power supply in our case ... discharge is the keyword. It is a sourceless circuit. The equations and the relations are a completely different story.

Yes for sure, otherwise how we would have ever got those relations.

Andy

Unfortunately Radio Shack doesn't sell super conducting inductors. Maybe a liquid Helium cooled super conductor would work.

I don't have the impression the Op wanted to add Active circuits to the capacitor.

Andy

On a sunny day (Wed, 14 Feb 2007 09:40:17 +0100) it happened Andrew Edge wrote in :

And states the current is _constant_. snip rest.

On a sunny day (Wed, 14 Feb 2007 11:54:45 +0000) it happened Eeyore wrote in :

I think it is safe to agree with that :-)

A few things to consider:

Electrolytics have very wide tolerances- -20 +200 % sometimes. So if a "10,000uF" capacitor works for you, that may only be because it's actually 18,553uF.

Lookup the cap vs. temperature curve. If your gadget has to work at

Do you really need 10mA? Can you run your chip at a lower clock rate? if it's CMOS, the current drain will go waaay down if you can slow down the clock.

On Feb 14, 3:48 am, Andrew Edge wrote: (snip\\0

You missed a post. He says he wants to run a microcontroller on the capacitor.

Doesn't get much more active than that. :-)

Lessee...the OP stated that the current is constant, not i=v/R.

Lessee...you asked for examples that disproved your statement, "The only way way you can get a current out of a capacitor is by using a resistor." That seems pretty absolute to me, and not a reference to the OP's question about the capacitor discharging at a constant current. I think we've given you several good examples. Take 'em or leave 'em.

As I re-read the OP's original posting, I think he's _very_ clear that the current discharging the capacitor is constant, and he posts the correct equations for calculating the discharge as a result. He does ask about exponential discharge, but that would be only if the load looks like R across the cap, and he's already stated it doesn't, by writing that it's a constant current.

Cheers, Tom