Cpacitor discharge

BTW don't you get more resistant, more stamina when you get more active. I think it may apply to circuits too. :-)

Andy

On a sunny day (Thu, 15 Feb 2007 09:47:02 +0100) it happened Andrew Edge wrote in :

No it is not, you need to buy a clue.

Good.

...

OK, Andy, a little homework assignment for you. Get a small light bulb, like a light from a 20-lamp Christmas tree string. Current should be about 50mA at 12V. Connect it across a capacitor, about

50,000 microfarads or larger. Connect a 12VDC power source and a scope across it also. The lamp should be lighting to nominally full brilliance. Disconnect the power and trigger the scope at the same time and report back what the voltage as a function of time looks like, down to a volt or so. How well does the discharge voltage curve match the predicted exponential? For the given values (50,000uF, 50mA@12V = 240 ohms), we'd predict that the initial slope will be 1V/ second and we'll reach 12/e = 4.41V at 12 seconds. How closely does the measurement match that? If the initial slope is x volts/second, and the intial voltage is y volts, do we reach y/e volts in y/x seconds, as expected for an exponentail discharge?

If you were serious about always being willing to learn, go do the experiment and see what you learn. Don't take my word for anything; don't take Jan's word; don't take John's word. Go make some measurements, and believe those measurements. They won't lie to you. With a little larger capacitor you could even do the measurement with a volt meter instead of a scope, if you don't have a scope available. But don't try the expriment with a 50uF cap and expect to get the same sort of results.

Cheers, Tom

Hi Tom ... Now why experiment. Didn't Bertrand Russel say Mathematics takes us into the region of absolute necessity, to which not only the actual word, but every possible word, must conform.

Andy

Couldn't possibly. We're alive, and have self-repair ability.

Non-living stuff just wears out. )-;

Thanks, Rich

Andrew,

You are taking positions that show some amount of inexperience on your part. You should investigate more detail before committing so strongly to a position, There us an old expression "Caution - Engage brain before putting mouth in gear".

Consider the circuit as shown here:

Now take a look at the discharge curve. Wow...its linear:

Get the LTSpice Circuit definition file here:

- mkaras

Hi. You missed the position I was taking in the thread in answer to the Op's question. Just read it carefully again.

Besides, without all those fancy 7 components and exoticly Pulsed Power supplies included in the circuit given above I could get you the linear voltage output your circuit gives with only 2 components plus a capacitor.

Careful though on your statements. Before telling anyone that a linear output is not an exponential watch out though on one point. I have sometimes used a linear output as an approximation to an exponential and it works excellently. Check on the Math of exponentials to see why.

Andy

So you are saying this circuit won't work, within any time limits, whatsoever:

/ +12 ----o o-----+--->|---+ | | [R] | | | +--->|---+ I -> ~ 10 mA | | ----- [1F] +---In|LM317|Out---+ | ----- | | Adj [120R] | | | | +---------+ | | | [R] | | Gnd -----------+---------------------------+

The op is discharging a cap through an active load which he says draws constant current. Such a circuit is shown above.

After the switch has been closed for sufficient time to fully charge the capacitor, it is opened. Will the discharge be constant current, or does the circuit suddenly lose the ability to regulate?

Ed

You don't really think your logic will matter to him, do you, Ed?

Note that he wrote, "For your information a Capacitor is an open circuit to a constant current." That would mean that in a loop comprising a constant current source, a capacitor, and a resistor, there would be no current through the resistor...

Sigh.

Cheers, Tom

OK Andrew: The full truth . . . you are an idiot. And sadly, without the ears to hear it or the humility to acknowledge it, you will charge forth continuing to prove it.

- mkaras

tube,

minimum

such

My favorite is from Physics 101.

Put a 365 pF variable capacitor (set to maximum C) in parallel with a 365 pF fixed capacitor. Charge the combination to 100 VDC. Adjust the variable capacitor to 10 pF.

1. What is the resultant voltage across the combination?
2. What is the stored energy before and after?
3. How much work did it take to move the capacitor plates?

Charge is moved, no resistance involved.

I suppose not.

Yup. I still have't figured out how to charge a capacitor from a constant current source using his gem of an idea.

Maybe it's a flux capacitor? :-)

Ed

Do you call it "All fluxed up" when it fails? ;-)

-- Service to my country? Been there, Done that, and I've got my DD214 to prove it. Member of DAV #85.

Michael A. Terrell Central Florida

You forgot. Capacitors have internal resistance. Wires connecting resistance. Quite a lot of resistance is involved.

Andy

YES ... Z(Capacitor) = 1/jwC

w=0 for a direct current source =>

Z = Infinity =>No current flows through circuit. Can it be any easier.

Got a better argument . Lets listen to it.

Hi Ed. The Op said he wass discharging through a Microcontroller lead, Not through a Voltage regulator chip.

Andy

You are beyond medical help.

Andy

I didn't forget anything, in contrast to you, who never knew anything.

PLONK

Oh sounds like you let go of something. Thank God. Explains a lot of things. Now get to the flushing part, which sounds just like your name.

Andy