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Since my last post on Engineering-Beta, I have composed a few simple simple circuits involving capacitors(super-caps specifically), resistors and LED's.

The following image is of a branch from a circuit I designed,which highlights the discharge of a capacitor through a resistor and LED. I will be using this as a reference to all my questions:

Circuit C1.3, Branch-B

The moment That I allow current to flow from this circuit, Voltage will begin to drop, and current would begin to drop also. I know that the addition of resistance into the circuit can slow the loss of charge at the capacitor...

I wrote a bunch of stuff regarding all this, but it seems trivial now. Basically, I'm dealing with a rate of change that is not constant, and I'm trying to find precision in my circuit, I would like to be able to calculate the time duration of the discharge but, I believe that:

"Calculus is concerned with things that do not change at a constant rate" -TheMathPage.com

Sum everything up, all I know right now is intermediate algebra, and from all my google searches about this topic, they keep talking about derivatives and calculus. So, how can I figure out the rate of discharge of a capacitor? Or have I answered my own question already?

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  • $\begingroup$ Check out electronics.stackexchange.com. That might be the stack exchange you're looking for... $\endgroup$
    – BobT
    Sep 29, 2017 at 22:42
  • $\begingroup$ Thanks a lot, been posting to this community and I knew that this wasn't going to get me the input I needed. $\endgroup$
    – Iam Pyre
    Sep 29, 2017 at 23:06
  • $\begingroup$ What is LED forward current? What is the LED part number? $\endgroup$
    – Mahendra Gunawardena
    Oct 1, 2017 at 1:20
  • $\begingroup$ @MahendraGunawardena well the forward current is 25mA max, but I set things up the way I did to get the most observation time. You see, if I use resistors that only restricted charge flow from the capacitor at 25mA, and consider that Q=CV, it wouldn't be much to observe, the LED would flash for about fractions of a second. The higher the charge capacity, and the higher the resistance, the longer the LED will be able to shine. I understand the "mechanics " of the concept of a capacitors discharge, but to be able to precisely determine the point in time when it will fully be discharged... $\endgroup$
    – Iam Pyre
    Oct 1, 2017 at 3:30
  • $\begingroup$ @MahendraGunawardena ..., as someone else has commented, I believe also that The precision of calculation lies in utilizing concepts of calculus to equate the varying rate of change for the quantities Q=charge, V=Voltage, and I=current. I'm not a mathematician, but the moment current is allowed to flow from the capacitor, I realize that Q will change, and Q=CV, thus V=C/Q, and V now changes, and I=V/R, so I changes, and so on and so on with every moment that passes. Based on this, the rate of discharge couldn't be something constant, thus Algebra itself wouldn't be able to provide a solution. $\endgroup$
    – Iam Pyre
    Oct 1, 2017 at 3:40

1 Answer 1

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Correct you need to find the discharge rate of the capacitor. Take a look at Calculation of Discharge Time. The equation constant current discharge equation 1. For you equation total capacitance is is 2000uf. Your V1 is 15mA(400)+2V

enter image description here

Below are some good references on this topic

References:

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  • $\begingroup$ This will not be a constant current discharge. The current will fall with time, until at the point where there is 2V left on the capacitor, the current goes to zero. $\endgroup$
    – Jack B
    Oct 2, 2017 at 17:53
  • $\begingroup$ your are correct, without datasheets and specifications this is good start. If you have any suggestion or a better answer feel free to contribute in the form of an answer. $\endgroup$
    – Mahendra Gunawardena
    Oct 2, 2017 at 23:14

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