# discharging capacitor through led [duplicate]

I want to discharge a capacitor (0.47μF) through a led (red led: 2V, 15mA) but I want the discharge remains for a while e.g. 60s or 30s. I charged the capacitor with 10V then I calculated the resistance's value (533Ωhm) to have 15mA current but the discharge was very quick.The led lighed up for just a moment. I know that if I put a bigger capacitor the discharge will remain for a longer time. But, I want the specific capacitor to use with 0.47μF capacitance and if it is possible to change the resistance.

So I try to use the function V(t)=Vo*e^(-t/RC) to calculate the RC.

The values for the function are: t=60s Vo=10V and V(t)=2V as a result RC=186,45s

So, if I put in the function C=0.47μF, the value of the R is calculated 79MΩhm but I can't succeed 15mA current with this resistance.

As a result the led never light up! Could you recommend a way to have the above capacitor but with a little bit longer discharge?

• The cap only has so much energy to light the LED with 15mA for that time. If you want longer time, supply more energy or use less energy per time. Feb 14, 2018 at 15:10

No. To quote a famous actor "It's physics Jim". Your formula has already told you that it is not possible There is nothing we can say which changes that.

You can gain a bit by using a more sensitive LED, some work good at 0.5mA (I thought you 15mA was way over the top anyway). But that is where it stops.

On the one hand, you want a resistor of 533 ohms.
On the other hand, you want a resistor of 79000000 ohms.

That's quite a span to overcome.

It seems that a small MOSfet transistor would be useful to span that difference. Its high-resistance gate has resistance larger than 79 Megohms to ground. And its low-resistance drain-to-source path is much lower than 533 ohms. So a circuit like this is possible:

simulate this circuit – Schematic created using CircuitLab When C1's input side rises to +10V, the MOSfet turns on, and 15 mA flows through R2 and the LED.
C1 will initially raise M1's gate voltage to +10V, but then start discharging through that very large R1, toward zero volt. After a time, M1's gate voltage will have dropped to about +2V, and M1 will start to turn off.

This is a conceptual circuit to show how the RC time constant of C1 x R1 could be used to stretch the ON period of a relatively high-current LED. In practice, that 79M resistor on the gate may be too large - the leakage current of that BS170 MOSfet may interfere.