I’m charging a high voltage capacitor to 200v. I’m looking for a way to turn on an LED when the cap isn’t taking any more power or when it has reached the full 200v. This will serve as a visual indicator. Similar to the way a disposable camera has an LED that comes on when the flash is ready.
Would it be possible to use an op-amp comparator to compare the input voltage (200v) against the charging voltage of the capacitor? It seems logical but voltages are somewhat high.
Best Answer
The charge on a cap is directly proportional to its voltage. So yes, comparing the voltage to a threshold just below the "full" limit should get you the signal you want.
200 V is high, so you need divide it down to a managable level. This is done with a resistor divider, which is a term you can look up and get lots of information on. For example, with a resistor divider of 1 MΩ and 10 kΩ, the voltage is divided by 101. That means 200 V in will result in just about 2 V out. Use whatever reasonable voltage reference you have as one input to a comparator and the divider output as the other. Adjust the divider so that you get a little more than the voltage reference when the input is 200 V.
Note that this will drain the 200 V somewhat. In the example above, it will drain it by 200 µA, which will dissipate 40 mW. Use higher resistances to drain less current, but not so high that the output impedance gets so high that it picks up too much noise. I'd say 1 MΩ to 10 MΩ for the top resistor is about right.