I have this case where my V1 supply voltage is constant at 14V.
The C1 capacitor now charges up to 14V and assume my capacitor C1 is 35V rated.
But I have this condition where I need to check my circuit in case of high voltage say.
This is the pulse I need to apply.
My questions :
- V=E.(1-e^((-T)/(R.C))) - This is the formula I use to calculate the voltage at the capacitor after 50us. If I plug in my values of R (2ohms - should in the circuit which represents the resistance mentioned in the pulse specification), C(123.2uF - taken considering ALL tolerances),E and T, I get a value around 20.937V.
I have a query. My capacitor is rated properly if I measure this voltage after 50us. Suppose I measure like instantaneously, say after 10us, my capacitor fails. Should I not measure after 10us? If not, why should I not measure after 10us?
If I measure the voltage before that 50us, the voltage is going to be higher than the capacitor rated voltage. It won't fail? Or is it very small time for the capacitor to handle or accumulate?
- After 50us, my input V1 voltage goes back to 14V. But now my capacitor is holding 20.937V. What happens to this 20.937V? Assume I have connected a load across the capacitor. Does the excess voltage go to the load / source? If I have a reverse protection diode between the source and the capacitor, then where would this 20.937V go when the input voltage has come back to nominal 14V?