# Finding the capacitance of a full wave rectifier

I am currently learning electrical engineering,and I'm having trouble in finding the capacitance of a full wave rectifier so that it would match the graphs below.

From the image below, the input voltage is a 10 V sine wave and the load of the rectifier is 2000 ohms. I am having trouble finding the capacitance and the only idea I have tried is looking when the input voltage is at 0. I would like any help possible. Thank you.

First of all, it cannot be a full-bridge rectifier. It's a half-bridge with one diode because it passes only a half cycle of AC voltage.

Anyway, to calculate the capacitor, you should consider the discharging rate of the capacitor's voltage between two successive peaks. According to the graph, a discharge happens from 11V to 6V in a time of 180 (all values approximately.)

So, the solution would be achieved by putting these values into the discharging equation of a capacitor with a time-constant of $$\T= R_{load}*C\$$. Initial value of capacitor is 11V, final value would be zero. So, 180 is the time when the capacitor's voltage reaches 6V. The value of the capacitor would be derived in this way.

Something like this: Something you should remember forever:

I * t = C * V. It's actually the definition of the capacitance, only the charge is presented as what flows over time. This is valid for linear discharge, to which exponential discharge is very close at the beginning.

So probably in the academy, it's not accurate enough, but in real life, it's way more useful.

For example, ask the same question differently:

"I am building this rectifier, I estimate my load to be around 2kR, what capacitor should I chose so the voltage wouldn't dip below certain level?"

Then the answer anyway can't be very accurate, right? Because you are limited to available capacitors, etc. But also you will take all kinds of margins. Anyway, this is a calculation you can do quickly and simply.

consider a variation of the other answer:

I = C * dV/dT

and rearrange

C = I * dT/dV = 5mA * 1/60 second / 1 volt