# How do you find the peak-to-peak ripple voltage of a full wave bridge rectifier?

The 32VRMS secondary of a transformer feeds a full wave rectifier and capacitor filter as shown in the circuit.

The peak-to-peak ripple of the output voltage of this circuit needs to be less than 1V when it delivers a current of 3A. A filter capacitor of about 30 000 uF is required. I need to find the frequency.

I am unsure of the formula for finding peak-to-peak ripple voltage of a full wave bridge rectifier.

Is this correct:

V(p-p) = I/(2*f*C)


Therefore, f = 50 Hz.

I am unsure because I have seen the equation:

V(p-p) = Vm/(2*f* R*C)


used.

Which equation is correct?

• Have you tried to simulate it? Is this homework? Commented Jun 22, 2020 at 7:59
• It's not homework, i am just confused which formula to use
– John
Commented Jun 22, 2020 at 8:09
• If you work at understanding the circuit from first principles you'll never have to remember the formulas or how to apply them. Commented Jun 22, 2020 at 8:57
• But i don't understand the circuit from first principles
– John
Commented Jun 22, 2020 at 9:05
• The first equation doesn't have any term denoting the load or the current drawn. A load which draws more current will definitely cause a bigger dip in the capacitor voltage. So first equation is probably wrong. In the second equation, there is an R in the denominator. A smaller R draws more current and hence more dip in the capacitor voltage. Hence the second equation is more likely correct. Derivations are usually given in most textbooks.
– AJN
Commented Jun 22, 2020 at 12:32