Half wave rectifier Ripple Voltage

[SteDolan]

Homework Statement

A capacitor of capacitance C is connected in parallel with a resistor of R ohms impedance. Calculate the size of the peak to peak voltage of the ripple produced relative to the peak output voltage V_p.

Homework Equations

Q=CV
dQ/dt = I
V=IR
V_c=V_pe^-t/RC (am i allowed to use this in an AC circuit?)
or V_c=-jI/R[itex]\omega[/itex]C

V = V_pcos([itex]\omega[/itex]t)

The Attempt at a Solution

Voltage drop across the resistor is just:

V = V_pcos([itex]\omega[/itex]t)

Voltage drop across the capacitor can be aproximated to:

V_c=V_p(1-e^-t/RC)

The alteration to the equation in 2 is to ensure the capacitor voltage is 0 at t=0.

At t=0

V_R - V_c = V_p

at t=T/2 or at 1/f

V_R - V_c = V_pe^-1/fRC

The ripple voltage will be the difference between these two:

V_Ripple= V_p(1-e^-/fRC)

 

[DaveE]

This old National Semiconductor Audio handbook has a great appendix that deals with these rectifier problems. Check it out.