# Rectifier circuit questions

Discussion in 'Homework Help' started by Raikiri, Mar 2, 2010.

1. ### Raikiri Thread Starter Member

Mar 22, 2009
16
0

I was wondering if my answers for this were correct?

For 1. I doubled the frequency since its a full-rectifying circuit, f = 160Hz so T = 1/f = 6.25ms , conduction angle is 30 degrees so that gives 1.04 ms, discharge time = 5.2 ms

2. dVload/dt = Iload/C and delta Vpp = discharge time x (Iload/C) so C = 41666uF

3. P= I^2 x t = 8^2 x 5.2ms = 0.33 W

Could someone confirm if I'm correct please? Thanks!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Provided the conduction angle in degrees relates to the rectified frequency (160Hz) rather than the fundamental supply frequency (80Hz) then your answers to parts 1. & 2. are correct.

Part 3. is incorrect.

You have an average load current of 8A at a DC voltage of around 14.5 V which is a good deal more than 0.33W.

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3. ### Raikiri Thread Starter Member

Mar 22, 2009
16
0
Ah right, so I'd use P = VI then and use the RMS voltage for the v, thanks tnk.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
With respect to the (DC) voltage it's a bit more complicated than just taking the RMS secondary voltage.

You have to work out the peak secondary AC voltage, then subtract two diode drops to find the peak voltage at the capacitor. You know the ripple [1V] so you can then find the average DC capacitor (load) voltage - which is the actual voltage you multiply by the average load current [8A] to find the load power.

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