# Full wave Rectification Problem!

Discussion in 'Homework Help' started by big_tobacco, Oct 25, 2009.

1. ### big_tobacco Thread Starter New Member

Mar 9, 2009
2
0
Hello,

I'm currently revising some past papers for an up and coming phase test. Basically the question reads:

"A full wave rectifier using ideal diodes is fed from a 50 (Hz) supply and has a smoothing capacitor of 1000 (microfarads). The rectifier should have an average DC output voltage of VL = 12 (V) when connected to a load of 3 (KiloOhms). Calculate the following....

A) The value of Vmax
B) The ripple voltage Vr
C) The percentage regulation

I can probabaly B & C (as I have obvious equations for both) but it's the first question I'm trying to solve.....fortunately I had my hair cut this morning otherwise I would have pulled it all out!!!!!!

2. ### Thav Member

Oct 13, 2009
82
0
If you can calculate Vr, I would imagine Vmax would be simply $V_{\mathrm{Max}} = \langle V \rangle + \frac{V_r}{2}$ with $\langle V \rangle$ representing the average value (12V) you're shooting for.

3. ### big_tobacco Thread Starter New Member

Mar 9, 2009
2
0

To be honest I haven't seen that equation, plus this answer is prior to knowing what Vr is.

I initially thought of rearranging

VL = Vm ℮ -T/CR

i.e. giving Vm = VL / ℮ -T/CR

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
One approach might be:

AT 50 Hz the full wave rectified component is 100Hz - corresponding to a 10ms interval.

Assume the capacitor discharges over this 10ms interval at 12V into the 3kΩ resistor at a constant rate - not quite true but near enough. The capacitor charging time is considered negligibly small - not quite true but near enough.

So the assumed constant capacitor discharge current is 12V/3k = 3mA

Using I=3mA=CdV/dt = CVr/dt where C=1000uF and dt=10ms

re-arrange to find the ripple Vr

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Sorry 12V/3k=4mA not 3mA!