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!!!!!!

Please help me and point me in the right direction (no pun intended)

 

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.

 

Hello, thanks for your response

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

 

Hello, thanks for your response

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

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

 

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

Sorry 12V/3k=4mA not 3mA!