Full Bridge Rectifier

Thread Starter


Joined Apr 29, 2011
Hey guys,
I have a homework question that is confusing me.

I have to design a full wave bridge rectifier with the given the following,
-1 phase rectifier
-Voltage Source (Vs) = 230V, 50Hz
-Transformer = 230V/115V, 1.5kVA
-Ripple Factor = 3%
-All components can be considered ideal (no voltage drop across diodes)

a) Average voltage across the load
b) average current supplied to the load
c) load Resistance (neglecting ripple)
d)min filter capacitance required at rectifier output
and some more stuff

What I figure so far,

Vs----->Transformer------------>Rectifier------------>Resistive Load//Capacitor
230V-->230>/115V, 1.5kVA---->103.5V,(300/23)A-->7.935ohm//?

The output voltage from the rectifier Vd = 0.9 x Vs = 0.9 x 115 = 103.5V
Is = S/V = 1.5kVA/115V = (300/23)A
R = V/I = 103.5/(300/23) = 7.935 ohm

My confusion is weather my calculations are rms or average (dc) values.
I have also been having difficulty finding the relevant formulas for this type of rectifier.

From what I understand, the voltage after the rectifier is dc, which is why Vs is multiplied by 0.9. This is confusing because I thought dc voltage should be o.636 times the peak voltage...maybe I skipped a step somewhere???

I have the same kind of problem with my average current.

thanks guys :)

Thread Starter


Joined Apr 29, 2011
ok I get why its 0.9Vs now.
Assuming Vs is in rms
Vpk=sqr(2) Vrms
Vdc = 2Vm/pi = [2sqr(2)Vrms]/pi = 0.9Vm.

Still a little unsure weather I'm calculating IL correctly


Joined Aug 6, 2009
With an infinitely large reservoir capacitor the DC output would be equal to the peak AC voltage, or 162Vdc. (Your factor of 0.9 would only work if there were no reservoir capacitor)

Are you supposed to fully load the transformer? It is not clear from the question. If you are, then the maximum allowable RMS current in the secondary coil is:
1.5kVA/115V = 13A
But the maximum DC output current will be less, due to the poor power factor causing by using a reservoir capacitor.

I think there may be some missing information in your question. Also, what definition of ripple factor are you using?
Last edited:

Thread Starter


Joined Apr 29, 2011
Ok, so are you saying the average dc output voltage will also be equal to the peak voltage because the capacitor holds the charge?

This is really all the information we are given:
Source power = 230V, 50Hz, 1 phase main grid supplied though a 230V/115V, 1.5kVA two winding transformer. RMS ripple factor of the voltage at the load terminals has been limited to 3%. Consider components ideal.

The ripple factor I think he is talking about here is, the % ac component of the total waveform at the output terminals. An equation from my Lecture notes gives it as:
RF,rms= V,ripple/Vd

what do you think?


Joined Aug 6, 2009
Ok, if the problem assumes the components are ideal then the capacitor really will be charged up to the peak AC voltage on each half-cycle, or 162.6V.

If we assume the ripple voltage waveform is symmetrical, then the average DC voltage is:
Vdc(average) = Vpeak - Vripple/2.........................(1)
Vpeak = Vrms * sqr(2)
Vripple = peak-to-peak ripple voltage

Vripple = 0.03 * Vdc............................(2)
So substitute (2) into (1) and rearrange to get Vdc in terms of Vpeak. I make it 160.2V.

So Vdc = 160.2V and Vripple is 3% of that, or 4.8Vp-p.

The capacitor is charged twice every cycle, or every 10ms, and the voltage across it changes by 4.8V during that time. If you are given the average DC current then you can find the capacitance, or vice versa, but the question doesn't seem to tell you either..?

Thread Starter


Joined Apr 29, 2011
Yeah that makes sense. Thanks.
We haven't been given the average DC current.
This question keeps getting more weird