Unregulated power supply

1. Aug 18, 2016

BigMan52

1. The problem statement, all variables and given/known data
Specify ratings and values for the transformer, bridge rectifier and smoothing capacitor for a 230 V unregulated power supply with an output voltage of 12 V, output current of 2 A and a ripple factor of <0.024

2. Relevant equations
V * I
V*sqrt 2
C= I/2f*Vr
Id=(1+sqrt2)*IL

3. The attempt at a solution
The transformer VA rating should exceed 12*2 so 24VA
The peak secondary voltage will be 12 * sqrt 2 = 17V
The value of the capacitor will be: 2/2*50*0.815 = 24,540 micro-farads
Bridge rectifier will need to be able to handle the peak forward current of: (1+sqrt2)*2=4.83amps
Power dissipation is 2*2=4watts (assuming 2 volt drop across the bridge)
Have I approached this is the correct way?

2. Aug 18, 2016

tech99

It looks as if your PSU will give 15 volts DC out. 17 volts peak less 2 volts for the rectifier.
Also could you clarify your capacitor calculation - not sure about it.

3. Aug 18, 2016

BigMan52

OK thank you I thought that would be the output for the DC.
This is how I calculated the capacitor value:
Using C= I/2f*Vr:
I= 2 amps
f= 2 x 50
Vr = ripple voltage which I calculated by: 0.024 * 12 =0.288 V (rms) converted to peak to peak by 0.288*sqrt2*2 = 0.815 V

4. Aug 19, 2016

BigMan52

Is that the correct methodology?

5. Aug 19, 2016

tech99

Looks correct to me.