Question

You have an ac voltage of 10 sin(60t). Design a full wave bridge rectifier which will give you an output voltage that varies a maximum between 8.5 and 8.55 V across the load. The load has a resistance of 1 kilo-Ohm.

Answer 1

As there are 10 V, for Vp1, that is the peak-voltage of the source:

$Vp1=10*\sqrt{2}=14.14 V$

Then, transformer's theory says that the relation of transformations is:

V1/V2=a

Where V1 is the voltage in the primary and V2 in the secondary.

V1=14.14 V

V2=8.55 V

a=1.65

Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:

8.5 -1.4=7.1 V

And this will be called VpL

Now we proceed to calculate the mean voltage:

$V_{mean}=VpL-(\frac{Vr}{2} )$

Where Vr is the ripple voltage, we asume that is 1 V

So, Vmean = 6.6 V

Then we have

Vmean/R= I mean

We have that R=1000 Ohm

Imedia=6.6 V/1000 Ohm

Imedia=6.6 mAmps

Finally, we can calculate the capacitor:

C=Q/Vr

C=Imean/(Vr*2f)

Where f is 60Hz

C=6.6mA/(1V*120)

C=5.5 uFarads

Therefore:

C=5.5 uFarads that works at 12 V