Full Wave Rectification with Smoothing Capacitor

Thread Starter


Joined Mar 17, 2011

I'm trying to do this circuit, but have a bit of trouble figuring out some values. Before the AC voltage enters the bridge rectifier Im putting in a 10:1 step down transformer in phase. I know I'll have to divide my Vin by 10

I was just asking to see what the formula for finding out what Vdc if I measured across the load. thanks!


Joined Dec 20, 2007
All transformers have their output voltage printed on them except yours.

Assume that the transformer has an output voltage of 10VAC. Then its is 10VAC only when it has its rated load. It might be 12VAC or more with a very light load.

The rectifiers operate at the peak voltage which is the root of 2 higher than the transformer's output voltage. If the transformer output is 10VAC thenthe peak is 14.14V. The rectifiers have a forward voltage of about 1V each at full load (because they conduct for only a moment so their peak current is very high) so they charge the filter capacitor to 12.14VDC.
If the filter capacitor value is too low for the amount of output current then it will have a ripple voltage that reduces the average DC output voltage.

Cheap transformers saturate their core a little which causes the peaks of the sine-wave to be compressed resulting in the peak voltage being less than the calculation that uses the root of 2.