# Full-Wave Bridge Rectifier

Discussion in 'Homework Help' started by Yep, Dec 9, 2010.

1. ### Yep Thread Starter New Member

Dec 9, 2010
3
0
Hey Folks,

New to the forum and electronics as it happens ha!

I have been given a circuit and a series of values to compute. However, I am struggling with two and hope you can help...

The circuit is a full-wave bridge rectifier with a sinusoidal input voltage of 20V (RMS) @ 50Hz and a load of 50 Ohms.

Not present in the attachment is the smoothing capacitor of which I've been giving varying values (500/1000/2000uF).

I need to calculate the ripple voltage for each capacitance to compare with my simulation values, but I'm not sure on how to do this???

Also, I am unsure of how to calculate the supply current (RMS)???

Can anyone help or point me in the right direction?

Thanks!!!!!!

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3. ### Fraser_Integration Member

Nov 28, 2009
142
5
I think the RMS voltage will be the same as before. This should be obvious as it's root mean SQUARE so doesnt matter that before the negative half cycle was well negative, and this time it is positive voltage!

4. ### Yep Thread Starter New Member

Dec 9, 2010
3
0
Thanks for the responses so far guys!

@ t06afre...

I have and referred to a number of text books. The solution I have use is:

Vripple = I / 2fC

However, this gives me ripple voltages roughly double the simulation values I retrieved from PSPICE. Should I be worried? Why would the values be so different?

@ Fraser Integration...

Anymore input?

Thanks again to anyone who reads this and has any suggestions

5. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,222
This may be a typical beginner fail. I did it ones at a school test my self once many years ago.
A period is defined a the time from pattern start to it repeat it self. In full wave rectifier this means one period will be half of the source AC voltage period, or the frequency is doubled, in respect to the AC source voltage frequency

6. ### Yep Thread Starter New Member

Dec 9, 2010
3
0
t06afre, beginner I am ha! Thank you, that makes perfect sense. The values will be similar now

Any thoughts on calculating the supply current?