Full Wave Rectifier

Shima · Mar 3, 2013

Hi there,

I appritiate if anyone can give me the respond to the two questions I have in the problem below.

Design a full wave rectifier power supply that has an input of 120 vrms, 60 HZ, and requires a max voltage output of 17V and min output voltage of 12V. The power supply will provide power to an electronic circuit that requires a constant current of 1A. determine the winding ratio and capacitor size assuming the diodes and transformer are ideal.

Here is the answer:

Vin = 120x1.41 = 169.7 V

2a = 169.7/17 = 10

a = 5 [ the formula for ratio is (a) = (V1 / V2) why do we use 2a )

R load = 12/1 = 12 ohm for worst case

C= (Vmax) /(delta V x f x RL )

Delta V = 17-12 =5

C = (17/ (5x 2 x 60x 12) ( where did the 2 came from.)

benbo · Mar 4, 2013

For the first question, I think the 169.7 is the p-p voltage, and for the full wave rectifier you plan using only the peak voltage so they divide it by 2. Is that a possiblity?

Apparently the ripple frequency for a full wave rectifier is twice the line frequency. So the f in the equation would be 120 or 2 x 60.

Shima · Mar 5, 2013

Thanks for the respond. You are correct the frequency for a full wave rectifier is twice the line frequency. I just not sure about the first question. If the voltage is line to line, then the max output voltage is also line to line. IN this case in capacitor formula, the max voltage needs to be converted to phase voltage as well. Thanks again.

benbo · Mar 5, 2013

[ If the voltage is line to line, then the max output voltage is also line to line./quote]
Not sure how to explain it, but if you think about the waveform for a full wave rectifier. The wave comes off the secondary of the transformer as p-p, but then the diodes basically flips the negative portion of the sine wave up - so you get a bunch of little bumps that are only half the p-p value. So what you end up after you smooth it out and regulate it is a DC voltage that is the same as the peak voltage.

DK PE · Mar 5, 2013

I guess count me in as confused by solution of a = 5. If I have a 1:10 turns ratio and 120 V RMS in, I get a sine wave out with a peak-peak voltage of 339/10 or ~ 34 volts. If I then run this into a FWR, I get a waveform with a zero - peak magnitude of 1/2 that or 17V and the frequency as stated is 120 Hz. With a textbook capacitor, maybe you wouldn't get too much droop but I don't know where to buy those.

I think this is a pretty poor problem. I looked in one good reference and their rule of thumb would be you would want ~ 19VRMS into the FWR in order to get 17VDC out. If you apply that rule of thumb you get a turns ratio of ~ 6.4. The key in my mind is you know the what the waveforms look like the magnitudes.

So in the end, I don't follow the a = 5 or I must be missing something.

benbo · Mar 5, 2013

You're right. To start with, I messed up the RMS conversion. But I oversimplified the design anyway. Since I'm obviosuly very rusty I googled it and came up with this, which agrees with what you say above. So I'm not sure where the 2 came from.

http://www.electronics-tutorials.ws/diode/diode_6.html