# Fullwave Rectifier

1. Nov 26, 2009

### exis

1. The problem statement, all variables and given/known data

A 50V peak AC signal at 50Hz is fed to a 1:1 turns ratio transformer full-wave rectifier circuit. The full-wave rectifier waveform is smoothed by the use of a capacitor C=47uF, R=4.3kohms.

I need to calculate the peak diode current (however before this i was asked to calculate the average voltage across the load resistor R and the peak-peak rippe voltage across R)

3. The attempt at a solution

Vrippe(peak-peak) = Vpeak/(2fRC) = 50/(2*50*4300*47*10^-6) = 2.47V
Vdc = Vpeak - Vripple(peak-peak)/2 = 50 - 2.47/2 = 48.76V

I reasoned that the maximum current through the diode occurs when Vin=Vpeak. Therefore I_max=Vpeak/R=50/4300=11.6mA.

I tried considering the current through the capacitor aswell but didnt know how to find the max current through it.

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2. Nov 27, 2009

### Redbelly98

Staff Emeritus
Note: are you to assume the diodes have zero voltage across them, or would they have 0.6 to 0.7 V across them?

Once you answer that, the key here is: You know the peak voltage across the resistor. You know R for the resistor. Find I for the resistor.

3. Nov 28, 2009

### exis

I assumed the diodes are ideal, therefore zero voltage drop across them. I found the peak diode current exactly the way you just said and the answer I got isn't correct. Peak voltage = 50V, R=4300ohms, therefore I_max = 11.6mA. The answer i should be getting is 127mA

4. Nov 28, 2009

### Redbelly98

Staff Emeritus
Weird. Perhaps you are also supposed to calculate a current going into the capacitor when the voltage increases from it's minimum to maximum values (i.e., the rising part of the ripple in the voltage).

5. Nov 28, 2009

### Phrak

I usually work these things by first guestimating then refining. The rough, peak-to-peak ripple is about 0.6 volts across the capacitor--very small.

The discharge curve is nearly a straight line and will intercept the rectified voltage curve very close to 90 degrees. About 81 degrees. From dv/dt, at this point on the curve, you can obtain the peak capacitor current.

I(t) = C dv/dt

6. Nov 28, 2009

### Phrak

btw, I get a combined peak current of about 126 mA

7. Dec 1, 2009

### exis

Phrak, how did you work it out exactly please?