1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Fullwave Rectifier

  1. Nov 26, 2009 #1
    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.

    Attached Files:

  2. jcsd
  3. Nov 27, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
  4. Nov 28, 2009 #3
    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
  5. Nov 28, 2009 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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).
  6. Nov 28, 2009 #5
    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
  7. Nov 28, 2009 #6
    btw, I get a combined peak current of about 126 mA
  8. Dec 1, 2009 #7
    Phrak, how did you work it out exactly please?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook