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: Diode Rectification

  1. Feb 2, 2010 #1
    1. The question.
    Suppose that the input frequency into a full-wave diode bridge rectifier is 60Hz, and suppose that the RC time constant of the network filter capacitor and the load resistance is 10ms.
    Estimate the time after the peak input voltage when the diode shuts off.

    The circuit looks like this:
    http://newton.ex.ac.uk/teaching/cdhw/Electronics2/PHY2003-C14.2.gif [Broken]

    2. Relevant equations
    I have no idea where to start. I do know that the ripple voltage equation for full wave rectification is:

    [tex]\Delta V = \frac{I_{load}}{fC}[/tex]


    3. The attempt at a solution
    I know that after the voltage hits it's peak value, the voltage decays at both the rate of discharge of the capacitor and the sinusoidal input. I also know that I have to find the time it takes for the decreasing rate to become dependent only on the discharge of the capacitor.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 2, 2010 #2

    The Electrician

    User Avatar
    Gold Member

    Suppose you tell us what you're supposed to solve?
     
  4. Feb 2, 2010 #3
    Sorry, lol.

    Question: Estimate the time after the peak input voltage when the diode shuts off.
     
  5. Feb 2, 2010 #4

    The Electrician

    User Avatar
    Gold Member

    Suppose that you have a capacitor charged up to some voltage, and suddenly apply a resistor load. The voltage across the capacitor will then discharge exponentially. Do you know how to derive the expression for the voltage across the capacitor? Then, having that expression, do you know how to derive the initial rate of change (slope) of that voltage?

    Does this give you any ideas?
     
  6. Feb 2, 2010 #5
    Yeah sorta. I knew that I had to do:

    [tex]V_{c} = V_{p}(1-e^{\frac{-t}{RC}})[/tex]
    [tex]\frac{dV_{c}}{dt} = \frac{-V_{p}e^{\frac{-t}{RC}}}{RC}[/tex]

    Should I just set this equal to the rate of change of the voltage of the input sine wave? I.e.:
    [tex]\frac{dV_{c}}{dt} = \frac{-V_{p}e^{\frac{-t}{RC}}}{RC} = -(2\pi f) V_{c}cos{(2\pi f t)}[/tex]
     
    Last edited: Feb 2, 2010
  7. Feb 2, 2010 #6

    The Electrician

    User Avatar
    Gold Member

    That's what I would do. Work it out and see if the number you get makes sense; it should be just a little past the peak of the sine wave.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook