1. Not finding help here? Sign up for a free 30min 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!

Diode Circuit

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    R = 1 kohm and Vs(t) is sinusoidal of (peak) amplitude 3 V. The diode is modelled by the series combination of an ideal diode and 0.7 V voltage source.
    For what percentage of time will the diode conduct?
    answer: 42.5

    8e13efe1b4.png

    2. Relevant equations


    3. The attempt at a solution
    4351a6abc0.jpg

    I'm getting 76%, the answer is meant to be 42.5%
     
  2. jcsd
  3. Dec 4, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Why don't you start with a sketch of the source voltage for one cycle. Then indicate the part or parts of the cycle where the diode can conduct.

    Hint: If you're getting a conduction percentage over 50% then you're saying that the diode can conduct during during at least part the negative half cycle of the AC waveform...does that seem reasonable?
     
  4. Dec 4, 2016 #3
    7ab41d21de.jpg

    Shaded part is where it can conduct
     
  5. Dec 4, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    First, you've labelled both the top and bottom of the voltage axis with 3 V. How is that possible? Shouldn't one of them be negative?

    Second, you're implying with the two shaded areas that the diode can conduct both forwards and backwards, but not for some band of voltages of either polarity near zero volts. Is that really what a diode does? If the diode were ideal with no forward voltage drop (so 0 V instead of 0.7 V), would the sine curve be entirely shaded and the diode conducting continuously for the whole cycle? What would be the difference between that diode and a piece of wire?
     
  6. Dec 4, 2016 #5
    I don't think my edited picture is showing up. Here it is:
    7ab41d21de.jpg
     
  7. Dec 4, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    It's better to add new material to the end of a post rather than overwriting or replacing previously published material, particularly if the previous material has already been commented on in later posts. Otherwise anyone who comes along later will have no idea what's happened and why the thread conversation refers to nonexistent things. You WILL get hit with infraction points if a moderator has to step in to fix things.

    So, your new diagram looks much better. How will you determine the portion of the cycle where the diode conducts? Can you determine the phase angle where it first turns on? (Note that you can use either time or angle to determine the fraction of the cycle. Angle is probably more straightforward, the angle going from 0 to 2 π over a whole cycle).
     
  8. Dec 4, 2016 #7
    5fd0ea761e.jpg

    Alright. I tried it with using t as the variable. Put this into my calculator. Im getting 48.62% which is wrong. Ho w do I do it using phase angle and ignoring the t?
     
  9. Dec 4, 2016 #8
    Here is how I got the t values:
    dea5e1d438.jpg
     
  10. Dec 4, 2016 #9

    gneill

    User Avatar

    Staff: Mentor

    Just solve for the angle when source voltage turns on the diode. So:

    ##3 sin(θ_1) = 0.7##

    Find ##θ_1##. Then find the corresponding angle ##θ_2## when it turns off. Hint: Symmetry is your friend here!

    upload_2016-12-4_19-6-0.png
     
  11. Dec 4, 2016 #10
    b8b02249ee.jpg
     
  12. Dec 4, 2016 #11
    omg. got the right answer. just realised my calculator was in degrees. derp
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Diode Circuit
  1. Diode circuit (Replies: 4)

  2. Diode circuit (Replies: 8)

  3. Diode circuit (Replies: 10)

  4. Diode Circuit (Replies: 3)

  5. Circuit with Diodes (Replies: 11)

Loading...