Ideal DC chopper average value

Discussion in 'Homework Help' started by DC55, Oct 3, 2010.

  1. DC55

    Thread Starter New Member

    Oct 2, 2010
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    I was hoping one of you kind souls out there could share some of your wisdom with me. I am taking a distance learning course on power electronics. I am sure this is an easy question but I am confused...so here goes

    The circuit is an ideal DC chopper. Question is at steady state, what is the avg value of Vr


    I think I could calculate the Duty ratio by 40us/100us = .4
    1) Would the average value of Vr be simply 10V*.4 = 4V? I think what is confusing to me is the term steady state for this case (switching on and off does not seem steady).
    2) Separately, I am trying to visualize what is going on in the circuit. When the switch is closed the current goes through the resistor and inductor. The resistor would have voltage drop of 10V and current of 10V/2ohm or 5A. The instant the switch was closed the inductor voltage would be 10V but would be quickly working it’s way to zero as it fully charges. Is the inductor the same polarity as the resistor? Or is it – 10V so the Vr would be 0V the instant the switch was closed. I know the polarity of the inductor change depending on charging or discharging.
    Thank you
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
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    In steady state ΔIon = ΔIoff , Von * ton = Voff * toff
    V = L * ΔI/Δt
    ΔIon = (10V * 40μs)/100μH = 4A
    So if we have steady state and we remember that in steady state, inductor acts as a short circuit.
    So for me average voltage is equal 0V becaues the net area under the voltage curve of an inductor must be equal 0 in steady state condition.
     
    DC55 likes this.
  3. Simsima

    New Member

    Oct 5, 2010
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    But, if the average voltage at steady-state equals to zero, then it's not working as dc chopper ???
     
  4. Ghar

    Active Member

    Mar 8, 2010
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    The word chopper is pretty ambiguous, it could be several things.

    That being said it's a pretty strange circuit... it's not really a DC output and it won't be sinusoidal either.
    It would be outputting some sort of AC voltage with 0 average. (Sinusoids have 0 average too)
     
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