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Half-wave bridge rectifier

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Waveform_halfwave_rectifier.png

    Hello all,
    Looking for some help regarding half wave bridge rectifiers. I'm having trouble grasping how to determine the average output voltage for the sin function.

    Let Vd=voltage drop across the diode. Let θ=angle conduction begins. Let Vs=denote the ac supplied voltage source with peak voltage of A. Hence, the ac source operates with the sinusoid Vs=Asinθ. Can someone perhaps explain to me the theory behind determining the average output voltage?

    I know that sinθ=Vd/A, and the angle of conduction occurs from θ to π-θ.

    But how does the textbook come up with this:

    Vo,average=(1/π)∫(Asinθ-Vd)dθ from θ ---> π-θ

    I dont really see how
     
  2. jcsd
  3. Oct 2, 2011 #2
    Never mind. I figured it out. However, can someone explain how the initial conducting angle is determined to be sin(theta)=Vd/A or in other words,
    the intial conducting angle is equal to the inverse sine of the voltage drop of the diode divided by the peak amplitude.

    My textbook just randomly states that, and it doesnt provide any proof.


    Also, occasionally they will take the integral from 0-->pi (ignoring the initial conducting angle) and other times they will go from the initial conducting angle to (pi - initial conducting angle).... Whats with that?
     
    Last edited: Oct 2, 2011
  4. Oct 2, 2011 #3
    Hello Sandy.Bridge,
    First of all, when calculating the average output for a rectifier, you want to average the voltage over one complete cycle. Therefore, for the graph you have provided, you want to integrate over 2π.

    For a more precise result, your limits of integration would be as follows:
    [itex]\overline{V_o}=\frac{1}{2\pi}\int^{\pi-\theta}_{\theta}[Asin(\theta)+V_d]d\theta[/itex]


    Note: if you were to integrate from 0->pi instead, the result would still be very accurate. Hence, it is reasonable to integrate from 0->pi to reduce the calculations..
    Regards.
     
    Last edited: Oct 2, 2011
  5. Oct 2, 2011 #4

    gneill

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    Staff: Mentor

    The diode "turns on" when the voltage across it meets or exceeds the required forward bias, Vd. This occurs when the input voltage reaches Vd. (A similar occurrence at the end of the cycle is when the input voltage drops to Vd and then the diode cuts off again). When the diode is cut off no current flows and no voltage is developed across the load.

    The input is at Vd when [itex] V_d = A\;sin(\theta)[/itex], or [itex] sin(\theta) = V_d/A[/itex].

    The average voltage is computed over an entire cycle. But note that for half-wave rectification all the "action" occurs while the diode is conducting, between the computed conduction angles. The rest of the time the output voltage will be zero. The integral can sum up the voltage through just the conduction phase, but it should still be divided by the period: [itex]2 \pi [/itex] radians.
     
  6. Oct 2, 2011 #5
    @gneil, is it okay to simplify calculations by calculating from 0->pi rather than theta->(pi-theta)?
     
  7. Oct 2, 2011 #6

    gneill

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    Sure, as long as long as the "tail" bits are zero they don't contribute to the sum. Writing it that way tends to underscore the domain over which the integral applies. In practice (in order to actually solve the integral) it would probably be split up into several parts anyways, covering regions where particular conditions hold or functions apply. Like here the bit from 0 to theta would simply be zero.
     
  8. Oct 4, 2011 #7
    Alright, so in reality it would make more sense to set the limits of integration from the initial conduction angle to pi-initial conduction angle, correct? When I use 0-->pi as the limits, I get an answer that is off a little bit. Not sure what happens there.
     
  9. Oct 4, 2011 #8

    gneill

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    It's probably off a little bit because your function being integrated doesn't "know" that it should be zero for the "tails", and instead keeps following the sine function into negative territory for those bits.
     
  10. Oct 4, 2011 #9
    Alright. Thanks. One last question: should I be concerned with the derivation of the conduction angle? My textbook merely states that sinθ=Vd/Vp, however, it doesn't really explain why.
     
  11. Oct 4, 2011 #10

    gneill

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    See post #4 in this thread.
     
  12. Oct 4, 2011 #11
    I always thought that the average of an integral was the following:
    [itex]\overline{f}=\frac{1}{\theta_2 - \theta_1}\int^{\theta_2}_{\theta_1} sin\theta d\theta[/itex]

    However, in doing that, it would not be dividing the integral by the period, it would be dividing by the difference of the angles, which would not be pi or 2pi.
     
  13. Oct 4, 2011 #12

    gneill

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    Staff: Mentor

    That's true. And you'd only be computing the average value of the function over that smaller interval, rather than over the natural period of the function.
     
  14. Oct 4, 2011 #13
    So essentially the textbook merely evaluates the integral where it is not equal to zero, but accounts for the entire period. That makes sense now that I thought about it. Sometimes it would be easier to see things such as this if the textbook stated that they did that (at least the first time) rather than not.
     
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