Help in Boost rectifier's simulation

Discussion in 'The Projects Forum' started by AbhimanyuSingh, May 31, 2011.

  1. AbhimanyuSingh

    Thread Starter New Member

    May 22, 2011
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    What is the supposed output of the first circuit I have attached here in this post? Isnt it a DC? But when I simulated the same circuit, I got a half rectified AC voltage at the output> My simulated circuits's image is also attached. Why is it so?
     
  2. #12

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    Nov 30, 2010
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    I don't know what "half rectified" means, but I can say you need a capacitor to the left of L1.

    Without the capacitor to store the peak voltage of V1, M1 will be turned on at random points of the sine wave from V1.
     
  3. AbhimanyuSingh

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    May 22, 2011
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    sorry, i mean "rectified" not "half rectified".
     
  4. #12

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    If your first drawing did not have the switch, the output would be rectified and filtered AC voltage. It is almost pure DC but there will be some peaks and their size will depend on the size of the load resistor and Co.

    In your second drawing, you used things like, "0 50 50 0 0 0 1000" which I do not understand. I can not comment on that.
     
  5. AbhimanyuSingh

    Thread Starter New Member

    May 22, 2011
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    "0 50 50 0 0 0 1000" is nothing but an AC source of amplitude 50 volts and frequency 50 Hz. And the output of my second circuit is the same rectified voltage similar to the one at the left of the inductor. the only difference is that it is scaled, depending on the duty ratio of the switch of boost converter. Which, I think is wrong.
     
  6. #12

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    Without a capacitor to the left of L1, you will have rectified AC pulses and the M1 switch will turn on at random times during the pulses.
     
  7. AbhimanyuSingh

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    May 22, 2011
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    But the switch M1 gets on at desired moments, ie at gate pulses. Whereas with a capacitor on left of L1 results in almost zero voltage (with little fluctuations) at right of capacitor and every point toward right of that.
     
  8. #12

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    With a capacitor on left of L1 results in almost zero AC voltage at right of capacitor, but the capacitor holds DC almost full peak voltage of V1.
     
  9. AbhimanyuSingh

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    May 22, 2011
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    So then, the output of the first schematic is not supposed to be a DC voltage?
     
  10. #12

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    True. The first schematic has no filter capacitor to store the peak voltage of V1 before the switch. Read post #4 again.
     
  11. AbhimanyuSingh

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    May 22, 2011
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    What is it supposed to be then, the output of the first schematic?
     
  12. #12

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    Read post #4 and post #6.
     
  13. AbhimanyuSingh

    Thread Starter New Member

    May 22, 2011
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    The literature i am reading says this circuit is an ac to dc converter. I am attaching a snapshot of that pdf file.
     
  14. #12

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    The first schematic is a representation of an unfiltered rectifier with the switch "S" representing the active part of a boost circuit. It might work if the IRF530 (M1) only conducts during the peaks of the rectified waveform. As you said in post #7, "the switch gets on at desired moments".

    The way the boost part works is that the switch (M1) turns on when voltage is available to send current through L1. When the switch turns off, L1 keeps pushing current through D5 for a short time to charge C1.

    When the sine wave from V1 is at zero volts there is no voltage to send current through L1 and M1.

    If M1 switches very fast, it will send the rectifier pulses through in tiny pieces, but if you use more than the tiniest amount of current, the outline of the tops of the pieces will look like the rectified pulses because there is no filter capacitor left of L1.

    As an AC to DC boost rectifier, I can tell you it will not work with the components you have listed, and that is why your results are not what you expected.
     
    Last edited: Jun 1, 2011
  15. AbhimanyuSingh

    Thread Starter New Member

    May 22, 2011
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    So, what componene t should I add to the circuit to get desired DC output? Simply adding a capacitor to the left of the inductor leaves me almost zero output.
     
  16. SgtWookie

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    Jul 17, 2007
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    D1-D4 are rated for 1A, 40v; which is less than your sine wave input.
    D5 is the "default" diode.
    You need to select diodes that have more realistic models.

    You should show L1's inductance as 10uH; much easier to read than 0.00001.
    Same with C1; 10uF is much easier to read than 0.00001
    The load "r" has a very low resistance; if V(r) exceeds 5v, then the ratings of D1 thru D4 will be exceeded.

    I re-created your simulation using more suitable diodes, decreased the frequency of the gate drive to 100kHz, decreased the duty cycle to 20% and increased the value of L1 to 50uH for compensation; this makes the simulation run in a more reasonable amount of time.
    The 5 Ohm load resistor has been replaced with a 1A current load, as your 5 Ohm load was not realistic for the other components originally used.
    C2 has been added as a filter capacitor between the bridge and L1.
    C2 has been increased to 100uF.

    See the attached schematic and simulation.
     
  17. #12

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    No fair Wookie. Anybody can make that circuit work if the change every part# and add a filter cap.
     
  18. SgtWookie

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    Hey, it's exactly the same yet completely different! :D
     
  19. #12

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    Actually, I feel like I didn't do a good job on this one because I didn't figure out his skill level until post #13. 4 instance, I was thinking that anybody working on a switching regulator would know not to use 40 volt recitfiers with a 50 volt supply, etc. I never checked the basics, and that's my fault.

    apologies.
     
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