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Question about Buck Convertor Operation

  1. May 20, 2016 #1
    Hi ,

    I have a query on Buck convertors , I have attached a picture that has the circuit and associated waveforms.

    http://i.imgur.com/ydQ20m8.png

    When you look at the inductor it has a constant flow of current through it with a changing magnitude but the same direction , but the waveform shows the voltage through it being positive and later on negative.If the voltage across the inductor was negative wouldn't that mean the direction of current through it would also change?

    *edit*

    I did some further reading , is it because voltage across an inductor is linked to the change in current , so the current could fall but still be positive and this would create a negative voltage across the inductor?

    Thanks
     
  2. jcsd
  3. May 20, 2016 #2

    berkeman

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    Staff: Mentor

    Are you familiar with the simple differential equation that relates the inductor current to the voltage across it? :smile:
     
  4. May 20, 2016 #3
    Hi ,

    I edited my original post

    "I did some further reading , is it because voltage across an inductor is linked to the change in current , so the current could fall but still be positive and this would create a negative voltage across the inductor?"

    Does this sound ok?
     
  5. May 20, 2016 #4

    berkeman

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    Staff: Mentor

    Yes, that is correct. The equation I was asking about is [tex]v(t) = L \frac{di(t)}{dt}[/tex]
    Are you familiar with how to interpret such equations?
     
    Last edited: May 20, 2016
  6. May 20, 2016 #5

    I do but I was a little rusty on circuit , I have a further question on Ico if you could help? When the switch is off wouldn't the capacitor discharge and have the current Ico flow towards the node at Vo?
     
  7. May 20, 2016 #6

    berkeman

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    Yes, I think that's what they are showing in the waveforms. How much it discharges depends on the load current and the inductor current.
     
  8. May 20, 2016 #7

    So immediately after the switch opens which direction would Ico be flowing?
     
  9. May 20, 2016 #8

    berkeman

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    It would be positive for the first part of the OFF cycle (when the inductor current is enough to maintain Vo across the load, and would reverse as the inductor current fell below Vo/Rload. Co is like a reservoir storage capacitor. The ripple current in and out of Co is an important part of the design of a buck converter (the cap must be able to handle the ripple current, and the ripple voltage you get across the load depends a lot on the DCR of Co).
     
  10. May 20, 2016 #9

    But if Ico changed direction then wouldn't also the voltage across the capacitor change direction too? And if so wouldn't that also change the voltage across the load if you follow the KVL rule?
     
  11. May 20, 2016 #10

    berkeman

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    The voltage on Co ripples about the nominal value. It does not go negative. For example, for a 5V output DC-DC converter, it's typical for you to get about 200mVpp of ripple on the 5V output due to the switching action. So if you measure the output with an oscilloscope, you see it rippling between 4.9V and 5.1V, at the frequency of the DC-DC switching action.
     
  12. May 21, 2016 #11
    Thanks , it's becoming clear now.I was wondering , I'm trying to derive the equation for the rms current through the switch but am having some trouble.

    I have to prove that it is root(k*Io^2 + delta(i)^2 /12 )

    Bit stuck on how to start
     
  13. May 21, 2016 #12
    This is my approach to the problem

    http://i.imgur.com/R3zxz8k.jpg

    When the switch is on for a period of k*Tp the current rises from Iin which is = to kIo the current when the switch is on rises by delta(i)

    Does this approach look ok? I'm not having much luck

    Thanks
     
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