How is the average Diode Current Derived in this Buck-Boost Converter?

Discussion in 'General Electronics Chat' started by king_falcon, Jan 27, 2013.

  1. king_falcon

    Thread Starter New Member

    Apr 21, 2011
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  2. gootee

    Senior Member

    Apr 24, 2007
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    It looks like they are assuming a constant load current. All of the load current has to come through the diode, obviously. It might be stored somewhere for a while, before getting to the load, but the diode only passes as much current as the load needs, since there's no where else for the current to go, or come from. That MEANS that the average diode current (average over time) must equal the load current. Otherwise there would be either too much or too little current for the load.

    So then you need the average diode current. They state that the diode current is equal to the inductor current during the Off times. The diode current amplitude is varying. But it is just triangle-shaped variations.

    To get the average of a time varying periodic signal's amplitude, you just integrate the amplitude over the time period, and then divide by the time period. That means you just take the area under the amplitude curve, over one period, and then divide by the period length.

    Looking at figure 4, the diode current is zero except for an interval of length delta x T (labeled at top of figure 4). During that interval, it jumps instantaneously from zero to Ilmax and then goes linearly back to zero. Integrate over one period T. That means just take the area. The only non-zero area is that triangle. It's area is 1/2 x base x height. The base is delta x T. The height is Ilmax. So its integral (area) is 1/2 x delta x T x Ilmax. Now divide by the period, T, to get the average for one period.
     
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  3. king_falcon

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    Apr 21, 2011
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    Hey.. Thanks a lot for that lovely detailed answer. But I got confused at the fact that why would we integrate or find the area of current? Wouldn't we get Charge (Q) when we integrate current?

    Oh wait a minute.. We do get charge when we find the area of the triangle, and since I=Q/time period, we get the average current.. Am I correct?
     
  4. gootee

    Senior Member

    Apr 24, 2007
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    I think that you ARE correct. Interestingly, I didn't quite catch that. I was only thinking about the general form for finding the Mean value of a function over an interval, which is 1/x times integral from 0 to x (or, more generally, 1/(x2-x1) times the integral from x1 to x2).

    Good work!

    "Conservation of Charge" would probably make that equation even more intuitively clear. All of the charge must be accounted for, since the inductor and the load are operating in series. So the total charge that passes through one of them must equal the total charge through the other one, eventually, at least for t = 0 to infinity. But since everything is periodic, it must also be true for one period.
     
    Last edited: Jan 28, 2013
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  5. king_falcon

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    Apr 21, 2011
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    Yup that's it. Thanks for the help.. :)
     
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