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BOOST SMPS/Converter in Discontinuous Conduction Mode

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A boost SMPS has the following design properties:

    Input Voltage - Vi = 6.0v
    Maximum Output Current - Io(max) = 0.5A
    switching frequency - F = 100khz
    Inductor = 500μH
    Capacitor = 1000μF with ESR 25mΩ

    Consider the boost SMPS operating in discontinuous conduction mode with a duty cycle of 0.5.

    i) Calculate the energy stored in the inductor at the end of the on-time.
    ii) Consider discontinuous conduction mode as in part i) when the boost SMPS has a 600Ω resistor connected at the output. Find the value of the output voltage that the SMPS will adopt.


    2. Relevant equations

    1/2CV^2 = Energy stored in a capacitor.
    1/2Li^2 = Energy stored in an inductor.
    ΔVc = 1/c∫Io.dt = 1/C * Io * δ/F voltage ripple
    Assume voltage drop across diode is negligible. I have attached the circuit diagram of a simple boost that we have been using for this course.


    3. The attempt at a solution

    After having given a lot of thought, I have an attempt but do not think it's correct. I would really appreciate some help with this.

    We are operating in discontinuous conduction mode, which means the current flowing when the transistor is in it off state drops to zero before it turns on again. Therefore, all the energy that was built up in the inductor is shipped to the capacitor. Hence, if I am able to work out the voltage ripple across the capacitor, then I can apply the equation E = 1/2CV^2 and obtain the energy that was also stored in the inductor.

    I am having some trouble working out this ΔVc. My initial approach was to use ΔVc = 1/c∫Io.dt = 1/c * Io * δ/F where δ is the duty cycle (0.5). This brings into question what should I use as Io ? The question gives Io(max) which is 0.5A but I do not think this is what I should be using?.
    Using this gives me an answer of 3nJ which seems far too small. I am also absolutely clueless about whether to take the ESR into account too.

    If anyone could shed some light on this I would be extremely grateful ! Some indication of whether my method is correct would be most appreciated too. Many, many thanks and I wish you all a good day :)
     

    Attached Files:

  2. jcsd
  3. Aug 26, 2013 #2
    Apologies for not stating earlier, this is my first course on power electronics. I am a second year undergraduate student.
     
  4. Aug 27, 2013 #3

    rude man

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    2nd year undergraduate? Pretty advanced electronics for that!

    Anyway - taking the no-load case first - with a circuit like this you should first ask yourself: what happens in the steady-state? And the 1st thing you notice is that the average curent thru L has to be zero. Why? because otherwise the capacitor would keep charging until infinite voltage!

    So we have 50% duty cycle - and it's obvious what the voltage across L is when Q is ON so we also know how the current behaves during that time. So question to you - how must the current behave when Q is OFF? Remember, the avg L current is zero and Vc must be some constant voltage (diode barely goes on any time). Think about this for a while.
     
  5. Aug 29, 2013 #4
    Hi, thanks a lot for your reply. This question has been bothering me for some time, however I think I've made some real progress now with the help of a few more resources.

    The current through the L can me modelled as a triangular wave. Increasing when Q is on and decreasing when Q is off. Since we are in discontinuous conduction mode, when Q is off the current through the L will fall to zero before Q goes back on again. Therefore if I can find the area under the portion of the IL graph for which it is rising, I can find the charge that is deposited to the capacitor at the end of the off time. Using V = Q/C I can then find the voltage, and finally with either E = 1/2CV^2 or 1/2QV I get the energy in the capacitor at the end of the off time. This is equivalent to the energy in the inductor at the end of the on time. So doing that gives me:

    With the help of some textbooks I found some more useful equations, which I use here

    IL(MAX) = VI*D*T/L
    where D is the duty cycle and T is the time period (= 1/f)

    This gives me 0.06A. Now finding the area underneath that triangular portion:

    1/2 * tON * 0.06 (1/2 Base * perpendicular height) tON is D*T

    This gives me 1.5 X 10-7 C

    Now using V = Q/C = 1.5 X 10-4

    Applying the energy equations and I've got 11.25 pJ.

    As I say I've used quite a few resources and hope I'm not in danger of a "too many cooks spoil the broth" type of situation ! I don't see anything (in terms of physics, anyway) wrong with the above.

    Your comments would be really interesting, as would anyone else's. I look forward to your reply.

    EDIT : I've attached a graph of the IL current just to be clear what I'm talking about.
     

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    Last edited: Aug 29, 2013
  6. Aug 29, 2013 #5

    rude man

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    I have to ask a quick question: does T have a reverse built-in diode? In other words, can current flow backwards when T is OFF? Power MOS transistors usually have such a diode built in. But we need to know if this one does.

    EDIT: never mind, I don't think it matters. Stay tuned!
     
    Last edited: Aug 29, 2013
  7. Aug 29, 2013 #6
    Hey, no worries. For the record, there is no reverse built in diode.
     
  8. Aug 29, 2013 #7

    rude man

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    OK, let's see if we can work this together.
    Part (a): let me use i for the L current, Q for the transistor and T for the half-period (= 50us).

    First off, we recognize that the output voltage Vc builds up without limit. Why? Because when Q is on, di/dt > 0 and when Q shuts off, the current built up in L must flow into C. There is no limiting mechanism for this.

    It took me a while to convince myself of this, and who knows, maybe it's still wrong. As I see it, as Vo and Vx build up, di/dt during Q-off gets larger in magnitude and is negative. So i, which went from zero to i = (Vi/L)T during Q-on, must return to zero in time Li/(Vx - Vi) after Q shuts off. So as Vx and Vc rise without limit, the time to reset i to zero gets shorter & shorter.

    So your graph is almost right except the fall time is shorter than the rise time unless Vx < 2Vi (i.e. at startup). Eventually, the fall time approaches zero. In reality, something, probably Q or C, would break down and stop this rise, possibly damaging components etc.

    Anyway, from all this, if you agree, it should be obvious what i is at the end of the Q-on half-cycle, and therefore the energy stored in L at that time.

    One thing I haven't taken into account is the capacitor's esr. Not sure how to handle this or if it's important.

    In part (b) we have a load so the output voltage will not increase indefinitely.
     
  9. Aug 31, 2013 #8
    Thanks a lot for you help rude man, I'll think about this some more in light of what you've said.
     
  10. Aug 31, 2013 #9

    rude man

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    OK.
    I've analyzed part (b) which indicated that, if the switcher is to run in discontinuous mode (L current = 0 within a part of each 10 us cycle) then the load resistor must be 800 ohms or greater. With 600 ohms the inductor current never ceases and the analysis is more difficult (haven't done that part).

    There's always the possibility I made a mistake somewhere of course. But we can discuss my M.O. anyway.
     
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