Buck converter problem

Discussion in 'Homework Help' started by PsySc0rpi0n, Feb 3, 2018.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hello once again.

    I have a problem to solve about buck converters. This was an exam problem but my knowledge about this subject is very small.
    I read the chapter about Buck converters from Ned Mohan book but most of it is about voltages and currents!

    The info that was given is:
    Vin = 80V
    Pout=100W
    f=150kHz
    L=0.4mH

    I was asked the following:
    a) Draw the Buck schematic and plot I_L wave form. Evaluate I_L and Iout.
    b) Evaluate the mathematical expression the gives you I_Lmax and I_Lmin.
    c) Evaluate Rout for the border condition between CCM and DCM (Continuous Conduction Mode and Discontinuous Conduction Mode)

    So, for the schematic this is what I have:
    buck_schematic.png
    Please ignore capacitor value because in exam problem we were told nothing about it. This was just to run the simulation.

    Now about I_L and Iout (I'll use 'out' instead of 'Load' just to follow the exam words), exam didn't said anything about in which conditions to evaluate the inductor and ouput currents. I mean if in switch ON or in switch OFF mode.

    So, for switch OFF mode, only D1, L1, C1 and R1 are considered and D1 is an ideal diode, so it's a short there.
    This said, I can write:
    V_L = VLoad

    And from Mohan's book, and because we consider this an ideal circuit, Pin = Pout

    But I can't find an expression for the inductor current that I can use with the information given by the problem!
    How can I solve this?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For Ro = 18Ω and Pin = Pout = 100W we have Io = 100W/80V = 1.25A.
    Hence the otput voltage is Vo = √(P*Ro) = 42.43V and the duty-cicle is D = Vo/Vin = 0.53

    Ton = 1/F *D = 3.533μs
    and Toff = 3.133μs.

    Additional from

    VL = L * dI/dt = L * ΔI/Δt

    we can find ΔIL

    ΔIL = VL/L *Δt = (80V - 42.43V)/400μH *3.533μs = 331.84mA

    From this we can find IL_min and IL_max

    IL_max = Io + ΔIL/2 = 2.52A
    IL_min = Io - ΔIL/2 = 2.19A


    And to be able to find Rout for the border condition between CCM and DCM you need to set IL_min = 0A and remember that Io is equal to the avrage inductor current.


    https://www.allaboutcircuits.com/te...ysis-of-four-dc-dc-converters-in-equilibrium/
    https://en.wikipedia.org/wiki/Buck_converter
     
    Last edited: Feb 3, 2018
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hum, ok, I'll ask some questions about your reply in the next posts.

    I tried something similar, but I got not match to your math.

    Po = Ro*Io² <=> Io = sqrt(Ro/Po) = 0.42A

    Why I couldn't do this? Is it because Ro in this case would be an equivalent resistance seen from the output of the circuit and not simply Ro = 18Ω?
     
  4. Jony130

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    Good catch. Yes, I'm stupid and wrongly assumed that Iin = Iout wich is not true.
    Vo is Vo = √(P*Ro) = 42.43V



    Io = sqrt(Po/Ro) = sqrt(100W/18Ω) = sqrt( 5.556A^2) = 2.357A
     
    Last edited: Feb 3, 2018
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

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    I'm sorry. I'm lost again.

    Why Vo = \sqrt(P*Ro)?

    Isn't
    P = R*I² <=> I = \sqrt(P/R)
    or
    P=V*I <=> I = P/V <=> V = P/I

    or am I missing something?
     
  6. Jony130

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    P = I*V and from the Ohm's law we have V = I*R

    P = I*V = I*I*R = I^2*R now solve for I^2

    I^2 = P/R ----> I = sqrt(P/R)


    Or we I = V/R

    P = I*V = V/R *V = V^2/R ---> solve for V^2

    V^2 = P*R ---> V = sqrt(P*R)
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, so I think you mistyped Vo equation in your post #4, because you typed sqrt(P*Ro) and I think it should be sqrt(P/Ro), no?


    So if I am correct about that mistyping:

    Pi=Po=Ro*Io² <=> I = √(Po/Ro) = √(100/18) = 2.36A

    Vo = Ro*Io = 18Ω*2.36A = 42.43V

    D = Vo/Vi = 0.53

    Edited;
    Ok, I just read your post better, and I can't understand why we don't get the same Vo. I understood your equations, but what did I did wrong?
     
    Last edited: Feb 3, 2018
  8. PsySc0rpi0n

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    Ok, I got it! I inverted a fraction in my post #3.

    Going to fix above post and post correct values!

    Well, you already did it all!

    Anyway, the math we have been doing here is for when the switch is ON or when it is OFF?
     
    Last edited: Feb 3, 2018
  9. Jony130

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    In steady state ( CCM mode ) this is always true
    ΔIon = ΔIoff and Von * ton = Voff * toff
    Where the Von, Voff represent the voltage across the inductor during ton/toff time.
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

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    @Jony130 in the link you provided in post #2 where we can read about Buck analysis, here from AAC, when the max and min inductor currents are evaluated, those equations shouldn't be:

    \Delta iL_{OFF} = -\frac{V_{o}}{L}\left ( 1-D \right )T

    instead of

    \Delta iL_{OFF} = -\frac{V_{o}}{L}\left ( 1-DT \right )

    I mean, 'T' should be out of the parenthesis, no?


    Edited;
    Also, why, when Imax and Imin are being evaluated, right after Figure 6, it's written (Imax - Imin)SwitchON = ..... and then it's written (Imin - Imax)SwitchOFF=.... ??? I mean, why for SwitccOFF, they swapped Imin with Imax in the expression?
     
    Last edited: Feb 4, 2018
  11. Jony130

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    Yep. Toff time is (1-D)*T


    During Ton time the inductor current rises from Imin to Imax. The coil store the energy in this phase. And during Toff the coli current is dropping from Imax to Imin. The inductor release the stored energy into the load. And the circuit is in stedy state only when ΔIon = ΔIoff (CCM)

    http://images.elektroda.net/9_1344534972.png
     
  12. PsySc0rpi0n

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    Ok, so this is following something like "the last value minus the first value" right? I mean for Ton, Imax is last value and Imin is first value. Same for Toff, right?
     
  13. Jony130

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    Do not worry much about this because we are comparing the absolute value only.
    ΔIon = ΔIoff


    ΔIon = (Vin - Vo)/L * DT

    ΔIoff = - Vo/L * (1 -D)T


    (Vin - Vo)/L * DT = Vo/L * (1 -D)T ( absolute value only)

    So if we solve this for Vo we will get

    Vo = D*Vin
     
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  14. MrAl

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    Jun 17, 2014
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    Hello there,

    I might be missing something here but i just have to interject a quick question here...

    Are you sure the output resistor R1 is really 18 ohms?
    I ask because this would make more sense if it was 16 ohms.

    No problem if it is really 18 ohms, but i just had to ask because for these problems we often assume ideal elements with Pin=Pout and when the duty cycle is a fractional 0.50 then the output voltage is exactly 1/2 of the input voltage, which means the output voltage would be 40 volts here and that would make R1=16 ohms to get 100 watts out.
     
  15. PsySc0rpi0n

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    Hello @MrAl ...

    Yes, Ro = 18Ω. We found D = 0.53 so Vo = 0.53*Vin = 42.4V.

    How do you find Po = 100W with Ro = 16Ω?
     
  16. Jony130

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    There is nothing special about this Ro = 16Ω. Simply as point out by MrAl to get nice round numbers (D = 0.5 and Vo = 40V, Po = 100W) the load resistor needs to be equal to 16Ω instead of 18Ω.
     
  17. MrAl

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    Jun 17, 2014
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    Hi,

    Well there is really though, because if we had the duty cycle shown which was 50 percent then for 100 watt output we'd need 16 ohms not 18.
    But he has modified the duty cycle now so 18 could be acceptable.

    It's like a puzzle, if one piece does not fit then we like to see what we are doing wrong.
    Check it over and see what i was talking about.
     
  18. MrAl

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    Hi,

    Ok then maybe there is no problem, but the original problem stated 50 percent not 53 percent so it looked like something was wrong.
    I get 42.43 to two decimal places and that matches yours good enough. And yes i get 53.03 percent duty cycle which is also close.

    With a 50 percent duty cycle (as stated originally) then we would need 16 ohms to get 100 watts out because 40^2/16=100.
     
  19. PsySc0rpi0n

    Thread Starter Well-Known Member

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    @MrAl ok, I got where this went wrong! The LTSpice circuit I posted in post #1 was from a different context, and I forgot to change that Duty Cycle to the one we evaluated! My bad. I'm sorry for that! The problem didn't stated any Duty Cycle. We had to evaluate it!

    Ok, now, the next problem is to add a fully controlled full wave rectifier to that circuit and evaluate some more parameters! I'll do this later today in a new thread!

    Thanks
    Psy
     
  20. MrAl

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    Hi,

    Ok sounds interesting. I'll try to keep an eye out for that new thread.
     
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