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Efficiency of Buck-Boost Converter

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a Buck-Boost converter. If in addition to the transistor on resistance [itex](R_{ON})[/itex], the converter diode has a voltage drop [itex](V_{D_0})[/itex], symbolically derive an expression for the efficiency, η of the converter, where η = 100* PO/PIN. Verify that that when [itex]R_{ON}[/itex] and [itex]V_{D_0}[/itex] are set to 0, the efficiency is 100%.

    2. Relevant equations
    [itex]<V_L>=0[/itex]
    [itex]<I_C>=0[/itex]
    [itex]D'=(1-D)[/itex], where D is the duty cycle

    3. The attempt at a solution
    Analysis of the two modes of Buck-Boost converter:
    IqvCYzc.jpg

    When [itex]<V_L>=0[/itex]
    [itex]\frac{DT_s(V_{IN}-R_{ON}I_L)+D'T_s(V_O+V_{D_0})}{T_s}=0[/itex]
    Equation 1: [itex]DV_{IN}-DR_{ON}I_L+D'V_O+D'V_{D_0}[/itex]

    When [itex]<I_C>=0[/itex]
    [itex]\frac{DT_s(\frac{-V_O}{R})+D'T_s(I_L-\frac{-V_O}{R})}{T_s}=0[/itex]
    [itex]-D\frac{V_O}{R}+D'I_L-D'\frac{V_O}{R}=0=\frac{-V_O}{R}+D'I_L[/itex]
    Equation 2: [itex]I_L=\frac{V_O}{D'R}[/itex]

    Plug 1 into 2:
    Equation 3: [itex]V_O=\frac{DV_{IN}+D'V_{D_0}}{\frac{DR_{ON}}{D'R}-D'}[/itex]

    Equation 4: [itex]P_{IN}=V_{IN}*I_{IN}=V_{IN}*I_L=V_{IN}\frac{V_O}{D'R}=\frac{V_{IN}(DV_{IN}+D'V_{D_0})}{D'R(\frac{DR_{ON}}{D'R}-D')}[/itex]

    Equation 5: [itex]P_O=\frac{V_O^2}{R}=\frac{(DV_{IN}+D'V_{D_0})^2}{R(\frac{DR_{ON}}{D'R}-D')^2}[/itex]

    Dividing equation 5 by Equation 4 yields Equation 6: [itex]\frac{D'(DV_{IN}+D'V_{D_0})}{V_{IN}(\frac{DR_{ON}}{D'R}-D')}[/itex]

    Setting [itex]R_{ON}=0[/itex] and [itex]V_{D_0}=0[/itex] in Equation 6 yields: [itex]\frac{D'(DV_{IN})}{-D'V_{IN}}[/itex]

    I need Equation 6 to equal 1 when [itex]R_{ON}[/itex] and [itex]V_{D_0}[/itex] are set to 0.
     
  2. jcsd
  3. Sep 29, 2016 #2

    donpacino

    User Avatar
    Gold Member

    Im having trouble following your thought process.

    I would go about it a different way.

    eff = 100 * Po / Pin

    So derive an expression for Po and Pin.

    since Roff is inf, you can assume no current is transfered during the off state of the transistor. So use your duty cycle to determine the current. P=IV so you now should have an expression for input power.

    As you stated earlier the inductor voltage and capacitor current averages will be zero. You can use this to derive an expression for your output power. You will then see that setting Vdo and Ro to zero will cause Po=Pin
     
  4. Sep 30, 2016 #3

    rude man

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    Homework Helper
    Gold Member

    You show the output as + but I assume you know the output will be -.
    If you want I'm willing to do the problem and compare answers if you manage to get an expression that = 100% when Ron and Vd are both zero.
     
  5. Sep 30, 2016 #4
    I figured out my mistake. My line for Equation 4 should say [itex]P_{IN}=V_{IN}*DI_L=...[/itex]
    After that it worked out.
     
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