Captain1024
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Homework Statement
Consider a Buck-Boost converter. If in addition to the transistor on resistance (R_{ON}), the converter diode has a voltage drop (V_{D_0}), symbolically derive an expression for the efficiency, η of the converter, where η = 100* PO/PIN. Verify that that when R_{ON} and V_{D_0} are set to 0, the efficiency is 100%.
Homework Equations
<V_L>=0
<I_C>=0
D'=(1-D), where D is the duty cycle
The Attempt at a Solution
Analysis of the two modes of Buck-Boost converter:
When <V_L>=0
\frac{DT_s(V_{IN}-R_{ON}I_L)+D'T_s(V_O+V_{D_0})}{T_s}=0
Equation 1: DV_{IN}-DR_{ON}I_L+D'V_O+D'V_{D_0}
When <I_C>=0
\frac{DT_s(\frac{-V_O}{R})+D'T_s(I_L-\frac{-V_O}{R})}{T_s}=0
-D\frac{V_O}{R}+D'I_L-D'\frac{V_O}{R}=0=\frac{-V_O}{R}+D'I_L
Equation 2: I_L=\frac{V_O}{D'R}
Plug 1 into 2:
Equation 3: V_O=\frac{DV_{IN}+D'V_{D_0}}{\frac{DR_{ON}}{D'R}-D'}
Equation 4: 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')}
Equation 5: P_O=\frac{V_O^2}{R}=\frac{(DV_{IN}+D'V_{D_0})^2}{R(\frac{DR_{ON}}{D'R}-D')^2}
Dividing equation 5 by Equation 4 yields Equation 6: \frac{D'(DV_{IN}+D'V_{D_0})}{V_{IN}(\frac{DR_{ON}}{D'R}-D')}
Setting R_{ON}=0 and V_{D_0}=0 in Equation 6 yields: \frac{D'(DV_{IN})}{-D'V_{IN}}
I need Equation 6 to equal 1 when R_{ON} and V_{D_0} are set to 0.