- #1
Captain1024
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Homework Statement
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%.
Homework Equations
[itex]<V_L>=0[/itex]
[itex]<I_C>=0[/itex]
[itex]D'=(1-D)[/itex], where D is the duty cycle
The Attempt at a Solution
Analysis of the two modes of Buck-Boost converter:
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.