Efficiency of Buck-Boost Converter

AI Thread Summary
The discussion focuses on deriving the efficiency expression for a Buck-Boost converter, factoring in the transistor's on-resistance (R_{ON}) and the diode's voltage drop (V_{D_0}). The efficiency is defined as η = 100 * P_O / P_IN, with attempts to simplify the equations for input and output power. Participants explore the implications of setting R_{ON} and V_{D_0} to zero, confirming that this results in 100% efficiency. The conversation includes troubleshooting and correcting equations to ensure accurate calculations. Ultimately, the goal is to derive a valid expression that demonstrates the efficiency under ideal conditions.
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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:
IqvCYzc.jpg


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.
 
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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 transferred 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
 
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.
 
I figured out my mistake. My line for Equation 4 should say P_{IN}=V_{IN}*DI_L=...
After that it worked out.
 
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