Efficiency of Buck-Boost Converter

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 5K views
Captain1024
Messages
44
Reaction score
2

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:
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.
 
on Phys.org
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 [itex]P_{IN}=V_{IN}*DI_L=...[/itex]
After that it worked out.