Answer :
We have derived an expression for the output voltage of a Buck-Boost converter that accounts for their presence.V_out = Duty Ratio * V_in / (1 - Duty Ratio).
To derive an expression for the output voltage of a Buck-Boost converter with non ideal switch and diode, we start with the basic Buck-Boost converter circuit and consider the voltage drops across the switch and diode.
The output voltage of a Buck-Boost converter is given by the equation (6-88):
V_out = Duty Ratio * V_in / (1 - Duty Ratio),
where Duty Ratio is the ratio of the ON time of the switch to the total switching period.
Now, let's consider the voltage drops across the non ideal switch (V_
sw) and diode (Vd). We assume that the switch has a small voltage drop when ON and the diode has a small voltage drop when conducting.
When the switch is ON, the voltage across the switch (Vsw) can be approximated as a small positive value, and when the diode is conducting, the voltage across the diode (Vd) can be approximated as a small positive value as well.
Therefore, we can modify the equation for the output voltage as follows:
V_out = (Duty Ratio * Vin - V_sw + V_d) / (1 - Duty Ratio).
This modified expression accounts for the voltage drops across the non ideal switch and diode. The voltage drop across the switch subtracts from the input voltage, while the voltage drop across the diode adds to the output voltage.
It is important to note that the exact values of V_sw and V_d depend on the characteristics of the specific switch and diode used in the circuit, and these values may vary based on operating conditions.
By considering the voltage drops across the non ideal switch and diode, we have derived an expression for the output voltage of a Buck-Boost converter that accounts for their presence.
Learn more about Buck-Boost converter from the given link
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