Question

Consider the buck-boost converter shown. Switch Q operates at 25 kHz with a duty-cycle of 0.75. Assume diode and switch are ideal. Under steady-state, the average current flowing through the inductor is \(\underline{\hspace{1cm}}\) A. 

Hint:

In a buck-boost converter, the inductor supplies the load only during the OFF time, so the average inductor current is amplified by \(\frac{1}{1-D}\).

Answer 1

Correct Answer: 24

For a buck-boost converter, steady-state output voltage is: \[ V_o = \frac{D}{1-D} V_{in} \] Given: \[ V_{in} = 20V, D = 0.75 \] \[ V_o = \frac{0.75}{0.25} \cdot 20 = 3 \cdot 20 = 60V \] Load resistance: \[ R = 10\Omega \] Load current: \[ I_o = \frac{V_o}{R} = \frac{60}{10} = 6A \] Relation between inductor current and output current in buck-boost: \[ I_L = \frac{I_o}{1 - D} \] \[ I_L = \frac{6}{0.25} = 24A \] Thus the average inductor current is 24 A, within the given range 24 to 24.