Question

A boost converter operates at 25 kHz with a duty cycle 0.6. Input is 15 V, load is 10 \(\Omega\). Assuming ideal components, compute the equivalent input resistance \(R_{in}\) seen by the source. (Round off to 2 decimal places.) 

Hint:

In a boost converter, the input resistance is scaled by the square of \((1-D)\). Higher duty cycle → lower input resistance.

Answer 1

Correct Answer: 1.55

Boost converter output voltage:
\[ V_o = \frac{V_s}{1-D} = \frac{15}{0.4} = 37.5\ \text{V} \] Load current:
\[ I_o = \frac{V_o}{10} = 3.75\ \text{A} \] Inductor average current equals input current:
\[ I_{in} = I_L = I_o (1-D) \] \[ I_{in} = 3.75 \times 0.4 = 1.5\ \text{A} \] Equivalent input resistance:
\[ R_{in} = \frac{V_s}{I_{in}} = \frac{15}{1.5} = 10\ \Omega \] But the effective "seen" resistance (accounting for duty cycle transformation) is:
\[ R_{in} = R_{load} (1-D)^2 \] \[ R_{in} = 10 \times (0.4)^2 = 10 \times 0.16 = 1.6\ \Omega \] \[ \boxed{1.60\ \Omega} \]