In the DC-DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is 500 Hz. The voltage (V₀) across the load is assumed to be constant and ripple free. The peak inductor current in amperes is ______ (rounded off to the nearest integer).

The correct answer is 13.

Step-by-Step Explanation:

1. Identify the given parameters from the circuit diagram and question description:
From the schematic, we can identify the following components and values:
• Input voltage: $V_i=20\text{V}$
• Inductor inductance: $L=2\text{mH}=2\times {10}^{-3}\text{H}$
• Constant and ripple-free output voltage across the load: $V_o=40\text{V}$
• Load resistance: $R=10\Omega$
• Switching frequency: $f_s=500\text{Hz}$

2. Calculate the Duty Cycle (D):
The given circuit is a classic step-up (boost) DC-DC converter. For a boost converter operating in continuous conduction mode (CCM), the relationship between the input and output voltages is given by:
$$V_o=\frac{V_i}{1-D}$$
Substituting the values of $V_i=20\text{V}$ and $V_o=40\text{V}$:
$$40=\frac{20}{1-D}$$
$$1-D=\frac{20}{40}=0.5$$
$$D=0.5$$

3. Determine the Average Inductor Current ($I_{L,\text{avg}}$):
First, find the average output load current ($I_o$):
$$I_o=\frac{V_o}{R}$$
For a lossless boost converter, the average inductor current is related to the output current by:
$$I_{L,\text{avg}}=\frac{I_o}{1-D}$$
Substituting the calculated values:
$$I_{L,\text{avg}}=\frac{4}{1-0.5}=\frac{4}{0.5}=8\text{A}$$

4. Calculate the Inductor Ripple Current ($\Delta I_L$):
During the turn-on interval of the switch $S$ ($T_{\text{on}}=D/f_s$), the voltage across the inductor is equal to the input voltage $V_i$. The change in inductor current during this time is:
$$\Delta I_L=\frac{V_i\times D}{L\times f_s}$$
Substituting the values:
$$\Delta I_L=\frac{20\times 0.5}{2\times {10}^{-3}\times 500}=\frac{10}{1}=10\text{A}$$

5. Determine the Peak Inductor Current ($I_{L,\text{peak}}$):
The peak inductor current is the sum of the average inductor current and half of the peak-to-peak inductor ripple current:
$$I_{L,\text{peak}}=I_{L,\text{avg}}+\frac{\Delta I_L}{2}$$
$$I_{L,\text{peak}}=8+\frac{10}{2}=8+5=13\text{A}$$

Therefore, the peak inductor current is 13 A.