Question

Consider the boost converter circuit shown. Assume that the semiconductor devices are ideal. In steady state, the inductor current rises linearly from \(0\) A to \(6\) A in the first \(10\ \mu\)s and then falls linearly from \(6\) A to \(0\) A in the next \(10\ \mu\)s of every switching cycle as shown. The load resistance \(R\) is \(10\ \Omega\) and the capacitance \(C\) is \(500\ \mu\)F.

Neglect the ripple in the output voltage. What is the input voltage \(V_{dc}\)?

• A. 10.0 V
• B. 15.0 V
• C. 7.5 V
• D. 12.5 V

Hint:

Find the duty ratio from the given on/off times, use \(V_{out}=V_{dc}/(1-D)\), and separately find \(V_{out}\) from the average diode current feeding the load resistor.

Answer 1

The Correct Option is C

Step 1: Get the duty ratio from the given waveform.
The inductor current rises for $10\ \mu$s (switch closed) and falls for $10\ \mu$s (switch open), out of a $20\ \mu$s cycle, so $D=10/20=0.5$.

Step 2: Apply volt-second balance on the inductor.
In steady state the average voltage across the inductor over one cycle is zero. While the switch is closed, $v_L=V_{dc}$; while open, $v_L=V_{dc}-V_{out}$ (with the diode forward biased and output ripple ignored). Setting the volt-seconds equal:
\[ V_{dc}\cdot T_{on} = (V_{out}-V_{dc})\cdot T_{off} \]
\[ V_{dc}(10) = (V_{out}-V_{dc})(10) \implies V_{dc}=V_{out}-V_{dc}\implies V_{out}=2V_{dc} \]

Step 3: Find the average capacitor charging behavior over one cycle.
The capacitor is charged by the diode current whenever the diode conducts (during the $10\ \mu$s OFF time, when $i_L$ falls from $6$ A to $0$ A), and is discharged by the constant load current $V_{out}/R$ throughout the whole cycle. In steady state, the net charge into the capacitor over a full cycle must be zero:
\[ \int_0^T i_D\,dt = \int_0^T \frac{V_{out}}{R}\,dt \]

Step 4: Evaluate the left-hand side.
The diode carries a triangular pulse from $6$ A to $0$ A over the $10\ \mu$s OFF interval, whose area is the area of a triangle:
\[ \int_0^T i_D\,dt = \frac12(6)(10\ \mu\text{s}) = 30\times10^{-6}\text{ A}\cdot\text{s} \]

Step 5: Evaluate the right-hand side and equate.
\[ \frac{V_{out}}{R}\times T = \frac{V_{out}}{10}\times20\times10^{-6} \]
Setting the two equal:
\[ 30\times10^{-6} = \frac{V_{out}}{10}\times20\times10^{-6} \implies V_{out} = \frac{30\times10}{20} = 15\text{ V} \]

Step 6: Solve for $V_{dc}$ using the ratio from Step 2.
\[ V_{dc} = \frac{V_{out}}{2} = \frac{15}{2} = 7.5\text{ V} \]
\[ \boxed{V_{dc}=7.5\text{ V}} \]