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: Identify the switching intervals.
In a boost converter, the inductor current rises while the switch \(S\) is ON (the inductor is directly across the source, diode reverse biased) and falls while \(S\) is OFF (the inductor current is forced through the diode to the output). Here the rise takes \(10\ \mu\)s and the fall takes \(10\ \mu\)s, so
\[ T_{on}=10\ \mu\text{s},\quad T_{off}=10\ \mu\text{s},\quad T=20\ \mu\text{s},\quad D=\frac{T_{on}}{T}=0.5 \]

Step 2: Recall the ideal boost voltage relation.
For a lossless boost converter,
\[ V_{out} = \frac{V_{dc}}{1-D} = \frac{V_{dc}}{1-0.5} = 2V_{dc} \]
This gives the ratio between input and output but not their absolute values yet, since we still need one more equation from the load.

Step 3: Find the average diode current.
The diode conducts only while \(S\) is OFF, which is exactly when \(i_L\) falls from \(6\) A to \(0\) A over \(10\ \mu\)s; during the ON time the diode is reverse biased and carries no current. So the diode current waveform is a triangular pulse (peak \(6\) A, base \(10\ \mu\)s) followed by \(10\ \mu\)s of zero, every \(20\ \mu\)s.

Step 4: Average this waveform over the full period.
\[ I_{D,avg} = \frac{1}{T}\left[\frac{1}{2}(6)(10\ \mu\text{s})+0\right] = \frac{30}{20} = 1.5\text{ A} \]

Step 5: Use the fact that the capacitor carries no net (average) current in steady state.
With the output ripple neglected, all of the average diode current flows into the load resistor:
\[ I_{D,avg} = \frac{V_{out}}{R} \implies 1.5 = \frac{V_{out}}{10} \implies V_{out} = 15\text{ V} \]

Step 6: Solve for \(V_{dc}\).
\[ V_{out}=2V_{dc} \implies V_{dc} = \frac{15}{2} = 7.5\text{ V} \]

Step 7: Check with power balance.
The average inductor current (which equals the average input current, since the inductor sits directly in the input path of a boost converter) for a triangular wave between \(0\) and \(6\) A is the average of the two extremes, \((0+6)/2=3\) A. Input power:
\[ P_{in} = V_{dc}\times I_{L,avg} = 7.5\times3 = 22.5\text{ W} \]
Output power:
\[ P_{out} = \frac{V_{out}^2}{R} = \frac{15^2}{10} = \frac{225}{10} = 22.5\text{ W} \]
Since \(P_{in}=P_{out}\), the ideal (lossless) converter assumption is satisfied, confirming \(V_{dc}=7.5\) V.

Step 8: Rule out the other options.
\(15.0\) V would be mistaking the output voltage itself for the input. \(10.0\) V and \(12.5\) V do not satisfy both the boost ratio \(V_{out}=2V_{dc}\) and the load power balance simultaneously.

Final Answer:
\[ \boxed{V_{dc}=7.5\text{ V}} \]