Question

The \(I-V\) characteristics of a diode with a knee voltage \(V_{knee} = 0.8\) volts is shown in Figure 1.

This diode is used in the circuit shown in Figure 2, in which an input signal of \(V_{in} = 100\sin(1000t)\) volts is applied. What is the voltage \(V_{out}\) (in volts) across the capacitor at steady state?

Assume that the capacitor is initially discharged and the reverse breakdown voltage of the diode is much greater than 100 volts.

• A. 99.2
• B. \(99.2\cos(1000t)\)
• C. \(99.2\sin(1000t)\)
• D. \(-99.2\)

Hint:

With no resistor to discharge it, the capacitor charges only up to the input's peak minus the diode's knee drop, and holds that value forever.

Answer 1

The Correct Option is A

Step 1: Write the loop equation while the diode conducts.
While the diode is forward biased, current flows from the source through the diode into the capacitor. Going around the loop, the source voltage equals the diode's forward drop plus the capacitor voltage:
\[ V_{in}(t) = V_{knee} + V_{out}(t) \quad \text{(while the diode conducts)} \]
So whenever the diode is ON, $V_{out}(t) = V_{in}(t) - 0.8$, meaning the capacitor voltage simply follows the source, offset down by the fixed 0.8 V knee drop.

Step 2: Find when the diode turns OFF.
The diode conducts only as long as the source is trying to push the capacitor voltage higher, that is, while $V_{in}(t)$ is increasing. The moment $V_{in}(t)$ starts to decrease (right after its peak), the loop equation above would require $V_{out}$ to decrease too, but a capacitor with no discharge resistor cannot lose charge on its own, its voltage cannot fall unless current is drawn out of it. Since the diode blocks any reverse current, it switches OFF at that instant and freezes the capacitor at whatever value it last held.

Step 3: Find the frozen value using the peak of the input.
The source $V_{in}(t) = 100\sin(1000t)$ reaches its maximum value of 100 V once per cycle. Just before that peak, the diode is still ON (since $V_{in}$ was still rising), so at the peak instant:
\[ V_{out} = V_{in,max} - V_{knee} = 100 - 0.8 = 99.2 \text{ V} \]
The instant the source starts falling back down, the diode opens the loop and this 99.2 V value gets locked onto the capacitor.

Step 4: Confirm the voltage stays fixed at this value forever after.
With the diode open and no resistor to discharge the capacitor, there is no route for charge to leave. Even though $V_{in}(t)$ keeps oscillating between $-100$ V and $+100$ V every cycle after, it never again pushes the diode back on in a way that raises the stored charge, so the capacitor voltage stays parked at 99.2 V, a pure DC value with no dependence on $t$.

Step 5: Final Answer.
$V_{out} = 99.2$ V, a constant, matching option (A). \[ \boxed{V_{out} = 99.2 \text{ V}} \]