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: Identify what kind of circuit this is.
A source in series with a diode, feeding a capacitor to ground with the output taken across the capacitor, and no resistor anywhere in the loop, is the classic peak detector (peak rectifier) circuit. The diode lets current flow into the capacitor only while the source voltage exceeds the capacitor voltage by the knee voltage, and once the capacitor is charged there is no path for it to discharge, so it holds its highest reached value forever.

Step 2: Track the first quarter cycle, when the capacitor is charging.
The capacitor starts discharged, at 0 V. As \(V_{in} = 100\sin(1000t)\) rises from 0, the diode stays off until \(V_{in}\) exceeds the capacitor voltage plus the knee voltage of 0.8 V. Once that happens the diode conducts and charges the capacitor, and it keeps conducting, tracking the source almost exactly (capacitor voltage staying 0.8 V behind the source) all the way up to the peak of the sine wave, since the source is still rising and demanding more charge.

Step 3: Find the capacitor voltage at the peak of the input.
The peak of \(V_{in} = 100\sin(1000t)\) is 100 V (at \(\omega t = \pi/2\)). At that instant the diode has been conducting continuously, dropping its fixed knee voltage of 0.8 V, so the capacitor charges up to:
\[ V_{out,peak} = V_{in,peak} - V_{knee} = 100 - 0.8 = 99.2 \text{ V} \]

Step 4: Show the capacitor cannot discharge after the peak.
Just after the peak, \(V_{in}\) starts to fall. Now the source voltage is less than the capacitor voltage plus 0.8 V, so the diode becomes reverse biased and switches off, since a diode only conducts when its anode is high enough relative to its cathode. With the diode off and no resistor in the circuit to bleed the charge away, the capacitor has absolutely no discharge path, so it is stuck holding 99.2 V indefinitely, through the rest of that cycle and every cycle after (the diode only turns on again if \(V_{in}\) tries to exceed 99.2 V by more than the knee voltage, which never happens since 100 V is the fixed peak).

Step 5: Final Answer.
At steady state, \(V_{out}\) is a constant DC value, not a function of time, equal to 99.2 V. \[ \boxed{V_{out} = 99.2 \text{ V}} \] This matches option (A).