Question

A signal \( V_{\text{in}}(t) \) shown is applied from \( t = 0 \) ms to \( t = 6 \) ms to the circuit shown. Given the initial voltage across the capacitor is 0.3 V, and that the diode is ideal, the open circuit voltage \( V_{\text{out}}(t) \) at \( t = 5 \) ms is _________

• A. 0.3 V
• B. 0.6 V
• C. 0.7 V
• D. 1.0 V

Hint:

In a circuit with an ideal diode and capacitor, the output voltage will charge up to the input voltage as long as the diode conducts (i.e., when \( V_{\text{in}}(t)>V_{\text{out}}(t) \)).

Answer 1

The Correct Option is D

Step 1: Understanding the circuit operation.
The circuit involves a diode and a capacitor. The initial voltage across the capacitor is 0.3 V. The diode is ideal, meaning it will conduct when the input voltage \( V_{\text{in}}(t) \) is greater than the voltage across the capacitor and block the current when \( V_{\text{in}}(t) \) is lower than the capacitor voltage. The capacitor charges through the diode when \( V_{\text{in}}(t) \) increases.
Step 2: Analyzing the input signal at \( t = 5 \) ms.
At \( t = 5 \) ms, the input voltage \( V_{\text{in}}(t) \) reaches 1.0 V (as shown in the graph), which is higher than the initial voltage across the capacitor (0.3 V). Since the diode is forward-biased when \( V_{\text{in}}(t)>V_{\text{out}}(t) \), the capacitor will charge up to match the input voltage. Thus, the output voltage at \( t = 5 \) ms will be 1.0 V, as the diode conducts fully, allowing the capacitor to charge up to the input voltage.
Step 3: Conclusion.
The correct answer is (D) 1.0 V.