In the circuit shown below, the voltage appearing across the diode D will be of the form: ____.
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Be careful! If the question asks for voltage across the Resistor, it's a standard rectifier (Positive halves). If it asks for voltage across the Diode, it's the "leftover" part of the signal (Negative halves).
Step 1: Understanding the Topic:
This question is part of "Semiconductor Electronics" and involves the study of a half-wave rectifier circuit. Crucially, it asks for the voltage across the diode, not across the load resistor. This is a common point of confusion for students. Step 2: Key Formulas and Approach:
Kirchhoff's Voltage Law (KVL) states:
\[ V_{input} = V_{diode} + V_{resistor} \]
When diode conducts ($V_{diode} \approx 0$), all voltage appears across the resistor.
When diode is off ($I = 0$, so $V_{resistor} = 0$), all voltage appears across the diode.
Step 3: Detailed Explanation:
Positive Half Cycle: The diode is forward-biased. In an ideal case, a forward-biased diode acts like a closed switch with zero resistance. Therefore, the voltage drop across it is $0 \text{ V}$.
Negative Half Cycle: The diode is reverse-biased. It acts like an open switch (infinite resistance). Because the circuit is "broken," no current flows through the resistor $R$. By KVL, the entire input voltage from the source must appear across the terminals of the open diode.
Resulting Waveform: The graph of $V_{diode}$ will show a flat line at zero during the positive halves of the AC input and will show the negative "humps" of the AC input during the negative halves.
This is exactly what is depicted in Plot (3).
Step 4: Final Answer:
The voltage across the diode consists of the negative half-cycles only.
