Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 393 of 494
In an experiment, a student connects a variable DC power supply to a silicon \( \text{P-N} \) diode in series with a resistor. When the forward voltage is set to \( 0.2\text{ V} \), the measured current is negligible. Why?
A
The diode has broken down and created an open circuit
B
The applied voltage is in reverse bias orientation across the terminals
C
The applied forward voltage is below the built-in barrier potential (\( \approx 0.7\text{ V} \))
D
The silicon semiconductor has transitioned into an insulator state
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: The applied forward voltage is below the built-in barrier potential (\( \approx 0.7\text{ V} \))
Concept:

A silicon diode requires a forward voltage greater than its built-in barrier potential (\( \approx 0.7\text{ V} \)) to reduce the depletion layer resistance and allow significant majority carrier conduction.

Formula:

$$V_{\text{applied}} = 0.2\text{ V} < V_{\text{barrier}}(\text{Si}) \approx 0.7\text{ V}$$

Solution:

  • At \( V_{\text{applied}} = 0.2\text{ V} \), the external field only partially offsets the built-in potential barrier.


  • The remaining potential barrier (\( \approx 0.5\text{ V} \)) continues to block the majority of diffusing charge carriers.


  • Consequently, only a negligible sub-threshold current flows through the circuit.


Why other options are incorrect:

  • Option A: Breakdown occurs under high reverse voltage, not low forward voltage.
  • Option B: The diode is forward-biased, but the voltage is below the threshold required for significant conduction.
  • Option D: The semiconductor material remains functional; its operating point is simply below the forward knee voltage.

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