Difficulty: Easy
Correct Answer: 0.7 V
Explanation:
Introduction / Context:
Understanding the forward conduction threshold of a pn-junction diode is fundamental in electronics. Designers often approximate a silicon diode as a switch that begins to conduct significantly after a certain forward voltage is reached. This question probes your grasp of that practical threshold, sometimes called the knee voltage or cut-in voltage.
Given Data / Assumptions:
Concept / Approach:
For a pn-junction diode, the Shockley relation ID = IS * (e^(VD/(n*VT)) - 1) shows current rises exponentially with forward voltage VD. In practice, engineers use a rule-of-thumb threshold where conduction becomes significant. For silicon diodes, this forward drop settles around 0.6–0.7 V at modest currents. Germanium diodes are typically ~0.2–0.3 V, and Schottky diodes ~0.2–0.4 V.
Step-by-Step Solution:
Recognize device type: standard silicon pn-junction.Recall practical knee/turn-on region: about 0.6–0.7 V for silicon.Select the closest standard value used in circuit calculations: 0.7 V.
Verification / Alternative check:
Datasheets of common silicon diodes (e.g., 1N4148, 1N400x) show typical forward drops near 0.7 V at tens to hundreds of milliamps. SPICE diode models also default to ~0.7 V as a convenient design estimate.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
0.7 V
Discussion & Comments