Question
A silicon p-n junction diode has a knee voltage of 0.7 V. It is connected in series with a resistor and a 5 V battery in forward bias. Find: (a) current through the diode, (b) voltage across the resistor, (c) power dissipated in the diode, (d) what changes if we reverse the battery.
Solution — Step by Step
In forward bias, the diode acts like a fixed voltage drop of V (knee voltage). Applying KVL around the loop:
Voltage across resistor: V.
Power dissipated in diode: .
In reverse bias, the diode allows only a tiny reverse saturation current (typically nA for silicon). For practical purposes, in reverse bias. So and the entire 5 V appears across the diode.
(a) mA, (b) V, (c) mW, (d) reverse: , V.
Why This Works
The ideal-diode-with-knee-voltage model treats a forward-biased diode as a constant 0.7 V drop (silicon) or 0.3 V (germanium). This linearizes the exponential I-V characteristic into a piecewise straight line — accurate enough for circuit analysis at the JEE/NEET level.
In reverse bias, the diode behaves like an open circuit (until breakdown). This asymmetric conduction is what makes the diode useful as a rectifier.
For half-wave rectifier output, and . For full-wave: and . These formulas are tested directly almost every NEET year.
Alternative Method
Using the ideal diode (zero knee voltage): mA. This overestimates by 0.7 mA. For JEE questions, always check whether “ideal” or “silicon (0.7 V)” is specified.
Common Mistake
Students write , ignoring the 0.7 V knee. Or worse, they subtract 0.7 V twice (once thinking it’s a battery, once as a drop). The diode is one element — one drop, applied once, in the direction of current flow.