For a basic pn-junction diode, under which terminal polarity does the junction conduct significant current (i.e., forward bias condition)?

Introduction / Context:
A pn junction is the fundamental building block of diodes, BJTs, and many IC devices. Understanding forward and reverse bias conditions is essential for predicting current flow and designing rectifiers, clamps, and limiters.

Given Data / Assumptions:

• Simple silicon pn-junction diode, not in breakdown.
• Room-temperature operation; typical forward drop around 0.6–0.8 V for silicon.
• No series resistance or special effects considered.

Concept / Approach:
Forward bias reduces the depletion region by making the p-side more positive than the n-side. This lowers the barrier potential, allowing majority carriers to cross the junction and creating significant conduction. Reverse bias widens the depletion region and blocks current except for a small leakage (until breakdown).

Step-by-Step Solution:
Apply polarity: V_p > V_n (p more positive).Barrier potential decreases, enabling carrier injection across the junction.Conduction increases exponentially with applied forward voltage.Therefore, forward conduction occurs when the p-type is more positive than the n-type.

Verification / Alternative check:
Use a DMM’s diode test: red lead (positive) on anode (p-side), black lead (negative) on cathode (n-side) shows a forward drop. Reversing leads shows OL (reverse-biased).

Why Other Options Are Wrong:
n more positive than p → reverse bias, blocking conduction.

Same potential or no potential → no forward bias; only leakage exists.

Zener breakdown is a special reverse-bias condition, not ordinary forward conduction.

Common Pitfalls:
Confusing anode/cathode polarity. Remember: forward bias → anode positive, cathode negative.

Final Answer:
the p-type material is more positive than the n-type material