Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
No. NCERT is clear on this. If you press two flat slabs together, the surface roughness is far larger than the atomic spacing (about 2 to 3 angstrom), so there is no continuous contact at the atomic level. Charge carriers cannot flow smoothly across such a gap. A real p-n junction is made from ONE crystal: you start with a p-type wafer and add a pentavalent impurity to part of it, turning that part into n-type. The boundary inside this single crystal is the metallurgical junction.
Two processes happen together. Diffusion: the n-side is crowded with electrons and the p-side is crowded with holes, so holes move p to n and electrons move n to p because of the concentration difference. This gives a diffusion current. Drift: as electrons leave the n-side they leave behind fixed positive donor ions, and as holes leave the p-side they leave behind fixed negative acceptor ions. These fixed ions set up an electric field that pushes carriers back. That field-driven motion is the drift current.
When an electron leaves the n-side it exposes a fixed positive ion; when a hole leaves the p-side it exposes a fixed negative ion. So the n-side near the junction becomes positive and the p-side near the junction becomes negative. This charge separation creates a potential difference across the junction, called the barrier potential (built-in potential V0). Its polarity opposes further crossing of carriers. Typical value is about 0.3 V for germanium and about 0.7 V for silicon.
No. At the start, diffusion current is large and drift current is small. As more carriers cross, the fixed-ion (space-charge) regions grow, the electric field gets stronger, so drift current rises. This continues until diffusion current exactly equals drift current. At that point the net current is zero and the junction is in equilibrium. So an isolated p-n junction carries NO net current on its own.
Diffusion current is due to majority carriers moving because of a concentration difference (holes p to n, electrons n to p). Drift current is due to the built-in electric field sweeping minority carriers the other way (n to p direction of field). At equilibrium these two are equal and opposite, so they cancel. Under bias (applied voltage) one dominates, which is how a diode conducts or blocks.
Consider Statements A and B: A. A zener diode is connected in reverse bias when used as a voltage regulator. B. The potential barrier of a p-n junction lies between 0.1 V to 0.3 V. Identify the correct answer.
The increase in the width of the depletion region in a p-n junction diode is due to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is one semiconductor crystal that is p-type on one side and n-type on the other, with a boundary in between. At that boundary carriers cross over and a built-in barrier forms.
Start with a p-type silicon wafer and add a small amount of pentavalent impurity to part of it, converting that part to n-type. Diffusion and drift then create the junction and its barrier potential.
About 0.3 V for germanium and about 0.7 V for silicon. It is the built-in voltage V0 that opposes further diffusion of carriers across the junction.
No. At equilibrium the diffusion current and drift current are equal and opposite, so the net current is zero until an external voltage (bias) is applied.
Surface roughness is much larger than atomic spacing, so there is no atomic-level contact. The gap behaves as a discontinuity and carriers cannot flow across it properly.
