Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
It is called depletion because it is depleted (emptied) of free (mobile) charge carriers. When holes from the p-side and electrons from the n-side diffuse across the junction, they recombine and cancel out. So this narrow zone has almost no free electrons or holes. What is left are only the fixed, immobile ion cores. Because it has no free carriers, its resistance is very high.
The n-side of the depletion region is positive and the p-side is negative. On the n-side, electrons leave and expose fixed positive donor ions, so it becomes positive. On the p-side, holes are filled by arriving electrons, exposing fixed negative acceptor ions, so it becomes negative. So the built-in electric field points from n-side to p-side, which is why the n-material is at a higher potential than the p-material.
The layer of fixed positive ions (n-side) and fixed negative ions (p-side) sets up an electric field across the junction. This charge separation creates a potential difference V0 across the depletion region, called the potential barrier or built-in potential. It opposes the further diffusion of majority carriers (holes from p, electrons from n). Diffusion stops when this barrier grows strong enough that the drift current exactly balances the diffusion current - that is equilibrium.
Yes. The potential barrier, barrier potential and built-in potential V0 all mean the same thing for an unbiased junction: the voltage difference that appears across the depletion region at equilibrium. Its typical value is about 0.3 V for germanium and about 0.7 V for silicon. Important NEET point: you cannot measure V0 with a voltmeter across the diode terminals, because a matching contact potential appears at the metal-semiconductor contacts and cancels it in an external circuit.
Because two opposite currents cancel. The concentration gradient drives a diffusion current of majority carriers across the junction. The built-in field drives a drift current of minority carriers in the opposite direction. At equilibrium these two are equal and opposite, so the net current is zero even though a potential barrier V0 exists across the depletion region.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a thin layer around the junction that is depleted of free charge carriers. It contains only fixed positive donor ions on the n-side and fixed negative acceptor ions on the p-side. Its typical width is a fraction of a micrometre.
The potential barrier V0 is about 0.7 V for a silicon junction and about 0.3 V for a germanium junction at room temperature. Silicon has a larger barrier because it has a larger energy gap.
It decreases in forward bias (effective barrier becomes V0 minus V) and increases in reverse bias (effective barrier becomes V0 plus V). So the depletion width shrinks in forward bias and widens in reverse bias.
No. When you connect a voltmeter, contact potentials appear at the metal-semiconductor junctions that exactly cancel V0. So no current flows and the voltmeter reads zero. V0 exists across the depletion region but is not available in the external circuit.
The field points from the n-side to the p-side (from positive ions to negative ions) across the depletion region. This field opposes the diffusion of majority carriers, which is why it forms a barrier.
