Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
| Dopant type | Pentavalent donor (As, Sb, P) | Trivalent acceptor (B, Al, In) |
| Majority carrier | Electrons | Holes |
| Minority carrier | Holes | Electrons |
| Carrier relation | ne >> nh | nh >> ne |
| Impurity level | Donor level just below conduction band | Acceptor level just above valence band |
| Net charge | Neutral | Neutral |
No. Both are electrically neutral. The 'n' and 'p' refer only to the majority carrier, not net charge. In n-type, extra free electrons come from donor atoms, but the fixed donor cores left behind carry equal positive charge, so the crystal stays neutral. Same idea for p-type: extra holes are balanced by fixed negative acceptor ions. The names describe carrier type, not overall charge.
n-type is made by adding a pentavalent (valency 5) impurity such as Arsenic (As), Antimony (Sb) or Phosphorus (P). These are called donors because they donate one extra electron. p-type is made by adding a trivalent (valency 3) impurity such as Boron (B), Aluminium (Al) or Indium (In). These are called acceptors because they create a hole that accepts an electron.
In n-type, electrons are the majority carrier and holes are the minority carrier. The reverse is true in p-type: holes are majority, electrons are minority. Even though one type dominates, the other still exists in a small amount and follows the mass action law: ne times nh = ni squared.
For equal doping level, an n-type sample has about the same number of majority electrons as a p-type sample has majority holes. But the currents are not equal, because electron mobility is higher than hole mobility. So for the same carrier concentration and same electric field, current in n-type is greater than current in p-type. This exact idea was asked in NEET 2021.
Each donor atom gives one nearly free electron, and each acceptor atom creates one hole. Since the dopant is added in ppm amounts that still hugely outnumber the intrinsic carriers ni, the majority carrier count is approximately equal to the number of doping atoms (ND for n-type, NA for p-type). The minority carrier is then found from ne times nh = ni squared.
For a p-type semiconductor, which of the following statements is true?
The electron concentration in an n-type semiconductor is the same as the hole concentration in a p-type semiconductor. An external electric field is applied across each of them. Compare the currents in them.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The dopant and the majority carrier. n-type uses a pentavalent donor (As, Sb, P) and its majority carriers are electrons. p-type uses a trivalent acceptor (B, Al, In) and its majority carriers are holes.
Yes, both are neutral. The mobile majority carriers are exactly balanced by the fixed charged ion cores of the dopant atoms.
In n-type the minority carriers are holes; in p-type the minority carriers are electrons. Their number follows ne times nh = ni squared.
n-type conducts more for the same carrier concentration and same field, because electron mobility is higher than hole mobility.
Donors (pentavalent, make n-type): Arsenic, Antimony, Phosphorus. Acceptors (trivalent, make p-type): Boron, Aluminium, Indium.
