How does doping effect the depletion layer in a semiconductor?

Answer 1

A semiconductor has an energy band (a range of energy levels) that is forbidden -- ideally void of charged particles at all temperatures. Practically, at low temperatures (T < 40 K for silicon), the probability of finding a free charge carrier outside the forbidden gap is nearly nil. When the temperature is increased, the probability of finding a free charge carrier outside the forbidden gap increases, but the net charge is still zero (negative charge exactly cancels positive charge). However, an intrinsic semiconductor (pure or undoped) is just a resistor of little importance (other materials are cheaper and better-controlled than a semiconductor).

When we introduce foreign atoms into a semiconductor (the process is called doping), we change its electrical properties -- it has a lot more free charge carriers than an intrinsic semiconductor, although again, the net charge is zero. The total charge of free carriers is balanced by immobile ions of equal and opposite total charge. For example, boron and indium will be used to dope silicon p-type; phosphorus and arsenic will be doping silicon n-type. I am quoting boron, phosphorus, and silicon as examples from hereon.

p-type doping is a process where a silicon atom in the lattice is replaced by a boron atom. A Boron atom has 3 electrons in the outer shell, compared with an electron occupancy of 4 for a silicon atom. So a Boron atom provides a vacancy for any free electrons to occupy with a little effort, when an electron chances to be nearby (the four boron-silicon covalent bonds needs 8 electrons to be stable, but only 7 are provided). The net charge of the material is still zero. More about from where the free electron is coming.

n-type doping is using a phosphorus atom to replace a silicon atom. A phosphorus atom has 5 electrons in the outer shell. So a phosphorus atom provides an electron that can be freed with a little effort (the four phosphorus-silicon covalent bonds only need 8 electrons to be stable, each atom needing only to contribute four electrons; the 9th electron will be loosely bound). The net charge of the material is still zero. Where can the electron go?

Magic happens when a p-type silicon is brought in contact with an n-type silicon to form a pn junction. The excess electron vacancies (holes) in p-Si now can exchange with the excess electrons in n-Si, but the net charge of the p-n silicon entity is still zero. However, microscopically, a depletion region is formed at the pn junction, where excess carriers can cross over to the other side. In the p-Si, excess electrons from the n-Si start filling up the holes (the lack of the 8th outer-shell electron to form the four stable boron-silicon covalent bonds) and negatively-charged boron atoms are formed. In the n-Si, excess holes from the p-Si start swallowing up the loosely-bound electrons (the 9th electron in the outer shell) of phosphorus atoms and positively-charged phosphorus atoms are formed. Once formed, and in the absence of an electric field, the depletion region now presents an energy barrier to any further carrier movement and a steady state results -- no net current in the pn junction.

Answer 2

Width of depletion layer is given by

x = (2*ebsylum*Vb).5/(qN)

x = width

Vb = potential barrier

q = charge of electron

N = doping concentration.

Thus increase in doping will reduce width of depletion layer.