Inhomogeneously Doped Semiconductors
Most recent answer: 03/07/2013
Q:
When we have doped a semiconductor inhomogeneously e.g. n(x) is linear and decreases from left to right first the same does happen with the holes i. e p(x) is linear but increases from left to right and of course because we have doped it with electrons holes have decreased? Second n(x),p(x) after relaxation time are linear parallel to x?Third we have a diffusion current because n(x),p(x) are initially not constant and after a while what happens to this current?If after relaxation time n(x),p(x)are constant then there is not a diffusion current or p(x),n(x) are less changing and diffusion current decreases?Initially the semiconductor is not charged ,at the same time when an electron goes to right because of the diffusion current we have a gradually increasing E field and creates an
opposite idrift until idiffusion and idrift are equal and then at relaxation time we ve no current but before we have a decreasing current?Because of E is there a charge? my teacher said that there isn't but how we have E and V without charge?Sorry if i was wearing but i cannot get it. The Milman and Grabel book does not explaining very analytically the same with Razavi. Are there any books more analytical about the physics of semiconductors and microelectronics i don't want scientific hard books but more descriptive.Thank you for your time.
- Kostas (age 19)
Patra Greece
- Kostas (age 19)
Patra Greece
A:
If the n doping changes linearly with x, the p concentration will not. The basic equilibrium relation is n*p= n02 where n0 is the concentration of electrons (and holes) in the undoped material at the operating temperature. So p(x)=n02/n(x). So it will increase from left to right but not linearly.
With regard to the diffusion current, you're right that some charges move to set up an electric field. Once equilibrium is reached, quickly, the drift current caused by the field then just cancels the diffusion current. You're right that there must be some charge density as the source of the field, but that density is typically a negligible fraction of n for n-doped materials. Thus it's usually an excellent approximation to treat the equilibrium concentrations as determined by the local doping and charge neutrality, even though the small violations of charge neutrality are essential to setting up the equilibrium fields.
I'm not sure what text would be good.
Mike W.
With regard to the diffusion current, you're right that some charges move to set up an electric field. Once equilibrium is reached, quickly, the drift current caused by the field then just cancels the diffusion current. You're right that there must be some charge density as the source of the field, but that density is typically a negligible fraction of n for n-doped materials. Thus it's usually an excellent approximation to treat the equilibrium concentrations as determined by the local doping and charge neutrality, even though the small violations of charge neutrality are essential to setting up the equilibrium fields.
I'm not sure what text would be good.
Mike W.
(published on 03/07/2013)