pn-junction
I have read at several places if we take exact same cubes of p and and n doped semiconductor material and form a junction, the depletion region(blue colored region in my drawing) will be asymmetric around the physical center as below:
This means for some reason the doping difference in two materials shows itself in such asymmetry.
If correct, how could it be explained in simple terms?
How can a charge neutral substance have a potential?
To say that a PN junction has built-in potential isn't to say that the PN junction has a potential relative to ground or infinity etc.
How can there be an inherent potential in a doped semi-conductor if it is charge neutral?
Charge has been separated within the PN junction and, thus, there is an electric field across the depletion region and an associated potential difference. A charged capacitor is neutral but there is a potential difference (voltage) across the dielectric. There are some similarities but...
However, I don't understand why we can't measure the voltage drop across the PN block with a voltmeter.
As explained, for example, here, the built-in potential is not readily measured with, e.g., a voltmeter. In other words, this question has been asked here several times (which means you are not the only one perplexed by this - most are at first) and there are good answers already available.
When ever you put two material together that have a different work function, they will form a junction which acts as a diode/rectifier. So the equivalent circuit diagram becomes "A' or "B" below:
simulate this circuit – Schematic created using CircuitLab
Either of those scenarios yields a device that isn't very useful, in fact it can't be used. Ideally you want to make sure that you contact to the device looks like a resistor (ideally a zero ohm resistor, or at very least a very low value resistor) and NOT a diode. I'm making the assumption here that you'll understand why having diodes in series with your transistor is a bad, bad thing.
The interesting aspect of this question is: Just HOW is a metal attached to a semiconductor such that is does not form a shottky barrier (i.e. a diode) which a metal semi-conductor system will usually form.
The background theory is very interesting but it requires an understanding of Quantum mechanics, semi-conductor theory (fermi levels etc.) and processing knowledge.
For a N-Type contact, implants are formed in such a way that at the surface the semiconductor is degenerate (i.e. Metal like) so a Metal, Metal contact does not have a barrier. Often these system will have silicides on them (refractory Si-Metal system)- to decrease the resistance) but that is not necessary here. As the current moves into the substrate it experiences less and less doping (the band edges are gently distorted) until the carriers make it into the semiconductor.
For an P-Type contact, there are Quantum mechanical effects that allows for a tunnelling across the barrier (formed by the band gap ~ 1.12 eV), the rest of the trip from the conduction band edge up to the metal work function is explained by the same mechanism as used in a N-Type device (degenerately doped). the Qm -effect is accomplished via a hyper abrupt metal - P material junction ( a few atoms deep at most).
Best Answer
It's simple enough — if you want to transfer some number of carriers from one side to the other in order to neutralize them, they must be more spread out (occupy a greater volume) on the lightly-doped side.
The doping level refers to the density of donor or acceptor atoms in the semiconductor matrix; light doping means a lower density.