Effect of doping on the width of depletion layer of PN-junction diode I've read that if there is very low doping in a pn-junction diode, the depletion region will be large because a large volume of depleted semiconductor is needed to generate enough electric field to balance the diffusion current. I have trouble understanding this.
How is it that a large volume of depleted semiconductor is needed to generate enough electric field to balance the diffusion current in case of low doping?
 A: Let our semiconductor be a crystal of a Group 14 element, for clarity.
When you replace a Group 14 atom with a Group 15 atom, it releases one extra electron to the conduction band, and remains with only 4 valence electrons (instead of the regular 5), and hence acquires a positive charge. This is how a N-type semiconductor is created.
When you replace a Group 14 atom with a Group 13 atom, it brings only 3 electrons instead of 4 allotted for valence band, and one state becomes vacant, i.e. a hole is formed. Likewise, if this “primordial hole”☺ drifts away, the atom remains with 4 (instead of its own 3) valence electrons, and hence acquires a negative charge. This is how a P-type semiconductor is created.
Update: another text about formation of N- and P-type semiconductors is at When recombination in PN junction occurs, which atom becomes an ion?
More concentrated is your N-type (P-type) semiconductor, higher charge density will have respective side of the depletion zone, and thinner will it be for the same voltage drop. You can also refer to Reverse bias P-N junction and Charge density in depletion layer for pn-junction
