Final answer:
In a p-n junction, increasing the p-doping from NA to 2NA will decrease the depletion region's width but increase the total amount of depletion charges due to the higher concentration of acceptor atoms.
Step-by-step explanation:
When considering two n+p junctions with different p-doping levels, the total amount of depletion charges will change in the second junction compared to the first one. If the p-doping is increased from NA to 2NA, the depletion region width will decrease because the higher concentration of acceptor atoms in the p-type semiconductor will more readily accept electrons from the n-type side. As a result, it will take a smaller volume to achieve charge neutrality. However, the total charge within the depletion region increases as there are more donor and acceptor ions exposed due to the higher initial doping concentration of the p-type semiconductor.
The balance of drift and diffusion currents set up the equilibrium condition in a p-n junction. By increasing the p-doping, you increase the charge density on the p-side, which increases the electric field intensity in the depletion region. However, if we assume that the n-type doping level remains constant, the overall change in the electric field will still depend on both the n and p-side charge concentration.