Using mass-action law to determine doping densities
I would like to understand the solution to part c) given below:
Find the electron and hole densities in silicon at $300$ K, for each of the following impurity densities:
a) Boron atoms at $5 \times 10^{21}\, \mathrm{m}^{-3}$;
b) Boron atoms at $2 \times 10^{22} \,\mathrm{m}^{-3}$ and Gallium atoms at $1.5 \times 10^{22} \,\mathrm{m}^{-3}$;
c) Boron atoms at $10^{23} \,\mathrm{m}^{-3}$ and Arsenic atoms at $10^{23}\,\mathrm{m}^{-3}$.
You may assume that the intrinsic carrier concentration of silicon is $_i = 1.5 \times 10^{16}\,\mathrm{m}^{-3}$.
a) Boron is an acceptor, so the hole density will be $ \approx 5 \times 10^{21}\,\mathrm{m}^{-3}$. The electron density will be $ = {_i}^2/ \approx 4.5 \times 10^{10}\,\mathrm{m}^{-3}$.
b) Boron and Gallium are both acceptors, so both contribute holes. Thus $ \approx 2 \times 10^{22} + 1.5 \times 10^{22} = 3.5 \times 10^{22}\,\mathrm{m}^{-3}$. The electron density will be $ = {_i}^2/ \approx 6.4 \times 10^9\,\mathrm{m}^{-3}$.
c) Boron atoms are acceptors and Arsenic atoms are donors. Since the dopant densities are the same, there are just as many electrons as there are holes. In this case, the Fermi level will be in the middle of the band gap and we would expect $ = = _i = 1.5 \times 10^{16}\,\mathrm{m}^{-3}$.
It was my interpretation that the mass-action law states that the intrinsic electron and hole density are identical, (basically a statement of charge neutrality); $n_i^2=np$, so this means that before any impurities are introduced there are already $1.5\times 10^{16}$ electrons and $1.5\times 10^{16}$ holes present in silicon. 565
Since we were told that the intrinsic concentration of silicon is $_i = 1.5 \times 10^{16}\,\mathrm{m}^{-3}$ and we now add $10^{23}$ holes and electrons, why is the answer to c) not $n=p=1.5\times 10^{16}\,\mathrm{m}^{-3}+10^{23}\,\mathrm{m}^{-3}\,$?
Edit:
Okay, so it seems I am totally misunderstanding what the mass-action law actually means. For the sake of understanding suppose that ${n_i}^2=16$. My interpretation of this statement is that in the absence of doping there are $n_i=n=p=4$ carriers (per unit volume) - there are intrinsically 4 electrons and 4 holes: this has to be the case as the sample is electrically neutral overall.
Now suppose that the sample is doped with donor density $n=8$, according to mass-action law, $p={n_i}^2/n=16/8=2$. While there were 4 holes to begin with, now there are just 2 holes (per unit volume)? But no charge has been added to the sample, it is still electrically neutral. When the donors ionize, the ions fixed into the lattice have a positive charge $N_D^+$; this has the same magnitude of charge as the 8 electrons added so that $n+N_D^+=0$. But according to mass-action law there are 2 holes leftover, so the sample gains an overall (net) positive charge through doping. Anyone got a problem? 531