In thermal equilibrium, the mass action law states that $$n_{h}n_{e}=n_{i}^{2},$$ where \$n_{h}\$ is the equilibrium concentration of holes and which equals to the equilibrium concentration of conduction electrons \$n_{e}\$ $$n_{h}=n_{e}=n_{i}.$$
After doping, for example with electrons donor, arsenic for example, we would have $$ N_{D}>>n_{e}$$
After doping, do we still have a thermodynamic equilibrium?
According to the wikipedia page on this topic where the mass action law is written as follows:
(where n is free electron concentration, and p the free hole concentration.)
$$ n_{}p_{}=n_{i}^{2} $$
In place of the above original $$ n_{h}n_{e}=n_{i}^{2} $$
The explanation continues with the extended equation of $$ n_{}p_{}=N_{c}N_{v}exp(-E_{g}/k_{B}T_{})=n_{i}^{2} $$
and ends with the statement:
"The above equation holds true even for lightly doped extrinsic semiconductors as the product np is independent of doping concentration."
So from that I'd venture to say that after doping, you would still have a thermodynamic equilibrium.
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