# Does carrier concentration at thermal equilibrium depend on doping concentration?

I came across a general equation at thermal equilibrium for carrier concentration that seems to be independent of doping concentration:

$$n_0= 2\left( \frac{2\pi m_n^* k_BT}{h^2} \right)^{3\over 2} \exp \left[ \frac{-(E_C-E_F)}{k_BT} \right]$$

which is an approximate result (for $E-E_F>3K_BT$) of the following integral:

$$n_0 = \int_{E_C}^{\infty} \underbrace{\frac{1}{1+\exp\left(\frac{E-E_F}{K_BT}\right)}}_{f(E)} \underbrace{\frac{8\sqrt{2}\pi}{h^3}m_n^* \sqrt{E-E_C}}_{N(E)} \cdot \mathrm{d}E$$

The first equation is supposed to be valid for both intrinsic and doped materials (under thermal equilibrium), but it has no apparent dependence on dopant concentration.

On the other hand, another famous equation states clearly that there's a dependance on dopant concentration:

$$n_0 = \frac{N_d-N_a}{2} + \sqrt{\left( \frac{N_d-N_a}{2} \right)^2 + n_i^2}$$

How are these two equations consistent? What am I missing here?

Related questions:

• Ok, yes you can have similar formula for doped. But for usual doped semiconductors at room temperature all donor electrons are excited, that is the region where $n$ is not dependent on $T$ and you get third formula. I am not sure whether the model predicts this behavior by taking into account finite number of donors, or it brakes down in this region. Back to your first question. The concentration of donors comes into first formula as Fermi level depends on donor concentration(see Neamen, Semiconductor Physics and Devices page 123). – sa101 Dec 6 '14 at 19:27