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I am puzzled with impurity ionization in Semi-conductors.

Suppose $N_d$ is the density of donor impurities and $n_d$ the density of electrons bound to the single impurity orbital with energy level $\varepsilon_d$.

Defining $\mu$ as the chemical potential of the semi-conductor electron gas (with $\varepsilon_d\geq\mu$), the density of bound electrons $n_d$ can be found as :

\begin{equation} n_d = \frac{N_d}{\frac{1}{2}e^{\frac{\varepsilon_d-\mu}{k_BT}} + 1} \end{equation}

The full ionization condition is then written as $\varepsilon_d-\mu\gg k_BT$ for which almost no electrons are bound to the impurity orbital. Note that even though $\varepsilon_d-\mu\gg k_BT$, impurity ionization is still physically achieved through thermal excitation from the impurity orbital $\varepsilon_d$ to the conduction band with energy $\varepsilon_c$ where $\varepsilon_c\geq\varepsilon_d$.

What I do not understand is :

Looking at the expression of $n_d$, low temperature seems to foster impurity ionization. However, at very low temperature, we know that incomplete ionization should arise. What is the correct picture?

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In your formula, $n_d/N_d$ gives you the ratio of the occupied to the total (occupied plus unoccupied) number of impurity states.

When considering effects of temperature changes, you must, in general, also include the change of the Fermi level $\mu$ with temperature which requires to take into account the effective densities if states and occupation of the valence and conduction bands of the semiconductor.(See, e.g., S.M. Sze, Physics of Semiconductor Devices,1969, Chapt. 4) When you do this, you will find that at very low temperatures, the Fermi level moves to an energy between the conduction band edge and the impurity energy level so that ratio $n_d/N_d$ becomes close to one signifying that most of the impurity energy levels are occupied and thus the impurities are not ionized. This behavior at low temperatures is usually accompanied by a decrease of the conduction electron concentration, which is also called "carrier freeze-out".

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  • $\begingroup$ Why not however, looking at the chemical potential for low donor impurity levels: $\mu = \mu_i + k_BT asinh(\frac{\Delta n}{2n_i})$ with $\Delta n = N_d$ suggests that as T decreases the chemical potential remains close to the intrinsic chemical potential $\mu_i$ which is right in the middle of the gap for identical valence and conduction density of states. $\endgroup$ Commented Mar 26, 2017 at 17:52
  • $\begingroup$ @RonanTarikDrevon - I don't know what you mean by "low donor impurity levels". Low in energy or in concentration? Further, it is unclear where your equation for the chemical potential comes from. The chemical potential stays the closer to the intrinsic level the higher the temperature and the lower the impurity concentrations are. If you let the impurity concentration go to zero, the chemical potential will approach the intrinsic level. At very low temperatures, the donors tend to be fully occupied by the neutralizing electrons and the chemical potential always moves above the donor level. $\endgroup$
    – freecharly
    Commented Mar 26, 2017 at 19:04
  • $\begingroup$ Low donor impurity in comparison with the intrinsic concentration indeed $\frac{N_d}{ n_i}=O(1)$. This is quite easy to demonstrate considering $p_vn_c=n_i^2$ and $n_c-p_v=N_d$ (hence full ionisation) and $n_c=n_ie^{\beta(\mu-mu_i)}$,$p_v=n_ie^{-\beta(\mu-mu_i)}$. It makes sense that as impurity tends to $0$ the chemical potential tends to its intrinsic value. However, the behaviour through the temperature is less obvious to me. $\endgroup$ Commented Mar 26, 2017 at 19:09
  • $\begingroup$ @Ronan Tarik Drevon - This will become clear to you when you look at the equations governing the position of the chemical potential as a function of temperature (and a corresponding graph) as given in the reference mentioned above. $\endgroup$
    – freecharly
    Commented Mar 26, 2017 at 19:13
  • $\begingroup$ You are answer is still right actually, since even with my formula for the chemical potential, $\mu$ increases with temperature T. This is due to the fact that the intrinsic concentration $n_i$ strongly decreases with T (stronger than linear) such that keeping $N_d$ constant makes it proportionately larger than $n_i$ hence increasing $\mu$ towards the conduction band edge. $\endgroup$ Commented Mar 26, 2017 at 19:19

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