Semiconductor chemical potential and the density of holes

In summary, the chemical potential in intrinsic semiconductors at T=0 is located at the middle of the energy gap. At a temperature of (3/4)(E_c+E_v), the chemical potential is equal to the average of the bottom of the conduction band and the top of the valence band. When doped with donors or acceptors, the chemical potential changes with temperature. In a direct band gap semiconductor doped with impurities, the density of holes at room temperature can be calculated using the given equations and values. However, it is unclear how to apply the equation for density of impurities in the problem.
  • #1
steroidjunkie
18
1

Homework Statement



[a) Show that chemical potential in intrinsic semiconductor at T = 0 lies exactly in the middle of the energy gap. At what temperature is the chemical potential equal ##(3/4)(E_c + E_v)##, where ##E_c## and ##E_v## are energies od the bottom of the conduction band and of the top of the valence band?
b) Explain how the chemical potential changes with temperature if we dope the matherial with:
##b_1##) donors;
##b_2##) acceptors?
c) Semiconductor with direct band gap is doped with impurities that give density of ##10^{23}## electron/##m^3##. Calculate density of holes at room temperature if energy band gap is 1 eV, and effective masses of charge carriers in conduction and valence bands are 0,25 and 0,4 mass of free electron, respectively.

Homework Equations



a) ##n=2(\frac{m_e k_B T}{2 \pi \hbar^2})^{\frac{3}{2}} e^{\frac{\mu-E_C}{k_B T}}##
##p=2(\frac{m_h k_B T}{2 \pi \hbar^2})^{\frac{3}{2}} e^{\frac{E_V-\mu}{k_B T}}##
##n_i=p_i##

c) ##T = 300 K##
## n = 10^{23}/m^3 ##
##E_g = 1 eV ##
## m_e^{*} = 0.25 m_e ##
## m_h^{*} = 0.25 m_e ##

The Attempt at a Solution



a) Equaling n to p gives out an equation for chemical potential:

## \mu=\frac{E_C + E_V}{2} + \frac{3}{4} k_B T \ln \frac{m_h}{m_e}##

and for ##T=0##:

## \mu=\frac{E_C + E_V}{2}##

because the second term vanishes.

The second part is just putting the ##(3/4)(E_c + E_v)## instead of ##\mu## in equation ## \mu=\frac{E_C + E_V}{2} + \frac{3}{4} k_B T \ln \frac{m_h}{m_e}## and solving equation:

##(3/4)(E_c + E_v) = \frac{E_C + E_V}{2} + \frac{3}{4} k_B T \ln \frac{m_h}{m_e}##

for T gives:

##T = \frac{E_C + E_V}{3 k_B \ln \frac{m_h}{m_e}} ##

b) and c) are the problem.
b) I know what happens with chemical potential if the semiconductor is doped, but I am clueless on proving it.
c) I found this equation in Kittel's "Introduction to solid state physics" (page 210) but I'm not sure if it's applicable to the problem:
## n = (n_0 N_d) e^\frac{E_d}{2 k_B T} ##, ## n_0 = 2 (\frac{m_e^* k_B T}{2 \pi \hbar^2})^{\frac{3}{2}} ##
 
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  • #2
where ##N_d## is the density of impurities, ##E_d## - the energy level of dopant, and ##m_e^*## - the effective mass of electron in conduction band.But I don't know what to do with it.
 

Related to Semiconductor chemical potential and the density of holes

1. What is a semiconductor chemical potential?

A semiconductor chemical potential is a measure of the energy required to add or remove an electron from a semiconductor material. It is also known as the Fermi level and is a key factor in determining the electronic properties of a semiconductor.

2. How is the semiconductor chemical potential related to the density of holes?

The semiconductor chemical potential is directly related to the density of holes in a semiconductor material. As the chemical potential increases, the density of holes also increases. This is because the chemical potential determines the likelihood of an electron occupying a specific energy level, and as the energy level increases, more electrons will occupy it, leaving behind holes.

3. What factors can affect the semiconductor chemical potential?

The semiconductor chemical potential can be affected by several factors, such as temperature, doping concentration, and bandgap energy. As temperature increases, the chemical potential decreases, and as doping concentration increases, the chemical potential shifts towards the energy levels of the dopant atoms. The bandgap energy also plays a role, as a larger bandgap leads to a higher chemical potential.

4. How is the density of holes calculated in a semiconductor?

The density of holes in a semiconductor can be calculated using the equation: p = Nv * e(-Ev-Ef)/kT, where Nv is the density of states in the valence band, Ev is the valence band energy, Ef is the Fermi level, k is the Boltzmann constant, and T is the temperature in Kelvin.

5. How does the density of holes affect the conductivity of a semiconductor?

The density of holes has a direct impact on the conductivity of a semiconductor. As the density of holes increases, the number of charge carriers available to conduct electricity also increases, leading to a higher conductivity. This is why doped semiconductors, which have a higher density of holes due to the presence of dopant atoms, are more conductive than pure semiconductors.

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