Electronic Devices & Circuits · Lecture 06

Semiconductors Current Carriers & Fermi Level

Semiconductor Physics

Dr. Mithun Mondal BITS Pilani, Hyderabad Campus
Lecture Recording
SECTION 01

Carrier Concentration in Intrinsic Semiconductors

  • \[\begin{aligned} n &=\int_{E_c}^{\infty} N(E) f(E) \mathrm{d} E \\ f(E) &=\mathrm{e}^{-\left(E-E_F\right) / k T} \quad \text{for}~ E \geq E_C, ~E-E_F>>k T \\ n &=\int_{E_c}^{\infty} \gamma\left(E-E_C\right)^{\frac{1}{2}} \mathrm{e}^{-\left(E - E_F\right) / k T} \mathrm{~d} E \\ n &=N_C \cdot \mathrm{e}^{-\left(E_c-E_F\right) / k T} \\ N_C &=2\left(\frac{2 \pi m_n k T}{h^2}\right)^{3 / 2}\left(1.60 \times 10^{-19}\right)^{3 / 2} \end{aligned}\]
    \(m_n\)Concentration of electrons in conduction band
  • \[\begin{aligned} N(E) & = \gamma \cdot (E_v-E)^{1/2} \end{aligned}\]
    is the maximum valence band energy, the density of states is If
  • \[\begin{aligned} 1-f(E) &=\frac{\mathrm{e}^{\left(E-E_F\right) / k T}}{1+\mathrm{e}^{\left(E-E_F\right) / k T}}=\mathrm{e}^{-\left(E_F-E\right) / k T} \quad E_F-E>>k \text { for } E \leq E_V \\ p &=\int_{-\infty}^{E_v} \gamma\left(E_V-E\right)^{1 / 2} \mathrm{e}^{-\left(E_F-E\right) / k T} \mathrm{~d} E \\ p &=N_v \cdot \mathrm{e}^{\left(E_F-E_v\right) / k T} \\ N_v &=2\left(\frac{2 \pi m_p k T}{h^2}\right)^{3 / 2}\left(1.60 \times 10^{-19}\right)^{3 / 2} \end{aligned}\]
    \(m_p\)For holes calculation:
  • \[\begin{aligned} n_i &=p_i \\ N_C \cdot \mathrm{e}^{-\left(E_c-E_F\right) / k T} &=N_V \cdot \mathrm{e}^{-\left(E_F-E_v\right) / k T} \\ \text{Taking logarithmic}~ \operatorname{ln} \frac{N_C}{N_V} &=\frac{E_C+E_V-2 E_F}{k T} \\ E_F &=\frac{E_C+E_V}{2}-\frac{k T}{2} \ln \frac{N_C}{N_V} \\ N_C &=N_V \quad (\text{considering same effective masses }) \\ E_F &=\frac{E_C+E_V}{2} \end{aligned}\]
    From the equation, at the centre of the forbidden energy band, Fermi level is presentFermi level of an intrinsic semiconductor:

  • Fermi level of an extrinsic semiconductor:

\[\begin{aligned} {\color{teal}{\textbf{N-type}}} & \\ E_F& =E_C+k T \ln \frac{N_C}{N_D} \\ N_D&=N_C \cdot \mathrm{e}^{-\left(E_C-E_F\right) / k T} \\ \end{aligned}\]
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SECTION 02

Problem-1

\[\begin{aligned} \text{For N-type material:}~E_F & = E_C - kT\operatorname{ln}\dfrac{N_C}{N_D} \\ \Rightarrow ~ \left(E_C - E_F\right) & = kT \operatorname{ln}\dfrac{N_C}{N_D}\\ \Rightarrow~0.3 & = 300k\operatorname{ln}\dfrac{N_C}{N_D} \qquad (1)\\ \text{Also}~E_C-E_{F1} & = 360k\operatorname{ln}\dfrac{N_C}{N_D} \qquad (2)\\ \Rightarrow~\dfrac{E_C-E_{F1}}{0.3} & = \dfrac{360}{300} \\ \Rightarrow~E_C-E_{F1} & = \dfrac{360}{300}\times 0.3 = 0.36~\mathrm{eV}~(\text{Ans}) \end{aligned}\]
SECTION 03

Problem-2

\[\begin{aligned} E_F & = E_v +kT \operatorname{ln} \dfrac{N_V}{N_A} \Rightarrow (E_F-E_V) = kT \operatorname{ln}\dfrac{N_V}{N_A} \\ T=300^{\circ}K \Rightarrow 0.3 & = 300k\operatorname{ln}\dfrac{N_V}{N_A} \qquad (1) \\ T=350^{\circ}K \Rightarrow (E_{F1}-E_V) & = 350k\operatorname{ln}\dfrac{N_V}{N_A} \qquad (2) \\ \text{From (1) and (2)} & \\ (E_{F1}-E_V) & = \dfrac{350}{300}\times 0.3=0.35~\mathrm{eV}\\ T=400^{\circ}K \Rightarrow (E_{F2}-E_V) & = 400k\operatorname{ln}\dfrac{N_V}{N_A} \qquad (3) \\ \Rightarrow~ \dfrac{E_{F2}-E_V}{0.3} & = \dfrac{400}{300} \Rightarrow~ (E_{F2}-E_V) = \dfrac{400}{300}\times 0.3 = 0.4~\mathrm{eV} \end{aligned}\]