Semiconducting Materials – Solved Problems

1. Calculate the intrinsic carrier concentration in germanium at 300 K. Given that m_e *=1.09 m₀, m_h *=0.31 m₀ and E_g = 0.68 eV.

Given

T = 300 K, k=1.38×10^-23 Jk^-1, h=6.626×10^-34 Js,
m₀ = 9.11×10^-31 Kg, E_g = 0.68 eV = 0.68×1.6×10^-19 J,
m_e *=1.09 m₀ = 9.929×10^-31 Kg, mp^* = 0.31
m₀ = 2.824×10^-31 Kg, n_i =?

Solution :-

\begin{aligned} \mathrm{n}_{\mathrm{i}} & =2\left(\frac{2 \pi \mathrm{kT}}{\mathrm{h}^{2}}\right)^{3 / 2}\left(\mathrm{~m}_{\mathrm{e}}^{*} \mathrm{~m}_{\mathrm{h}}^{*}\right)^{3 / 4} \exp -\left(\frac{\mathrm{E}_{\mathrm{g}}}{2 \mathrm{kT}}\right) \\ \mathrm{n}_{\mathrm{i}} & \left.=2\left[\frac{2 \times 3.14 \times 1.38 \times 10^{-23} \times 300}{\left(6.626 \times 10^{-34}\right)^{2}}\right]^{3 / 2}\left(9.929 \times 2.824 \times 10^{-62}\right)^{3 / 4} \times \mathrm{e}^{-\left[\frac{0.68 \times 1.6 \times 10^{-19}}{2 \times 1.38 \times 10^{-23} \times 300}\right.}\right] \\ & =2\left[\frac{2.599 \times 10^{-20}}{6.626 \times 6.626 \times 10^{-68}}\right]^{3 / 2} \times\left(3.853 \times 10^{-46}\right) \times \frac{1}{\mathrm{e}^{13.14}} \\ & =2\left[5.9197 \times 10^{46}\right]^{3 / 2} \times 3.853 \times 10^{-46} \times \frac{1}{508896.53} \\ \mathrm{n}_{\mathrm{i}} & =2.18 \times 10^{19} / \mathrm{m}^{3} \end{aligned}

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2. Mobilities of electrons and holes in a sample of intrinsic germanium at 300 Kare 0.32 m² V^-1 s⁻¹ and 0.15 m² V^-1 s⁻¹ respectively. If the conductivity of the specimen is 2Ω⁻¹ m⁻¹, compute the carrier concentration.

Given

σ_i = 2.0 Ω^-1 m^-1, μ_e = 0.32 m² V^-1 s^-1. μ_h = 0.15m^² V^-1 s^-1 e=1.602×10^-19 C, n_i = ?

Solution:-

\begin{aligned} \sigma_{\mathrm{i}} & =\mathrm{n}_{\mathrm{i}} \mathrm{e}\left(\mu_{\mathrm{e}}+\mu_{\mathrm{h}}\right) \\ \mathrm{n}_{\mathrm{i}} & =\frac{\sigma_{\mathrm{i}}}{\mathrm{e}\left(\mu_{\mathrm{e}}+\mu_{\mathrm{h}}\right)} \\ & =\frac{2.0}{1.602 \times 10^{-19} \times(0.32+0.15)} \\ \mathrm{n}_{\mathrm{i}} & =2.656 \times 10^{19} / \mathrm{m}^{3} \end{aligned} 

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3. The following data are given for intrinsic germanium at 300 K n_i=2.4×10¹⁹/m³,μ_e=0.39 m² V^-1 s⁻¹,μ_h=0.19 m² V^-1 s⁻¹. Calculate the conductivity of the sample.

Given

μ_e = 0.39 m² V^-1 s^-1, μ_h=0.19m² V^-1 s^-1, e=1.602×10^-19 C, σ_i =?

Solution:-

σ_i = n_ie(μ_e + μ_h)

=2.4×10¹⁹ × 1.602 × 10^-19 (0.39+0.19)

σ_i = 2.22Ω^-1 m^-1

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4. The Hall coefficient of certain silicon specimen was found to be −7.35×10⁻⁵ m³ C⁻¹. The -ve sign indicates it is n type of semiconductor. Calculate the density and mobility of charge carrier if σ=200Ω⁻¹ m⁻¹.

Given

R_H = 7.35×10^-5 m³ C^-1, σ = 200Ω^-1 m^-1

Solution:-

\begin{aligned} \mathrm{n}_{\mathrm{e}} & =\frac{-1}{\mathrm{eR}_{\mathrm{H}}}=\frac{-1}{1.602 \times 10^{-19}\left(-7.35 \times 10^{-5}\right)} \\ \mathrm{n}_{\mathrm{e}} & =8.492 \times 10^{22} / \mathrm{m}^{3} \\ \therefore \mu_{\mathrm{e}} & =\frac{\sigma}{\mathrm{n}_{\mathrm{e}} \mathrm{e}} \\ & =\frac{200}{8.492 \times 10^{22} \times 1.602 \times 10^{-19}} \\ \mu_{\mathrm{e}} & =0.0147 \mathrm{~m}^{2} \mathrm{v}^{-1} \mathrm{~s}^{-1} \end{aligned}

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5. A Silicon plate of thickness 1 mm, breadth 10 mm and length 100 mm is placed in a magnetic field of 0.5Wb/m² acting perpendicular to it’s thickness. If 10^-2 A current flows along it’s length. Calculate the Hall voltage developed if the Hall coefficient is 3.66×10^-4 m³ / coulomb.

Given

R_H= 3.66×10^-4 m³ / C,I=10^-2 A, B=0.5 Wb/m², t=1mm, V_H=?

Solution:-

\begin{aligned} R_{H} & =\frac{V_{H} t}{I_{x} B} \\ V_{H} & =\frac{R_{H} I_{x} B}{t} \\ & =\frac{3.66 \times 10^{-4} \times 10^{-2} \times 0.5}{1 \times 10^{-3}} \\ V_{H} & =1.83 \mathrm{mV} \end{aligned}

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6. A sample of n type semiconductor has a resistivity of 10⁻³ ohm –m and a Hall coefficient of 10⁻⁴ m³/C. Assuming only electrons as charge carriers, determine the electron density and mobility.

Given

R_H = 10^-4 m³ / C, ρ = 10^-3 ohm/m, n=? μ=?

Solution:-

\begin{aligned} \mathrm{R}_{\mathrm{H}} & =\frac{1}{\mathrm{ne}} \\ \mathrm{n} & =\frac{1}{\mathrm{eR}_{\mathrm{H}}}=\frac{1}{1.602 \times 10^{-19} \times 10^{-4}} \\ \mathrm{n} & =6.24 \times 10^{22} / \mathrm{m}^{3} \\ \sigma & =\mathrm{ne} \mu \\ \rho & =\frac{1}{\sigma}=\frac{1}{\mathrm{ne} \mu} \\ \mu & =\frac{1}{\mathrm{ne} \rho}=\frac{1}{6.24 \times 10^{22} \times 1.602 \times 10^{-19} \times 10^{-3}} \\ \mu & =0.1 \mathrm{~m}^{2} / \mathrm{Vs} \end{aligned}

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7. Find the resistance of an intrinsic germanium rod 1 cm long, 1 mm wide and 1 mm thick at 300 K.

For germanium n_i = 2.5×10¹⁹ / m³

μ_e = 0.39 m² V^-1 s^-1

μ_e =  0.19 m² V^-1 s^-1 at 300 K

Given data

Intrinsic carrier concentration n_i = 2.5×10¹⁹ / m³
Electron mobility μ_e = 0.39 m² V^-1 s^-1
Hole mobility μ_e =  0.19 m² V^-1 s^-1

Solution:-

We know that the electrical conductivity of an intrinsic semiconductor (Germanium)

σ = n_i e(μ_e + μ_h)

Substituting the given values, we have

σ = 2.5×10¹⁹ × 1.6×10^-19 (0.39+0.19)

σ = 2.32 Ω^-1 m^-1

Resistance \mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}} \text { or } \mathrm{R}=\frac{1}{\sigma \mathrm{A}}\left(\therefore \sigma=\frac{1}{\rho}\right)

where,

l → length of the rod = 1cm =1×10^-2 m
A → Area of cross-section (width x thickness)
A → (1×10^-3 m)(1×10^-3 m)

\begin{aligned} & =\frac{1 \times 10^{-2}}{2.32 \times\left(1 \times 10^{-3} \times 1 \times 10^{-3}\right)} \\ & =4310 \Omega \end{aligned}

Resistance of germanium = 4310Ω

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8. For an intrinsic semiconductor with a band gap of 0.7 eV, determine the position of E_F at T=300 K if m_h^∗=6 m_e^∗.

Given Data

\mathrm{E}_{\mathrm{g}}=0.7 \mathrm{eV}\left(\right. or) 1.12 \times 10^{-19} Joules, \mathrm{T}=300 \mathrm{~K}, \frac{\mathrm{m}_{\mathrm{h}}^{*}}{\mathrm{~m}_{\mathrm{e}}^{*}}=6

Solution:-

We know Fermi energy of an intrinsic semiconductor

\begin{aligned} \mathrm{E}_{\mathrm{F}} & =\frac{\mathrm{E}_{\mathrm{g}}}{2}+\frac{3 \mathrm{~K}_{\mathrm{B}} \mathrm{T}}{4} \log _{\mathrm{e}}\left(\frac{\mathrm{m}_{\mathrm{h}}^{*}}{\mathrm{~m}_{\mathrm{e}}^{*}}\right) \\ \mathrm{E}_{\mathrm{F}} & =\frac{1.12 \times 10^{-19}}{2}+\frac{3 \times 1.38 \times 10^{-23} \times 300}{4} \log _{\mathrm{e}} 6 \\ & =5.6 \times 10^{-20}+5.5634 \times 10^{-21} \\ \mathrm{E}_{\mathrm{F}} & =6.15634 \times 10^{-20} \mathrm{~J} \text { (or) } \mathrm{E}_{\mathrm{F}}=\frac{6.15634 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV} \\ \mathrm{E}_{\mathrm{F}} & =0.3847 \mathrm{eV} \end{aligned}

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9. Find the intrinsic resistivity of Ge at room temperature 300K if the carrier density is 2.15×10¹³/cm³.

Given Data

Mobility of electron μ_e = 3900 cm² / Vs
Mobility of hole μ_n = 1900 cm² / Vs
Mobility density n_i = 2.15×10¹³ /cm³

Solution:-

We know that σ_i = e(μ_e + μ_h) n_i

Substituting the given values, we have

\begin{aligned} \sigma_{\mathrm{i}} & =1.6 \times 10^{-19} \times(3900+1900) \times 2.15 \times 10^{13} \\ & =1.6 \times 10^{-19} \times 5800 \times 2.15 \times 10^{13} \\ \sigma_{\mathrm{i}} & =2.32 \times 10^{-2} \mathrm{ohm} / \mathrm{cm} \end{aligned}

Intrinsic resistivity \rho_{\mathrm{i}}=\frac{1}{\sigma_{\mathrm{i}}}=\frac{1}{2.32 \times 10^{-2}} \rho_{\mathrm{i}}=43 \Omega \mathrm{cm}