Semiconductor Electronics: Materials, Devices and

14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is $\stackrel{?}{A+A}$ = $\stackrel{?}{A}$

The truth table for the same is given as:

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are $\stackrel{?}{A}$ and $\stackrel{?}{B}$  , as shown in the following figure

Above is given the inputs for the last NOR gate.

Hence, the output for the circuit can be written as:

Y = $\stackrel{?}{A+B}$ = $\stackrel{?}{\stackrel{?}{A}.\stackrel{?}{B}}$   = A.B

The truth table for the same can be written as:

This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

14.10 In a p-n junction diode, the expression for current is given as:

$\mathit{I}=\mathit{}{\mathit{I}}_{0\mathit{}}\mathit{e}\mathit{x}\mathit{p}\left(\frac{\mathit{e}\mathit{V}}{2{\mathit{k}}_{\mathit{B}}\mathit{T}}-1\right)$

Where, $I_0$ = Reverse saturation current = 5 $*{10}^{-12}$ A

T = Absolute temperature = 300 K

${\mathit{k}}_{\mathit{B}}$ = Boltzmann constant = 8.6  $*{10}^{-5}$  eV/K = 8.6 $*{10}^{-5}*1.6*{10}^{-19}$ J/K = 1.376 $$ $*{10}^{-23}J/K$

V = voltage across the diode

e = charge of an electron = 1.6 $*{10}^{-19}$ C

Forward voltage, V = 0.6V

Current, $I=5*{10}^{-12}*exp\left(\frac{1.6*{10}^{-19}*0.6}{1.376*{10}^{-23}*300}-1\right)$ = $5*{10}^{-12}*\mathrm{e}\mathrm{x}\mathrm{p}?\left(22.256\right)$

= 0.02315 A

For forward voltage, V = 0.7V, we can write

Current, $I\text{'}=5*{10}^{-12}*exp\left(\frac{1.6*{10}^{-19}*0.7}{1.376*{10}^{-23}*300}-1\right)$ = $5*{10}^{-12}*\mathrm{e}\mathrm{x}\mathrm{p}?\left(26.132\right)$

= 1.117 A

Hence increase in the current, ΔI = I'-I = 1.117 – 0.02315 = 1.0934 A

Dynamic resistance =  $\frac{Changeinvoltage}{ChangeinCurrent}$ = $\frac{0.7-0.6}{1.0934}=0.091\Omega$

If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as $I_0$ will be equal in both cases. Therefore the dynamic resistance in the reverse bias will be $\frac{1}{0}=\infty$

14.9 Energy gap of the given intrinsic semiconductor,  $E_g$  = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

$n_i=n_{0}exp\left(-\frac{E_g}{2k_{B}T}\right)$

Where,  $k_B$ = Boltzmann constant = 8.62 $*{10}^{-5}$
 eV/K

T = Temperature

$n_0$ = constant

Initial temperature, $T_1$
 = 300 K

The intrinsic carrier-concentration at this temperature can be written as

$n_{i1}=n_{0}exp\left(-\frac{E_g}{2k_B*300}\right)$ ……(1)

Final temperature,  $T_2$ = 600 K

The intrinsic carrier-concentration at this temperature can be written as

$n_{i2}=n_{0}exp\left(-\frac{E_g}{2k_B*600}\right)$ ……(2)

The ratio between conductivity at 600K and at 300 K is equal to the ratio between the respective intrinsic carrier concentration at these temperatures.

Therefore,

$\frac{n_{i2}}{n_{i1}}$ =  $\frac{n_{0}exp\left(-\frac{E_g}{2k_B*600}\right)}{n_{0}exp\left(-\frac{E_g}{2k_B*300}\right)}$

= exp $\frac{E_g}{2k_B}\left[\frac{1}{300}-\frac{1}{600}\right]$

= exp $\frac{E_g}{2k_B}\left[\frac{1}{300}-\frac{1}{600}\right]$

= exp $\frac{1.2}{2*8.62*{10}^{-5}}\left[\frac{2-1}{600}\right]$  = exp(11.6) = 109 $\frac{{}_{}}{}$ $*{10}^5$

Therefore, the ratio between two conductivities is 1.09 $*{10}^5$

$\mathit{I}=\mathit{}{\mathit{I}}_{0\mathit{}}\mathit{e}\mathit{x}\mathit{p}\left(\frac{\mathit{e}\mathit{V}}{2{\mathit{k}}_{\mathit{B}}\mathit{T}}-1\right)$

14.8 Number of silicone atoms, $n_S$ = 5 $*{10}^{28}$ / $m^3$

Number of arsenic atoms, $n_A$ = 5 $*{10}^{22}$ / $m^3$

Number of indium atoms, $n_I$ = 5 $*{10}^{20}$ / $m^3$

Number of thermally generated atoms, $n_i$ = 1.5 $*{10}^{16}$ / $m^3$

Hence number of electrons, $n_e$ = 5 $*{10}^{22}-$ 1.5 $*{10}^{16}$ = 4.99 $*{10}^{22}$

Number of holes = $n_H$

In thermal equilibrium, the concentration of electrons and holes in a semiconductor are related as:

$n_e{n}_H$ = $n_i^2$

Therefore, $n_H$ = $\frac{n_i^2}{n_e}$ = $\frac{{(1.5*{10}^{16})}^2}{4.99*{10}^{22}}$ = 4.51 $*{10}^9$

Since the number of electrons are more (4.99 $*{10}^{22})$ than the number of holes (4.51 $*{10}^9)$ , the material is an n-type semiconductor.