CBSE Class 12 Physics Chapter 26: Semiconductor Electronics NCERT Solutions

We are given the following information:

• Number of silicon atoms, $N = 5 \times 10^{28}$ atoms/m³
• Number of Arsenic atoms (donor impurity), $n_{As} = 10^{22}$ atoms/m³
• Number of Indium atoms (acceptor impurity), $n_{In} = 5 \times 10^{20}$ atoms/m³
• Intrinsic carrier concentration, $n_i = 1.5 \times 10^{16}$ m^-3

Arsenic (As) is a pentavalent impurity, so it acts as a donor, contributing free electrons. Indium (In) is a trivalent impurity, so it acts as an acceptor, creating holes.

The net concentration of donor atoms is $N_D = n_{As} = 10^{22}$ m^-3.

The net concentration of acceptor atoms is $N_A = n_{In} = 5 \times 10^{20}$ m^-3.

Since $N_D > N_A$, the semiconductor will be n-type, and the majority charge carriers will be electrons.

The concentration of electrons ($n_e$) in an n-type semiconductor is approximately equal to the concentration of donor atoms, provided the doping concentration is much higher than the intrinsic carrier concentration ($N_D \gg n_i$).

So, the electron concentration is approximately:

$n_e \approx N_D = 10^{22}$ m^-3

We can verify this by noting that $10^{22}$ is indeed much larger than $1.5 \times 10^{16}$.

Now, we use the mass action law for semiconductors in thermal equilibrium, which states that the product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration:

$$n\_e\; n\_h\; =\; n\_i^2$$

We can now calculate the hole concentration ($n_h$):

$$n\_h\; =\; \backslash frac\{n\_i^2\}\{n\_e\}$$

$$n\_h\; =\; \backslash frac\{(1.5\; \backslash times\; 10^\{16\}\; \backslash text\{\; m\}^\{-3\})^2\}\{10^\{22\}\; \backslash text\{\; m\}^\{-3\}\}$$

$$n\_h\; =\; \backslash frac\{2.25\; \backslash times\; 10^\{32\}\; \backslash text\{\; m\}^\{-6\}\}\{10^\{22\}\; \backslash text\{\; m\}^\{-3\}\}$$

$$n\_h\; =\; 2.25\; \backslash times\; 10^\{10\}\; \backslash text\{\; m\}^\{-3\}$$

Therefore, the number of electrons is approximately $10^{22}$ m^-3, and the number of holes is approximately $2.25 \times 10^{10}$ m^-3.

Since the concentration of electrons ($10^{22}$ m^-3) is significantly greater than the concentration of holes ($2.25 \times 10^{10}$ m^-3), the material is an n-type semiconductor.

We are given:

• Energy gap, $E_g = 1.2$ eV
• Boltzmann constant, $k_B = 8.62 \times 10^{-5}$ eV/K
• Initial temperature, $T_1 = 300$ K
• Final temperature, $T_2 = 600$ K
• The intrinsic carrier concentration is given by: $$n\_i\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\}\; \backslash right]$$

The conductivity ($\sigma$) of an intrinsic semiconductor is directly proportional to the intrinsic carrier concentration ($n_i$) and the sum of electron and hole mobilities ($\mu_e + \mu_h$). Since the semiconductor is intrinsic, $n_e = n_h = n_i$. The conductivity is given by $\sigma = n_i e (\mu_e + \mu_h)$.

We are told that hole mobility ($\mu_h$) is much smaller than electron mobility ($\mu_e$) and is independent of temperature. Also, $\mu_e$ is assumed to be independent of temperature for this calculation.

Therefore, the conductivity is approximately proportional to the intrinsic carrier concentration:

$$\backslash sigma\; \backslash propto\; n\_i$$

We need to find the ratio of conductivity at $T_2 = 600$ K to conductivity at $T_1 = 300$ K. This ratio will be equal to the ratio of the intrinsic carrier concentrations at these temperatures:

$$\backslash frac\{\backslash sigma\_2\}\{\backslash sigma\_1\}\; =\; \backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}$$

Let's calculate $n_{i1}$ at $T_1 = 300$ K:

$$n\_\{i1\}\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_1\}\; \backslash right]$$

And $n_{i2}$ at $T_2 = 600$ K:

$$n\_\{i2\}\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_2\}\; \backslash right]$$

Now, we find the ratio $\frac{n_{i2}}{n_{i1}}$:

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; \backslash frac\{n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_2\}\; \backslash right]\}\{n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_1\}\; \backslash right]\}$$

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_2\}\; -\; \backslash left(-\backslash frac\{E\_g\}\{2k\_B\; T\_1\}\backslash right)\; \backslash right]$$

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; \backslash exp\; \backslash left[\; \backslash frac\{E\_g\}\{2k\_B\}\; \backslash left(\backslash frac\{1\}\{T\_1\}\; -\; \backslash frac\{1\}\{T\_2\}\backslash right)\; \backslash right]$$

Substitute the given values:

$$E\_g\; =\; 1.2\; \backslash text\{\; eV\}$$

$$k\_B\; =\; 8.62\; \backslash times\; 10^\{-5\}\; \backslash text\{\; eV/K\}$$

$$T\_1\; =\; 300\; \backslash text\{\; K\}$$

$$T\_2\; =\; 600\; \backslash text\{\; K\}$$

First, calculate the term $\frac{E_g}{2k_B}$:

$$\backslash frac\{E\_g\}\{2k\_B\}\; =\; \backslash frac\{1.2\; \backslash text\{\; eV\}\}\{2\; \backslash times\; 8.62\; \backslash times\; 10^\{-5\}\; \backslash text\{\; eV/K\}\}\; =\; \backslash frac\{1.2\}\{1.724\; \backslash times\; 10^\{-4\}\}\; \backslash text\{\; K\}\; \backslash approx\; 6950\; \backslash text\{\; K\}$$

Now, calculate the term $\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$:

$$\backslash frac\{1\}\{T\_1\}\; -\; \backslash frac\{1\}\{T\_2\}\; =\; \backslash frac\{1\}\{300\; \backslash text\{\; K\}\}\; -\; \backslash frac\{1\}\{600\; \backslash text\{\; K\}\}\; =\; \backslash frac\{2\; -\; 1\}\{600\; \backslash text\{\; K\}\}\; =\; \backslash frac\{1\}\{600\; \backslash text\{\; K\}\}$$

Now, substitute these values back into the exponent:

$$\backslash frac\{E\_g\}\{2k\_B\}\; \backslash left(\backslash frac\{1\}\{T\_1\}\; -\; \backslash frac\{1\}\{T\_2\}\backslash right)\; =\; (6950\; \backslash text\{\; K\})\; \backslash times\; \backslash left(\backslash frac\{1\}\{600\; \backslash text\{\; K\}\}\backslash right)\; \backslash approx\; 11.58$$

So, the ratio of intrinsic carrier concentrations is:

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; \backslash exp(11.58)$$

Using a calculator, $\exp(11.58) \approx 106900$.

Therefore, the ratio between the conductivity at 600 K and that at 300 K is approximately:

$$\backslash frac\{\backslash sigma\_\{600K\}\}\{\backslash sigma\_\{300K\}\}\; =\; \backslash frac\{\backslash sigma\_2\}\{\backslash sigma\_1\}\; =\; \backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; \backslash approx\; 106900$$

The ratio of conductivity at 600 K to that at 300 K is approximately 106900.