CBSE Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits NCERT Solutions

Solution:

Given data:

• Number of silicon atoms, $N\; =\; 5\; \backslash times\; 10^\{28\}$ atoms/m³
• Concentration of Arsenic atoms (pentavalent impurity), $n\_\{As\}\; =\; 10^\{22\}$ atoms/m³
• Concentration of Indium atoms (trivalent impurity), $n\_\{In\}\; =\; 5\; \backslash times\; 10^\{20\}$ atoms/m³
• Intrinsic carrier concentration, $n\_i\; =\; 1.5\; \backslash times\; 10^\{16\}$ m^-3

Arsenic is a pentavalent impurity, which acts as a donor and contributes free electrons. Indium is a trivalent impurity, which acts as an acceptor and creates holes. Since the concentration of Arsenic atoms ($10^\{22\}$ m^-3) is significantly higher than that of Indium atoms ($5\; \backslash times\; 10^\{20\}$ m^-3), the material will predominantly exhibit n-type characteristics.

The concentration of donor electrons ($n\_\{As\}$) is much larger than the intrinsic carrier concentration ($n\_i$). Therefore, the electron concentration ($n\_e$) in the n-type semiconductor is approximately equal to the concentration of donor atoms:

$n\_e\; \backslash approx\; n\_\{As\}\; =\; 10^\{22\}$ m^-3

We can also consider the difference between donor and acceptor concentrations for a more precise calculation, but given the large difference, $n\_e\; \backslash approx\; n\_\{As\}$ is a good approximation. The source calculation used $n\_e\; =\; n\_\{As\}\; -\; n\_i$, which is also valid as $n\_\{As\}\; \backslash gg\; n\_i$.

Using the relation $n\_e\; \backslash approx\; n\_\{As\}\; -\; n\_i$:

$n\_e\; \backslash approx\; 10^\{22\}\; -\; 1.5\; \backslash times\; 10^\{16\}\; \backslash approx\; 10^\{22\}$ m^-3

In thermal equilibrium, the product of electron and hole concentrations is given by:

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

Now, we can calculate the hole concentration ($n\_h$):

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

$$n\_h\; =\; \backslash frac\{(1.5\; \backslash times\; 10^\{16\})^2\}\{10^\{22\}\}\; =\; \backslash frac\{2.25\; \backslash times\; 10^\{32\}\}\{10^\{22\}\}\; =\; 2.25\; \backslash times\; 10^\{10\}$$ m^-3

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

Since the concentration of electrons ($n\_e$) is much greater than the concentration of holes ($n\_h$), the material is an n-type semiconductor.

Solution:

Given:

• Energy gap, $E\_g\; =\; 1.2$ eV
• Intrinsic carrier concentration dependence on temperature: $n\_i\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\}\; \backslash right]$
• Boltzmann constant, $k\_B\; =\; 8.62\; \backslash times\; 10^\{-5\}$ eV/K
• Initial temperature, $T\_1\; =\; 300$ K
• Final temperature, $T\_2\; =\; 600$ K

The conductivity ($\backslash sigma$) of an intrinsic semiconductor is given by:

$$\backslash sigma\; =\; n\_i\; e\; (\backslash mu\_e\; +\; \backslash mu\_h)$$

where $n\_i$ is the intrinsic carrier concentration, $e$ is the electronic charge, $\backslash mu\_e$ is the electron mobility, and $\backslash mu\_h$ is the hole mobility.

We are given that hole mobility ($\backslash mu\_h$) is much smaller than electron mobility ($\backslash mu\_e$) and is independent of temperature. Therefore, we can approximate the conductivity as being primarily dependent on electron mobility and intrinsic carrier concentration:

$\backslash sigma\; \backslash approx\; n\_i\; e\; \backslash mu\_e$

Since $e$ and $\backslash mu\_e$ are assumed to be constant with respect to temperature (as $\backslash mu\_h$ is independent and $\backslash mu\_e$ is not specified to be dependent), the ratio of conductivities at two different temperatures ($T\_2$ and $T\_1$) will be equal to the ratio of their intrinsic carrier concentrations:

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

Let's calculate the intrinsic carrier concentrations at $T\_1\; =\; 300$ K and $T\_2\; =\; 600$ K:

At $T\_1\; =\; 300$ K:

$$n\_\{i1\}\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_1\}\; \backslash right]\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{1.2\}\{2\; \backslash times\; (8.62\; \backslash times\; 10^\{-5\})\; \backslash times\; 300\}\; \backslash right]$$

At $T\_2\; =\; 600$ K:

$$n\_\{i2\}\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{E\_g\}\{2k\_B\; T\_2\}\; \backslash right]\; =\; n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{1.2\}\{2\; \backslash times\; (8.62\; \backslash times\; 10^\{-5\})\; \backslash times\; 600\}\; \backslash right]$$

Now, we find the ratio $\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}$:

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; \backslash frac\{n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{1.2\}\{2k\_B\; \backslash times\; 600\}\; \backslash right]\}\{n\_0\; \backslash exp\; \backslash left[\; -\backslash frac\{1.2\}\{2k\_B\; \backslash times\; 300\}\; \backslash right]\}\; =\; \backslash exp\; \backslash left[\; -\backslash frac\{1.2\}\{1200\; k\_B\}\; +\; \backslash frac\{1.2\}\{600\; k\_B\}\; \backslash right]$$

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; \backslash exp\; \backslash left[\; \backslash frac\{1.2\}\{600\; k\_B\}\; \backslash left(\; -1\; +\; 2\; \backslash right)\; \backslash right]\; =\; \backslash exp\; \backslash left[\; \backslash frac\{1.2\}\{600\; k\_B\}\; \backslash right]$$

Substitute the value of $k\_B$:

$$\backslash frac\{1.2\}\{600\; k\_B\}\; =\; \backslash frac\{1.2\}\{600\; \backslash times\; (8.62\; \backslash times\; 10^\{-5\})\}\; =\; \backslash frac\{1.2\}\{0.05172\}\; \backslash approx\; 23.199$$

So, the ratio is:

$$\backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; e^\{23.199\}$$

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

$$\backslash frac\{\backslash sigma\_\{600K\}\}\{\backslash sigma\_\{300K\}\}\; =\; \backslash frac\{n\_\{i2\}\}\{n\_\{i1\}\}\; =\; e^\{23.199\}$$

Calculating the value:

$$e^\{23.199\}\; \backslash approx\; 1.079\; \backslash times\; 10^\{10\}$$

The ratio between the conductivity at 600 K and that at 300 K is approximately $1.079\; \backslash times\; 10^\{10\}$.