In an intrinsic semiconductor, the energy gap E_g is 1.2eV. Its hole mobility is very much smaller than electron mobility and is independent of temp. What is the ratio b/w conductivity at 600K and that at 300K?

Assume temp. dependence of intrinsic carrier conc. n_i is expressed as

n_i=n_oexp[-E_g/2k_bT]

where n_o is a constant and k_b=8.62*10^-5eVK^-1 is Boltzmann constant.

To determine the ratio of conductivity in an intrinsic semiconductor at two different temperatures, we need to consider how the intrinsic carrier concentration changes with temperature and how this affects conductivity. The conductivity (σ) of an intrinsic semiconductor can be expressed as:

Understanding Conductivity in Intrinsic Semiconductors

The conductivity is given by the formula:

σ = q(nμ_n + pμ_p)

Where:

• σ = conductivity
• q = charge of an electron (approximately 1.6 x 10^-19 C)
• n = concentration of electrons
• p = concentration of holes
• μ_n = electron mobility
• μ_p = hole mobility

In intrinsic semiconductors, the concentration of electrons (n) is equal to the concentration of holes (p), so we can simplify the equation to:

σ = q n (μ_n + μ_p)

Temperature Dependence of Carrier Concentration

Given that the intrinsic carrier concentration is expressed as:

n_i = n_o exp[-E_g / (2k_b T)]

Where:

• n_o = a constant
• E_g = energy gap (1.2 eV in this case)
• k_b = Boltzmann constant (8.62 x 10^-5 eV/K)
• T = temperature in Kelvin

We can calculate the intrinsic carrier concentration at two different temperatures, 600 K and 300 K.

Calculating Carrier Concentration at Different Temperatures

First, let's find the intrinsic carrier concentration at 300 K:

n_i(300K) = n_o exp[-(1.2 eV) / (2 * 8.62 x 10^-5 eV/K * 300 K)]

Now, calculating the exponent:

n_i(300K) = n_o exp[-(1.2) / (0.05172)]

This results in:

n_i(300K) = n_o exp[-23.19]

Next, for 600 K:

n_i(600K) = n_o exp[-(1.2 eV) / (2 * 8.62 x 10^-5 eV/K * 600 K)]

Calculating the exponent for 600 K:

n_i(600K) = n_o exp[-(1.2) / (0.10304)]

This results in:

n_i(600K) = n_o exp[-11.65]

Finding the Ratio of Conductivities

Now, we can find the ratio of the intrinsic carrier concentrations:

n_i(600K) / n_i(300K) = exp[-11.65] / exp[-23.19]

This simplifies to:

n_i(600K) / n_i(300K) = exp[11.54]

Now, we can find the ratio of conductivities:

σ(600K) / σ(300K) = (n_i(600K) * (μ_n + μ_p)) / (n_i(300K) * (μ_n + μ_p))

Since the hole mobility is much smaller than the electron mobility and is independent of temperature, we can approximate:

σ(600K) / σ(300K) ≈ n_i(600K) / n_i(300K)

Thus, the ratio of conductivities becomes:

σ(600K) / σ(300K) = exp[11.54]

Final Thoughts

In conclusion, the ratio of conductivity at 600 K to that at 300 K is approximately equal to exp[11.54], which indicates a significant increase in conductivity with temperature due to the exponential increase in intrinsic carrier concentration. This illustrates the crucial role temperature plays in semiconductor behavior, particularly in intrinsic materials.