Silicon Sample Doping Analysis

The question asks us to determine the displacement of the Fermi energy level (\(E_F\)) relative to the intrinsic Fermi energy level (\(E_i\)) in a Silicon (Si) sample doped with Arsenic atoms. We are given the doping concentration and need to calculate the value of \(E_F - E_i\).

Semiconductor Type Identification

Silicon (Si) is a Group IV element. Arsenic (As) is a Group V element. When a Group V element like Arsenic is added to a Group IV semiconductor like Silicon, it acts as a donor impurity. This means it contributes free electrons to the conduction band, making the semiconductor n-type.

• In an n-type semiconductor, the concentration of electrons (n) is much higher than the concentration of holes (p).
• The Fermi energy level (\(E_F\)) in an n-type semiconductor lies closer to the conduction band edge (\(E_C\)) and is above the intrinsic Fermi energy level (\(E_i\)).
• Therefore, the displacement \(E_F - E_i\) is expected to be positive.

Key Parameters for Silicon

To calculate the Fermi level displacement, we need the following standard parameters for Silicon at room temperature (assumed to be 300 K):

• Doping Concentration (\(N_D\)): \(10^{17} \text{ atoms/cm}^3\) (given). Since Arsenic is a donor, \(N_D\) is the donor concentration. In a heavily doped n-type semiconductor where all donors are ionized, the electron concentration \(n \approx N_D\). So, \(n = 10^{17} \text{ cm}^{-3}\).
• Intrinsic Carrier Concentration (\(n_i\)): For Silicon at 300 K, a commonly accepted value is \(1.5 \times 10^{10} \text{ cm}^{-3}\).
• Thermal Voltage (\(k_BT/q\)): At 300 K, the thermal voltage (\(V_T\)) is approximately 0.0259 eV. Here, \(k_B\) is Boltzmann's constant and \(T\) is the absolute temperature. So, \(k_BT = 0.0259 \text{ eV}\).

Fermi Level Displacement Formula

For an n-type semiconductor, the relationship between the Fermi energy level (\(E_F\)), intrinsic Fermi energy level (\(E_i\)), electron concentration (n), and intrinsic carrier concentration (\(n_i\)) is given by:

Since the electron concentration \(n\) in a sufficiently doped n-type semiconductor is approximately equal to the donor concentration (\(N_D\)), we can substitute \(n \approx N_D\):

The displacement of \(E_F\) relative to \(E_i\) is therefore:

Step-by-Step Calculation

Now, let's substitute the values into the formula:

• \(N_D = 10^{17} \text{ cm}^{-3}\)
• \(n_i = 1.5 \times 10^{10} \text{ cm}^{-3}\)
• \(k_BT = 0.0259 \text{ eV}\)

First, calculate the ratio \(\frac{N_D}{n_i}\):

Next, calculate the natural logarithm of this ratio:

Finally, multiply by \(k_BT\):

Rounding this to three significant figures, we get 0.407 eV.

Result Interpretation

The calculated displacement \(E_F - E_i\) is approximately 0.407 eV. Since the value is positive, it confirms that the Fermi level is indeed above the intrinsic Fermi level, which is characteristic of an n-type semiconductor. This displacement indicates how much the Fermi energy has shifted due to the doping, moving closer to the conduction band edge.

Therefore, the displacement of \(E_F\) relative to \(E_i\) is Positive, 0.407 eV.