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Question

A semiconductor uses P-type substrates with $N_A = 5 \times 10^{15}cm^{-3}$ in an n-MOS device having depletion region width of 0.415 $\mu$m. What is the depletion charge in magnitude?

The correct answer is
$3.32\times10^{-8} C/cm^2$

Depletion Charge Calculation for n-MOS

The question asks for the magnitude of the depletion charge per unit area in a P-type semiconductor substrate used in an n-MOS device.

Depletion Charge Formula

The depletion charge ($Q_d$) per unit area in a uniformly doped P-type semiconductor is given by the formula:

$ |Q_d| = q N_A x_d $

Where:

  • $q$ is the magnitude of the elementary charge ($1.602 \times 10^{-19} C$).
  • $N_A$ is the acceptor concentration in the P-type substrate.
  • $x_d$ is the depletion region width.

Input Parameters and Unit Conversion

Given values are:

  • Acceptor concentration, $N_A = 5 \times 10^{15} cm^{-3}$.
  • Depletion region width, $x_d = 0.415 \mu m$.

We need to ensure consistent units. Convert the depletion width from micrometers ($\mu m$) to centimeters ($cm$):

$ x_d = 0.415 \mu m = 0.415 \times 10^{-4} cm $

Magnitude Calculation

Substitute the values into the formula:

$ |Q_d| = (1.602 \times 10^{-19} C) \times (5 \times 10^{15} cm^{-3}) \times (0.415 \times 10^{-4} cm) $

Calculate the result:

$ |Q_d| = (1.602 \times 5 \times 0.415) \times (10^{-19 + 15 - 4}) C/cm^2 $

$ |Q_d| = (8.01 \times 0.415) \times 10^{-8} C/cm^2 $

$ |Q_d| \approx 3.324 \times 10^{-8} C/cm^2 $

Result Comparison

The calculated magnitude of the depletion charge is approximately $3.32 \times 10^{-8} C/cm^2$. This matches Option B.

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