Before You Begin

Question #96051

Difficulty: 8/10 | 🟢 Free

The Question

Consider a 1D n-type semiconductor region ( $x \ge 0$ ) in thermal equilibrium at temperature $T$ . The hole concentration is negligible ( $p \approx 0$ ). The electric potential profile is established as $V(x) = - \frac{k_B T}{e} \ln\left(1 + \frac{x}{L}\right)$ , where $k_B$ is the Boltzmann constant, $e$ is the elementary charge, and $L$ is a characteristic length. The electron concentration at $x=0$ is $n_0$ , and the permittivity of the semiconductor is $\epsilon$ . Assuming complete ionization of donors and that all quantities are in standard SI units, calculate the required gradient of the donor doping concentration, $\frac{dN_d}{dx}$ , at $x=0$ to maintain this steady state.

Options

A.
$-\frac{n_0}{L}$
B.
$-\frac{n_0}{L} - \frac{2 \epsilon k_B T}{e^2 L^3}$
C.
$-\frac{n_0}{L} + \frac{2 \epsilon k_B T}{e^2 L^3}$
D.
$-\frac{n_0}{L} + \frac{\epsilon k_B T}{e^2 L^3}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Law of Mass Action|Work-Energy Theorem Validity|Mechanical Energy Conservation

Why This Question Is Tricky

Examiner Levers

Negative Thinking

Uses "NOT", "EXCEPT", or "INCORRECT" to flip the logic gate.

Difficulty Boost : +2

Extraneous Data

Provides unnecessary variables (Red Herrings) to test conceptual focus.

Difficulty Boost : +3

Common Mistakes

This question tests the rigorous application of Poisson's equation in conjunction with the Boltzmann relation for a semiconductor in thermal equilibrium.

The primary trap lies in the assumption of quasi-neutrality ( $N_d \approx n$ ) everywhere.

Students falling for this trap effectively ignore the $\frac{d^2 V}{dx^2}$ term in Poisson's equation, which is equivalent to assuming a linear potential profile (constant electric field) where the net space charge is zero.

By doing so, they simply differentiate $n(x)$ to find the gradient, leading to the incorrect answer $-\frac{n_0}{L}$ .

The problem demands recognizing that the logarithmic nature of the potential creates a non-linear field, requiring a specific non-zero space charge density that must be supplied by the doping gradient.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Semiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
Concept	Law of Mass Action
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free