What Is a Semiconductor With Example: Exam Walkthrough & Material Selection

The Core Concept: What Is a Semiconductor With Example?

A semiconductor is a solid material whose electrical conductivity falls between that of a conductor (like copper) and an insulator (like glass). This intermediate conductivity is governed by the material's bandgap—the energy difference between the valence band (where electrons are bound to atoms) and the conduction band (where electrons are free to move). By introducing specific impurities (doping), we can precisely control whether the material conducts via negative electrons (N-type) or positive holes (P-type).

To answer what is semi conductor with example in a practical sense: pure Silicon (Si) is an intrinsic semiconductor with very low conductivity at room temperature. However, if we dope it with Phosphorus (which has five valence electrons compared to Silicon's four), we create an N-type extrinsic semiconductor. The extra electron from each Phosphorus atom becomes a free charge carrier, drastically increasing conductivity without requiring high heat or voltage. This principle is the foundation of every diode, transistor, and microchip on your workbench.

Exam Problem Statement: N-Type Silicon Conductivity

Step-by-Step Algebraic Solution & Sanity Check

To solve this, we apply the Mass Action Law and the Drift Conductivity Equation. Because Phosphorus is a Group V element, it acts as a donor, making this an N-type semiconductor.

Step 1: Determine Carrier Concentrations ($n$ and $p$)

At room temperature, we assume complete ionization of the dopants. Therefore, the majority carrier concentration (electrons, $n$) is approximately equal to the donor concentration:

$n \approx N_d = 1.0 \times 10^{16} \text{ cm}^{-3}$

To find the minority carrier concentration (holes, $p$), we use the Mass Action Law ($n \cdot p = n_i^2$):

$p = \frac{n_i^2}{n} = \frac{(1.0 \times 10^{10} \text{ cm}^{-3})^2}{1.0 \times 10^{16} \text{ cm}^{-3}}$

$p = \frac{1.0 \times 10^{20} \text{ cm}^{-6}}{1.0 \times 10^{16} \text{ cm}^{-3}} = 1.0 \times 10^{4} \text{ cm}^{-3}$

Step 2: Calculate Total Conductivity ($\sigma$)

The general formula for conductivity includes both electrons and holes:

$\sigma = q(n\mu_n + p\mu_p)$

Let's check the magnitude of the two terms inside the parenthesis before multiplying by $q$:

• Electron term: $n\mu_n = (10^{16})(1200) = 1.2 \times 10^{19}$
• Hole term: $p\mu_p = (10^{4})(400) = 4.0 \times 10^{6}$

Because $1.2 \times 10^{19} \gg 4.0 \times 10^{6}$, the minority carriers contribute negligibly to conductivity. We can safely simplify the equation to $\sigma \approx q \cdot n \cdot \mu_n$.

Now, plug in the values with explicit units:

$\sigma = (1.602 \times 10^{-19} \text{ C}) \times (1.0 \times 10^{16} \text{ cm}^{-3}) \times (1200 \text{ cm}^2 / (\text{V}\cdot\text{s}))$

$\sigma = (1.602 \times 10^{-3} \text{ C}\cdot\text{cm}^{-3}) \times (1200 \text{ cm}^2 / (\text{V}\cdot\text{s}))$

$\sigma = 1.9224 \text{ (C/s)} \cdot \text{V}^{-1} \cdot \text{cm}^{-1}$

Since $1 \text{ C/s} = 1 \text{ Ampere (A)}$, and $1 \text{ A/V} = 1 \text{ Siemens (S)}$ or $\Omega^{-1}$:

$\sigma = 1.92 \text{ S/cm}$ (or $1.92 \ (\Omega\cdot\text{cm})^{-1}$)

The Trap: Temperature Dependence and Intrinsic Swamping

The most common trap in semiconductor exams—and in real-world circuit design—is assuming that $n \approx N_d$ holds true at all temperatures. It does not.

The intrinsic carrier concentration ($n_i$) is highly temperature-dependent, scaling exponentially with temperature according to the relation $n_i \propto T^{3/2} e^{-E_g / 2kT}$. If the silicon wafer in our problem were heated to 200°C (473K), $n_i$ would rise from $10^{10}$ to roughly $10^{16} \text{ cm}^{-3}$.

When $n_i$ approaches $N_d$, the thermally generated intrinsic carriers swamp the dopant atoms. The material loses its extrinsic N-type characteristics and reverts to intrinsic behavior. In a practical aerospace or automotive sensor operating at high ambient temperatures, this causes massive leakage currents, thermal runaway, and total device failure. According to Georgia State University's HyperPhysics, this intrinsic transition temperature is the hard operational limit for standard silicon devices.

Decision Path: Picking the Right Semiconductor for High-Temp Apps

If your design requires a semiconductor to operate reliably above 150°C, standard Silicon is the wrong choice. Use this decision tree to select the correct material based on bandgap energy ($E_g$) and thermal limits.

FAQ: Verifying Your Answers and Common Pitfalls

How do I verify the conductivity answer independently in a lab?

You cannot accurately measure semiconductor conductivity with a standard multimeter; the contact resistance of the probes will dwarf the wafer's resistance. Instead, use a Four-Point Probe (4PP) station. By forcing a known constant current through the outer two probes and measuring the voltage drop across the inner two probes, you eliminate lead and contact resistance from the calculation. The sheet resistance is calculated as $R_s = (\pi / \ln 2) \times (V/I)$, and conductivity is derived from the sheet resistance and the known wafer thickness.

Why did we ignore hole mobility in the final calculation?

In an N-type semiconductor where $N_d \gg n_i$, the concentration of electrons outnumbers holes by a factor of $10^{12}$ in our specific problem. Even though holes are present, their contribution to the total current is mathematically insignificant (adding 0.0000001% to the final answer). In exams, showing that you evaluated the minority term and consciously dropped it demonstrates rigorous engineering judgment, which is exactly what MIT's OpenCourseWare device physics materials emphasize for full credit.

What if the problem asks for resistivity instead of conductivity?

Resistivity ($\rho$) is simply the reciprocal of conductivity. If $\sigma = 1.92 \text{ S/cm}$, then $\rho = 1 / 1.92 = 0.52 \ \Omega\cdot\text{cm}$. Always check the requested units before boxing your final answer.