What Is an Example of a Semiconductor? Solved Conductivity Problem

The Core Question: What Is an Example of a Semiconductor?

The most definitive example of a semiconductor is Silicon (Si). In its pure, intrinsic crystalline form, silicon acts as an insulator at absolute zero but conducts a small amount of current at room temperature. By intentionally introducing impurities (doping), we can precisely tune its conductivity to build the diodes, transistors, and integrated circuits that power everything from a simple 2N2222 BJT to the complex 40nm silicon die inside an ESP32-WROOM-32 microcontroller.

Other common examples include Germanium (Ge), used in early transistors and modern RF/solar applications, and Gallium Arsenide (GaAs), favored for high-frequency RF amplifiers and high-efficiency solar cells. But how do we mathematically prove that a material is a semiconductor rather than a conductor or an insulator? We calculate its electrical conductivity and verify that it falls in the narrow gap between the two extremes.

Practice Problem Walkthrough: Proving Silicon is a Semiconductor

Which Theorem/Method Applies and Why?

We use the Drift Conductivity Equation combined with the Law of Mass Action (charge neutrality). Conductivity in a semiconductor depends on both the concentration of charge carriers (electrons and holes) and their mobility. Because Phosphorus is a Group V element, it acts as a donor, creating an n-type semiconductor where electrons are the majority carriers.

Step-by-Step Solution

Step 1: Identify majority and minority carrier concentrations.
Since the doping concentration ($N_d = 10^{16}$ cm$^{-3}$) is vastly greater than the intrinsic concentration ($n_i = 1.5 \times 10^{10}$ cm$^{-3}$), we assume complete ionization at 300 K. The majority electron concentration ($n$) is approximately equal to the donor concentration:

$n \approx N_d = 1 \times 10^{16}$ cm$^{-3}$

Step 2: Calculate the minority hole concentration ($p$).
Using the Law of Mass Action ($n \cdot p = n_i^2$):

$p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{10})^2}{1 \times 10^{16}}$

$p = \frac{2.25 \times 10^{20}}{1 \times 10^{16}} = 2.25 \times 10^4$ cm$^{-3}$

Step 3: Apply the conductivity formula.
The general formula is $\sigma = q(n\mu_n + p\mu_p)$. Because $n \gg p$ ($10^{16}$ vs $10^4$), the hole contribution is negligible. We simplify to:

$\sigma \approx q \cdot n \cdot \mu_n$

Step 4: Execute the algebra and track units.
$\sigma = (1.6 \times 10^{-19} \text{ C}) \times (1 \times 10^{16} \text{ cm}^{-3}) \times (1350 \text{ cm}^2/\text{V}\cdot\text{s})$
$\sigma = (1.6 \times 10^{-3}) \times 1350$
$\sigma = 2.16 \, (\Omega \cdot \text{cm})^{-1}$

Step 5: Convert to standard SI units (Siemens per meter, S/m) for comparison.
Since $1 \text{ m} = 100 \text{ cm}$, we multiply by 100 to convert from $(\Omega \cdot \text{cm})^{-1}$ to $(\Omega \cdot \text{m})^{-1}$ (or S/m):

$\sigma_{SI} = 2.16 \times 100 = 216 \text{ S/m}$

Answer Sanity Check

To verify our answer, we check the order of magnitude against known material classes:

• Conductors (e.g., Copper): $\sigma \approx 5.9 \times 10^7$ S/m
• Semiconductors (e.g., Doped Si): $\sigma \approx 10^{-4}$ to $10^4$ S/m
• Insulators (e.g., Glass): $\sigma \approx 10^{-12}$ S/m

Our calculated value of 216 S/m sits squarely in the semiconductor range. It is roughly 270,000 times less conductive than copper, but trillions of times more conductive than glass. The math proves Silicon is a semiconductor.

How to Verify Your Answer Independently

If you were holding this physical wafer on a lab bench, you wouldn't use a standard multimeter to verify this calculation. The contact resistance of the probes would ruin the measurement. Instead, you would use a Four-Point Probe station.

A four-point probe passes a known current through the outer two pins and measures the voltage drop across the inner two pins. Because the voltmeter draws virtually zero current, the voltage drop across the probe-to-silicon contact resistance is eliminated. According to standard BYU Cleanroom measurement protocols, the sheet resistance ($R_s$) is calculated as $R_s = 4.532 \times (V/I)$ for a standard thin wafer. You would then convert sheet resistance to bulk conductivity using the known thickness of the wafer, confirming our theoretical 2.16 $(\Omega \cdot \text{cm})^{-1}$ result.

Frequently Asked Questions

What is an example of a semiconductor used in everyday electronics?

Beyond raw silicon wafers, the most common practical examples of semiconductors are the discrete components and ICs on your workbench. The 1N4007 rectifier diode, the 2N7000 N-channel MOSFET, and the NE555 timer IC are all everyday examples of semiconductor devices. In modern IoT projects, the ESP32-WROOM-32 module contains a highly complex semiconductor die integrating CPUs, WiFi, and Bluetooth radios onto a single piece of doped silicon.

Why is silicon the most common example of a semiconductor instead of germanium?

While Germanium was used in the very first transistors, Silicon dominates today for two primary physical reasons. First, Silicon has a wider bandgap (1.12 eV vs 0.67 eV for Ge), meaning it remains stable and functional at much higher operating temperatures (up to 150°C or more) without suffering from thermal runaway caused by intrinsic carrier generation. Second, Silicon readily forms a high-quality, stable insulating oxide layer (Silicon Dioxide, SiO2) when heated in oxygen. This native oxide is the foundational reason the MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor) exists, enabling the entire modern VLSI microchip industry. For a deeper physics breakdown, Georgia State University's HyperPhysics provides excellent bandgap comparison tables.

Is an example of a semiconductor always a solid material?

No. While crystalline solids like Silicon, GaAs, and Silicon Carbide (SiC) are the standard examples, the field of organic semiconductors utilizes carbon-based polymers and small molecules that exhibit semiconductor properties. These are used in OLED displays and flexible solar panels. Additionally, certain liquid solutions and ionic melts can exhibit semiconductor-like bandgaps and charge transport mechanisms, though these are largely confined to advanced chemical research rather than consumer electronics.

How does temperature affect a semiconductor example like silicon?

Semiconductors exhibit a negative temperature coefficient of resistance. As the temperature of a silicon die increases, more valence electrons gain enough thermal energy to jump the bandgap into the conduction band, creating additional electron-hole pairs. This increases the intrinsic carrier concentration ($n_i$), which in turn increases conductivity (and lowers resistance). This is the exact opposite of a copper wire, where increased temperature causes more lattice scattering, increasing resistance. This thermal behavior is why high-power semiconductor devices (like IGBTs in motor drives or MOSFETs in switching power supplies) require robust heatsinking to prevent thermal runaway and catastrophic failure.