All Exams Test series for 1 year @ ₹349 only
Question

For semiconductor, with both electrons and holes as carriers, the resistivity can be expressed as:

The correct answer is \(\rm \frac{1}{q(\mu_n n + \mu _p p)}\)

Understanding Semiconductor Resistivity

Resistivity ($\rho$) is a fundamental property of a material that measures how strongly it resists the flow of electric current. Its unit is ohm-meter ($\Omega \cdot \text{m}$). In semiconductors, electrical conduction occurs due to the movement of charge carriers, which can be both electrons and holes.

The reciprocal of resistivity is conductivity ($\sigma$), which measures how easily current flows through a material. The relationship is given by:

\(\rho = \frac{1}{\sigma}\)

The unit of conductivity is siemens per meter ($\text{S}/\text{m}$), or $\Omega^{-1} \cdot \text{m}^{-1}$.

Electrical Conductivity in Semiconductors

In a semiconductor where both electrons and holes are present as charge carriers, the total electrical conductivity is the sum of the conductivity due to electrons and the conductivity due to holes. This is because both types of carriers contribute to the overall current flow.

Conductivity due to Electrons ($\sigma_n$)

Electrons are negatively charged carriers. Their contribution to conductivity depends on their concentration and how easily they move through the material (their mobility). The formula for electron conductivity is:

\(\sigma_n = n q \mu_n\)

Where:

  • \(n\) is the concentration of electrons (number of electrons per unit volume).
  • \(q\) is the magnitude of the elementary charge (\(1.602 \times 10^{-19}\) C).
  • \(\mu_n\) is the mobility of electrons. Mobility ($\mu$) indicates how fast a carrier drifts in an electric field. Its unit is usually $\text{cm}^2/\text{V} \cdot \text{s}$ or $\text{m}^2/\text{V} \cdot \text{s}$.

Conductivity due to Holes ($\sigma_p$)

Holes are positively charged carriers. Similar to electrons, their contribution to conductivity depends on their concentration and mobility. The formula for hole conductivity is:

\(\sigma_p = p q \mu_p\)

Where:

  • \(p\) is the concentration of holes (number of holes per unit volume).
  • \(q\) is the magnitude of the elementary charge.
  • \(\mu_p\) is the mobility of holes.

Total Conductivity ($\sigma$)

The total conductivity of the semiconductor is the sum of the conductivity due to electrons and the conductivity due to holes:

\(\sigma = \sigma_n + \sigma_p\)

Substituting the formulas for \(\sigma_n\) and \(\sigma_p\):

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

We can factor out the elementary charge \(q\):

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

Or, written slightly differently as often seen:

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

Deriving Resistivity from Conductivity

Since resistivity ($\rho$) is the reciprocal of conductivity ($\sigma$), we can find the expression for resistivity by taking the reciprocal of the total conductivity formula:

\(\rho = \frac{1}{\sigma} = \frac{1}{q (\mu_n n + \mu_p p)}\)

Analyzing the Options

Let's compare our derived formula for semiconductor resistivity with the given options:

  1. \(q (\mu_n n + \mu_p p)\): This is the expression for conductivity (\(\sigma\)), not resistivity ($\rho$).
  2. \(q (\mu_n n - \mu_p p)\): This formula involves a difference, which is incorrect as both carriers contribute positively to conductivity.
  3. \(\rm \frac{q}{(\mu_n n + \mu _p p)}\): This formula has \(q\) in the numerator, which is incorrect.
  4. \(\rm \frac{1}{q(\mu_n n + \mu _p p)}\): This formula exactly matches our derived expression for resistivity ($\rho$).

Therefore, the correct expression for the resistivity of a semiconductor with both electrons and holes as carriers is \(\rm \frac{1}{q(\mu_n n + \mu _p p)}\).

Symbol Meaning Unit (SI)
\(\rho\) Resistivity \(\Omega \cdot \text{m}\)
\(\sigma\) Conductivity \(\text{S}/\text{m}\)
\(n\) Electron concentration \(\text{m}^{-3}\)
\(p\) Hole concentration \(\text{m}^{-3}\)
\(q\) Elementary charge magnitude C
\(\mu_n\) Electron mobility \(\text{m}^2/\text{V} \cdot \text{s}\)
\(\mu_p\) Hole mobility \(\text{m}^2/\text{V} \cdot \text{s}\)

Revision Table: Semiconductor Properties

Property Description Dependence
Resistivity (\(\rho\)) Resistance to current flow Inverse of conductivity, depends on carrier concentration and mobility
Conductivity (\(\sigma\)) Ease of current flow Sum of electron and hole conductivity, depends on \(n, p, \mu_n, \mu_p\)
Carrier Concentration (\(n, p\)) Number of charge carriers per unit volume Temperature, doping level
Mobility (\(\mu_n, \mu_p\)) Ease of movement in electric field Temperature, scattering mechanisms (impurities, lattice vibrations)

Additional Information: Charge Carrier Behavior

In semiconductors, charge carriers move under the influence of an electric field (drift) and due to concentration gradients (diffusion). Resistivity and conductivity are primarily related to drift current, which is proportional to the electric field and the total charge density and drift velocity of carriers.

The drift velocity (\(v_d\)) of a carrier is proportional to the electric field (\(E\)), with the proportionality constant being the mobility ($\mu$):

\(v_d = \mu E\)

The current density (\(J\)) due to drift is given by the charge density multiplied by the drift velocity. For electrons:

\(J_n = (-nq) v_{dn} = (-nq) (-\mu_n E) = n q \mu_n E\)

For holes:

\(J_p = (pq) v_{dp} = (pq) (\mu_p E) = p q \mu_p E\)

The total current density is the sum:

\(J = J_n + J_p = (n q \mu_n + p q \mu_p) E\)

We know that current density is also related to conductivity and electric field by Ohm's Law in differential form:

\(J = \sigma E\)

Comparing the two expressions for \(J\), we get:

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

So, the total conductivity is:

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

And finally, resistivity is the reciprocal of this conductivity:

\(\rho = \frac{1}{\sigma} = \frac{1}{q (n \mu_n + p \mu_p)}\)

This detailed derivation confirms the formula for semiconductor resistivity when both electrons and holes are present.

Was this answer helpful?

Important Questions from Semiconductors

  1. In which one of the following devices, the light energy is converted into the electrical energy?

  2. The majority charge carriers in a p-type semiconductor are

  3. The thyristor is turned off when the anode current falls below-

  4. In P-type semiconductor, the majority carriers are-

  5. What is the forbidden energy gap in a pure conductor?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App