A conducting line on an IC chip is 2.8 mm long and has a rectangular cross-section of 1 μm x 4 μm.. A current of 5 mA produces a voltage drop of 100 mV across the line. If the electron mobility is 500 cm2/V-s, the electron concentration is
4.38 x 1021 cm-3
To solve this problem, we'll calculate the electron concentration in the conducting line based on the given parameters. The important parameters given are:
- Length of the conducting line, \( L = 2.8 \, \text{mm} = 2.8 \times 10^{-3} \, \text{m} \)
- Cross-sectional area, \( A = 1 \, \mu\text{m} \times 4 \, \mu\text{m} = 4 \times 10^{-12} \, \text{m}^2 \)
- Current, \( I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \)
- Voltage drop, \( V = 100 \, \text{mV} = 0.1 \, \text{V} \)
- Electron mobility, \( \mu_e = 500 \, \text{cm}^2/\text{V-s} = 500 \times 10^{-4} \, \text{m}^2/\text{V-s} \)
The resistance \( R \) of the conducting line can be calculated using Ohm's Law:
R = \frac{V}{I} = \frac{0.1}{5 \times 10^{-3}} = 20 \, \Omega
Using the formula for resistance in terms of resistivity \( \rho \), we have:
R = \rho \cdot \frac{L}{A}
Substituting the values, we get:
20 = \rho \cdot \frac{2.8 \times 10^{-3}}{4 \times 10^{-12}}
Solving for \( \rho \):
\rho = \frac{20 \cdot 4 \times 10^{-12}}{2.8 \times 10^{-3}} = \frac{80 \times 10^{-12}}{2.8 \times 10^{-3}} = 28.57 \times 10^{-9} \, \Omega \cdot m
The conductivity \( \sigma \) is related to resistivity by:
\sigma = \frac{1}{\rho} = \frac{1}{28.57 \times 10^{-9}} \, \text{S/m}
Calculating \( \sigma \):
\sigma = 35 \times 10^{6} \, \text{S/m}
Using the formula for conductivity in terms of electron mobility and concentration:
\sigma = n \cdot e \cdot \mu_e
Where \( e \) is the electron charge, \( e = 1.6 \times 10^{-19} \, \text{C} \). Solve for the electron concentration \( n \):
n = \frac{\sigma}{e \cdot \mu_e} = \frac{35 \times 10^{6}}{1.6 \times 10^{-19} \cdot 500 \times 10^{-4}}
Calculate \( n \):
n = \frac{35 \times 10^{6}}{8 \times 10^{-23}} = 4.38 \times 10^{21} \, \text{cm}^{-3}
Therefore, the electron concentration is \( 4.38 \times 10^{21} \, \text{cm}^{-3} \), which corresponds to option A.
Similar Questions
Consider the following statements :
If an electric field is applied to an n-type semiconductor bar, the electrons and holes move in opposite directions due to their opposite charges. The net current is
1. due to both electrons and holes with electrons as majority carriers.
2. the sum of electron and hole currents.
3. the difference between electron and hole current.Which of these statements is/are correct ?
Consider the following :
1. In semiconductors, the mobility of electrons is more than that of holes.
2. In semiconductors, the resistivity increases with the increase in temperature.
3. Metal has positive TCR.
4. In metals, thermal conductivity is inversely proportional to electronic conductivity at a particular temperature.Which of the following statements are correct ?
The Debye length is a characteristic length for semiconductor. Arrange the doping density given below in such a way that Debye length changes from maximum to minimum.
(A) N = 1015 cm–3
(B) N = 1017 cm–3
(C) N = 3 × 1015 cm–3
(D) N = 1016 cm–3
(E) N = 5 × 1016 cm–3Choose the most appropriate answer from the options given below :
Match the following :
List – I List – II a. Laplace equation i. \(q\mu_n\left(\varepsilon\cdot n+\dfrac{kT}{q}\dfrac{\partial n}{\partial x}\right)\) b. Continuity equation ii. \(G_n-\dfrac{n_p-n_{p0}}{\tau_n}+n_p\mu_n\dfrac{\partial\varepsilon}{\partial x}+\mu_n\varepsilon\dfrac{\partial n_p}{\partial x}+D_n\dfrac{\partial^{2}n_p}{\partial x^{2}}\) c. Current density equation iii. \(\dfrac{\partial^{2}\psi}{\partial x^{2}}+\dfrac{\partial^{2}\psi}{\partial y^{2}}+\dfrac{\partial^{2}\psi}{\partial z^{2}}=\dfrac{qN_D}{\epsilon}\) d. Poisson’s equation iv. \(\nabla^{2}\psi=0\) Choose the correct answer from the codes given below:
For non-degenerate semiconductors the product of the majority and minority carrier concentration is fixed as
(a) \(ni ^ {2}\) (b) \(N_cN_v\exp\left(\dfrac{Eg}{kT}\right)\) (c) \(N_cN_v\exp\left(-\dfrac{Eg}{kT}\right)\) (d) \(N_v\exp\left(-\dfrac{Eg}{kT}\right)\)
Which of the above are correct ?
The electron density of a n-type semiconductor is proportional to :
The ionised concentration for donors is given by :
Important Questions from Carriers in Semiconductors
The velocity with which electrons are emitted in the photoemission process
The process of adding impurities to a pure semiconductor is called
Mobility and conductivity are related by which of the following equations?
How many electrons are there in the valence shell of a pure semiconductor?
In a pure silicon, what is the time for an electron to drift $1\mu m$ in an electric field of 100 V/cm?
Assume electron mobility of $1350 \text{ cm}^2/V-s$
