Minority Carrier Distribution in Inverse Active Mode MCQ Quiz - Objective Question with Answer for Minority Carrier Distribution in Inverse Active Mode - Download Free PDF
Last updated on Apr 15, 2025
Latest Minority Carrier Distribution in Inverse Active Mode MCQ Objective Questions
Top Minority Carrier Distribution in Inverse Active Mode MCQ Objective Questions
Consider the recombination process via bulk traps in a forward-biased pn homojunction diode. The maximum recombination rate is Umax. If the electron and the hole capture cross-sections are equal, which one of the following is FALSE?
Answer (Detailed Solution Below)
Minority Carrier Distribution in Inverse Active Mode Question 1 Detailed Solution
Download Solution PDFConcept:
A forward-biased p-n homojunction diode is shown below:
The maximum recombination rate (Umax) is given by:
Where,
ni: intrinsic concentration.
VTH: thermal velocity of carriers.
σ0 = Capture cross-section.
V: applied forward voltage.
VT: Thermal voltage.
Nt: probability density of energy state Et which is present between EC and Ev
Observations:
- Umax ∝ ni, so option 1 is correct.
- Maximum recombination (Umax) occurs at the center of the depletion region. ∴ Option 2 is wrong.
- Umax ∝ eV, ∴ Option 3 is correct.
- Umax ∝ VTH, ∴ Option 4 is correct.
Minority Carrier Distribution in Inverse Active Mode Question 2:
At T = 300 K, an n-type silicon sample contains a donor concentration Nd = 1016 cm-3 and intrinsic concentration ni = 1.5 × 1010 cm-3. The minority carrier hole lifetime is found to be τp0 = 20 μ sec.
What will be the lifetime of the majority carrier electrons?
Answer (Detailed Solution Below)
Minority Carrier Distribution in Inverse Active Mode Question 2 Detailed Solution
Concept:
Mass-action law states that:
n0 = concentration electrons
p0 = concentration of holes
ni = Intrinsic carrier concentration
Since the electrons and holes recombine in pairs, the minority carrier recombination rate will be equal to the majority carrier recombination rate, i.e.
τ = Recombination lifetime
Calculation:
Given,
Nd = 1016 cm-3
Nd ≫ ni
The electron concentration is:
n0 = Nd = 1016 cm-3
Using mass-action law, the hole concentration (minority concentration) is obtained as:
Now, the hole recombination rate in thermal equilibrium is:
The recombination rate of majority carriers will be the same as that of minority carrier holes i.e.
∴ The lifetime of the majority carrier electron will be:
τn0 = 8.89 × 106 sec
Minority Carrier Distribution in Inverse Active Mode Question 3:
Consider the recombination process via bulk traps in a forward-biased pn homojunction diode. The maximum recombination rate is Umax. If the electron and the hole capture cross-sections are equal, which one of the following is FALSE?
Answer (Detailed Solution Below)
Minority Carrier Distribution in Inverse Active Mode Question 3 Detailed Solution
Concept:
A forward-biased p-n homojunction diode is shown below:
The maximum recombination rate (Umax) is given by:
Where,
ni: intrinsic concentration.
VTH: thermal velocity of carriers.
σ0 = Capture cross-section.
V: applied forward voltage.
VT: Thermal voltage.
Nt: probability density of energy state Et which is present between EC and Ev
Observations:
- Umax ∝ ni, so option 1 is correct.
- Maximum recombination (Umax) occurs at the center of the depletion region. ∴ Option 2 is wrong.
- Umax ∝ eV, ∴ Option 3 is correct.
- Umax ∝ VTH, ∴ Option 4 is correct.
Minority Carrier Distribution in Inverse Active Mode Question 4:
An n-types GaAs semiconductor is doped with Nd = 1016 cm-3 and Na = 0 at T = 300 K. The electron mobility is 8500 cm2/V-sec and hole mobility is 400 cm2/V-sec for GaAs. The minority carrier lifetime is τp0 = 2 × 10-7 sec. If a uniform generation rate g’ = 2 × 1021 cm-3 sec-1 is incident on the semiconductor, then the steady state increase in conductivity will be _______ (Ω – cm)-1.
Answer (Detailed Solution Below) 0.55 - 0.65
Minority Carrier Distribution in Inverse Active Mode Question 4 Detailed Solution
Concept:
The conductivity for a semiconductor at thermal equilibrium in given as:
σ = e(μn n + μp p) ---(1)
n = Concentration of electrons
p= Concentration of holes
The amount of exess carriers generated will increase the conductivity by increasing the number of charge carriers, i.e.
Concentration of electron = n + δn
Concentration of holes = p + δp
The conductivity now will be:
σnew = e(μn (n + δn) + μp (p + δp)) ---(2)
The change (increase) in conductivity (Δσ) is calculated as:
Δσ = eμnδn + eμpδp
Since the excess electron and holes are always generated in pairs, i.e. δp = δn, the conductivity can be written as:
Δσ = e(μn + μp) δp 'or'
Δσ = e(μn + μp) δn
Calculation:
The generation rate is the ratio of the excess carriers generated with the minority carrier lifetime, i.e.
The excess hole concentration will be:
δp = g’ τp
The change in conductivity ‘or’ the conductivity is increased by:
= 0.57 (Ω cm)-1