Diffusion Capacitance Questions and Answers

This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Diffusion Capacitance”.

1. On which of these does the diffusion capacitance of a diode not depend upon?
a) Forward current
b) Dynamic conductance
c) Doping concentration
d) Reverse resistance
View Answer

Answer: d
Explanation: C_D = \(\frac{εA}{W}\) which means C_D ∝ \(\frac{1}{W}\). Thus, C_D ∝ \(\frac{1}{\sqrt{Doping}}\) concentration
C_D = τg = \(\frac{τ}{r} = \frac{τI_F}{ηV_T}\)
where τ = carrier life time, g = dynamic conductance, I_F = forward current.

2. If the diffusion capacitance is directly proportional only to the lifetime of holes in N side then ________
a) The diode is a p⁺n junction diode
b) The diode is a pn⁺ junction diode
c) The diode is a p⁺n⁺ diode
d) The diode can have any type of pn junction
View Answer

Answer: a
Explanation: Diffusion capacitance is proportional to the carrier lifetime of injected minority carriers / excess minority carriers. Since it is proportional to the lifetime of holes in N side, this means the N side is the minority and P side is the majority. Hence it is a p⁺n diode.

3. Diffusion capacitance is larger than transition capacitance.
a) True
b) False
c) Both are same
d) Depends on doping concentrations
View Answer

Answer: b
Explanation: Diffusion capacitance occurs in a forward biased diode, transition capacitance is easy to see in reverse bias. C_D > C_T for a forward bias junction. In reverse bias though, C_D may be neglected compared to C_T.

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4. In a diode, the change in voltage being applied across it is 2V. The change in minority carriers outside the depletion region is 1.2×10^-8. Find diffusion capacitance.
a) 6 pF
b) 6 μF
c) 1.2 nF
d) 6nF
View Answer

Answer: d
Explanation: Diffusion capacitance = \(\frac{dQ}{dV}\)
C_D = \(\frac{1.2 x 10^{-8}}{2}\)
C_D = 6 nF.

5. The minority charge concentration can be represented as Q = V² – 33V. If the voltage being applied is now 5V, find the diffusion capacitance.
a) 1μF
b) 0.1μF
c) 10μF
d) 0.9μF
View Answer

Answer: b
Explanation: Q = – \(\frac{V^{-9}}{9}\)
C_D = \(\frac{dQ}{dV}\) = V^-10 = 1.024 x 10^-7
C_D = 0.1μF.

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6. Consider two almost similar diodes, whose diffusion capacitances are C1 and C2, and doping concentrations of 1020 /cubic centimeter and 1016/cubic centimeter. Find \(\frac{C1}{C2}\).
a) 100
b) 0.01
c) 10000
d) 0.0001
View Answer

Answer: b
Explanation: C ∝ \(\frac{1}{\sqrt{Doping}}\)
\(\frac{C1}{C2} = \sqrt{\frac{10^{16}}{10^{20}}} = \sqrt{\frac{1}{10^4}} = \frac{1}{100}\)
\(\frac{C1}{C2}\) = 0.01.

7. If the relative permittivity of a diode remains constant, its area is doubled and its doping concentration is quadrupled. What is its new diffusion capacitance, of originally it was C_D?
a) 4C_D
b) 2C_D
c) 3C_D
d) C_D
View Answer

Answer: a
Explanation: C_D = \(\frac{εA}{W}\) ∝ A.\(\sqrt{Doping}\)
C_D‘ = \(\frac{ε2A}{W}\) ∝ 2A.2 2A.\(\sqrt{4Doping}\) = 4 C_D
C_D‘ = 4C_D.

8. Consider a step graded diode, which has built in voltage 0.7V, and depletion width 2μm. Depletion width is W2 when a reverse bias voltage of 11.3 V is applied. Find W2?
a) 8.28μm
b) 8μm
c) 4.14 μm
d) 9.88 μm
View Answer

Answer: a
Explanation: \(\frac{W1}{W2} = \sqrt{\frac{V1}{V2}} = \sqrt{\frac{0.7}{12}}\)
W2 = 4.14*2 = 8.28μm.

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