# Analog Circuits Questions and Answers – Diffusion Capacitance

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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

Explanation: CD = $$\frac{εA}{W}$$ which means CD ∝ $$\frac{1}{W}$$. Thus, CD ∝ $$\frac{1}{\sqrt{Doping}}$$ concentration
CD = τg = $$\frac{τ}{r} = \frac{τI_F}{ηV_T}$$
where τ = carrier life time, g = dynamic conductance, IF = 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

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

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

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

Explanation: Diffusion capacitance = $$\frac{dQ}{dV}$$
CD = $$\frac{1.2 x 10^{-8}}{2}$$
CD = 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

Explanation: Q = – $$\frac{V^{-9}}{9}$$
CD = $$\frac{dQ}{dV}$$ = V-10 = 1.024 x 10-7
CD = 0.1μF.

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

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 CD?
a) 4CD
b) 2CD
c) 3CD
d) CD

Explanation: CD = $$\frac{εA}{W}$$ ∝ A.$$\sqrt{Doping}$$
CD‘ = $$\frac{ε2A}{W}$$ ∝ 2A.2 2A.$$\sqrt{4Doping}$$ = 4 CD
CD‘ = 4CD.

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

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

Sanfoundry Global Education & Learning Series – Analog Circuits.

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