Analog Circuits Questions and Answers – High Frequency Response


This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “High Frequency Response”.

1. We cannot use h-parameter model in high frequency analysis because ____________
a) They all can be ignored for high frequencies
b) Junction capacitances are not included in it
c) Junction capacitances have to be included in it
d) AC analysis is difficult for high frequency using it
View Answer

Answer: b
Explanation: The effect of smaller capacitors is considerable in high frequency analysis of analog circuits, and hence they cannot be ignored. Instead of h-parameter model, we use π-model.

2. Consider a CE circuit, where trans-conductance is 50mΩ-1, diffusion capacitance is 100 pF, transition capacitance is 3 pF. IB = 20μA. Given base emitter dynamic resistance, rbe = 1000 Ω, input VI is 20*sin(107t). What is the short circuit current gain?
a) 30
b) 35
c) 40
d) 100
View Answer

Answer: b
Explanation: AI = IL/IB
IL = -gmVb’e
Vb’e = Ib rb’e/(1+jωCrb’e)
C = CD + CT = 103pF
Vb’e = 20μ.1k/(1+j.107.103.10-12.1000)
AI = IL/IB = 50m.1k/(1+j.107.103.10-12.1000)
AI = 35 (approx).

3. Given that transition capacitance is 5 pico F and diffusion capacitance is 80 pico F, and base emitter dynamic resistance is 1500 Ω, find the β cut-off frequency.
a) 7.8 x 106 rad/s
b) 8.0 x 106 rad/s
c) 49.2 x 106 rad/s
d) 22.7 x 106 rad/s
View Answer

Answer: a
Explanation: The frequency in radians is calculated by
ωβ = 1/C.rbe
ωβ = 7.8 x 106.

4. For given BJT, β=200. The applied input frequency is 20 Mhz and net internal capacitance is 100 pF. What is the CE short circuit current gain at β cut-off frequency?
a) 200
b) 100
c) 141.42
d) 440.2
View Answer

Answer: c
Explanation: The current gain for the CE circuit is A = \(\frac{β}{\sqrt{1+(\frac{f}{f_β})^2}}\)
At f = fβ, A = \(\frac{β}{\sqrt{2}}\)
Hence A = 141.42.

5. Given that β=200, input frequency is f = 20Mhz and short circuit current gain is A = 100. What is the unity gain frequency?
a) 2308 Mhz
b) 2000 Mhz
c) 2508 Mhz
d) 3000 Mhz
View Answer

Answer: a
Explanation: A = \(\frac{β}{\sqrt{1+(20Mhz/fβ)^2}}\)
1 + (20/f)2 = 4
20/f = 1.732
fβ = 11.54 Mhz
Unity gain frequency = βfβ = 200 x 11.54Mhz = 2308 Mhz.

6. Gain bandwidth frequency is GBP= 3000 Mhz. The cut-off frequency is f=10Mhz. What is the CE short circuit current gain at the β cutoff frequency?
a) 212
b) 220
c) 300
d) 200
View Answer

Answer: a
Explanation: fT = 3000Mhz
βfβ = 3000Mhz
β = 3000/10 = 300
A = \(\frac{β}{\sqrt{2}}\) = 212.13.

7. Which of the statement is incorrect?
a) At unity gain frequency the CE short circuit current gain becomes 1
b) Unity gain frequency is the same as Gain Bandwidth Product of BJT
c) Gain of BJT decreases at higher frequencies due to junction capacitances
d) β- cut-off frequency is one where the CE short circuit current gain becomes β/2
View Answer

Answer: d
Explanation: At unity gain frequency the current gain is 1 is a correct statement. The same frequency is fT = βfβ which is the gain bandwidth product of BJT. Gain of BJT at high frequency decreases due to the junction capacitance. However, at β cut-off frequency, current gain becomes \(\frac{β}{\sqrt{2}}\).

8. Given a MOSFET where gate to source capacitance is 300 pF and gate to drain capacitance is 500 pF. Calculate the gain bandwidth product if the transconductance is 30 mΩ-1.
a) 5.98 Mhz
b) 4.9 Mhz
c) 6.5Mhz
d) 5.22Mhz
View Answer

Answer: a
Explanation: Gain bandwidth product for any MOSFET is fT = gm/2π(Cgs+Cgd)
Thus GBP is approximately 5.9 Mhz.

9. In an RC coupled CE amplifier, when the input frequency increases, which of these are incorrect?
a) Reactance CSH decreases
b) Voltage gain increases
c) Voltage gain decreases due to shunt capacitance
d) An RC coupled amplifier behaves like a low pass filter
View Answer

Answer: b
Explanation: When frequency increases, shunt reactance decreases. The voltage drop across shunt capacitance decreases and net voltage gain decrease. RC coupled amplifier acts as a low pass filter at high frequencies.

Sanfoundry Global Education & Learning Series – Analog Circuits.


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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn