Analog Circuits Questions and Answers – Comparison of Amplifier Classes

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on the “Comparison of Amplifier Classes”.

1. Which of the following amplifier class have the highest linearity and lowest distortion?

a) Class A

b) Class B

c) Class C

d) Class B push-pull

Answer: a

Explanation: Class A amplifier has the highest linearity and the lowest distortion. The amplifying element is always conducting and close to the linear portion of its transconductance curve. The point where the device is almost off is not at a zero signal point and hence its distortions compared to other classes are less.

2. Which of the following letter is not used to represent a class?

a) D

b) E

c) C

d) K

Answer: d

Explanation: There is no amplifier called Class K. There are only A, B, C, D,E/F, G, H, S.

3. Which of the following letter is not used to represent a class?

a) I

b) H

c) G

d) S

Answer: a

Explanation: There is no amplifier called Class I. There are only A, B, C, D, E/F, G, H, S.

4. Which of the following class has the poorest linearity

a) Class A

b) Class B

c) Class C

d) Class AB

Answer: c

Explanation: Class C amplifiers have high efficiency but have the poorest linearity since they only take less than 180° oscillations. They are suitable for amplifying constant envelope signals.

5. Which of the following amplifier cannot be used for audio frequency amplification?

a) Class A

b) Class C

c) Class AB

d) Class B push-pull

Answer: b

Explanation: Class C amplifiers cannot be used for audio frequency amplifiers because of their high distortion.

6. Which of the following amplifier is less efficient than others?

a) Class C

b) Class B

c) Class A

d) Class AB

Answer: c

Explanation: Class A amplifiers are the least efficient of all. A maximum of 25% theoretical efficiency is obtainable, 50% for when using a transformer or with induced coupling. This wastes power, as well as increases the cost, and requires higher-rated output devices.

7. Which of the following amplifier is designed to operate in digital pulses?

a) Class D

b) Class C

c) Class AB

d) Class B

Answer: a

Explanation: Class D amplifiers use a form of PWM to control the output devices. The conduction angle varies with the pulse width and doesn’t depend on the input directly. The analog signal is converted into a stream of pulses representing the signal using a modulation technique.

8. Which of the following class have a theoretical efficiency of 50%?

a) Class A

b) Class C

c) Class AB

d) Class D

Answer: a

Explanation: Class A amplifier has a theoretical efficiency of 50%. 50% of the energy supplied is a waste.

9. Which of the following class have a theoretical efficiency of 78.5%?

a) Class A

b) Class D

c) Class C

d) Class B

Answer: d

Explanation: Class B amplifier has a theoretical efficiency of 78.5% which is higher than Class A while Class D theoretically has an efficiency of 100%.

10. Which of the following amplifier is most suited for making tuning circuits?

a) Class A

b) Class B

c) Class C

d) Class D

Answer: c

Explanation: Class C is the most suitable amplifier type for tuning circuits and radio frequency amplification. It employs filtering and hence the final signal is completely acceptable. Class C amplifiers are quite efficient than other types.

Analog Circuits Questions and Answers – Features of Power Amplifier

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Features of Power Amplifier”.

1. The use of amplifier in a circuit is to _____________ for input signal.

a) Provide a phase shift

b) Provide strength

c) Provide frequency enhancement

d) Make circuit compatible

View Answer

Answer: b

Explanation: The only use of amplifier in a circuit is to provide strength to signal. This may refer to an increase in current, voltage or power of the output w.r.t the input being applied.

2. The unwanted characteristics of amplifier output apart from the desired output is collectively termed as ___________

a) Inefficiency

b) Damage

c) Fault

d) Distortion

View Answer

Answer: d

Explanation: The unwanted characteristics of amplifier output apart from desired output is collectively termed as distortion. This should be avoided.

3. Unit of power rating of a transistor is expressed in ___________

a) Watts

b) KWh

c) W/s

d) Wh

View Answer

Answer: a

Explanation: Power rating is the maximum power allowable to dissipate by a transistor beyond this point transistor may behave unlikely. This is expressed in watts.

4. Which device was used for the amplification of audio signals before the invention of power amplifiers?

a) Diode

b) Op-amp

c) Vacuum tubes

d) SCR

View Answer

Answer: c

Explanation: Before the invention of power amplifier vacuum tubes are used for audio signal amplification which consumes large space and costly.

5. Power amplifier directly amplifies ___________

a) Voltage of signal

b) Current of the signal

c) Power of the signal

d) All of the mentioned

View Answer

Answer: d

Explanation: Power amplifier increases voltage as well as current. Increase in voltage or current is small compared to normal amplifiers. But power amplification has occurred ie. Voltage x current is more.

6. Input stage of power amplifier is also called ___________

a) First op

b) Beginning stage

c) Front end

d) Normal stage

View Answer

Answer: c

Explanation: Input stage of the power amplifier is also called the front end.

7. Transistor in power amplifier is ___________

a) An active device

b) A passive device

c) A op-amp

d) A voltage generating device

View Answer

Answer: a

Explanation: Transistor is an active device since transistor contains voltage sources which are necessary for amplification.

8. For a perfect power amplifier output power rating will be ________ if the output impedance is halved.

a) Halved

b) Squared

c) Doubled

d) Square rooted

View Answer

Answer: c

Explanation: In the equation of power output for the power amplifier, the power is proportional to the square of the current and inversely proportional to the resistance. If the impedance is halved then power is doubled.

9. Which of the following audio speaker will be hard to be driven by a power amplifier?

a) 4ohm

b) 8ohm

c) 12ohm

d) 2ohm

View Answer

Answer: d

Explanation: If the resistance of the audio amplifier is less, the output power of the transistor will be high since output current is increasing. Hence to drive 2ohm speaker amplifier needs double power that for 4ohm speaker.

10. The power rating of the amplifier is 100watts then the transistor can only operate at ___________

a) Power higher than 100w

b) Power lower than 100w

c) Power near to 100w

d) Power lower than 200W

View Answer

Answer: b

Explanation: The power rating is 100 W, and that is the maximum allowable power usage of a transistor, beyond which it may damage. If the power is less than 100W, the circuit operates. Near to 100W, the power may also be higher than 100W, hence that option is incorrect.

Analog Circuits Questions and Answers – Effects of Feedback

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

1. What is reverse transmission factor?

a) Ratio of output by input signal

b) Ratio of feedback by input signal

c) Ration of feedback by output signal

d) Ratio of input by feedback signal

View Answer

Answer: c

Explanation: In feedback systems, the feedback signal is in proportion with the output signal.

XF ∝ XO

XF = βXO, where β is the feedback factor or reverse transmission factor.

2. Return ratio for a circuit is 220. What is the amount of feedback, correct up to 2 decimal places?

a) 2.34 dB

b) – 46.84 dB

c) – 46.88 dB

d) 46.88 dB

View Answer

Answer: c

Explanation: Return ratio is Aβ, where β=feedback factor, and A=open loop gain.

Amount of feedback is AF/A = 1 + Aβ

In decibels, amount of feedback = -20log10(1+Aβ) = -20log10(221) = -46.88 decibels.

3. Consider the given diagram. Loop gain is 19. Consider closed loop gain is 50. Find the output without any feedback when input is 5.

analog-circuits-questions-answers-effects-feedback-q3

a) 1000

b) 500

c) 5000

d) 50000

View Answer

Answer: c

Explanation: Feedback factor, β=10.

AF = A/(1+Aβ) = 50

A = 20*50 = 1000

Output is thus 5000.

4. Consider an open loop circuit with lower cutoff frequency 5kHz and upper cutoff frequency 20kHz. If negative feedback is applied to the same, choose correct option stating the new cutoff frequencies.

a) Lower cutoff = 5kHz, Upper cutoff = 20kHz

b) Lower cutoff = 2kHz, Upper cutoff = 18kHz

c) Lower cutoff = 2kHz, Upper cutoff = 25kHz

d) Lower cutoff = 10kHz, Upper cutoff = 25kHz

View Answer

Answer: c

Explanation: Negative feedback decreases lower cutoff frequencies and increases the higher cutoff frequency.

fHF = fH(1+Aβ)

fLF = fL/(1+Aβ)

Total bandwidth is thus increased.

5. Find the relative change in gain with negative feedback given that return ratio is 24, and feedback factor is 3, when the change in open loop gain is 2.

a) 1

b) 1.6

c) 0.1

d) 0.01

View Answer

Answer: d

Explanation: AF = A/(1+Aβ)

Aβ = 24

A = 8

Relative change in gain = dAF/AF = dA/A(1+Aβ)

dAF/AF = 2/8*25 = 0.01.

6. Relative change of gain of feedback amplifier is 0.05. Also, loop gain is 9. Find desensitivity?

a) 50

b) 10

c) 20

d) 1/9

View Answer

Answer: b

Explanation: We can simply use the ratio of 0.1 to find the answer.

Loop gain Aβ = 9

1+Aβ = 10

Desensitivity = 1/S = 1+Aβ = 10.

7. Circuit P has desensitivity 20, circuit Q has sensitivity 0.1 and circuit R has desensitivity 40. Which of the following is more stable in gain?

a) Circuit P

b) Circuit Q

c) Circuit R

d) All circuits are equally stable in gain

View Answer

Answer: c

Explanation: Greater desensitivity indicates better stability in gain. More desensitivity means gain becomes smaller, but stable.

For circuit Q, desensitivity = 1/S = 10

Circuit R has higher desensitivity, hence most stable.

8. Consider the open loop response. An unknown feedback is applied. Choose the correct output of the new system from the following.

analog-circuits-questions-answers-effects-feedback-q8

a) Output response of Increased frequency distortion

analog-circuits-questions-answers-effects-feedback-q8a

b) Output response of Decreased frequency distortion

analog-circuits-questions-answers-effects-feedback-q8b

c) Output response of Decreased frequency distortion

analog-circuits-questions-answers-effects-feedback-q8c

d) Output response of No change in frequency distortion.

analog-circuits-questions-answers-effects-feedback-q8d

View Answer

Answer: b

Explanation: Output should have increased Bandwidth and decreased frequency distortion. If bandwidth increases, the distortion cannot increase since it’s a case of negative feedback. Also, the distortion cannot remain the same.

9. Consider the total harmonic distortion of a closed loop system is 5%. Distortion without feedback is 10%. Calculate the sensitivity of closed loop system.

a) 0.5

b) 0.2

c) 0.6

d) 0.1

View Answer

Answer: a

Explanation: DHF = DH/1+Aβ

1 + Aβ = DH/DHF = 10/5 = 2

Sensitivity = 12 = 0.5.

10. For the system shown gain with feedback is 200. Find feedback factor.

analog-circuits-questions-answers-effects-feedback-q10

a) 0.41

b) 5

c) 0.0041

d) It can be any real number

View Answer

Answer: c

Explanation: AF = A/1+Aβ

A = A1 x A2 = 1200

Thus 1 + Aβ = 1200/200 = 6

Aβ = 5

β = 5/A = 0.0041.

Analog Circuits Questions and Answers – Feedback Connection Types

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Feedback Connection Types”.

1. Which of these doesn’t refer to a series-shunt feedback?

a) Voltage in and Voltage out

b) Current in and Voltage out

c) Voltage Controlled Voltage Source

d) Series voltage feedback

View Answer

Answer: b

Explanation: In a series shunt feedback network, feedback is connected in series with signal source but in shunt with the load. Error voltage from feedback network is in series with the input. Voltage fed back from output is proportional to output voltage, hence parallel or shunt connected. The current in and voltage out connection refers to a shunt-shunt connection.

2. In the following diagram, shaded portions are named A and B.

analog-circuits-questions-answers-feedback-connection-types-q2

What are A and B?

a) A = Current sampling network, B = Voltage sampling network

b) A = Current mixing network, B = Voltage sampling network

c) A = Shunt mixing network, B = Current sampling network

d) A = Voltage mixing network, B = Current sampling network

View Answer

Answer: c

Explanation: When feedback network is in shunt with load, then output voltage appears as input to feedback. In above case, output current appears as the feedback input, hence B is a current sampling network. Also, feedback network is in shunt with the signal source, hence it’s called shunt mixing or current mixing.

3. Given that a feedback network is shunt-series, and output load is 10kΩ, what is the output voltage across it given that transfer gain is 10, source current is 20mA and feedback current is 10mA?

a) 1V

b) 2V

c) 10V

d) 20V

View Answer

Answer: c

Explanation: RL = 10kΩ

IF = βIL

IL = IF/β = 10/10 = 1mA

VL = ILRL = 10V.

4. Consider the circuit shown.

analog-circuits-questions-answers-feedback-connection-types-q4

What is the type of sampling observed?

a) Shunt-Series feedback

b) Series-Series feedback

c) Shunt-Shunt feedback

d) Series-Shunt feedback

View Answer

Answer: d

Explanation: The feedback network is connected directly to output node, so voltage sampling occurs. However, it’s not connected directly to the input node. Hence it’s series mixing at the input. Voltage sampling is a shunt network.

5. Consider a voltage series feedback network, where amplifier gain = 100, feedback factor = 5. For the basic amplifier, input voltage = 4V, input current=2mA. Find the input resistance of the network.

a) 1.002kΩ

b) 1002kΩ

c) 2kΩ

d) 2000kΩ

View Answer

Answer: b

Explanation: RI = VI / II = 4/2m = 2kΩ

RIF = RI(1+A.β) = 2k(1+500) = 1002kΩ.

6. In which network is the unit of the feedback factor Ω?

a) Shunt-shunt feedback

b) Shunt-series feedback

c) Series-series feedback

d) Series-shunt feedback

View Answer

Answer: c

Explanation: In series-series feedback, the output is current sampled, that is it is in series with the load. Also, input is a voltage mixer, which is in series with signal source. So feedback factor

Β = VF/IL in Ohms.

7. A circuit can have more than one type of feedback.

a) True

b) False

View Answer

Answer: a

Explanation: In any circuit, the feedback depends on the configuration of resistor network and presence of capacitances. Consider a collector to base bias circuit, in which base resistance causes voltage shunt feedback. However, presence of an emitter resistance provides a second feedback of current series type.

8. Consider given circuit.

analog-circuits-questions-answers-feedback-connection-types-q8

What is the feedback configuration?

a) Current series feedback

b) Current shunt feedback

c) Voltage series feedback

d) Voltage shunt feedback

View Answer

Answer: a

Explanation: The resistance R4 is the feedback network resistance. There is no bypass capacitor being used. The resistance is not directly connected to either the input node, or output node. Hence it’s a current series feedback.

9. Consider the circuit shown below.

analog-circuits-questions-answers-feedback-connection-types-q9

Consider A: Current-shunt feedback

B: Current-series feedback

C: Voltage-shunt feedback

D: Voltage-series feedback

Which of the above are present?

a) A and B

b) A only

c) B only

d) A and D

View Answer

Answer: a

Explanation: Resistor R5 causes global feedback. It is connected to the input node, causing shunt mixing but not to output node, meaning current sampling. Hence it’s a current shunt feedback. Resistors R6 and R7 are neither connected to input nor the output, causing series mixing and current sampling, hence causing current series feedback.

10. In a feedback network, input voltage is 14V, feedback voltage is 6V and source voltage is 20V.

β is in ohms. What is its configuration?

a) Shunt-Shunt feedback

b) Shunt-Series feedback

c) Series-Series feedback

d) Series-Shunt feedback

View Answer

Answer: c

Explanation: Given that input is 14V, feedback is 6V and source is 20 V, we can see

VI = VS – VF, which is voltage mixing. Also β is in ohms that is voltage/current. Since output of feedback is voltage and input is current, the output has current sampling. Thus, configuration is a series-series feedback / current – series feedback.

Analog Circuits Questions and Answers – Cascode and Darlington Amplifier

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Cascode and Darlington Amplifier”.

1. Which of these are incorrect about Darlington amplifier?

a) It has a high input resistance

b) The output resistance is low

c) It has a unity voltage gain

d) It is a current buffer

Answer: d

Explanation: A Darlington amplifier has a very high input resistance, low output resistance, unity voltage gain and a high current gain. It is a voltage buffer, not a current buffer.

2. Consider the circuit shown below where hfe=50

analog-circuits-questions-answers-cascode-darlington-amplifier-q2

Calculate the input resistance of the network.

a) 255 kΩ

b) 13 MΩ

c) 5 MΩ

d) 250 kΩ

Answer: b

Explanation: The load for the first transistor in the figure is the input resistance of the second.

RE1 = (1+hfe)5k = 255kΩ

Net input resistance, RI = (1+hfe)RE1=(1+hfe)25k = 13005k = 13MΩ.

3. Given the following circuit

analog-circuits-questions-answers-cascode-darlington-amplifier-q3

It is given that hfe=55, hie=1kΩ, hoe=25μΩ-1. Calculate the net current gain and the voltage gain of the network.

a) AI=192.6, Av=220

b) AI=1, AV=220

c) AI=192.6, AV=1

d) AI=192.6, AV=55

Answer: c

Explanation: AI=A1xA2

AI = [1+hfe/1+hoehfeRE]x[1+hfe] AI = 51×51/(1+25x50x10x10-3) = 192.6.

4. In a Darlington pair, the overall β=15000.β1=100. Calculate the collector current for Q2 given base current for Q1 is 20 μA.

analog-circuits-questions-answers-cascode-darlington-amplifier-q4

a) 300 mA

b) 298 mA

c) 2 mA

d) 200mA

Answer: b

Explanation: IB = 20 μA

IC = β.IB = 15000 x 20μ = 300 mA

IC1 = β1.IB = 100.20μ = 2mA

IC2 = 300 – 2 = 298mA.

5. Darlington amplifier is an emitter follower.

a) True

b) False

Answer; a

Explanation: Darlington pair is an emitter follower circuit, in which a darling pair is used in place of a single

transistor. It also provides a large β as per requirements.

6. What is the need for bootstrap biasing?

a) To prevent a decrease in the gain of network

b) To prevent an increase in the input resistance due to the biasing network

c) To prevent a decrease in the input resistance due to the presence of multiple BJT amplifiers

d) To prevent a decrease in the input resistance due to the biasing network

Answer: b

Explanation: A bootstrap biasing network is a special biasing circuit used in a Darlington amplifier to prevent the decrease in input resistance due to the biasing network being used. Capacitors and resistors are added to the circuit to prevent it from happening.

7. Consider a Darlington amplifier. In the self-bias network, the biasing resistances are 220kΩ and 400 kΩ. What can be the correct value of input resistance if hfe=50 and emitter resistance = 10kΩ.

a) 141 kΩ

b) 15 MΩ

c) 20 MΩ

d) 200 kΩ

Answer: a

Explanation: R’ = 220k||400k = 142 kΩ

RI = (1+hfe)2RE = 26MΩ

RI’ = 26M||142k = 141.22 K.

8. What is a cascode amplifier?

a) A cascade of two CE amplifiers

b) A cascade of two CB amplifiers

c) A cascade of CE and CB amplifiers

d) A cascade of CB and CC amplifiers

Answer: c

Explanation: A cascode amplifier is a cascade network of CE and CB amplifiers, or CS and CG amplifiers.

It is used as a wide-band amplifier.

9. Consider the figure shown.

analog-circuits-questions-answers-cascode-darlington-amplifier-q9

Given that gm1 = 30mΩ-1 and gm2 = 50mΩ-1, α1 = 1.1, α2 = 1.5 what is the transconductance of the entire network?

a) 80 mΩ-1

b) 75 mΩ-1

c) 33 mΩ-1

d) 55 mΩ-1

Answer: d

Explanation: The above circuit is a cascode pair. For this circuit, the overall transconductance is

gm = α1gm2

gm = 1.1 gm2 = 55mΩ-1.

10. Find the transconductance of the network given below, provided that gm1 = 30mΩ-1. VT = 25mV, VBias > 4V.

analog-circuits-questions-answers-cascode-darlington-amplifier-q10

a) 30mΩ-1

b) 10mΩ-1

c) 1mΩ-1

d) 20mΩ-1

Answer: a

Explanation: For a MOSFET cascode amplifier, the net transconductance in the above network shown is equal to the transconductance of MOSFET M1 that is equal to 30mΩ-1.

11. In the given circuit, hfe = 50 and hie = 1000Ω, find overall input and output resistance.

analog-circuits-questions-answers-cascode-darlington-amplifier-q11

a) RI=956Ω, RO=1.6 kΩ

b) RI=956 kΩ, RO=2 kΩ

c) RI=956 Ω, RO=2 kΩ

d) RI=900Ω, RO=10 kΩ

Answer: c

Explanation: RO = RC = 2kΩ

Input resistance = hie||50k||40k = 0.956 kΩ.

Analog Circuits Questions and Answers – Cascaded Amplifier

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

1. For any cascaded amplifier network, which of these are incorrect?

a) Cascading increases gain

b) Overall input resistance is equal to the input resistance of the first amplifier

c) The overall output resistance is less than the lowest output resistance in all amplifiers used

d) Loading effect occurs

View Answer

Answer: c

Explanation: In cascading, the output of one amplifier is connected to the input of another amplifier. It is used to increase gain while obtaining desired values of input and output resistances. Overall input resistance is the same as input resistance of the first amplifier and net output resistance is the same as output resistance of the last (nth) amplifier in the network. When amplifiers are connected in cascade, then loading effect does occur.

2. Consider the circuit

analog-circuits-questions-answers-cascaded-amplifier-q2

Loading effect occurs when ______________

a) R2 is small

b) A1 is small

c) R2 is large

d) A2 is large

View Answer

Answer: a

Explanation: Loading effect refers to the effect of the input resistance of the nth amplifier in the net load resistance of the n-1th amplifier. The decrease of voltage gain of A1 above, due to smaller input resistance R2 of A2 is called loading effect. If the amplifier has larger input resistance, that is if R2 is large, no loading effect occurs. Such amplifiers are called non-interacting amplifiers.

3. Consider the circuit shown.

analog-circuits-questions-answers-cascaded-amplifier-q3

Find internal voltage gain of the network given gm = 50mΩ-1 and β=100.

a) 100

b) -90

c) 90

d) 95

View Answer

Answer: c

Explanation: Both are CE amplifiers without a bypass capacitor.

For this, voltage gain is AV = -gmRL’/1+gmRE

A2 = -50 x 5/1 + 50 x 0.2 = -22.7

RL1 = 5k||RI2

RI2 = rπ + (1+β)RE = 100/50m + 101 x 0.2k = 2k + 20.2k = 22.2k

RL1 = 4.08kΩ

A1 = -50×4.08/1+ 50×1 = – 4

Gain = A’= A1 x A2 = 90.8.

4. Cascading increases lower cut-off frequencies.

a) True

b) False

View Answer

Answer: a

Explanation: Considering each amplifier has lower cut-off frequency fL1 then net lower cut-off frequency for a network of N cascaded amplifiers is fL = fL121/N√−1

For N>=2, 21/N−−−−√−1 < 1, thus fL > fL1.

5. 6 similar amplifiers are cascaded, with lower cut-off frequency 100Hz. Bandwidth is B1=10 kHz. What is the higher cut-off frequency of the cascaded network?

a) 4000 Hz

b) 1667 Hz

c) 3642 Hz

d) 3000 Hz

View Answer

Answer: c

Explanation: fL1 = 50 Hz

fL = 50/ 21/6−−−√−1 = 142 Hz

B1 = 10kHz

B2 = B1 21/N−−−−√−1 = 3.5 Khz

fH – fL = 3500

fH = 3642 Hz.

6. It is provided that the lower cut-off frequency of an individual amplifier is 25Hz, find the net cut-off frequency of a cascaded network of 8 similar amplifiers.

a) 200 Hz

b) 83 Hz

c) 100 Hz

d) 25 Hz

View Answer

Answer: b

Explanation: fL1 = 25Hz

fL = fL121/N√−1=2521/8√−1=250.0905√

fL = 83 Hz.

7. Given that the higher cut-off frequency of the cascaded network of 6 amplifiers is 2Mhz, find the higher cut-off frequency of one amplifier, if all amplifiers are similar.

a) 5.7 Mhz

b) 0.33 Mhz

c) 12 Mhz

d) 64 Mhz

View Answer

Answer: a

Explanation: fH = 2Mhz

fH = fH1 21/N−−−−√−1

fH1 = 2Mhz/√21/6−−−√−1 = 5.71 Mhz.

8. The lower and upper cutoff frequencies of an amplifier are unknown. If originally, individual BW of such an amplifier is B1, and now the bandwidth of the cascaded network of 10 such amplifiers is B2, find B2/B1.

a) 0.26

b) 3.84

c) Insufficient data

d) 5

View Answer

Answer: a

Explanation: To calculate net bandwidth B2 = B1 x 21/N−−−−√−1

B2/B1 = 21/N−−−−√−1

B2/B1 = 0.26.

9. Provided a cascade multistage amplifier network, their pole frequencies obtained are f1=10Mhz, f2=12Mhz, f3=20Mhz, f4=16Mhz. What is the approximate higher cutoff frequency of the cascaded network?

a) 3.4 Mhz

b) 8 Mhz

c) 5 Mhz

d) 6 Mhz

View Answer

Answer: a

Explanation: The net frequency

1/fH = 1/f1 + 1/f2 + 1/f3 + 1/f4 and so on for multiple stages.

1/fH = 0.295833

fH ≈ 3.4 Mhz.

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 as. 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 = β1+(ffβ)2√

At f = fβ, A = β2√

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) 2300 Mhz

b) 2000 Mhz

c) 2500 Mhz

d) 3000 Mhz

View Answer

Answer: a

Explanation: A = β1+(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 = β2√ = 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 β2√.

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.