Analog Circuits Questions and Answers – Distortion in Amplifier-1

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Distortion in Amplifier-1”.

1. The problem in which output signal is not an exact reproduction of output signal in amplifier is collectively called __________

a) Thermal runaway

b) Phase error

c) Distortion

d) Biasing error

Answer: c

Explanation: The deviation of output from an exact copy of input signal with amplification is collectively known as the distortion of the amplifier. They are of a different kind.

2. Which of the following is not a reason for distortion in amplifier output?

a) Incorrect biasing level

b) Sinusoidal input

c) Non- linear amplification

d) Large input signal

Answer: d

Explanation: Incorrect biasing level that is, if biasing level is not properly managed improper gain may lead to distortion. Non-linear amplification is a common reason for distortion in case of transistor. If the input signal is large then output exceed maximum peak of output that can be provided by an amplifier.

3. Amplitude distortion is due to ___________

a) Shift in Q-point

b) Change in input

c) Linear amplification

d) Small input signal

Answer: a

Explanation: If we incorrectly design our amplifier and a change in Q-point occurs, then distortion in the amplifier is observed. Also, if we apply too large an input signal, it may end up causing distortion.

4. If output of amplifier exceeds maximum allowable value, ___________ is occurs in output waveform.

a) Clipping

b) Clamping

c) Rectifying

d) Rounding

Answer: a

Explanation: If amplifier output is beyond maximum value it cannot display voltage further than maximum value. This constitute clipping. This maximum output value depends on the source voltage(VCC) of the amplifier, and can’t exceed the value.

5. Flat tops in the output signal is due to

a) Frequency distortion

b) Amplitude distortion

c) Phase distortion

d) Harmonic distortion

Answer: b

Explanation: If amplifier output is beyond maximum value it cannot display voltage further than maximum value. This constitute clipping and creates flat tops in the output wave. This is due to amplitude distortion in the amplifier.

6. Frequency distortion occurs when _______ is varied with frequency.

a) Amplitude

b) Amplification

c) Distortion

d) Output

Answer: b

Explanation: Due to abnormalities of the transistor level of amplification varies with a frequency which constitutes frequency distortion.

7. Phase distortion can also be called as _________

a) Frequency distortion

b) Amplitude distortion

c) Delay distortion

d) Harmonic distortion

Answer: c

Explanation: Another name of phase distortion is delay distortion. It is called so, because it associated with a delay.

8. The distortion caused by multiple frequencies in output is called _________

a) Amplifier distortion

b) Harmonic distortion

c) Phase distortion

d) None of the mentioned

Answer: b

Explanation: The distortion happened due to the presence of harmonic frequencies in output is known as harmonic distortion.

9. Harmonic distortion is caused by nonlinearities of _________

a) Voltage divider circuit

b) Resistive elements only

c) Passive elements

d) Active elements

Answer: d

Explanation: Harmonic distortion is caused by Active elements in the circuit.

10. Which of the following components in a transistor circuit is really responsible for harmonic distortion?

a) Capacitor

b) Resistor

c) Transistor

d) Inductance

Answer: c

Explanation: Harmonic distortion is caused by Active elements in the circuit. Hence transistor is causing harmonic distortion.

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.

Analog Circuits Questions and Answers – Low Frequency Response and Miller Effect Capacitance

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Low Frequency Response and Miller Effect Capacitance”.

1. In Miller’s theorem, what is the constant K?

a) Total voltage gain

b) Internal voltage gain

c) Internal current gain

d) Internal power gain

View Answer

Answer: b

Explanation: The constant K=V2/V1, which is the internal voltage gain of the network.

Thus resistance RM=R/1-K

RN=R/1-K-1.

2. When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?

a) C1

b) C2

c) Both are equal

d) Insufficient data

View Answer

Answer: a

Explanation: Given R1>R2

R/1-K > R/1-K-1, and so 1-K-1>1-K

Thus K2>1, K>1, K<-1 (correct)

Thus, C1=C(1-K) and C2=C(1-K-1)

Hence C1>C2

3. Find net voltage gain, given hfe = 50 and hie = 1kΩ.

analog-circuits-questions-answers-low-frequency-miller-effect-capacitance-q3

a) 27.68

b) -22

c) 30.55

d) -27.68

View Answer

Answer: d

Explanation: Apply millers theorem to resistance between input and output.

At input, RM=100k/1-K = RI

Output, RN=100k/1-K-1 ≈ 100k

Internal voltage gain , K = -hfeRL’/hie

K = – 50xRc||100k/1k = – 50x4x100/104= – 192

RI = 100k/1+192 = 0.51kΩ

RI’ = RI||hie = 0.51k||1k = 0.51×1/1.51 = 0.337kΩ

Net voltage gain = K.RI’/RS+RI’ = – 192 x 0.337/2k + 0.337k = -27.68.

4. Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node.

a) 0.67 pF

b) 1.34pF

c) 0.44pF

d) 2.2pF

View Answer

Answer: a

Explanation: C1=C(1-K), C2=C(1-K-1)

C1=2pF

C2=4pF

C1/C2=1/2=1-K/1-K-1

K = -2

C1 = C(1+2) = 3C

C = C1/3 = 2/3pF = 0.67 pF.

5. Consider an RC coupled amplifier at low frequency. Internal voltage gain is -120. Find the voltage gain magnitude, when given that collector resistance = 1kΩ, load = 9kΩ, collector capacitance is 0. is 0.1μF, and input frequency is 20Hz.

a) 120

b) 12

c) 15

d) -12

View Answer

Answer: c

Explanation: AV = -120

fL = 1/2πCC(RC+RL) = 1/2π*0.001 = 1000/2π = 159.15Hz

AV’ = |AV|1√+(fLf)2

AV’ = 120/8.02 ≈ 15.

6. Find the 3-dB frequency given that the gain of RC coupled amplifier is 150, the low frequency voltage gain is 100 and the input frequency is 50Hz.

a) 50.8 Hz

b) 55.9 Hz

c) 60Hz

d) 100Hz

View Answer

Answer: b

Explanation: AVM = 150

AVL = 100

f = 50Hz

100 = 1501√+(fL50)2

1+f2/2500 =1.52

f2 = 2500*1.25 = 3125

f = 55.90 Hz.

7. Given collector resistance = 2kΩ, load resistance = 5kΩ, collector capacitance = 1μF, emitter capacitance = 20μF, collector current = 2mA, source resistance = 2kΩ. If the effect of blocking capacitor is ignored, find the applicable cut-off frequency.

a) 22.73 Hz

b) 612 Hz

c) 673Hz

d) 317 Hz

View Answer

Answer: b

Explanation: RC = 2kΩ, RL = 5kΩ, CC = 1μF, CB = 10μF, CE = 20μF, RS = 2 kΩ

hie = 1kΩ, IC = 2mA

fL1 = 1/2πCC(RC+RL) = 22.73 Hz

fL2 = gm/2πCE = IC/2πCEVT = 612 Hz

Since fL2 > 4fL1, hence fL2 is the correct answer.

8. Consider the circuit shown.

analog-circuits-questions-answers-low-frequency-miller-effect-capacitance-q8

hfe = 50, hie = 1000Ω. Find magnitude of voltage gain at input frequency 10Hz.

a) 100

b) 133

c) 166

d) 220

View Answer

Answer: b

Explanation: Net load = 10k||10k = 5kΩ = RL’

AVM = -hfeRL’/hie = -50×5/1 = -250

fL = 1/2πCC(RC+RL) = 15.9 Hz

AVL = AVM1√+(fLf)2 = 133.

9. What is the phase shift in RC coupled CE amplifier at lower 3dB frequency?

a) 180°

b) 225°

c) 270°

d) 100°

View Answer

Answer: b

Explanation: Total phase shift = 180°+ tan-1(fL/f)

At 3dB frequency fL/f = 1

Total phase shift = 180° + 45° = 225°.

10. Consider that the phase shift of an RC coupled CE amplifier is 260°. Find the low frequency gain when the voltage gain of the transistor is -150.

a) 100

b) 26

c) 40

d) 55

View Answer

Answer: b

Explanation: 180° + tan-1(fL/f) = 260°

fL/f = tan(80) = 5.67

A = 1501√ + 5.672 = 26.05.

Analog Circuits Questions and Answers – Effect of Various Capacitors on Frequency Response – 2

 This set of Analog Circuits Mcqs focuses on “Effect of Various Capacitors on Frequency Response – 2”.

1. Ignoring early effect, if R1 is the total resistance connected to the base and R2 is the total resistance connected at the collector, what could be the approximate input pole of a simple C.E. stage?

a) 1 / [R1 * (Cµ(2+gm*R2) + Cπ)]

b) 1 / [R1 * (Cµ(1+2*gm*R2) + Cπ)]

c) 1 / [R1 * (Cµ(1+gm*R2) + Cπ)]

d) 1 / [R1 * (Cµ(1-gm*R2) + Cπ)]

View Answer

Answer: c

Explanation: The input pole can be approximately calculated by observing the input node. The input node is the node where the base of the B.J.T. is connected to the input voltage. The product of total resistance and capacitance connected at that particular node is R1 * Cin and Cin is Cµ(1+gm*R2) + Cπ- the inverse of this product gives us the input pole. Thus the correct option is 1 / [R1 * (Cµ(1+gm*R2) + Cπ)].

2. Ignoring early effect, if R2 is the total resistance at the collector, what could be the approximate output pole of a simple C.E. stage?

a) 1 / [R2 * (Ccs + Cµ*(1 + 2/gm*R2))]

b) 1 / [R2 * (Ccs – Cµ*(1 + 1/gm*R2))]

c) 1 / [R2 * (Ccs + Cµ*(1 – 1/gm*R2))]

d) 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))]

View Answer

Answer: d

Explanation: The output pole can be approximately calculated by observing the output node. For a C.E. stage, the output node is the node where the Collector of the B.J.T. is connected to the output measuring device. The product of total resistance and capacitance connected at that particular node is R2 * Cout and Cout is (Ccs + Cµ*(1 + 1/gm*R2). The inverse of this product gives us the output pole. Thus the correct option is 1 / [R2 * (Ccs + Cµ*(1 + 1/gm*R2))].

3. If the load resistance of a C.E. stage increases by a factor of 2, what happens to the high frequency response?

a) The 3 db roll off occurs faster

b) The 3 db roll off occurs later

c) The input pole shifts towards origin

d) The input pole becomes infinite

View Answer

Answer: a

Explanation: If the load resistance increases by a factor of 2, the output pole decreases since it’s inversely proportional to the load resistance. Hence the C.E. stage experiences a faster roll off due to the pole.

4. During high frequency applications of a B.J.T., which of the following three stages do not get affected by Miller’s approximation?

a) C.E.

b) C.B.

c) C.C.

d) Follower

View Answer

Answer: b

Explanation: During the C.B. stage, the capacitance between the base and the collector doesn’t suffer from Miller approximation since the input is applied to the emitter of the B.J.T. There are no capacitors connected between two nodes having a constant gain. Hence the C.B. stage doesn’t get affected by miller approximation.

5. Ignoring early effect, if C1 is the total capacitance tied to the emitter, what is the input pole of a simple C.B. stage?

a) 1/gm * C1

b) 2/gm * C1

c) gm * C1

d) gm * 2C1

View Answer

Answer: a

Explanation: The resistance looking into the emitter of the B.J.T. is 1/gm. The capacitance connected to the input node is C1 (as mentioned). The inverse product of these two provides us the input pole of the C.B. stage.

6. Ignoring early effect, if R1 is the total resistance connected to the collector; what is the output pole of a simple C.B. stage?

a) 1/[R1 * (Ccs + Cµ)]

b) 1/[R1* (Ccs + 2*Cµ)]

c) 1/[R1 * (2*Ccs + Cµ)]

d) 1/[R1 * 2*(Ccs + Cµ)]

View Answer

Answer: a

Explanation: The output pole is calculated, approximately, by the inverse product of the total resistance and the capacitance connected at the output node. We find that the total resistance connected to the output node is R1 while the total capacitance is Ccs + Cµ. In absence of early effect, 1/[R1 * (Ccs + Cµ)] becomes the output pole.

7. If early effect is included, and R1 is the total resistance connected at the collector. What is the output pole of a simple C.B. stage?

a) 1/[(R1 || ro) * 2(Ccs + Cµ)]

b) 1/[(R1 || ro) * (Ccs + Cµ)]

c) 1/[(R1 || ro) * (2*Ccs + Cµ)]

d) 1/[(R1 || ro) * 2*(Ccs + 2*Cµ)]

View Answer

Answer: b

Explanation: The output pole is calculated, approximately, by the inverse product of the total resistance and the capacitance connected at the output node. We find that the total resistance connected to the output node is R1 in parallel with ro, due to early effect, while the total capacitance is C2 ie Ccs + Cµ. Thus, the correct option is 1/[(R1 || ro) * (Ccs + Cµ)].

8. In a simple follower stage, C2 is a parasitic capacitance arising due to the depletion region between the collector and the substrate. What is the value of C2?

a) 0

b) Infinite

c) Ccs

d) 2*Ccs

View Answer

Answer: a

Explanation: During the high frequency response, the capacitor between the collector and the substrate gets shorted to A.C. ground at both of its terminals. Hence, C2=0. The answer would have been Ccs for any other stage of B.J.T.

9. For a cascode stage, with input applied to the C.B. stage, the input capacitance gets multiplied by a factor of ____

a) 0

b) 1

c) 3

d) 2

View Answer

Answer: d

Explanation: The small signal gain, of the C.B. stage, in a cascode stage is approximately equal to the ratio of the transconductances of the two B.J.T.’s. Since they are roughly same, the gain is 1. Miller multiplication leads to multiplying the capacitance, between base and collector, by a factor of (1 + small signal gain) which is 2. Hence, the correct option is 2.

10. If the B.J.T. is used as a follower, which capacitor experiences Miller multiplication?

a) Cπ

b) Cµ

c) Ccs

d) Cb

View Answer

Answer: a

Explanation: We find that the input is given to the base of the B.J.T. while the output is sensed at the collector of the B.J.T. We observe that the only capacitance connected between two nodes- where there is an amplification unit between the nodes, is Cπ. Hence, the correct option is Cπ.

11. If 1/h12 = 10 for a C.E. stage- what is the value of the base to collector capacitance, after Miller multiplication, at the output side?

a) 1.1Cµ

b) 1.2Cµ

c) 2.1Cµ

d) 2.2Cµ

View Answer

Answer: a

Explanation: At the output side of a C.E. stage, Cµ gets multiplied by a factor of (1+1/Av) where Av is the voltage gain. 1/h12 is nothing but Av. Hence, the value changes to 1.1Cµ.

12. If 1/h12 = 4, for a C.E. stage- what is the value of the base to collector capacitance, after Miller multiplication, at the input side?

a) 4Cµ

b) 5Cµ

c) 6Cµ

d) 1.1Cµ

View Answer

Answer: c

Explanation: The capacitor, Cµ, gets multiplied by a factor of (1 + Av), at the input side of a C.E. stage. 1/h12 is equal to Av since h12 is the reverse voltage amplification factor. Hence, the final value becomes 5Cµ.

13. The transconductance of a B.J.T.is 5mS (gm) while a 2KΩ (Rl) load resistance is connected to the C.E. stage. Neglecting Early effect, what is the Miller multiplication factor for the input side?

a) 21

b) 11

c) 20

d) 0

View Answer

Answer: b

Explanation: The Miller multiplication factor for the input side of a C.E. stage is (1+Av). Now, Av is the small signal low frequency gain of the C.E. stage which is gm*RL=10. Hence, the Miller multiplication factor is 11.

Analog Circuits Questions and Answers – Effect of Various Capacitors on Frequency Response – 1

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Effect of Various Capacitors on Frequency Response – 1”.

1. During high frequency applications of a B.J.T., which parasitic capacitors arise between the base and the emitter?

a) Cje and Cb

b) Ccs

c) Cb

d) Ccs and Cb

View Answer

Answer: a

Explanation: There are two capacitors which arise between bases and emitter. One is Cje due to depletion region associated between base and emitter. Cb is another capacitor which arises due to the accumulation of electrons in the base which further results into the concentration gradient within the base of the transistor.

2. During high frequency applications of a B.J.T., which parasitic capacitors arise between the collector and the emitter?

a) No capacitor arises

b) Ccs

c) Cb

d) Ccs and Cb

View Answer

Answer: a

Explanation: The emitter and the collector are far away from each other when the B.J.T. is being constructed. Hence, we find that they don’t share a common junction where charges can accumulate. Thus, no such parasitic capacitors appear.

3. During high frequency applications of a B.J.T, which parasitic capacitors arise between the collector and the base?

a) Cje and Cb

b) Ccs

c) Cπ

d) Cµ

View Answer

Answer: d

Explanation: Only one capacitor up between the base and the collector. This is due to the depletion region present between the base and the collector region.

4. Which parasitic capacitors are present at the collector terminal of the B.J.T.?

a) Cje and Cb

b) Ccs and Cµ

c) Cb

d) Ccs and Cb

View Answer

Answer: b

Explanation: There are two capacitors attached to the collector terminal. The collector-base junction provides a depletion capacitance (Cµ) while the collector substrate junction provides a certain capacitance (Ccs).

5. Which parasitic capacitors do not affect the frequency response of the C.E. stage, of the B.J.T.?

a) Cje and Cb

b) Ccs and Cµ

c) Cb and Cµ

d) No parasitic capacitor gets deactivated

View Answer

Answer: d

Explanation: While observing the frequency response of a C.E. stage, we find that all the parasitic capacitances of the B.J.T. end up slowing the speed of the B.J.T. The frequency response of this stage is affected by all the parasitic capacitors.

6. Which parasitic capacitors don’t affect the frequency response of the C.B. stage of the B.J.T.?

a) None of the parasitic capacitances

b) All the parasitic capacitances

c) Some of the coupling capacitors

d) Ccs and Cb

View Answer

Answer: b

Explanation: All the parasitic capacitors of a B.J.T. affect the C.B. stage. None of the parasitic capacitors gets deactivated and they end up behaving as a pole during the frequency response of the C.B. stage.

7. Which parasitic capacitors don’t affect the frequency response of the C.C. stage of the B.J.T.?

a) Ccs

b) Ccs and Cb

c) Cb

d) Ccs and Cµ

View Answer

Answer: a

Explanation: In the follower stage, the load is present at the emitter. The parasitic capacitors present between the collector and the substrate i.e. Cµ gets deactivated. This is observed from the small signal analysis where both the terminals of this capacitor get shorted to A.C. ground.

8. If the transconductance of the B.J.T increases, the transit frequency ______

a) Increases

b) Decreases

c) Doesn’t get affected

d) Doubles

View Answer

Answer: a

Explanation: The transit frequency is directly proportional to the transconductance of the B.J.T. Hence, the correct option is increases. Since it hasn’t been mentioned that whether the transconductance has been doubled or not, we cannot conclude the option “doubles” as an answer.

9. If the total capacitance between the base and the emitter increases by a factor of 2, the transit frequency __________

a) reduces by 2

b) increases by 2

c) reduces by 4

d) increases by 4

View Answer

Answer: a

Explanation: The transit frequency is almost inversely proportional to the total capacitance between the base and the emitter of the B.J.T. Hence, the transit frequency will approximately reduce by 2 and the correct option becomes reduces by 2.

10. Which effect plays a critical role in producing changes in the frequency response of the B.J.T.?

a) Thevenin’s effect

b) Miller effect

c) Tellegen’s effect

d) Norton’s effect

View Answer

Answer: a

Explanation: The miller effect results in a change in the capacitance seen between the base and the collector. This is why, it affects the frequency response of the B.J.T. deeply by changing the poles and affecting the high frequency voltage gain stage.

11. If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the input side gets multiplied?

a) 1 + gmRl

b) 1 – gmRl

c) 1 + 2*gmRl

d) 1 – 2*gmRl

View Answer

Answer: a

Explanation: The low frequency gain of the C.E. stage is gmRl. By the application of miller effect, we find that the capacitor between the base and the collector, looking into the input of the C.E. stage, will be increased by a factor of 1 + gmRl.

12. If a C.E. stage has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side gets multiplied?

a) 1 + 1/gmRl

b) 1 – 1/gmRl

c) 1 + 2/gmRl

d) 1 – 2/gmRl

View Answer

Answer: a

Explanation: The low frequency response of the C.E. stage is gmRl. By the application of miller effect, we find that the capacitance between the base and the collector, looking from the output side, will be increased by a factor of 1 + 1/gmRl. Hence, the correct option is 1 + 1/ gmRl.

13. If a C.E. stage with early effect has a load Rl and transconductance gm, what is the factor by which the capacitance between the base and the collector at the output side, gets multiplied?

a) 1 + 2/gm*(Rl || ro)

b) 1 – 1/gm*(Rl || ro)

c) 1 + 1/gm*(Rl || ro)

d) 1 – 2/gm*(Rl || ro)

View Answer

Answer: c

Explanation: If the early effect is considered, the low frequency response of the C.E. stage becomes gm*(Rl || ro). Thereby, miller approximation shows that the capacitance between the base and the collector, looking from the output side, will be increased by a factor of 1 + 1/gm*(Rl || ro). Hence the correct option is 1 + 1/ gm*(Rl || ro).

14. For a high frequency response of a simple C.E. stage with a transconductance of gm, what is Cin?

a) Cµ(1 + gm*R2) – Cπ

b) Cµ(1 + gm*R2) + Cπ

c) Cµ(1 – 2*gm*R2) + Cπ

d) Cµ(1 + 2*gm*R2) – Cπ

View Answer

Answer: b

Explanation: The input capacitance is an equivalent of the base to emitter capacitance in parallel to the miller approximation of the base to collector capacitance. Due to miller approximation, the base to collector capacitance becomes Cµ(1+gm*R2) while the base to emitter capacitance is Cπ. Capacitors get added, when in parallel and thus Cµ(1+gm*R2) + Cπ is correct.

15. For a high frequency response of a simple C.E. stage with a transconductance of gm, what is Cout?

a) Ccs – Cµ*(2 + 1/gm*R2)

b) Ccs + Cµ*(1 + 2/gm*R2)

c) Ccs – Cµ*(1 + 1/gm*R2)

d) Ccs + Cµ*(1 + 1/gm*R2)

View Answer

Answer: d

Explanation: We have a capacitor from the collector to substrate, Ccs, which comes in parallel to the miller approximation of the capacitance from base to collector. The miller approximation defines the latter as Cµ*(1 + 1/gm*R2). Since capacitors gets added, when in parallel, the correct option is Ccs + Cµ*(1+ 1/gm*R2).

Analog Circuits Questions and Answers – General Frequency Consideration

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

1. The frequency f1 and f2 from the below picture are respectively called ___________

analog-circuits-questions-answers-general-frequency-consideration-q1

a) lower cut-off frequency and upper cut-off frequency

b) upper cut-off frequency and lower cut-off frequency

c) left frequency, right frequency

d) there is no specific name

View Answer

Answer: a

Explanation: The frequencies are called lower cut-off frequency and upper cut-off frequency. At these frequencies power of the signal becomes half of its original value, 3dB less than the maximum and they are also called 3dB cut-off frequencies.

2. Bandwidth of amplifier is __________

a) Difference between upper cut-off frequency and lower cut-off frequency

b) Sum of upper cut-off frequency and lower cut-off frequency

c) Average of upper cut-off frequency and lower cut-off frequency

d) Independent to cut off frequency

View Answer

Answer: a

Explanation: 3dB bandwidth of an amplifier is the difference between upper cut-off frequency and lower cut-off frequency. The unity gain bandwidth is the difference between frequencies where gain is 1.

3. At 3dB cut-off frequency the voltage gain will be __________

a) 100% of maximum gain

b) 70.7% of maximum gain

c) 80.7% of maximum gain

d) 47.5% of maximum gain

View Answer

Answer: b

Explanation: 3dB cut-off frequency is the frequency at which the power becomes half of its maximum value. That is the voltage gain becomes 0.707 times maximum voltage gain.

4. At 3dB cut-off frequencies power will be __________

a) Half of maximum value

b) Quarter of maximum value

c) 70.7% of maximum value

d) Same as maximum value

View Answer

Answer: a

Explanation: 3dB cut-off frequency is the frequency at which the power becomes half of its maximum value. That is the voltage gain becomes 0.707 times maximum voltage gain. The importance of 3dB points is that for a human ear, it will not notice the fall in power of the signal up to 50% of maximum power.

5. A voltage amplifier has a voltage gain of 100. What will be gain at 3dB cut-off frequencies?

a) 70.7

b) 80.7

c) 45.7

d) 50

View Answer

Answer: a

Explanation: 3dB cut-off frequency is the frequency at which the power becomes half of its maximum value. That is the voltage gain becomes 0.707 times maximum voltage gain. Therefore if voltage gain is 100 then gain at 3dB frequencies will be

100 x 0.707 = 70.7.

6. What is the roll-off rate of single order filter?

a) 20dB/decade

b) 5dB/octave

c) 40dB/decade

d) 10dB/octave

View Answer

Answer: a

Explanation: At lower and higher frequencies the gain is decreasing. This fall or decrease in gain is known as roll-off rate. It is commonly specified in dB/octave or dB/decade. For a single order filter the roll-off rate is 6dB/octave or 20dB/decade.

7. -6dB is equivalent to __________ power gain.

a) 0.5

b) 0.25

c) 0.75

d) 0.8

View Answer

Answer: b

Explanation: The relation between dB and power gain is

dB = 10 log(power gain)

That is power gain for -6dB is

10(-6 /10) = 0.251.

8. Voltage gain of 1,00,000 is equivalent to __________

a) 10dB

b) 1000dB

c) 100dB

d) 50dB

View Answer

Answer: c

Explanation: dB = 20 log(voltage gain)

Therefor 100000 voltage gain is equivalent to

20 log(1,00,000)dB = 100 dB.

9. If the output power from an audio amplifier is measured at 100W when the signal frequency is 1kHz, and 1W when the signal frequency is 10kHz. Calculate the dB change in power.

a) -10dB

b) -20dB

c) -30dB

d) 15dB

View Answer

Answer: b

Explanation: The initial power gain in dB = 10 log (output power)

= 10 log(100) = 20dB

The final power gain in dB = 10 log(output power)

= 10 log(1) = 0 dB

So change in power = final power – initial power

= 0-20 = -20dB.

10. If an electronic system produces a 48mV output voltage when a 12mV signal is applied, calculate the decibel value of the systems output voltage gain.

a) 12dB

b) 6dB

c) 20dB

d) 4dB

View Answer

Answer: a

Explanation: Gain of the system is = output voltage/input voltage

= 48/12 = 4

Gain in dB = 20 log(voltage gain)

= 20 log(4) = 12.04 dB

Analog Circuits Questions and Answers – Collector Feedback Configuration

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

1. The Collector feedback configuration is better than __________

a) Fixed Bias Configuration

b) Voltage divider configuration

c) C.E. configuration

d) C.B. configuration

View Answer

Answer: a

Explanation: The fixed bias circuit has been seen to offer low stability with respect to change in ICO. The Voltage divider bias provides the most stable biasing mechanism. Hence, the collector feedback configuration is better than the fixed bias configuration while C.E. and C.B. are not biasing stages.

2. The Collector feedback is done by connecting a resistor from the collector to the __________

a) Emitter

b) Base

c) Supply voltage

d) Bias voltage

View Answer

Answer: b

Explanation: The collector feedback configuration is done by connecting a resistor from the collector to the base voltage. This is done to stabilize the biasing voltage against thermal runaway.

3. The Collector feedback configuration helps to stabilize __________

a) Bias voltage

b) Collector voltage

c) Bias current

d) Collector current

View Answer

Answer: d

Explanation: The collector feedback configuration is used to stabilize the Collector current. The Collector current is seen to increase at a sincere rate which can harm the transistor by thermal runaway. Stabilizing this current is necessary during biasing a transistor for proper application.

4. The Collector feedback helps to evade __________

a) Inverse Active mode

b) Pinch Off

c) Thermal Runaway

d) Breakdown

View Answer

Answer: c

Explanation: The increase in the Collector current is primarily due to ICO. A sudden increase in ICO can increase the Collector current and this will increase the temperature of the device. This is called thermal runaway and duee to high current gain of the transistor, the transistor can get destroyed due to thermal runaway. The Collector feedback configuration helps to evade this phenomenon.

5. Due to the Collector feedback mechanism, the transistor remains always remains in the __________

a) Active mode

b) Saturation mode

c) Inverse Active

d) Cut-off

View Answer

Answer: a

Explanation: The collector feedback configuration helps to keep the transistor in the active region. This is done because the bias voltage can get changed if the input and the bias voltage is superposed. With the introduction of this feedback mechanism, the transistor always stays biased in the active region.

6. What are the effects on the output voltage if the Collector resistance increases in a Collector feedback configuration?

a) Not much effect

b) Bias voltage reduces

c) Bias voltage increases

d) Bias voltage doubles

View Answer

Answer: a

Explanation: The increase in collector resistance doesn’t have much impact on the output voltage since the bias voltage is kept stable by the feedback operation. After all, one of the important applications of feedback is increasing stability and the output voltage is kept stable by this configuration.

7. What kind of configuration is this?

analog-circuits-questions-answers-collector-feedback-configuration-q7

a) Collector feedback

b) Base Bias

c) Self Bias

d) No bias

View Answer

Answer: d

Explanation: The transistor isn’t biased since the voltage drop from the base to the collector is 0. The transistor action won’t get be manifested.

8. If the current gain of the transistor is β, what is the stability factor pertaining to IC and ICO?

analog-circuits-questions-answers-collector-feedback-configuration-q8

a) β+1/ {1 + β * R1/(R1 + β)}

b) β+1

c) β+1/ {1 + β * R2/(R1 + β)}

d) β+1/ {1 + (R1 + β)}

View Answer

Answer: a

Explanation: The stability factor is determined by calculating the change in IC with respect to the change in ICO. Hence, we can simply apply the method of K.V.L. and derive a relation between these two currents. After differentiation, we’ll get the stability factor as β+1/ {1 + β * R1/(R1 + β)}.

9. To keep the B.J.T. in the active region, what should be the relation between R1 and R2?

analog-circuits-questions-answers-collector-feedback-configuration-q8

a) R1 >> R2

b) R1 << R2

c) R1 = R2

d) No such relation

View Answer

Answer: b

Explanation: The feedback resistance should be much smaller than the collector resistance since we need to reduce the sensitivity of the collector current to the current gain. Typically, R2 should be lower than R1 by a factor of β.

10. From the base bias to the collector feedback configuration, the stability facto S reduces by a factor of __________

a) 1 + R1/(R1 + β)

b) 1 + β * R1/(R1 + β)

c) β * R1/(R1 + β)

d) R1/(R1 + β)

View Answer

Answer: b

Explanation: This can be simply observed from the stability factors of both cases. The correct factor is 1 + β * R1/(R1 + β).

11. What is the stability factor against VBE for the collector feedback configuration?

a) β / (Rc * (1 + β)

b) -β / (RB + Rc * (1 + β)

c) β / (RB + Rc * (1 + β)

d) -β / (Rc * (1 + β)

View Answer

Answer: c

Explanation: This is easily derived from the Collector feedback configuration by using the method of K.V.L. deriving a relation between the collector current and the base emitter voltage. It should be noted that the stability against VBE increases in comparison to the base bias comparison.

12. What is the condition of stability of the following circuit?

analog-circuits-questions-answers-collector-feedback-configuration-q12

a) Highly stable

b) Poorly stable

c) Marginally stable

d) Unstable

View Answer

Answer: b

Explanation: The Collector feedback configuration does make the biasing stable but here, the collector and the base resistances are same. Hence, we conclude that the circuit is poorly stable. The base resistance should be substantially lower than the collector resistance.

13. If β = 100, what should be the ratio of the collector to base resistance for achieving insensitivity to β?

analog-circuits-questions-answers-collector-feedback-configuration-q12

a) 110

b) 90

c) 20

d) 140

View Answer

Answer: d

Explanation: The Base resistance should be substantially lower than the Collector resistance by a factor of β. 110 is a good choice but provided 140 is present as an option, a better choice is a factor of 140.

14. Why are we worried about β during the Collector feedback configuration?

a) To maintain a stable q-point irrespective of β

b) To increase the gain

c) To decrease the output impedance

d) To maintain a stable input impedance

View Answer

Answer: a

Explanation: If we change the device, β changes. But we want to keep the q-point stable so that circuit if represented as a black box, would provide the same characteristics and not be highly dependent on the transistor. Hence, we want to make the circuit insensitive to β.

15. The stability factors change from npn to pnp transistor.

a) True

b) False

View Answer

Answer: b

Explanation: The stability factors are independent of whether the transistor is of npn or pnp type. It is only dependent on β and the impedance connected to the terminals of the transistor.uation of a hyperbola.

Analog Circuits Questions and Answers – Hybrid Equivalent Model

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Hybrid Equivalent Model”.

1. Which of the following statement is incorrect?

a) Output of CE amplifier is out of phase with respect to its input

b) CC amplifier is a voltage buffer

c) CB amplifier is a voltage buffer

d) CE amplifier is used as an audio (low frequency) amplifier

View Answer

Answer: c

Explanation: The output of the CE amplifier has a phase shift of 180o with respect to the input. The CC amplifier has AV≅1, thus it is a voltage buffer. However, the CB amplifier has a large voltage gain, and its current gain AI≅1, thus it is a current buffer. CE amplifier has an application has an audio amplifier.

2. Consider the following circuit. __________ provides DC isolation. _____________ prevents a decrease in voltage gain. _____________ is used to control the bandwidth.

analog-circuits-questions-answers-hybrid-equivalent-model-q2

a) C3, C1, C4

b) C4, C1, C2

c) C2, C3, C2

d) C4, C3, C2

View Answer

Answer: b

Explanation: Capacitor C3 and C4, are the blocking capacitor and coupling capacitor respectively, both providing DC isolation to biasing circuit. Capacitor C1 is the emitter bypass capacitor, to prevent decrease in voltage gain by avoiding negative feedback. Capacitor C2 is the shunt capacitor, used to control the bandwidth, wherein the bandwidth is inversely proportional to C2.

3. Given hfe = 60, hie=1000Ω, hoe = 20μ Ω–, hre = 2 * 10-4. Find the current gain of the BJT, correct up to two decimal points.

analog-circuits-questions-answers-hybrid-equivalent-model-q3

a) – 58.44

b) -59.21

c) – 60.10

d) – 60.00

View Answer

Answer: a

Explanation: Current gain, AI = – hf / (1 + hoRL’) where RL’ = 2kΩ||4kΩ

RL’ = 1.33kΩ.

Thus AI = – 60 / (1 + 0.0266) = -58.4453.

4. Consider the circuit. Given hfe = 50, hie = 1200Ω. Find voltage gain.

analog-circuits-questions-answers-hybrid-equivalent-model-q4

a) – 278

b) -277.9

c) – 300

d) – 280

View Answer

Answer: a

Explanation: Voltage gain = AV = -hfeRL’/hie

RL’ = 20k||10k = 6.67kΩ

AV = -50 * 6.67k/1.2k = -277.9 ≅ – 278.

5. Given that IB = 5mA and hfe = 55, find load current.

analog-circuits-questions-answers-hybrid-equivalent-model-q5

a) 28mA

b) 280mA

c) 2.5A

d) 2A

View Answer

Answer: b

Explanation: In given circuit, which is an emitter follower, current gain = 1 + hfe

IL = IB (1+hfe)

IL = 5mA(56) = 280 mA.

6. Consider the following circuit, where source current = 10mA, hfe = 50, hie = 1100Ω, then for the transistor circuit, find output resistance RO and input resistance RI.

analog-circuits-questions-answers-hybrid-equivalent-model-q6

a) RO = 0, RI = 21Ω

b) RO = ∞, RI = 0Ω

c) RO = ∞, RI = 21Ω

d) RO = 10, RI = 21Ω

View Answer

Answer: c

Explanation: Since hoe is not given, we can consider it to be small; i.e 1/hoe is neglected, open circuited. Hence output resistance RO = ∞.

Input resistance = hie/(1 + hfe) = 1100/51 ≅ 21Ω.

7. For the given circuit, input resistance RI = 20Ω, hfe = 50. Output resistance = ∞. Find the new values of input and output resistance, if a base resistance of 2kΩ is added to the circuit.

analog-circuits-questions-answers-hybrid-equivalent-model-q7

a) RI = 20Ω, RO = ∞

b) RI = 20Ω, RO = 2kΩ

c) RI = 59Ω, RO = ∞

d) RI = 59Ω, RO = 2kΩ

View Answer

Answer: c

Explanation: RI = 20k = hie/(1+hfe) = hie/51

hie =1020 Ω

Hence, after adding base resistance, RI’= (hie+RB)/(1+hfe) = (1020+2000) / 51 ≅ 59Ω

There is no change in output resistance or current gain due to an extra base resistance. RO’ = ∞.

8. Consider its input resistance to be R1. Now, the bypass capacitor is attached, so that the new input resistance is R2. Given that hie = 1000Ω and hfe = 50, find R1-R2.

analog-circuits-questions-answers-hybrid-equivalent-model-q8

a) 112.2Ω

b) 0Ω

c) 110Ω

d) 200Ω

View Answer

Answer: a

Explanation: For the circuit, CE amplifier without bypass capacitor, input resistance, R1=hie + (1+hfe)RE

R1 = 1000 + 51*2.2 = 1000 + 112.2 = 1112.2Ω

With a bypass capacitor attached, input resistance, R2 = hie = 1000Ω

Thus R1 – R2 = 112.2Ω.

9. Given that for a transistor, hie = 1100Ω, hfe = 50, hre = 2*10-4 and hoe = 2μΩ-1. Find CB h-parameters.

a) hfb = 1, hib = 22, hob = 3μΩ-1, hrb = -1.5×10-4

b) hfb = -0.98, hib = -21.56, hob = 0.03μΩ-1, hrb = 1.5×10-4

c) hfb = -0.98, hib = 21.56, hob = 0.03μΩ-1, hrb = -1.5×10-4

d) hfb =1, hib = -21.56, hob = 0.03μΩ-1, hrb = -2×10-4

View Answer

Answer: c

Explanation: hfb = -hfe/(1+hfe) = -50/51= -0.98

hib = hie/(1+hfe) = 21.56Ω

hob = hoe/(1+hfe) = 0.03 μΩ-1

hrb = (hiehoe/1+hfe) – hre = -1.5×10-4.

10. If source resistance in an amplifier circuit is zero, then voltage gain (output to input voltage ratio) and source voltage gain (output to source voltage ratio) are the same.

a) True

b) False

View Answer

Answer: a

Explanation: When a source resistance RS is present, the voltage gain with respect to source becomes

AVS = AVRI’/(RS+RI’), where AV is voltage gain with respect to transistor input. However when RS=0 then AVS = AV.

Analog Circuits Questions and Answers – Characteristics of Amplifier

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

1. The state amplifier has no input is not called ______________

a) Zero signal condition

b) Non-signal condition

c) Quiescent condition

d) Empty-signal condition

View Answer

Answer: d

Explanation: The state at which amplifier has zero input signal is called zero signal condition, Non-signal condition, quiescent condition. There is nothing named empty-signal condition.

2. Which of the following is not considered for quiescent operating point?

a) DC collector-emitter voltage

b) DC collector current

c) DC base current

d) DC input voltage

View Answer

Answer: d

Explanation: The quiescent point is the operating point of an amplifier where the DC condition of amplifier is constant. For that we have to make sure that DC collector-emitter voltage, DC collector current, DC base current are constant.

3. Which of the following resistor is not involving in biasing the circuit shown below?

analog-circuits-questions-answers-characteristics-amplifier-q2

a) R1

b) R2

c) RC

d) RL

View Answer

Answer: d

Explanation: R1, R2, RC are, used to bias the circuit while RL is used as a load resistor. R1, R2 are used as a voltage divider. RC is used to control collector current.

4. Which of the following statements is most correct to explain role of biasing circuit in the implementation of a transistor circuit?

a) It is used provide proper voltage to every component in the circuit

b) It is used to ensure maximum power is obtained out of the circuit

c) It is used to provide the quiescent collector current

d) It is used to provide proper and stable functional environment to all quiescent point parameters

View Answer

Answer: d

Explanation: The basic function of biasing is to maintain amplifier in quiescent condition. The amplifier will properly work only if the quiescent condition is stable.

5. What is the role of input capacitance in the transistor amplifying circuit?

a) To prevent input variation from reaching output

b) To prevent DC content in the input from reaching transistor

c) There isn’t any role for input capacitance

d) To increase input impedance

View Answer

Answer: b

Explanation: The input capacitance, as its name indicates is used to prevent DC offset voltages in the input. It also prevents the transistor bias voltage to be fed back to input generating circuit.

6. What is the role of emitter bypass capacitance in the transistor amplifying circuit?

a) To prevent damage of emitter resistance from variation in voltage

b) To prevent emitter from over voltage

c) To increase gain

d) To increase load to transistor circuit

View Answer

Answer: c

Explanation: When an emitter resistance is added to the amplifier circuit, in common emitter mode, voltage gain is reduced and input impedance increases. When we need to obtain higher gain, we add a capacitance in parallel to the emitter resistance, called emitter bypass capacitance, and voltage gain does not decrease.

7. Which of the following is actually not a function of emitter bypass capacitor?

a) Increase gain

b) Lower the impedance of emitter resistance

c) Provide a low reactance path

d) Help emitter resistance to withstand voltage variation

View Answer

Answer: d

Explanation: The emitter bypass capacitor is not meant for reducing loading effect of emitter resistance. It is to increase gain. It provides a low reactive path to the AC signal without changing the quiescent point.

8. What is the role of emitter resistance in the transistor amplifying circuit?

a) To prevent thermal runaway

b) To prevent increase in gain

c) To lower the output impedance

d) To increase gain

View Answer

Answer: a

Explanation: Thermal runaway is the increase in the collector current without an increase in input due to heating of semiconductor material which in turn reduce the resistance thus increases current. The emitter resistor decreases effective input voltage decrease when collector current increases and thus it reduces collector current itself.

9. Which of the following is not true regarding the output capacitor in the transistor biasing circuit?

a) To pass AC signal

b) To stop DC signal

c) To couple the amplifier to load or next amplifier

d) There is no importance for an output capacitance

View Answer

Answer: d

Explanation: The output capacitor or output coupling capacitor is provided to pass AC signal and to block DC signal. It also helps to couple the amplifier to load or next amplifier.

10. Which of the following is the best biasing method for transistor bias?

a) emitter bias

b) voltage divider bias

c) fixed bias

d) collector feedback bias

View Answer

Answer: b

Explanation: Voltage divider bias is more stable because the biased voltage will not change. It is best to use voltage divider bias for accuracy.

Analog Circuits Questions and Answers – Transistor Switching Network

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Transistor Switching Network”.

1. At saturation, which of these is not true for a BJT?

a) The collector current IC cannot increase further

b) The base current IB, cannot increase further

c) The collector-to-emitter voltage, VCE is due to the non-zero internal resistance of BJT

d) VCE(saturation) is the minimum voltage drop between C and E

View Answer

Answer: b

Explanation: At saturation, the collector-to-emitter voltage is the minimum drop possible occurring due to the non-zero internal resistance of the BJT. Since it cannot decrease further, the current IC cannot increase further. The BJT is said to be saturated. However, the base current, IB, can keep increasing with the input voltage and hence, in saturation, the relation IC = βIB is not satisfied.

2. For a transistor in saturation, which is true?

a) IC = βIB

b) IC > βIB

c) IC < βIB

d) IC = (β+1)IB

View Answer

Answer: c

Explanation: At saturation, collector current remains constant. However, the base current increases with the input voltage being applied and hence BJT cannot satisfy the relation IC = βIB. In the saturation region, βIB > IC is the correct relation.

3. Given that the BJT is completely saturated, what is the overdrive?

a) Overdrive = 1

b) Overdrive < 1

c) Overdrive > 1

d) Overdrive > 0

View Answer

Answer: c

Explanation: Overdrive during BJT saturation is the ratio of its normal β and its forced β. Forced β is the ratio of IC(sat) and IB when BJT is in saturation. Since in saturation IC is constant and IB increases thus, IC/IB decreases and the forced β is less than the normal β. Hence the overdrive > 1. In hard/strong saturation, β>>1.

4. Consider the graph of IC vs VI shown below for a transistor. Find the correct relation for region 3 in the diagram.

analog-circuits-questions-answers-transistor-switching-network-q4

a) IC = IC(sat) and VCE = VCE(sat)

b) IC = IC(sat) and VCE = VCC

c) IC = βIB and VCE = VCE(sat)

d) IC = βIB and VCE = VCC

View Answer

Answer: a

Explanation: Region 3 in the above is the saturation region in which IC remains constant with respect to the input voltage and the voltage VCE is the saturation voltage, almost zero. At this point, the transistor acts as an ON switch.

5. What is the ON resistance of a transistor?

a) RON = VCEsat/βIB

b) RON = VCEsat + VA/ICsat

c) RON = VCEsat/(β+1)IB

d) RON = VCEsat/ICsat

View Answer

Answer: d

Explanation: In the saturation region, we consider that the transistor acts as an ON switch. In this region, both collector-to-emitter voltage as well current are constant and do not change. The ON resistance is the ratio of this saturation voltage to saturation current.

6. For the graph which depicts collector current, find the ON time.

analog-circuits-questions-answers-transistor-switching-network-q6

t1 = 1ms

t2 = 2ms

t3 = 4ms

t4 = 6ms

t5 = 16ms

t6 = 18ms

a) 3ms

b) 1ms

c) 2ms

d) 5ms

View Answer

Answer: a

Explanation: On time is the time taken by BJT to change from the OFF state to the ON state. It is the sum of the delay time and the rise time. Rise time is the time taken by current to increase from 10% to 90% of saturation and delay time is time taken by current to increase from 0 to 10% of the saturation.

ton = td + tr = t2 – t1 + t3 – t2 = 2 – 1 + 4 – 2 = 3ms.

7. Find the storage time for the current variation shown below.

analog-circuits-questions-answers-transistor-switching-network-q6

t1 = 2ms

t2 = 3ms

t3 = 4ms

t4 = 6ms

t5 = 19ms

t6 = 20ms

a) 1ms

b) 13ms

c) 3ms

d) 2ms

View Answer

Answer: b

Explanation: Storage time is the time taken by IC to decrease from ICsat to 90% of ICsat.

TS = t5 – t4 = 19 – 6 = 13ms.

8. Which of these relations is true always for the BJT as a switch?

a) Off time >> On time

b) Off time = Storage time – Rise time

c) Off time << On time

d) Off time = Storage time + Delay time

View Answer

Answer: a

Explanation: Off time for a BJT is larger than its ON time. Off time=Storage time + Fall time.

Often Storage time is larger than fall/delay/rise time and hence OFF time is quite large than ON time.

9. How is BJT used as a faster switch?

a) By operating it in the saturation and cut-off region

b) By operating it in the active and cut-off region

c) By using it in strong saturation

d) By decreasing its ON resistance

View Answer

Answer: b

Explanation: If BJT is to act as a switch with negligible power dissipation, then BJT is operated in the cut-off and saturation region, as in the TTL family. When BJT has to be operated as a fast switch, then it is operated in the active and cut-off region, as in the ECL family.

Analog Circuits Questions and Answers – Transistor Bias Configuration

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Transistor Bias Configuration”.

1. BJT is biased to _________

a) Work as a switch

b) Prevent thermal runaway

c) Increase DC collector current

d) Operate it in the saturation region

View Answer

Answer: b

Explanation: A BJT is biased to operate in the active region, to work as an amplifier. It is not biased in the cut-off or saturation region to work as a switch. Also, biasing is done to maintain a stable collector current so that the operating point does not change. This also prevents thermal runaway.

2. For the given circuit, β = 150.

analog-circuits-questions-answers-transistor-bias-configuration-q2

Fig. 1 has stability factor S1

If the above circuit is changed to below circuit

analog-circuits-questions-answers-transistor-bias-configuration-q2a

Fig. 2 having stability factor S2

Choose the correct option which is having better stability with S1 and S2 Values.

a) S2 = 151, S1 = 150. Circuit 1 has better stability

b) S2 = 100, S1 = 10 Circuit 2 has better stability

c) S2 = 151, S1 = 10.3 Circuit 1 has better stability

d) S2 = 151, S1 = 10.3 Circuit 2 has better stability

View Answer

Answer: c

Explanation: The above circuits are a collector to base bias circuits, where stability factor

S=1+β1+βRCRC+RB

Thus, in the question, S1 =1+1501+150∗10110=15114.63 = 10.32

However, in circuit 2, which is a transformer coupled amplifier, RC = 0

Thus, S2 = 1+150 = 151.

3. Consider the biasing circuit shown. The β for the circuit is large. R3 = 1kΩ, R4 = 2kΩ. The stability factor varies between 10 and 11. Find the maximum and minimum values of R2.

analog-circuits-questions-answers-transistor-bias-configuration-q3

a) Minimum = 16.36kΩ, Maximum = 20kΩ

b) Minimum = 16.36kΩ, Maximum = 18kΩ

c) Minimum = 10 kΩ, Maximum = 20kΩ

d) Minimum = 6 kΩ, Maximum = 10kΩ

View Answer

Answer: a

Explanation: Circuit is a self bias circuit.

Base resistance = RB = R1*R2 / (R1+R2)

Since β is large, stability factor, S = 1 + RB/RE = 1 + RB/R3

1 + RB/R3 = 10

RB/R3 = 9 => RB = 9k => R2 = 16.36 kΩ

For S = 11

RB/R3=10 => RB = 10k => R2 = 20kΩ.

4. Choose the incorrect option according to self bias circuit?

a) Voltage gain increases

b) Stability factor is independent of collector resistance

c) BJT can be used in either of the three configurations

d) Excellent stability in collector current is achieved

View Answer

Answer: d

Explanation: In a self bias circuit, due to emitter resistance a negative feedback exists. This decreases voltage gain. Also, stability factor S does not depend on collector resistance, only on base and emitter resistance and β, if required. S=(1+β)(RB+RE)RB+(1+β)RE.

S is least in self bias circuit, hence excellent stability is achieved.

5. In the circuit given, the two Si transistors are similar. Given β=50, Vcc=12V, I1=5mA. Find I?

analog-circuits-questions-answers-transistor-bias-configuration-q5

a) 5 mA

b) 4.807 mA

c) 4.5 mA

d) 5.2 mA

View Answer

Answer: b

Explanation: The circuit is a current mirror circuit. Both transistors are similar. I1= IREF = 5 mA

Β is not large so 2/β is not negligible.

Thus current I = I1/(1 + 2/β) = 5mA/(1 + 2/50) = 4.807 mA.

6. Given Vout = 5V, β=100, I1=10mA, R1=100KΩ. Find the output resistance.

analog-circuits-questions-answers-transistor-bias-configuration-q5

analog-circuits-questions-answers-transistor-bias-configuration-q6

a) 8 kΩ

b) 8.163 kΩ

c) 7.582 kΩ

d) 8.4 kΩ

View Answer

Answer: b

Explanation: The circuit is a current mirror, whose output resistance, ROUT = (VA + VCE)/IOUT

Here, IOUT = I1 / (1+ 2/β) = 10mA / (1 + 2/100) = 9.8 mA

VCE = VOUT = 5V

VA = Early voltage = 75 V (obtained from the graph)

ROUT = (VA + VCE)/ IOUT = (75 + 5) / 9.8 * 10-3

ROUT = 80,000/9.8 = 8.163 kΩ.

7. Why is self bias circuit not used in IC amplifier?

a) To reduce power losses

b) To reduce area used on the chip

c) Stability factor reduces in the IC

d) Voltage gain is reduced

View Answer

Answer: b

Explanation: Self biased circuits are not preferred in IC amplifiers because they need large resistances R1 and R2, since then S will be smaller and stability will be more. However, using large resistances in ICs means a requirement of larger chip area, so to reduce this area requirement, we use current mirror circuits instead.

8. Considering all transistors to be similar and β is very large, when I1 = 10 mA, find current I2.

analog-circuits-questions-answers-transistor-bias-configuration-q8

a) 10 mA

b) 50 mA

c) 25 mA

d) 20 mA

View Answer

Answer: c

Explanation: Let current through the transistor Q1/Q2 be IC. Since both are similar, we can say that,

IC = I1/2

Similarly, current through transistors Q3 to Q7 is assumed to be IC’, where IC’ = I2/5

Since all transistors are similar. IC = IC’

I2 = 5IC’ = 5 IC = 2.5*I1.

9. Consider the following circuit, where the transistors are similar Si transistors. Given I1 = 2mA, I2 = 1μA, Vcc = 12 V, find R1 and R2.

analog-circuits-questions-answers-transistor-bias-configuration-q9

a) R1 = 10kΩ, R2 = 6kΩ

b) R1 = 5.6kΩ, R2 = 20kΩ

c) R1 =18kΩ, R2 = any value

d) R1 = 18kΩ, R2 = 5.6kΩ

View Answer

Answer: d

Explanation: R2 = (Vcc – VBE)/I1 = (12 – 0.7)/2m = 5.65 kΩ – Using KVL

Also, I2*R1 = VT ln(I1/I2) = 0.026 * ln(2)

Thus, R1 = 0.026*ln(2) / 1μ = 18 kΩ.

10. Widlar current source was introduced to obtain a smaller output current.

a) True

b) False

View Answer

Answer: a

Explanation: In a current mirror circuit, to obtain lower values of output current, the resistance values required to increase a lot, which becomes difficult to manufacture on an IC. Instead, we use a widlar current source, where an emitter resistance is also present, affecting the output current.

Analog Circuits Questions and Answers – Operating Point of Transistor

 This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Operating Point of Transistor”.

1. Reverse saturation current of a common emitter transistor is __________

a) Collector current when emitter is open circuited and base-collector junction is reverse biased

b) Emitter current when collector is open circuited and base-collector junction is reverse biased

c) Base current when emitter circuit is open circuited and emitter-collector junction is reverse biased

d) Collector current when base circuit is open circuited

View Answer

Answer: a

Explanation: Reverse saturation current is the collector current when emitter is open circuited and base-collector junction is reverse bias mode. In this mode of operation collector-base junction act as a reverse biased diode. The current in this reverse biased junction is known as reverse saturation current.

2. Reverse collector saturation current ICBO is __________

a) Collector current when emitter current is zero

b) Collector current when base current is zero

c) Same as reverse saturation current

d) Collector current when either emitter or base current is zero

View Answer

Answer: A

Explanation: Reverse collector current ICBO is collector-base current when emitter is open. This is same as reverse saturation current in ideal but have slight difference in practical.

3. Reverse collector saturation is greater than Reverse saturation current because of reverse collector saturation __________

a) Is ideal

b) Doesn’t include leakage current

c) They are same

d) Include avalanche multiplication current which is caused by the collision in collector junction

View Answer

Answer: d

Explanation: One of the reasons why reverse collector current exceeds the reverse saturation current is the introduction of avalanche multiplication current in the base collector junction. This happens when high energy electron collides in the lattice it creates more number of electron and thus a greater current. Another major reason is the presence of surface leakage currents flowing in the reverse collector saturation.

4. Reverse collector saturation is greater than Reverse saturation current because reverse collector saturation __________

a) Is ideal

b) Consist leakage current flowing through junction and surface

c) Doesn’t include avalanche multiplication current opposing collector current

d) They are same

View Answer

Answer: b

Explanation: One of the reasons why reverse collector current exceeds the reverse saturation current is the introduction of avalanche multiplication current in the base collector junction. This happens when high energy electron collides in the lattice it creates more number of electron and thus a greater current. Another major reason is the presence of surface leakage currents flowing in the reverse collector saturation.

5. Which of the following statement about a common base transistor is true?

a) Very low input impedance

b) Very low output Impedance

c) Current gain is greater than unity

d) Voltage gain is very low

View Answer

Answer: a

Explanation: Common base transistor has very low input resistance (20Ω). It also has very high output resistance. Its current gain is less than unity and it has a medium voltage gain.

6. Which of the following statement about a common emitter transistor is true?

a) Very high input resistance

b) High output resistance

c) Current gain is less than unity

d) Voltage gain is very low

View Answer

Answer: b

Explanation: Common emitter transistor has high output resistance (about 40k). It has low input resistance (about 1k). Current gain is high (20 to a few hundreds). Voltage gain is medium.

7. Which of the following statement about a common collector transistor is true?

a) Very low input impedance

b) Very high output impedance

c) Unity current gain

d) Unity voltage gain

View Answer

Answer: d

Explanation: Common collector configuration has high input impedance and low output impedance. The current gain is high but voltage gain is low, almost equal to unity.

8. Which of the following configuration is used as input stage of multistage amplifier?

a) Common base configuration

b) Common emitter configuration

c) Common collector configuration

d) All configurations are equally suited

View Answer

Answer: a

Explanation: Since input resistance is low and output resistance is high common base configuration is used as an input stage of multistage amplifier. Common emitter configuration is used for audio signal amplification. Common collector is used for impedance matching.

9. Which of the following configuration is used for audio signal amplification?

a) Common base configuration

b) Common emitter configuration

c) Common collector configuration

d) All configurations are equally suited

View Answer

Answer: b

Explanation: Common base configuration is used as input stage of multistage amplifier since it has low input resistance and high output resistance. Since voltage gain is high, common emitter configuration is used for audio signal amplification. Common collector is used for impedance matching since the voltage gain is unity.

10. Which of the following configuration is used for impedance matching?

a) Common base configuration

b) Common emitter configuration

c) Common collector configuration

d) All configurations are equally suited

View Answer

Answer: c

Explanation: Common base configuration is used as input stage of multistage amplifier since it has low input resistance and high output resistance. Since voltage gain is high, common emitter configuration is used for audio signal amplification. Common collector is used for impedance matching since the voltage gain is unity.