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The correct choice is (a) COLLECTOR current when emitter current is zero

For 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.

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

The best I can explain: 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.

Correct option is (d) Include avalanche multiplication CURRENT which is caused by the collision in collector junction

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.

Right CHOICE is (a) Very low input impedance

For 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.

Correct choice is (b) Consist leakage current flowing through junction and surface

For 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.

The correct answer is (a) COMMON base configuration

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

Right OPTION is (d) Unity voltage gain

The 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.

Correct CHOICE is (b) High output RESISTANCE

The best I can explain: COMMON emitter transistor has high output resistance (about 40K). It has LOW input resistance (about 1k). Current gain is high (20 to a few hundred). Voltage gain is medium.

The correct answer is (b) Common EMITTER configuration

Best 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.

Right option is (c) Common collector configuration

The best 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.

Right OPTION is (a) Minimum = 16.36kΩ, Maximum = 20kΩ

For explanation I would say: 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.

The CORRECT choice is (b) Prevent THERMAL runaway

Easiest 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.

Correct choice is (b) 8.163 kΩ

To explain I would say: 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.

Correct choice is (B) 4.807 mA

Best 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.

Right answer is (d) EXCELLENT stability in COLLECTOR current is achieved

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. \(S = \frac{(1+β)(R_B +R_E)}{R_B + (1+β)R_E}\).

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

Right option is (B) To reduce area used on the chip

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.

The CORRECT answer is (d) R1 = 18kΩ, R2 = 5.6kΩ

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.

Correct choice is (C) 25 mA

For explanation I would say: 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.

The correct option is (b) The base current IB, cannot increase further

For explanation I would say: 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.

The correct CHOICE is (c) IC < βIB

Easiest 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.

Right ANSWER is (c) Overdrive > 1

The best I can explain: 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.

Right choice is (a) IC = IC(sat) and VCE = VCE(sat)

For explanation I would say: 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.

Correct OPTION is (a) Off time >> On time

The 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.

The CORRECT choice is (d) DC INPUT voltage

Best 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.

Right choice is (b) By operating it in the active and cut-off region

The EXPLANATION is: 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.

Correct option is (d) RL

To explain: 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.

The correct answer is (d) Empty-signal CONDITION

To explain: 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.

Correct option is (d) It is used to provide proper and stable functional environment to all quiescent POINT parameters

For 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.

The correct OPTION is (b) To prevent DC content in the input from reaching transistor

Best 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.

The correct CHOICE is (c) To increase gain

The 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.

Correct OPTION is (d) Help emitter RESISTANCE to withstand VOLTAGE variation

Best 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.

The correct answer is (a) To prevent THERMAL runaway

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.

Right OPTION is (d) There is no IMPORTANCE for an output capacitance

Easy 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.

The correct CHOICE is (B) voltage divider BIAS

For explanation I would say: Voltage divider bias is more stable because the biased voltage will not CHANGE. It is best to use voltage divider bias for accuracy.

Right ANSWER is (c) CB amplifier is a voltage buffer

The best I can explain: The OUTPUT of the CE amplifier has a phase shift of 180^o 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 that has an audio amplifier.

The correct choice is (b) C4, C1, C2

Easiest 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.

Correct OPTION is (a) – 58.44

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

RL’ = 1.33kΩ.

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

Correct choice is (c) RO = ∞, RI = 21Ω

The EXPLANATION is: 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.

Correct option is (a) – 278

The best I can EXPLAIN: Voltage GAIN = AV = -hfeRL’/hie

RL’ = 20k||10k = 6.67kΩ

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

The CORRECT choice is (b) 280mA

To elaborate: In GIVEN circuit, which is an emitter FOLLOWER, current gain = 1 + hfe

IL = IB (1+hfe)

IL = 5mA(56) = 280 mA.

Right option is (a) 112.2Ω

Easiest 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.

The correct answer is (c) RI = 59Ω, RO = ∞

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’ = ∞.

Right OPTION is (c) hfb = -0.98, hib = 21.56, hob = 0.03μΩ^-1, HRB = -1.5×10^-4

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.

Correct option is (a) True

Easy 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.

The correct choice is (a) Fixed BIAS Configuration

Easy 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.

Correct option is (C) Thermal Runaway

Best 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 due to the high current gain of the TRANSISTOR, the transistor can get destroyed due to thermal runaway. The Collector FEEDBACK configuration helps to evade this phenomenon.

The CORRECT OPTION is (d) Collector CURRENT

The explanation is: 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.

Correct choice is (a) Active mode

For 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.