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The correct answer is (a) Cje and Cb

For EXPLANATION: There are two CAPACITORS that 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.

The correct choice is (a) No capacitor arises

The BEST I can explain: 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.

Right answer is (B) Ccs and Cµ

To explain: 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).

Correct choice is (a) Ccs

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.

The correct answer is (a) Increases

The EXPLANATION is: 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.

Correct CHOICE is (a) reduces by 2

Best 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.

The CORRECT OPTION is (a) Thevenin’s effect

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.

The correct choice is (a) 1 + gmRl

Best 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.

Right OPTION is (a) 1 + 1/gmRl

The best I can EXPLAIN: 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.

The correct choice is (c) 1 + 1/gm*(Rl || RO)

The 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).

Correct option is (a) The 3 db roll off occurs faster

Easy 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.

Correct choice is (b) Cµ(1 + GM*R2) + Cπ

For explanation I would say: 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.

Correct answer is (C) 1 / [R1 * (Cµ(1+gm*R2) + Cπ)]

To EXPLAIN I would say: 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)].

Right answer is (a) 1/[R1 * (Ccs + Cµ)]

The 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.

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

The best 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))].

Correct answer is (b) C.B.

Easy 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.

The correct OPTION is (b) 1/[(R1 || RO) * (Ccs + Cµ)]

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)].

Right choice is (a) 0

The explanation is: 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.

Right choice is (a) 1.1Cµ

The BEST 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.

The correct answer is (a) Cπ

Easiest 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.

Correct option is (C) 6Cµ

To explain I would say: 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.

The correct OPTION is (a) C1

Easiest explanation: Given R1>R2

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

Thus K^2>1, K>1, K<-1 (correct)

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

HENCE C1>C2.

The correct answer is (b) 11

For explanation I would say: 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.

Correct OPTION is (d) -27.68

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.

The correct choice is (b) Internal VOLTAGE GAIN

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.

The correct OPTION is (c) 15

The best I can explain: AV = -120

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

AV’ = \(\frac{|AV|}{\SQRT{1}} + (\frac{f_L}{f})^2\)

AV’ = 120/8.02 ≈ 15.

Correct choice is (a) 7.8 x 10^6 rad/s

For EXPLANATION I would SAY: The FREQUENCY in radians is CALCULATED by

 ωβ = 1/C.rbe

ωβ = 7.8 x 10^6.

Right answer is (B) 55.9 Hz

Easiest explanation: AVM = 150

AVL = 100

f = 50Hz

100 = \(\frac{150}{\SQRT{1}}+(\frac{f_L}{50})^2\)

1+f^2/2500 = 1.5^2

f^2 = 2500*1.25 = 3125

f = 55.90 Hz.

Correct ANSWER is (b) 35

Best 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.10^7.103.10^-12.1000)

AI = IL/IB = 50m.1k/(1+j.10^7.103.10^-12.1000)

AI = 35 (approx).

The correct option is (b) Junction CAPACITANCES are not included in it

The explanation: The effect of smaller CAPACITORS is considerable in high frequency analysis of ANALOG circuits, and hence they cannot be ignored. Instead of h-parameter model, we USE π-model.

The CORRECT choice is (c) 141.42

To elaborate: The CURRENT gain for the CE CIRCUIT is A = \(\frac{β}{\sqrt{1+(\frac{f}{f_β})^2}}\)

At f = fβ, A = \(\frac{β}{\sqrt{2}}\)

Hence A = 141.42.

Correct answer is (d) β- cut-off frequency is one where the CE short CIRCUIT CURRENT gain becomes β/2

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

Correct answer is (b) Voltage GAIN increases

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.

Correct answer is (B) All the parasitic capacitances

Best 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.