Electronic Circuits I Set 4 MCQs

Q1 | How many h-parameters of a transistor are dimensionless?

four
two
three
one

Answer: two

Explaination:
(i) h11 = v1/i1; for v2 = 0 (output short circuited)

Q2 | The values of h-parameters of a transistor in CE arrangement are arrangement.

same as for cb
same as for cc
different from that in cb
similar to no

Answer: different from that in cb

Explaination:
the values of h-parameter in ce arrangement:

Q3 | If the load resistance of a C.E. stage increases by a factor of 2, what happens to the high frequency response?

the 3 db roll off occurs faster
the 3 db roll off occurs later
the input pole shifts towards origin
the input pole becomes infinite

Answer: the 3 db roll off occurs faster

Explaination:
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.

Q4 | During high frequency applications of a B.J.T., which of the following three stages do not get affected by Miller’s approximation?

c.e.
c.b.
c.c.
follower

Answer: c.b.

Explaination:
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.

Q5 | Ignoring early effect, if C1 is the total capacitance tied to the emitter, what is the input pole of a simple C.B. stage?

1/gm * c1
2/gm * c1
gm * c1
gm * 2c1

Answer: 1/gm * c1

Explaination:
the resistance looking into the

Q6 | 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?

0
infinite
ccs
2*ccs

Answer: 0

Explaination:
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.

Q7 | For a cascode stage, with input applied to the C.B. stage, the input capacitance gets multiplied by a factor of

0
1
3
2

Answer: 2

Explaination:
the small signal gain, of the

Q8 | If the B.J.T. is used as a follower, which capacitor experiences Miller multiplication?

cπ
cµ
ccs
cb

Answer: cπ

Explaination:
we find that the input is given to the base of the b.j.t. while the output is

Q9 | 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?

4cµ
5cµ
6cµ d) 1.1cµ

Answer: 6cµ d) 1.1cµ

Explaination:
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.

Q10 | 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?

21
11
20

Answer: 11

Explaination:
the miller multiplication factor for the input side of a c.e. stage is (1+av).

Q11 | For what type of signals does a transistor behaves as linear device?

small signals only
large signals only
both large and small signal
no signal

Answer: small signals only

Explaination:
the small variation in the total voltage and current due to an application of signal moves the point up and down just by a bit and that whole up and down dynamics of the operating point from its dc value point

Q12 | Which of these are incorrect about Darlington amplifier?

it has a high input resistance
the output resistance is low
it has a unity voltage gain
it is a current buffer

Answer: it is a current buffer

Explaination:
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.

Q13 | In a Darlington pair, the overall β=15000.β1=100. Calculate the collector current for Q2 given base current for Q1 is 20 μA.

300 ma
298 ma
2 ma
200ma

Answer: 298 ma

Explaination:
ib = 20 μa

Q14 | What is the need for bootstrap biasing?

to prevent a decrease in the gain of network
to prevent an increase in the input resistance due to the biasing network
to prevent a decrease in the input resistance due to the presence of multiple bjt amplifiers
to prevent a decrease in the input resistance due to the biasing network

Answer: to prevent an increase in the input resistance due to the biasing network

Explaination:
a bootstrap biasing network is a special biasing circuit used in 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.

Q15 | 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.

141 kΩ
15 mΩ
20 mΩ
200 kΩ

Answer: 141 kΩ

Explaination:
r’ = 220k

Q16 | What is a cascode amplifier?

a cascade of two ce amplifiers
a cascade of two cb amplifiers
a cascade of ce and cb amplifiers
a cascade of cb and cc amplifiers

Answer: a cascade of ce and cb amplifiers

Explaination:
a cascode amplifier is a cascade network of ce and cb amplifiers, or cs and cg amplifiers.

Q17 | 1, α2 = 1.5 what is the transconductance of the entire network?

80 mΩ-1
75 mΩ-1
33 mΩ-1
55 mΩ-1

Answer: 55 mΩ-1

Explaination:
the above circuit is a cascode pair. for this circuit, the overall transconductance is

Q18 | In the given circuit, hfe = 50 and hie = 1000Ω, find overall input and output resistance.

ri=956Ω, ro=1.6 kΩ
ri=956 kΩ, ro=2 kΩ
ri=956 Ω, ro=2 kΩ
ri=900Ω, ro=10 kΩ

Answer: ri=956 Ω, ro=2 kΩ

Explaination:
ro = rc = 2kΩ

Q19 | A Differential Amplifier should have collector resistor’s value (RC1 & RC2) as

5kΩ, 5kΩ
5Ω, 10kΩ
5Ω, 5kΩ
5kΩ, 10kΩ

Answer: 5kΩ, 5kΩ

Explaination:
the values of collector current will be equal in differential amplifier (rc1=rc2).

Q20 | A Differential Amplifier amplifies

input signal with higher voltage
input voltage with smaller voltage
sum of the input voltage
none of the mentioned

Answer: none of the mentioned

Explaination:
the purpose of differential amplifier is to amplify the difference between two signals.

Q21 | If output is measured between two collectors of transistors, then the Differential amplifier with two input signal is said to be configured as

dual input balanced output
dual input unbalanced output
single input balanced output
dual input unbalanced output

Answer: dual input balanced output

Explaination:
when two input signals are applied to base of transistor, it is said to be dual input. when both collectors are at same dc potential with respect to ground, then it is said to be balance output.

Q22 | A differential amplifier is capable of amplifying

dc input signal only
ac input signal only
ac & dc input signal
none of the mentioned

Answer: ac & dc input signal

Explaination:
direct connection between stages removes the lower cut off frequency imposed by coupling capacitor; therefore it can amplify both ac and dc signal.

Q23 | In ideal Differential Amplifier, if same signal is given to both inputs, then output will be

same as input
double the input
not equal to zero
zero

Answer: zero

Explaination:
in ideal amplifier, output voltage

Q24 | Find IC, given VCE=0.77v, VCC=10v, VBE=0.37v and RC=2.4kΩ in Dual Input Balanced Output differential amplifier

0.4ma
0.4a
4ma
b) d) 4a

Answer: 4ma

Explaination:
substitute the values in collector to emitter voltage equation, vce= vcc+ vbe-rc ic

Q25 | Obtain the collector voltage, for collector resistor (RC) =5.6kΩ, IE=1.664mA and VCC=10v for single input unbalanced output differential amplifier

0.987v
0.682v c) 0.555v
d) none of the mentioned

Answer: 0.682v c) 0.555v

Explaination:
substitute the given values in collector voltage equation,