210+ Electronic Circuits 1 Solved MCQs

1.

Which of the following condition is true for cut-off mode?

A. the collector current is zero
B. the collector current is proportional to the base current
C. the base current is non zero
D. all of the mentioned
Answer» A. the collector current is zero
Explanation: the base current as well as the collector current are zero in cut-off mode.
2.

Which of the following is true for the cut- off region in an npn transistor?

A. potential difference between the emitter and the base is smaller than 0.5v
B. potential difference between the emitter and the base is smaller than 0.4v
C. the collector current increases with the increase in the base current
D. the collector current is always zero and the base current is always non zero
Answer» B. potential difference between the emitter and the base is smaller than 0.4v
Explanation: both collector and emitter current are zero in cut-off region.
3.

Which of the following is true for a typical active region of an npn transistor?

A. the potential difference between the emitter and the collector is less than 0.5 v
B. the potential difference between the emitter and the collector is less than 0.4 v
C. the potential difference between the emitter and the collector is less than 0.3 v
D. the potential difference between the emitter and the collector is less than 0.2 v
Answer» C. the potential difference between the emitter and the collector is less than 0.3 v
Explanation: most commonly used transistors have vce less than 0.4 v for the active region.
4.

Which of the following is true for the active region of an npn transistor?

A. the collector current is directly proportional to the base current
B. the potential difference between the emitter and the collector is less than 0.4 v
C. all of the mentioned
D. none of the mentioned
Answer» C. all of the mentioned
Explanation: the base current and the collector current are directly proportional to each other and the potential difference between the collector and the base is always less than 0.4 v.
5.

Which of the following is true for a npn transistor in the saturation region?

A. the potential difference between the collector and the base is approximately 0.2v
B. the potential difference between the collector and the base is approximately 0.3v
C. the potential difference between the collector and the base is approximately 0.4v
D. the potential difference between the collector and the base is approximately 0.5v
Answer» D. the potential difference between the collector and the base is approximately 0.5v
Explanation: the commonly used npn transistors have a potential difference of around 0.5v between he collector and the base.
6.

The potential difference between the base and the collector Vcb in a pnp transistor in saturation region is                  

A. -0.2 v
B. -0.5v
C. 0.2 v
D. 0.5 v
Answer» B. -0.5v
Explanation: the value of vcb is -0.5v for a pnp transistor and 0.5v for an npn transistor.
7.

For a pnp transistor in the active region the value of Vce (potential difference between the collector and the base) is

A. less than 0.3v
B. less than 3v
C. greater than 0.3v
D. greater than 3v
Answer» A. less than 0.3v
Explanation: for a pnp transistor vce is less
8.

Which of the following is true for a pnp transistor in active region?

A. cb junction is reversed bias and the eb junction is forward bias
B. cb junction is forward bias and the eb junction is forward bias
C. cb junction is forward bias and the eb junction is reverse bias
D. cb junction is reversed bias and the eb junction is reverse bias
Answer» A. cb junction is reversed bias and the eb junction is forward bias
Explanation: whether the transistor in npn or pnp, for it be in active region the eb junction must be reversed bias the cb junction must be forward bias.
9.

Which of the following is true for a pnp transistor in saturation region?

A. cb junction is reversed bias and the eb junction is forward bias
B. cb junction is forward bias and the eb junction is forward bias
C. cb junction is forward bias and the eb junction is reverse bias
D. cb junction is reversed bias and the eb junction is reverse bias
Answer» B. cb junction is forward bias and the eb junction is forward bias
Explanation: whether the transistor in npn or pnp, for it be in saturation region the eb junction must be forward bias the cb junction must be forward bias.
10.

For the circuit shown, find the quiescent point.

A. none of the below
B. (4v, 10ma)
C. (10v, 3ma)
D. (3ma, 10v)
Answer» C. (10v, 3ma)
Explanation: we know, ie=vee/re=30/10kΩ=3ma ic=α ie =ie =3ma
11.

What is the DC characteristic used to prove that the transistor is indeed biased in saturation mode?

A. ic = βib
B. ic > βib
C. ic >> βib
D. ic < βib
Answer» D. ic < βib
Explanation: when in a transistor is driven into saturation, we use vce(sat) as another linear parameter. in, addition when a transistor is biased in saturation mode, we have ic < βib. this characteristic used to prove that the transistor is indeed biased in saturation mode.
12.

The feature of an approximate model of a transistor is

A. it helps in quicker analysis
B. it provides individual analysis for different configurations
C. it helps in dc analysis
D. ac analysis is not possible
Answer» A. it helps in quicker analysis
Explanation: the small signal model helps in quicker ac analysis of a transistor. the approximate model is applicable for all the configurations. the dc analysis is not obtained by using a small signal model of transistor.
13.

005mmhos, hre=0. Find the output impedance if the lad resistance is 5kΩ.

A. 5kΩ
B. 4kΩ
C. 20kΩ
D. 15kΩ
Answer» B. 4kΩ
Explanation: ro=i/hoe=1/0.005m
14.

A transistor has hie =2kΩ, hoe=25µmhos and hfe=60 with an unbypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance?

A. 90kΩ and 50kΩ respectively
B. 33kΩ and 45kΩ respectively
C. 6kΩ and 40kΩ respectively
D. 63kΩ and 40kΩ respectively
Answer» D. 63kΩ and 40kΩ respectively
Explanation: as the emitter is unbypassed, the input resistance ri=hie+(1+hfe)re
15.

A transistor has hie =1KΩ and hfe=60 with an bypassed emitter resistor Re=1kΩ. What will be the input resistance and output resistance?

A. 90kΩ and 50kΩ respectively
B. 33kΩ and 45kΩ respectively
C. 6kΩ and 40kΩ respectively
D. 63kΩ and 40kΩ respectively
Answer» D. 63kΩ and 40kΩ respectively
Explanation: as the emitter is bypassed, the input resistance ri=hie
16.

In the given circuit, find the equivalent resistance between A and B nodes.

A. 100kΩ
B. 50kΩ
C. 40kΩ
D. 60kΩ
Answer» B. 50kΩ
Explanation: rab=ro
17.

Which of the following acts as a buffer?

A. cc amplifier
B. ce amplifier
C. cb amplifier
D. cascaded amplifier
Answer» A. cc amplifier
Explanation: the voltage gain of a common collector amplifier is unity. it is then used as a buffer. the cc amplifier is also called as an emitter follower. though there is no amplification done, the output will be stabilised.
18.

Which of the following is true?

A. cc amplifier has a large current gain
B. ce amplifier has a large current gain
C. cb amplifier has low voltage gain
D. cc amplifier has low current gain
Answer» B. ce amplifier has a large current gain
Explanation: the ce amplifier has high current and voltage gains. the cc amplifier has unity voltage gain which cannot be regarded as high. the common base amplifier has a unity current gain and high voltage gain.
19.

In CB configuration, the value of α=0.98A. A voltage drop of 4.9V is obtained across the resistor of 5KΩ when connected in collector circuit. Find the base current.

A. 0.01ma
B. 0.07ma
C. 0.02ma
D. 0.05ma
Answer» C. 0.02ma
Explanation: here, ic=4.9/5k=0.98ma α = ic/ie .so,
20.

Which of the following is the correct relationship between base and emitter current of a BJT?

A. ib = β ie
B. ib = ie
C. ib = (β + 1) ie
D. ie = (β + 1) ib
Answer» D. ie = (β + 1) ib
Explanation: for a bjt, the collector current
21.

For best operation of a BJT, which region must the operating point be set at?

A. active region
B. cutoff region
C. saturation region
D. reverse active region
Answer» A. active region
Explanation: operating point for a bjt must always be set in the active region to ensure proper functioning. setting up of q-point in any other region may lead to reduced functionality.
22.

From the given circuit, using a silicon transistor, what is the value of IBQ?

A. 47.08 ma
B. 47.08 ua
C. 50 ua
D. 0 ma
Answer» B. 47.08 ua
Explanation: consider the bjt to be in saturation. then ic=12-0.2/2.2k=5.36 ma and ib=12-0.8/240k=0.047 ma ibmin=icsat/β=5.09/50=0.1072ma which is
23.

From the given circuit, using a silicon BJT, what is the value of VCEQ?

A. 7 v
B. 0.7 v
C. 6.83 v
D. 7.17 v
Answer» C. 6.83 v
Explanation: consider the bjt to be in saturation. then ic=12-0.2/2.2k=5.36 ma and ib=12-0.8/240k=0.047 ma ibmin=icsat/β=5.09/50=0.1072ma which is
24.

From the given circuit, using a silicon BJT, what is the value of VBC?

A. 6.13 v
B. -6.13 v
C. 7 v
D. -7 v
Answer» B. -6.13 v
Explanation: consider the bjt to be in saturation. then ic=12-0.2/2.2k=5.36 ma and ib=12-0.8/240k=0.047 ma ibmin=icsat/β=5.09/50=0.1072ma which is
25.

From the given circuit, using silicon BJT, what is the value of the saturation collector current?

A. 5 ma
B. 5.36 ma
C. 5.45 ma
D. 10.9 ma
Answer» B. 5.36 ma
Explanation: to obtain an approximate answer, under saturation the bjt is on and hence acts like a short circuit. however, ideally a drop exists for the transistor which is a fixed value. for an exact answer, if the bjt is a silicon transistor, then drop vce = 0.2v and current is 12-0.2/2.2=5.36 ma.
26.

In the given circuit, what is the value of IC if the BJT is made of Silicon?

A. 2.01 ma
B. 2.01 ua
C. 10.05 ma
D. 10.05 ua
Answer» A. 2.01 ma
Explanation: consider the bjt to be in saturation. then ic=20-0.2/2k=9.9 ma and ib=20-0.8/430k=0.044 ma
27.

In the given circuit, using a silicon BJT, what is the value of VCE?

A. 20 v
B. 15.52 v
C. 14.98 v
D. 13.97 v
Answer» B. 15.52 v
Explanation: consider the bjt to be in saturation. then ic=20-0.2/2k=9.9 ma and ib=20-0.8/430k=0.044 ma
28.

In the given circuit, what is the value of VE when using a silicon BJT?

A. 2.01 v
B. 0.28 v
C. 0 v
D. 2.28 v
Answer» D. 2.28 v
Explanation: consider the bjt to be in saturation. then ic=20-0.2/2k=9.9 ma and ib=20-0.8/430k=0.044 ma
29.

In the given circuit using a silicon BJT, what is the value of saturation collector current?

A. 10 ma
B. 8.77 ma
C. 6.67 ma
D. 5 ma
Answer» C. 6.67 ma
Explanation: to obtain an approximate answer, under saturation the bjt is on and hence acts like a short circuit. however, ideally a drop exists for the transistor which is a fixed value. for an exact answer, if the bjt is a silicon transistor, then drop vce = 0.2v and current is 20-0.2/2.2=9.9 ma.
30.

What is Stability factor?

A. ratio of change in collector current to change in a current amplification factor
B. ratio of change in collector current to change in base current
C. current amplification factor
D. ratio of base current to collector current
Answer» A. ratio of change in collector current to change in a current amplification factor
Explanation: stability factor is defined as the rate at which collector current changes when base to emitter voltage changes, keeping base
31.

The base current for a BJT remains constant at 5mA, the collector current changes from 0.2mA to 0.3 mA and beta was changed from 100 to 110, then calculate the value of S.

A. 0.01m
B. 1m
C. 100m
D. 25m
Answer» A. 0.01m
Explanation: since the current in the above case, remains constant, therefore stability factor is 0.01 as it is defined as the ratio of change in collector current to change in beta. s=change in collector current/change in beta=0.1ma/10=0.01m.
32.

For a n-p-n transistor, the collector current changed from 0.2mA to 0.22mA resulting a change of base emitter voltage from 0.8v to 0.8005V. What is the value of Stability factor?

A. 0
B. 0.25
C. 0.04
D. 0.333
Answer» C. 0.04
Explanation: change in vbe = 0.0005v change in collector current = 0.02ma
33.

There are two transistors A and B having ‘S’ as 25 and 250 respectively, on comparing the value of S, we can say B is more stable than A.

A. true
B. false
Answer» B. false
Explanation: more the value of s, lesser the stability, since a has lesser s value the
34.

For a fixed bias circuit having Ic = 0.3mA and In=0.0003mA, S is

A. 100
B. 0
C. 11
D. 111
Answer» C. 11
Explanation: for fixed bias s=1+beta beta=ic/ib=10
35.

For a fixed bias circuit having RC=2Kohm and VCC=60V, IB=0.25mA and S=101, find Vce.

A. 12v
B. 10v
C. 5v
D. 2.5v
Answer» B. 10v
Explanation: s = 1 + beta,
36.

For an ideal transistor having a fixed bias configuration, what will be the value of Beta?

A. 0
B. 2
C. -1
D. 1
Answer» C. -1
Explanation: s = 1 + beta s = 0
37.

The temperature changes do not affect the Stability.

A. true
B. false
Answer» B. false
Explanation: the temperature changes the value of beta which in turn changes the stability of the transition. the temperature changes affect the mobility of the charge carries which results in a change of the current parameters affecting stability.
38.

Comparing fixed and collector to base bias which of the following statement is true?

A. fixed bias is more stable
B. collector to base bias is more stable
C. both are the same in terms of stability
D. depends on the design
Answer» B. collector to base bias is more stable
Explanation: for fixed bias circuit, s = 1+beta, more the beta, lesser the stability for collector to base bias s = (1+beta)/(1+beta(rc/rc+rb))
39.

The compensation techniques are used to                   

A. increase stability
B. increase the voltage gain
C. improve negative feedback
D. decrease voltage gain
Answer» B. increase the voltage gain
Explanation: usually, the negative feedback is used to produce a stable operating point.
40.

Compensation techniques refer to the use of                   

A. diodes
B. capacitors
C. resistors
D. transformers
Answer» A. diodes
Explanation: compensation techniques refer to the use of temperature sensitive devices such as thermistors, diodes, transistors, sensistors etc to compensate variation in currents. sometimes for excellent bias and thermal stabilization, both stabilization and compensation techniques are used.
41.

In a silicon transistor, which of the following change significantly to the change in IC?

A. vce
B. ib
C. vbe
D. b) ie
Answer» C. vbe
Explanation: for germanium transistor, changes in ico with temperature contribute more serious problem than for silicon transistor. on the other hand, in a silicon transistor, the changes of vbe with
42.

What is the compensation element used for variation in VBE and ICO?

A. diodes
B. capacitors
C. resistors
D. transformers
Answer» A. diodes
Explanation: a diode is used as the compensation element used variation in vbe and ico. the diode used is of the same material and type as that of transistor. hence, the voltage across the diode has same temperature coefficient as vbe of the transistor.
43.

The expression for IC in the compensation for instability due to ICO variation                   

A. βi+βio+βico
B. βi+βio
C. βio+βico
D. βi+βico
Answer» A. βi+βio+βico
Explanation: in this method, diode is used for the compensation in variation of ico. the diode used is of the same material and type as that of transistor. hence, the reverse saturation current io of the diode will increase with temperature at the same rate as the transistor collector saturation current ico.
44.

Which of the following has a negative temperature coefficient of resistance?

A. sensistor
B. diode
C. thermistor
D. capacitor
Answer» C. thermistor
Explanation: the thermistor has a negative temperature coefficient of resistance. it
45.

Which of the following has a negative temperature coefficient of resistance?

A. capacitor
B. diode
C. thermistor
D. sensistor
Answer» D. sensistor
Explanation: the sensistor has a positive temperature coefficient of resistance. it is a temperature sensitive resistor. it is a heavily doped semiconductor. when voltage is decreased, the net forward emitter voltage decreases. as a result the collector current decreases.
46.

Increase in collector emitter voltage from 5V to 8V causes increase in collector current from 5mA to 5.3mA. Determine the dynamic output resistance.

A. 20kΩ
B. 10kΩ
C. 50kΩ
D. 60kΩ
Answer» B. 10kΩ
Explanation: ro=∆vce/∆ic
47.

The output resistance of CB transistor is given by                    

A. ∆vcb/∆ic
B. ∆vbe/∆ib
C. ∆vbe/∆ic
D. ∆veb/∆ie
Answer» A. ∆vcb/∆ic
Explanation: the ratio of change in collector base voltage (∆vcb) to resulting change in collector current (∆ic) at constant emitter current (ie) is defined as output resistance.
48.

The negative sign in the formula of amplification factor indicates                   

A. that ie flows into transistor while ic flows out it
B. that ic flows into transistor while ie flows out it
C. that ib flows into transistor while ic flows out it
D. that ic flows into transistor while ib flows out it
Answer» A. that ie flows into transistor while ic flows out it
Explanation: when no signal is applied, the ratio of collector current to emitter current is called dc alpha, αdc of a transistor. αdc=- ic/ie. it is the measure of the quality of a transistor. higher is the value of α, better is the transistor in the sense that collector current approaches the emitter current.
49.

In the given situation for n-channel JFET, we get drain-to-source current is 5mA. What is the current when VGS = – 6V?

A. 5 ma
B. 0.5a
C. 0.125 a
D. 0.5a
Answer» C. 0.125 a
Explanation: ids = idss(1-vgs/vp)2 when vgs = 0, idss = ids = 5ma when vgs = -6v, ids = 5ma(1 + 4)2 ids = 5 x 25 = 125 ma.
50.

7 BIASING BJT SWITCHING CIRCUITSJFET - DC LOAD LINE AND BIAS POINT, VARIOUS BIASING METHODS OF JFET - JFET BIAS CIRCUIT DESIGN

A. rd < 6kΩ
B. rd > 6kΩ
C. rd > 4kΩ
D. rd < 4kΩ
Answer» A. rd < 6kΩ
Explanation: in given circuit, vgs = -5v vds = vdd – idsrd
51.

Consider the circuit shown. VDS=3 V. If IDS=2mA, find VDD to bias circuit.

A. -30v
B. 30v
C. 33v
D. any value of voltage less than 12 v
Answer» C. 33v
Explanation: vds = vdd – ids(10k + 5k) 3 = vdd – 2(15)
52.

To bias a e-MOSFET                        

A. we can use either gate bias or a voltage divider bias circuit
B. we can use either gate bias or a self bias circuit
C. we can use either self bias or a voltage divider bias circuit
D. we can use any type of bias circuit
Answer» A. we can use either gate bias or a voltage divider bias circuit
Explanation: to bias an e-mosfet, we cannot use a self bias circuit because the gate to source voltage for such a circuit is zero.
53.

Consider the following circuit. IDSS = 2mA, VDD = 30V. Find R, given that VP = – 2V.

A. 10kΩ
B. 4kΩ
C. 2kΩ
D. 5kΩ
Answer» B. 4kΩ
Explanation: idss = 2ma
54.

Process transconductance parameter is 40μA/V2. Find drain to source current in saturation.

A. 0.10 ma
B. 0.05ma
C. – 0.05ma
D. – 50a
Answer» C. – 0.05ma
Explanation: isd = k’w(vsg –
55.

Consider the following circuit. Given that VDD = 15V, VP = 2V, and IDS = 3mA, to bias the circuit properly, select the proper statement.

A. rd < 6kΩ
B. rd > 6kΩ
C. rd > 4kΩ
D. rd < 4kΩ
Answer» A. rd < 6kΩ
Explanation: in given circuit, vgs = -5v vds = vdd – idsrd
56.

Consider the following circuit. Process transconductance parameter = 0.50 mA/V2, W/L=1, Threshold voltage = 3V, VDD = 20V. Find the operating point of circuit.

A. 20v, 25ma
B. 13v, 22ma
C. 12.72v, 23.61ma
D. 20v, 23.61ma
Answer» C. 12.72v, 23.61ma
Explanation: ids = [k’w/l(vgs – vt)2]/2 vgs = 20 x 35 / 55 = 12.72 v
57.

Which of the following relation is true about gate current?

A. ig=id+is
B. id=ig
C. is= ig
D. ig=0
Answer» D. ig=0
Explanation: the fet physical structure which contains silicon dioxide provides infinite resistance. hence no current will flow through the gate terminal.
58.

For a fixed bias circuit the drain current was 1mA, what is the value of source current?

A. 0ma
B. 1ma
C. 2ma
D. 3ma
Answer» C. 2ma
Explanation: we know that for an fet same current flows through the gate and source terminal, hence source current=1ma.
59.

For a fixed bias circuit the drain current was 1mA, VDD=12V, determine drain resistance required if VDS=10V?

A. 1kΩ
B. 1.5kΩ
C. 2kΩ
D. 4kΩ
Answer» C. 2kΩ
Explanation: vds=vdd-id rd
60.

Which of the following equation brings the relation between gate to source voltage and drain current in Self Bias?

A. vgs=vdd
B. vgs=-id rs
C. vgs=0
D. vgs=1+id rs
Answer» B. vgs=-id rs
Explanation: vrs=id rs
61.

For a self-bias circuit, find drain to source voltage if VDD=12V, ID=1mA, Rs=RD=1KΩ?

A. 1v
B. 2v
C. 10v
D. 5v
Answer» C. 10v
Explanation: vds=vdd-id (rd+rs)
62.

Find the gate voltage for voltage divider having R1=R2=1KΩ and VDD=5V?

A. 1v
B. 5v
C. 3v
D. 2.5v
Answer» D. 2.5v
Explanation: vg = r2×vdd/r1+r2
63.

Find the gate to source voltage for voltage divider having R1=R2=2KΩ and VDD=12V, ID=1mA and RS=4KΩ?

A. 3v
B. 2v
C. 0v
D. 1v
Answer» B. 2v
Explanation: vg = r2×vdd/r1+r2
64.

What will happen if values of Rs increase?

A. vgs increases
B. vgs decreases
C. vgs remains the same
D. vgs=0
Answer» B. vgs decreases
Explanation: increasing values of rs result in lower quiescent values of id and more negative values of vgs.
65.

What is the current flowing through the R1 resistor for voltage divider (R1=R2=1KΩ, VDD=10V)?

A. 5ma
B. 3ma
C. 1ma
D. 2ma
Answer» A. 5ma
Explanation: ir1=ir2 =vdd/r1+r2
66.

The h-parameters analysis gives correct results for                      

A. large signals only
B. small signals only
C. both large and small
D. not large nor small signals
Answer» B. small signals only
Explanation: every linear circuit is associated with h –parameters. when this linear circuit is terminated with load rl, we can find input impedance, current gain, voltage gain, etc in terms of h-parameters.
67.

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

A. small signals only
B. large signals only
C. both large and small signal
D. no signal
Answer» A. small signals only
Explanation: 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 can be approximated to be along a straight line. whole analysis can be done with same assumption of linearity with the limit of signal being in the same vicinity of the dc operating point.
68.

How many h-parameters are there for a transistor?

A. two
B. three
C. four
D. five
Answer» C. four
Explanation: a transistor has four h- parameters –
69.

The dimensions of hie parameters are

A. mho
B. ohm
C. farad
D. ampere
Answer» B. ohm
Explanation: hie = vbe/ib; common emitter input impedance
70.

The hfe parameter is called                in CE arrangement with output short circuited.

A. voltage gain
B. current gain
C. input impedance
D. output impedance
Answer» B. current gain
Explanation: hfe in ce arrangement is given as
71.

What happens to the h parameters of a transistor when the operating point of the transistor changes?

A. it also changes
B. does not change
C. may or may not change
D. nothing happens
Answer» A. it also changes
Explanation: it is very difficult to get exact values of h parameters for a particular transistor. it is because these parameters are subject to considerable variation unit to unit variation, variation due to change in temperature and variation due to the operating point.
72.

If temperature changes, h parameters of a transistor            

A. also change
B. does not change
C. remains same
D. may or may not change
Answer» A. also change
Explanation: it is very difficult to get exact values of h parameters for a particular transistor. it is because these parameters are subject to considerable variation unit to unit variation, variation due to change in temperature and variation due to the operating point.
73.

In CE arrangement, the value of input impedance is approximately equal to            

A. hie
B. hib
C. hoe
D. hre
Answer» A. hie
Explanation: hie = vbe/ib; common emitter input impedance
74.

How many h-parameters of a transistor are dimensionless?

A. four
B. two
C. three
D. one
Answer» B. two
Explanation: (i) h11 = v1/i1; for v2 = 0 (output short circuited)
75.

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

A. same as for cb
B. same as for cc
C. different from that in cb
D. similar to no
Answer» C. different from that in cb
Explanation: the values of h-parameter in ce arrangement:
76.

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
Answer» A. the 3 db roll off occurs faster
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.
77.

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
Answer» B. c.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.
78.

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
Answer» A. 1/gm * c1
Explanation: the resistance looking into the
79.

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
Answer» A. 0
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.
80.

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
Answer» D. 2
Explanation: the small signal gain, of the
81.

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

A.
B.
C. ccs
D. cb
Answer» A. cπ
Explanation: we find that the input is given to the base of the b.j.t. while the output is
82.

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
Answer» C. 6cµ
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.
83.

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
Answer» B. 11
Explanation: the miller multiplication factor for the input side of a c.e. stage is (1+av).
84.

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. it is a current buffer
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.
85.

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

A. 300 ma
B. 298 ma
C. 2 ma
D. 200ma
Answer» B. 298 ma
Explanation: ib = 20 μa
86.

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. to prevent an increase in the input resistance due to the biasing network
Explanation: 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.
87.

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. 141 kΩ
Explanation: r’ = 220k
88.

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. a cascade of ce and cb amplifiers
Explanation: a cascode amplifier is a cascade network of ce and cb amplifiers, or cs and cg amplifiers.
89.

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. 55 mΩ-1
Explanation: the above circuit is a cascode pair. for this circuit, the overall transconductance is
90.

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

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. ri=956 Ω, ro=2 kΩ
Explanation: ro = rc = 2kΩ
91.

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

A. 5kΩ, 5kΩ
B. 5Ω, 10kΩ
C. 5Ω, 5kΩ
D. 5kΩ, 10kΩ
Answer» A. 5kΩ, 5kΩ
Explanation: the values of collector current will be equal in differential amplifier (rc1=rc2).
92.

A Differential Amplifier amplifies

A. input signal with higher voltage
B. input voltage with smaller voltage
C. sum of the input voltage
D. none of the mentioned
Answer» D. none of the mentioned
Explanation: the purpose of differential amplifier is to amplify the difference between two signals.
93.

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

A. dual input balanced output
B. dual input unbalanced output
C. single input balanced output
D. dual input unbalanced output
Answer» A. dual input balanced output
Explanation: 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.
94.

A differential amplifier is capable of amplifying

A. dc input signal only
B. ac input signal only
C. ac & dc input signal
D. none of the mentioned
Answer» C. ac & dc input signal
Explanation: direct connection between stages removes the lower cut off frequency imposed by coupling capacitor; therefore it can amplify both ac and dc signal.
95.

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

A. same as input
B. double the input
C. not equal to zero
D. zero
Answer» D. zero
Explanation: in ideal amplifier, output voltage
96.

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

A. 0.4ma
B. 0.4a
C. 4ma
D. b) d) 4a
Answer» C. 4ma
Explanation: substitute the values in collector to emitter voltage equation, vce= vcc+ vbe-rc ic
97.

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

A. 0.987v
B. 0.682v
C. 0.555v
D. none of the mentioned
Answer» B. 0.682v
Explanation: substitute the given values in collector voltage equation,
98.

In a Single Input Balanced Output Differential amplifier, given VCC=15v, RE = 3.9kΩ, VCE=2.4 v and re=250Ω. Determine Voltage gain

A. 26
B. 56
C. 38
D. 61
Answer» A. 26
99.

For the difference amplifier which of the following is true?

A. it responds to the difference between the two signals and rejects the signal that are common to both the signal
B. it responds to the signal that are common to the two inputs only
C. it has a low value of input resistance
D. the efficacy of the amplifier is measured by the degree of its differential signal to the preference of the common mode signal
Answer» A. it responds to the difference between the two signals and rejects the signal that are common to both the signal
Explanation: all the statements are not true except for the fact that it responds only when there is difference between two signals only.
100.

The problem with the single operational difference amplifier is its

A. high input resistance
B. low input resistance
C. low output resistance
D. none of the mentioned
Answer» B. low input resistance
Explanation: due to low input resistance a large part of the signal is lost to the source’s internal resistance.
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