Electronic Devices and Circuits OHM Institute BJT (PYQs)

UNIT BJT
3 PREVIOUS GATE QUESTIONS WITH SOLUTIONS

6. If the transistor in the figure is in saturation, then,

ONE MARK QUESTIONS
(GATE-01)
1. The break down voltage of a transistor with its base
C
open is BVCEO and that with emitter open is BVCBO,
then (GATE-95)
B dc denotes the
(a) BVCEO = BVCBO
dc current gain
(b) BVCEO > BVCBO IB
(c) BVCEO < BVCBO
(d) BVCEO is not related to BVCBO E
(a) IC is always equal to dcIB
2. A BJT is said to be operating in the saturation region
if (GATE-95) (b) IC is always equal to -dcIB
(a) Both junctions are reverse biased (c) IC is greater than or equal to dcI dc
(b) Base-emitter junction is reverse biased and base (d) IC is less than dc I B
collector junction is forward biased
(c) Base-emitter junction is forward biased and 7. For a BJT, the common-base current gain  = 0.98
base-collector junction reverse biased and the collector base junction reverse bias saturation
(d) Both the junctions are forward biased current IC0 = 0.6A. This BJT is connected in the
common emitter mode and operated in the active
3. The Ebers-Moll model is applicable to (GATE-95)
region with a base drive current IB = 20A. The
(a) Bipolar junction transistors
collector current IC for this mode of operation is
(b) NMOS transistors
(GATE-11)
(c) Unipolar junction transistors
(a) 0.98 mA
(d) Junction field – effect transistors
(b) 0.99 mA
4. The Early-Effect in a bipolar junction transistor is (c) 1.0 mA
caused by (GATE-95) (d) 1.01 mA
(a) Fast – turn – on.
8. An increase in the base recombination of a BJT will
(b) Fast – turn – off.
increase (GATE-14) (Set2)
(c) Large collector-base reverse bias.
(a) the common emitter dc current gain 
(d) Large emitter-base forward bias.
(b) the breakdown voltage BVCE0
5. In a bipolar transistor at room temperature, if the (c) the unity-gain cut-off frequency fT
emitter current is doubled the voltage across its base- (d) the trans conductance gm
emitter junction (GATE-96)
(a) doubles 9. In the circuit shown in the figure, the BJT has a
(b) halves current gain () of 50. For an emitter-base voltage VEB
(c) increase by about 20 mV = 600 mV, the emitter-collector voltage VEC (in Volts)
(d) Decrease by about 20 mV is _____. (GATE-15) (Set-3)

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Electronic Devices and Circuits OHM Institute BJT (PYQs)
3V (a) the effective base width increases and common-
emitter current gain increases
(b) the effective base width increases and common
emitter current gain decreases
(c) the effective base width decreases and common-

600k
60
500  emitter current gain increases
(d) the effective base width decreases and common-
emitter current gain decreases

10. If the base width in a bipolar junction transistor is 14. Consider the circuit shown in figure. Assume base to
doubled, which one of the following statements will emitter voltage VBE = 0.8V and common base current
be TRUE? (GATE-15) (Set-3) gain () of transistor is unity.
(a) Current gain will increase.
(b) Unity gain frequency will increase. +18V
(c) Emitter-base junction capacitance will increase.
(d) Early voltage will increase. 44 k 4 k

11. The Ebers - Moll model of a BJT is valid

(GATE-16) (Set-2)
(a) only in active mode 16 k 2 k
(b) only in active and saturation modes
(c) only in active and cut-off modes
(d) in active, saturation and cut-off modes
The value of the collector to emitter voltage VCE (in
12. For a narrow base PNP BJT, the excess minority volts) is _____.
carrier concentrations (nE for emitter, pB base, nC (GATE-17) (Set 2)
for collector) normalized to equilibrium minority
carrier concentrations (nE0 for emitter, pB0 for base, nC0 15. Which one of the following options describes
for collector) in the quasi-neutral emitter, base and correctly the equilibrium band diagram at T = 300 K
collector regions are shown below. Which one of the of a Silicon pnn+ p++ configuration shown in the
following biasing modes is the transistor operating in? figure? (GATE-19)
(GATE-17) (Set 1)
p n n p 
EC
Carrier Concentration
Normalized Excess

p B 105
pB 0 nC
nC 0 EV
(a) EF
nE
1
nE 0
Emitter (p) Base (N) Collector (P)
X and Y axes are not to scale
(a) Forward active (b) Saturation
(c) Inverse active (d) Cutoff EC
EF
13. An npn bipolar junction transistor (BJT) is operating
(b)
in the active region. If the reverse bias across the base-
collector junction is increased. Then,
(GATE-17) (Set 2) EV

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Electronic Devices and Circuits OHM Institute BJT (PYQs)

EC TWO MARK QUESTIONS

1. Which of the following effects can be caused by a rise

EF
(c) in the temperature? (GATE-90)
EV
(a) increase in MOSFET current (IDS)
(b) increase in BJT current (IC)
(c) decrease in MOSFET current (IDS)
(d) decrease in BJT current (IC)
EC
2. In a transistor having finite , the forward bias across
the base emitter junction is kept constant and the
(d) EF
reverse bias across the collector base junction is
increased. Neglecting the leakage across the collector
EV
base junction and the depletion region generations
current, the base current will ___________ (increase/
decrease/remains constant)
16. The correct circuit representation of the structure (GATE-92)
shown in the figure is (GATE-19)
B E C 3. Match the following. (GATE-94)
List-I:
n++ A. The current gain of a BJT will be increased
p+ B. The current gain of a BJT will be reduced
n++ C. The break-down voltage of a BJT will be reduced
n List-II:
1. The collector doping concentration is increased
n+
2. The base width is reduced.
C 3. The emitter doping concentration to base doping
concentration ratio is reduced
(a) B
4. The base doping concentration is increased
E keeping the ratio of the emitter doping
concentration to base doping concentration
C constant
5. The collector doping concentration is reduced
(b) B
(a) A-2, B-3, C-1
E (b) A-2, B-5, C-1
(c) A-2, B-3, C-4
C
(d) A-4, B-3, C-1
(c) B
4. In a bipolar-transistor at room temperature, if the
E emitter current is doubled, the voltage across its base-
emitter junction (GATE-97)
C
(a) Doubles
(d) B (b) Halves
(c) Increases by about 20 mV
E
(d) Decreases by about 20 mV
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Electronic Devices and Circuits OHM Institute BJT (PYQs)
5. For a BJT circuit shown, assume that the ‘’ of the 9. Consider two BJT’s biased at the same collector
transistor is very large and V BE = 0.7 V. The mode current with area A1 = 0.2 m  0.2m and A2 =
300m  300m. Assuming that all other device
of operation of the BJT is (GATE-06)
parameters are identical, kT/q = 26mV, the intrinsic
10 k carrier concentration is 11010cm-3 and q = 1.610-
19
C, the difference between the base–emitter voltages
R1 (in mV) of the two BJT ’s (i.e., VBE1 –VBE2) ___.
+ (GATE-14) (Set 4)
+ 10 V
2V 10. An npn BJT having reverse saturation current IS = 10-
R2 1 k  15
 A is biased in the forward active region with VBE =
700 mV. The thermal voltage (VT) is 25 mV and the
current gain () may vary from 50 to 150 due to
manufacturing variations. The maximum emitter
(a) Cut-off
current (in A) is _____.
(b) Saturation (GATE-15) (Set-3)
(c) Normal active
(d) Reverse active 11. The injected excess electron concentration profile in
the base region of an npn BJT, biased in the active
6. The DC current gain () of a BJT is 50. Assessing region, is linear, as shown in the figure. If the area of
that, the emitter junction efficiency is 0.995, the base the emitter-base junction is 0.001 cm2, n = 800
cm2/(V-s) in the base region and depletion layer
transport factor is (GATE-07) widths are negligible, then the collector current Ic (in
(a) 0.980 (b) 0.985 mA) at room temperature is ___. (Given: thermal
(c) 0.990 (d) 0.995 voltage VT = 26mV at room temperature, electronic
charge q = 1.6  10-19 C) (GATE-16) (Set-3)
7. In a uniformly doped BJT, assume that NE, NB and NC IB
are the emitter, base and collector doping in
n P n
atoms/cm3 respectively. If the emitter injection Excess electron
1014cm3 profile
efficiency of the BJT is close to unity which one of IE IC
the following is true? (GATE-10)
(a) N E  N B  N C
0
(b) N E  N B and N B  N C
0.5 m
(c) N E  N B and N B  N C
(d) N E  N B  N C 12. The base of an npn BJT T1 has linear doping profile
N B (x) as shown below. The base of another npn BJT
8. For a BJT, the common-base current gain  = 0.98 T2 has a uniform doping N B of 1017 cm 3 . All other
and the collector base junction reverse bias saturation parameters are identical for both the devices.
current ICO = 0.6 A . This BJT is connected in the Assuming that the hole density profile is the same as
that of doping, the common-emitter current gain of T2
common emitter mode and operated in the active is (EC-GATE-20)
region with a base drive current IB = 20 A . The
1017(cm-3)
collector current IC for this mode of operation is
NB(x)
(GATE-11)
(a) 0.98 mA
(b) 0.99 mA 1014(cm-3)
(c) 1.0 mA Collector
Emitter Base
(d) 1.01 mA 0 W
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Electronic Devices and Circuits OHM Institute BJT (PYQs)
(a) approximately 2.0 times that of T1 14. Which of the following statements is/are true for a
(b) approximately 0.3 times that of T1 BJT with respect to its DC current gain  ?
(c) approximately 2.5 times that of T1 (EC-GATE-24)
(d) approximately 0.7 times that of T1
(a) Under high-level injection condition in forward
(e) approximately 0.5 times that of T1 (Not given in
active mode,  will decrease with increase in the
the actual exam)
magnitude of collector current.
13. Select the CORRECT statement(s) regarding (b) Under low-level injection condition in forward
semiconductor devices (EC-GATE-22) active mode, where the current at the emitter-
(a) Electrons and holes are of equal density in an base junction is dominated by recombination-
intrinsic semiconductor at equilibrium generation process,  will decrease with
(b) Collector region is generally more heavily doped
increase in the magnitude of collector current.
than Base region in a BJT
(c) Total current is spatially constant in a two (c)  will be lower when the BJT is in saturation
terminal electronic device in dark under steady region compared to when it is in active region.
state condition (d) A higher value of  will lead to a lower value of
(d) Mobility of electrons always increases with the collector-to-emitter breakdown voltage.
temperature in Silicon beyond 300 K

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Electronic Devices and Circuits OHM Institute BJT (PYQs)

ANSWER KEY
One Mark Questions
1 c 2 d 3 a 4 c 5 c

6 d 7 d 8 b 9 2 10 d

11 d 12 c 13 c 14 6 15 a

16 a

Two Mark Questions

1 b&c 2 Decrease 3 a 4 c 5 b

6 b 7 b 8 d 9 380.3 10 1475.2

11 6.656 12 e 13 a, c 14 a, c, d

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Electronic Devices and Circuits OHM Institute BJT (Solutions)

SOLUTIONS
ONE MARK QUESTIONS 2I E1 eVBE 2 / VT
  V / V  e VBE 2  VBE1  / VT
I E1 e BE1 T
1. Answer: (c)  VBE 2  VBE1  VT ln(2) 18 mV
Solution:
1/n
  VBE  18mV
1
BVCEO = BVCBO  
 6. Answer: (d)
For  = 50, n = 6, BVCEO = 0.52 BVCBO Solution:
Current gain is reduced in saturation below its value
2. Answer: (d)
in active.
Solution:
Ic = dcIB in active.
When BJT enters saturation, Ic is less than dc IB

7. Answer: (d)
Solution:
A BJT is said to be operating in the saturation region
  
if base emitter and base collector junction are forward   
 1  
biased.
1
3. Answer: (a) 0.98 
1
1
Solution: 
Two diodes connected back-to-back is the
   49
representation of Ebers model. Ebers-model is
applicable to BJT’s I CEO  (1  ) ICBO
= (1+49) 0.6
= 30A
I C   I B  ICEO
= 49  20  30
 1.01mA

4. Answer: (c)
8. Answer: (b)
Solution:
Early effect in BJT’s is nothing but variation of base Solution:
1/ n
width in accordance with the reverse biased voltage 1
BVCEO  BVCBO  
across collector to base junction. 
5. Answer: (c)
As base recombination increases, IB increases.
Solution:
Hence, IC decreases (since IE = IB + IC). Hence, 
IE2 I ES  e VBE 2 /VT1  1
 1
I E1 I ES  e VBE1 / VT1  1 decreases. Hence,   increases hence BVCEO
  
 I E 2  2 I E1 increases.

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Electronic Devices and Circuits OHM Institute BJT (Solutions)
9. Answer: 2 12. Answer: (c)
Solution: Solution:
At C - B junction, due to FB a lot of holes are injected
into the base from collector. At E-B junction, due to
RB very few holes are injected into the base from
Emitter
(OR)
At collector – Base junction, ratio of Excess minority
carrier concentration to equilibrium minority carrier
concentration is in order of 105 (very high). This is
VEB = VE – VB
possible when the junction is forward bias (injection).
 VB = VE – VEB = 3 – 0.6
= 2.4V At emitter – base junction, ratio of Excess minority
V 2.4V carrier concentration to equilibrium minority carrier
IB = B   40 A
60K 60K concentration is in order of 1 (negligible). This is
IC  IB  50  40A  2mA possible when the junction is reverse Bias (no
VC  500IC  500  2mA  1V injection).
VEC = VE – VC = 3 – 1 = 2V
13. Answer: (c)
10. Answer: (d) Solution:
Solution:
If RB across the Base – collector junction increases
Early voltage is directly proportional to base width
 Effective Base width decreases
 As Base width increases, early voltage increases
So, re-combinations in Base decreases
11. Answer: (d) So,  increases, so  increases
Solution:
Ebers-moll model of BJT is valid in active, saturation 14. Answer: 6
and cut-off modes
Solution:
 = 1      IB = 0
16
VB  18 
16  44
16
 18  = 4.8 V
60
VE = 4.8 – 0.8 = 4V
 VBE VBC
 1  VBE
 4
I E  IS  e VT
e VT
 e VT
 1 IE   2mA
   F   2k
IC = IE = 2mA
1  VBE  1  VBE 
I B  IS  e VT  1   e VT  1  VC = 18 – (4k  2m) = 10 V
 F   R  
   VCE = 10 – 4 = 6 V
 VBE VBC
 1  VBC

IC  IS  e VT
E VT
 e VT
 1  15. Answer: (a)
  R  
    16. Answer: (a)
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Electronic Devices and Circuits OHM Institute BJT (Solutions)
6. Answer: (b)
TWO MARK QUESTIONS
Solution:
50
1. Answer: (b) & (c)  = /(1 + ) = = 0.9804
Solution: 1  50
IC = (1 + )ICBO + IB.  = T 
As T, ICBO & 
T = = 0.9804/0.995 = 0.985
IDS decrease with increase in Temp. 

2. Answer: Decrease 7. Answer:(b)

Solution: Solution:
As reverse bias to JC increases, IC effective base 1

width decreases, so recombination in base decreases, D E N B WB
1 . .
so IB. D B N E WE
For   1, NE >> NB.
3. Answer: (a)
For uniform doped BJT, NE >> NB & NB > NC.
Solution:
1. If base width is reduced recombination in base
8. Answer: (d)
decreases so IC and hence . Solution:
2. If emitter to base doping is  emitter injection
IC I b  1    ICBO
efficiency decreases so , 


4. Answer: (c) 1 
Solution: IC 1.01mA
Emitter current means diode current
VT
2I2 I (eV2  1) 9. Answer: 380.3
= 0 V VT
I1 I0 (e 1  1) Solution:
(V2 – V1) = VT ln2 IC  ISeVBE /VT
= 1  0.026  0.693 IS  area of the transistor.
= 0.018 VBE1  VBE 2 VBE1  VBE 2
IC1 IS1 VT A1 VT
= 18 Mv  e  e
IC2 IS2 A2
5. Answer: (b) IC1  IC2
Solution:
A 
Assuming active, VBE1  VBE2  VT n  2 
2  VBE  A1 
IC  I E 
R2  300  300 
 0.026n  
2  0.7  0.2  0.2 
 = 1.3 mA  380.3mV
1k
VC  VCC  IC R1
10. Answer: 1475.2
 10V  1.3mA 10k  Solution:
VC = – 1.3 V Given,
VB = 2V Is = 10–15A, VBE = 700 mV,
VBC = 3.3 V > 0.5 V VT = 25mV
Since, VC is less than VB, transistor is not in active. It 0.7
will operate in saturation. I C  IS .e VBE / VT  1015.e 0.025  1.4463mA

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Electronic Devices and Circuits OHM Institute BJT (Solutions)
1 12. Answer: (e) (In the official key it is MTA)
1
 1    Solution:
IC = IE  .I E  IE    .I C  .IC
1     1 D N W G
 B E E  E
D E N B WB G B
For IE to be maximum,  should be minimum which
G E is same for both transistors.
is 50.
WB
1
 1  50 
IE,max    .IC
GB 
DB  N  x .dx
B

 50  0

 1  50   2 G B1
 
  1.4463mA 1 G B2
 50 
 1.47523mA 1  14 1 
G B1   10 WB   10 WB 
17

1475.23 A DB  2 
1
11. Answer: 6.656
G B2 
DB
1017 WB 
Solution: 1017
1014 
2 2  1  0.5
dn dn 
IC  AeD n  Ae n VT 1 1017
2
dx dx
So, the current gain T2 is approximately 0.5 times that
19  1014  0  of T1 (Not given in the actual exam)
IC  0.001  1.6  10  800  0.026   4 
 0.5  10 
13. Answer: (a, c)
IC = 6.656mA
14. Answer: (a, c, d)

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