BJT DC Biasing - Fixed Bias & Emitter Bias Questions and Answers

This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “BJT DC biasing – Fixed Bias and Emitter Bias”.

1. Which of the following is the correct relationship between base and emitter current of a BJT?
a) I_B = β I_E
b) I_B = I_E
c) I_B = (β + 1) I_E
d) I_E = (β + 1) I_B
View Answer

Answer: d
Explanation: For a BJT, the collector current I_C = βI_B and I_E = I_C + I_B
Hence, I_E = (β + 1) I_B.

2. 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
View Answer

Answer: a
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.

3. From the given circuit, using a silicon transistor, what is the value of I_{B_Q}?

a) 47.08 mA
b) 47.08 uA
c) 50 uA
d) 0 mA
View Answer

Answer: b
Explanation: Consider the BJT to be in saturation. Then I_C=12-0.2/2.2k=5.36 mA
And I_B=12-0.8/240k=0.047 mA
I_BMIN=I_CSAT/β=5.09/50=0.1072mA which is greater than above I_B.
Hence transistor is in the active region.
Thus I_C=βI_B.
V_BE=0.7V
I_B=12-0.7/240=47.08μA

4. From the given circuit, using a silicon BJT, what is the value of V_{CE_Q}?

a) 7 V
b) 0.7 V
c) 6.83 V
d) 7.17 V
View Answer

Answer: c
Explanation: Consider the BJT to be in saturation. Then I_C=12-0.2/2.2k=5.36 mA
And I_B=12-0.8/240k=0.047 mA
I_BMIN=I_CSAT/β=5.09/50=0.1072mA which is greater than above I_B.
Hence transistor is in the active region.
Thus I_C=βI_B.
V_BE=0.7V
I_B=12-0.7/240=47.08μA
I_C=50×47.08=2.354 mA
V_CE=V_CC-I_CR_C=12-2.354*2.2=12-5.178=6.83V.

5. From the given circuit, using a silicon BJT, what is the value of V_BC?

a) 6.13 V
b) -6.13 V
c) 7 V
d) -7 V
View Answer

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6. 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
View Answer

Answer: b
Explanation: To obtain an approximate answer, undersaturation the BJT is ON and hence acts as 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 V_CE = 0.2V and current is 12-0.2/2.2=5.36 mA.

7. In the given circuit, what is the value of I_C if the BJT is made of Silicon?

a) 2.01 mA
b) 2.01 uA
c) 10.05 mA
d) 10.05 uA
View Answer

Answer: a
Explanation: Consider the BJT to be in saturation. Then I_C=20-0.2/2k=9.9 mA
And I_B=20-0.8/430k=0.044 mA
I_BMIN=I_CSAT/β=5.09/50=0.198mA which is greater than above I_B.
Hence transistor is in the active region.
Thus I_C=βI_B.
V_BE=0.7V
I_B=20-0.7/430=44.88μA
I_C=50×44.88=2.24 mA.

8. In the given circuit, using a silicon BJT, what is the value of V_CE?

a) 20 V
b) 15.52 V
c) 14.98 V
d) 13.97 V
View Answer

Answer: b
Explanation: Consider the BJT to be in saturation. Then I_C=20-0.2/2k=9.9 mA
And I_B=20-0.8/430k=0.044 mA
I_BMIN=I_CSAT/β=5.09/50=0.198mA which is greater than above I_B.
Hence transistor is in the active region.
Thus I_C=βI_B.
V_BE=0.7V
I_B=20-0.7/430=44.88μA
I_C=50×44.88=2.24 mA
V_CE=20-2.24*2=15.52V.

9. In the given circuit, what is the value of V_E when using a silicon BJT?

a) 2.01 V
b) 0.28 V
c) 0 V
d) 2.28 V
View Answer

Answer: d
Explanation: Consider the BJT to be in saturation. Then I_C=20-0.2/2k=9.9 mA
And I_B=20-0.8/430k=0.044 mA
I_BMIN=I_CSAT/β=5.09/50=0.198mA which is greater than above I_B.
Hence transistor is in the active region.
Thus I_C=βI_B.
V_BE=0.7V
I_B=20-0.7/430=44.88μA
I_C=50×44.88=2.24 mA
V_CE=20-2.24*2=15.52V
V_E=I_ER_E=(1+β)I_BR_E=51*44.88*1=2.28V.

10. 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
View Answer

Answer: c
Explanation: To obtain an approximate answer, under saturation the BJT is ON and hence acts as 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 V_CE = 0.2V and current is 20-0.2/2.2=9.9 mA.

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