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BJT - Biasing & Stabilization MCQs for Electronics and Communication Engineering (ECE) Exam
It covers all Important Questions with answers on BJT - Biasing & Stabilization for the Electronics and Communication Engineering (ECE) exam. The questions are based on important topics. Details about the questions:- Topic: BJT - Biasing & Stabilization
- Type of Questions: MCQs with solutions
- Number of Questions: 50
- You can attempt them on EduRev to score high in Electronics and Communication Engineering (ECE) exam.
Given that the collector power dissipation is 300 mW, what is the value of collector current for the collector to emitter voltage = 12 V?- a)50 mA
- b)0 mA
- c)25 mA
- d)100 mA
Correct answer is option 'B'. Can you explain this answer?
Given that the collector power dissipation is 300 mW, what is the value of collector current for the collector to emitter voltage = 12 V?
a)
50 mA
b)
0 mA
c)
25 mA
d)
100 mA
 | Sagarika Patel answered |
P=VCE.IC => 300mW = (12V)IC => IC=300/12 mA = 25 mA.
For the circuit shown below, the collector to emitter voltage is
- a)1.25 V
- b)3.40 V
- c)-2.30 V
- d)0 V
Correct answer is option 'B'. Can you explain this answer?
For the circuit shown below, the collector to emitter voltage is
a)
1.25 V
b)
3.40 V
c)
-2.30 V
d)
0 V
 | Sanya Agarwal answered |
Let the transistor be operating under active region.
Which of the following configuration is used for audio signal amplification- a)Common base configuration
- b)Common emitter configuration
- c)Common collector configuration
- d)All configurations are equally suited
Correct answer is option 'B'. Can you explain this answer?
Which of the following configuration is used for audio signal amplification
a)
Common base configuration
b)
Common emitter configuration
c)
Common collector configuration
d)
All configurations are equally suited
 | Ritika Sarkar answered |
Common base configuration is used as input stage of multistage amplifier. Since voltage gain is high, common emitter configuration is used for audio signal amplification. Common collector is used for impedance matching.
Which of the following statement about a common collector transistor is true- a)Very low input impedance
- b)Very high output impedance
- c)Very low current gain
- d)Low voltage gain
Correct answer is option 'D'. Can you explain this answer?
Which of the following statement about a common collector transistor is true
a)
Very low input impedance
b)
Very high output impedance
c)
Very low current gain
d)
Low voltage gain
 | Bijoy Nair answered |
common collector configuration has high input impedance and low output impedance. The current gain is high but voltage gain is low.
From the given circuit, what is the value of the saturation collector current?
- a)5 mA
- b)5.45 mA
- c)4.54 mA
- d)10.9 mA
Correct answer is option 'B'. Can you explain this answer?
From the given circuit, what is the value of the saturation collector current?
a)
5 mA
b)
5.45 mA
c)
4.54 mA
d)
10.9 mA
 | Rajveer Saha answered |
Saturation collector current = VCC/RC = 5.45 mA.
For a BJT, what is typically the shape of the power dissipation curve, if it’s plotted on the output characterisics?- a)Parabola
- b)Linear
- c)Hyperbola
- d)Circular
Correct answer is option 'C'. Can you explain this answer?
For a BJT, what is typically the shape of the power dissipation curve, if it’s plotted on the output characterisics?
a)
Parabola
b)
Linear
c)
Hyperbola
d)
Circular
 | Surbhi Chopra answered |
Power Dissipation in a BJT is given by P=VCE.IC. This is in the form of k=xy which is the equation of a hyperbola.
In a BJT, if the collector-base junction is forward-biased and the base-emitter junction is forward-biased, which region is the BJT operating in?- a)Saturation region
- b)Active region
- c)Cutoff region
- d)Reverse active region
Correct answer is option 'A'. Can you explain this answer?
In a BJT, if the collector-base junction is forward-biased and the base-emitter junction is forward-biased, which region is the BJT operating in?
a)
Saturation region
b)
Active region
c)
Cutoff region
d)
Reverse active region
 | Xena Das answered |
Operating Regions of a BJT
A Bipolar Junction Transistor (BJT) is a three-layer semiconductor device that consists of two pn junctions. These junctions are the collector-base (CB) junction and the base-emitter (BE) junction. Depending on the biasing conditions of these junctions, the BJT can operate in different regions. The operating regions of a BJT are as follows:
1. Cutoff Region:
When both the collector-base junction and the base-emitter junction are reverse-biased, the BJT is said to be in the cutoff region. In this region, the BJT is in an off state, and no current flows through it.
2. Active Region:
In the active region, the collector-base junction is reverse-biased, while the base-emitter junction is forward-biased. This is the region where the BJT operates as an amplifier. The current flowing through the BJT is controlled by the base current, and the collector current is proportional to the base current. The BJT operates in an active mode where it can amplify signals.
3. Saturation Region:
When both the collector-base junction and the base-emitter junction are forward-biased, the BJT is said to be in the saturation region. In this region, the BJT is fully turned on, and the collector current is at its maximum value. The BJT operates as a closed switch, allowing a large current to flow from the collector to the emitter.
4. Reverse Active Region:
In the reverse active region, the collector-base junction is forward-biased, while the base-emitter junction is reverse-biased. This region is rarely used in practical applications and is mainly used for testing purposes. In this region, the BJT operates with reversed current flow, and the collector current is controlled by the base current.
Answer:
Based on the given biasing conditions (collector-base junction forward-biased and base-emitter junction forward-biased), the BJT is operating in the saturation region. In this region, the BJT acts as a closed switch, allowing a large current to flow from the collector to the emitter. This biasing configuration is commonly used in digital applications and for switching purposes.
A Bipolar Junction Transistor (BJT) is a three-layer semiconductor device that consists of two pn junctions. These junctions are the collector-base (CB) junction and the base-emitter (BE) junction. Depending on the biasing conditions of these junctions, the BJT can operate in different regions. The operating regions of a BJT are as follows:
1. Cutoff Region:
When both the collector-base junction and the base-emitter junction are reverse-biased, the BJT is said to be in the cutoff region. In this region, the BJT is in an off state, and no current flows through it.
2. Active Region:
In the active region, the collector-base junction is reverse-biased, while the base-emitter junction is forward-biased. This is the region where the BJT operates as an amplifier. The current flowing through the BJT is controlled by the base current, and the collector current is proportional to the base current. The BJT operates in an active mode where it can amplify signals.
3. Saturation Region:
When both the collector-base junction and the base-emitter junction are forward-biased, the BJT is said to be in the saturation region. In this region, the BJT is fully turned on, and the collector current is at its maximum value. The BJT operates as a closed switch, allowing a large current to flow from the collector to the emitter.
4. Reverse Active Region:
In the reverse active region, the collector-base junction is forward-biased, while the base-emitter junction is reverse-biased. This region is rarely used in practical applications and is mainly used for testing purposes. In this region, the BJT operates with reversed current flow, and the collector current is controlled by the base current.
Answer:
Based on the given biasing conditions (collector-base junction forward-biased and base-emitter junction forward-biased), the BJT is operating in the saturation region. In this region, the BJT acts as a closed switch, allowing a large current to flow from the collector to the emitter. This biasing configuration is commonly used in digital applications and for switching purposes.
In typical circuits, the stabilization factor S = (ΔICICO) is- a)< 1
- b)> 1
- c)= 1
- d)<< 1
Correct answer is option 'D'. Can you explain this answer?
In typical circuits, the stabilization factor S = (ΔICICO) is
a)
< 1
b)
> 1
c)
= 1
d)
<< 1
 | Kajal Yadav answered |
For better stability of the circuit, stability factor S must be as small as possible.
What is the region on the output characteristics for VCE < VCEsat called?- a)Active region
- b)Cutoff region
- c)Saturation region
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
What is the region on the output characteristics for VCE < VCEsat called?
a)
Active region
b)
Cutoff region
c)
Saturation region
d)
None of the mentioned
 | Pranjal Datta answered |
The region below VCE < VCEsat is called the saturation region.
Which of the following is not a valid type of BJT?- a)PNP
- b)NPN
- c)PPN
- d)NNP
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not a valid type of BJT?
a)
PNP
b)
NPN
c)
PPN
d)
NNP
 | Akshara Kapoor answered |
PPN is not a valid type of BJT
PNP and NPN:
- PNP and NPN are the two most common types of bipolar junction transistors (BJTs) used in electronic circuits.
- PNP transistors have a layer of N-type semiconductor sandwiched between two layers of P-type semiconductor, while NPN transistors have a layer of P-type semiconductor sandwiched between two layers of N-type semiconductor.
PPN:
- PPN is not a valid type of BJT.
- BJT transistors are characterized by their structure of three layers of semiconductor material, which determine the type of BJT (PNP or NPN).
- The structure of PPN does not follow the typical configuration of a BJT.
MNP:
- MNP is also not a valid type of BJT.
- The correct types of BJT are PNP and NPN, which are widely used in electronic circuits for amplification and switching purposes.
In a pnp-BJT, when the E-B junction is forward biased and no voltage is applied across C-B junction, what happens to the width of the depletion region in the E-B junction?- a)Increases
- b)Decreases
- c)Remains same
- d)Can’t be determined
Correct answer is option 'B'. Can you explain this answer?
In a pnp-BJT, when the E-B junction is forward biased and no voltage is applied across C-B junction, what happens to the width of the depletion region in the E-B junction?
a)
Increases
b)
Decreases
c)
Remains same
d)
Can’t be determined
 | Ravi Singh answered |
On application of a forward bias voltage across E-B junction, the width of the depletion region decreases.
An enhancement type NMOS transistor with V1 = 0.7 V and source terminal grounded and 1.5 V applied to the gate. In what region does the device operate for VD = 0.5 V?- a)Cut off region
- b)Saturation region
- c)Triode region
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
An enhancement type NMOS transistor with V1 = 0.7 V and source terminal grounded and 1.5 V applied to the gate. In what region does the device operate for VD = 0.5 V?
a)
Cut off region
b)
Saturation region
c)
Triode region
d)
None of the above
| | Xena Das answered |
Explanation:
The given transistor is an enhancement type NMOS transistor with V1 = 0.7 V and a source terminal grounded. The gate voltage, Vg = 1.5 V, and the drain voltage, Vd = 0.5 V. We need to determine the region of operation for this transistor.
Cut-off Region:
In the cut-off region, the transistor is off and no current flows between the drain and source terminals. This occurs when the gate voltage is less than the threshold voltage (Vt) of the transistor. Since Vg = 1.5 V is greater than Vt, the transistor is not in the cut-off region.
Saturation Region:
In the saturation region, the transistor is fully on and acts as a closed switch between the drain and source terminals. This occurs when the gate voltage is greater than the threshold voltage (Vt) and the drain-source voltage (Vds) is also greater than the gate-source voltage (Vgs) minus the threshold voltage (Vt). In this case, Vds = 0.5 V and Vgs - Vt = 1.5 V - 0.7 V = 0.8 V, which means Vds is less than Vgs - Vt. Therefore, the transistor is not in the saturation region.
Triode Region:
In the triode region, the transistor operates as a linear amplifier. This occurs when the gate voltage is greater than the threshold voltage (Vt) and the drain-source voltage (Vds) is less than the gate-source voltage (Vgs) minus the threshold voltage (Vt). In this case, Vds = 0.5 V and Vgs - Vt = 1.5 V - 0.7 V = 0.8 V, which means Vds is less than Vgs - Vt. Therefore, the transistor is in the triode region of operation.
Conclusion:
Based on the given parameters, the transistor operates in the triode region for Vd = 0.5 V.
The given transistor is an enhancement type NMOS transistor with V1 = 0.7 V and a source terminal grounded. The gate voltage, Vg = 1.5 V, and the drain voltage, Vd = 0.5 V. We need to determine the region of operation for this transistor.
Cut-off Region:
In the cut-off region, the transistor is off and no current flows between the drain and source terminals. This occurs when the gate voltage is less than the threshold voltage (Vt) of the transistor. Since Vg = 1.5 V is greater than Vt, the transistor is not in the cut-off region.
Saturation Region:
In the saturation region, the transistor is fully on and acts as a closed switch between the drain and source terminals. This occurs when the gate voltage is greater than the threshold voltage (Vt) and the drain-source voltage (Vds) is also greater than the gate-source voltage (Vgs) minus the threshold voltage (Vt). In this case, Vds = 0.5 V and Vgs - Vt = 1.5 V - 0.7 V = 0.8 V, which means Vds is less than Vgs - Vt. Therefore, the transistor is not in the saturation region.
Triode Region:
In the triode region, the transistor operates as a linear amplifier. This occurs when the gate voltage is greater than the threshold voltage (Vt) and the drain-source voltage (Vds) is less than the gate-source voltage (Vgs) minus the threshold voltage (Vt). In this case, Vds = 0.5 V and Vgs - Vt = 1.5 V - 0.7 V = 0.8 V, which means Vds is less than Vgs - Vt. Therefore, the transistor is in the triode region of operation.
Conclusion:
Based on the given parameters, the transistor operates in the triode region for Vd = 0.5 V.
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)None of thye mentioned
Correct answer is option 'A'. Can you explain this answer?
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)
None of thye mentioned
| | Partho Saha answered |
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.
For the given transistor, what is the correct sequence of the pins from left to right?
- a)ECB
- b)BCE
- c)CEB
- d)The device is not a transistor
Correct answer is option 'B'. Can you explain this answer?
For the given transistor, what is the correct sequence of the pins from left to right?
a)
ECB
b)
BCE
c)
CEB
d)
The device is not a transistor
| | Kritika Gupta answered |
With the curved side facing away the correct lead sequence is base, collector, emitter.
Which of the following statement about a common base transistor is true- a)Very low input impedance
- b)Very low output Impedance
- c)Current gain is greater than unity
- d)Voltage gain is very low
Correct answer is option 'A'. Can you explain this answer?
Which of the following statement about a common base transistor is true
a)
Very low input impedance
b)
Very low output Impedance
c)
Current gain is greater than unity
d)
Voltage gain is very low
| | Prisha Sen answered |
Common base transistor has very low input resistance (20Ω). It also has very high output resistance. Its current gain is less than unity and it has a medium voltage gain.
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