Configuration of BJT MCQ Quiz - Objective Question with Answer for Configuration of BJT - Download Free PDF

Emitter Current in BJT (Bipolar Junction Transistor):

Definition: The emitter current (I_E) in a Bipolar Junction Transistor (BJT) is the total current flowing out of the emitter terminal. It is the sum of the base current (I_B) and the collector current (I_C).

Important Parameters:

• Collector Current (I_C): The current flowing through the collector terminal.
• Base Current (I_B): The current flowing through the base terminal.
• Emitter Current (I_E): The total current flowing out of the emitter terminal, given by:

Formula:

The emitter current can be expressed as:

I_E = I_C + I_B

Additionally, the common base short circuit current gain, denoted by α, relates the collector current to the emitter current:

α = I_C / I_E

Rearranging the formula, we can write:

I_E = I_C / α

Given Data:

• Collector Current, I_C = 1 mA
• Common Base Short Circuit Current Gain, α = 1 (unity)

Solution:

Using the formula:

I_E = I_C / α

Substitute the given values:

I_E = 1 mA / 1

I_E = 1 mA

Hence, the emitter current is 1 mA.

Correct Option Analysis:

The correct option is:

Option 3: 1 mA

This option is correct because the emitter current is calculated to be 1 mA based on the given collector current and the common base short circuit current gain (α = 1). The result directly matches the provided data and the relationship between these parameters in the BJT.

Additional Information

To further understand the analysis, let’s evaluate why the other options are incorrect:

Option 1: 1.5 mA

This option suggests that the emitter current is 1.5 mA. However, the given parameters (I_C = 1 mA and α = 1) clearly show that the emitter current is 1 mA. A value of 1.5 mA would imply a different value of α, which is not consistent with the problem statement.

Option 2: 2 mA

This option proposes an emitter current of 2 mA. This value is incorrect because, with α = 1, the emitter current must equal the collector current. An emitter current of 2 mA would require a collector current greater than 1 mA, which contradicts the given data.

Option 4: 0.5 mA

This option suggests an emitter current of 0.5 mA. This is incorrect because, with α = 1, the emitter current cannot be less than the collector current. A value of 0.5 mA for the emitter current would imply either an incorrect value of α or collector current, which is not the case here.

Conclusion:

Understanding the relationship between emitter current, collector current, and the common base current gain (α) is essential in analyzing BJTs. In this case, the emitter current is directly equal to the collector current due to the unity gain (α = 1). This simplicity highlights the importance of correctly applying the fundamental formulas of BJT operation.

Explanation:

The high-frequency gain of a common-emitter amplifier is mainly affected by the collector-base junction capacitance and the emitter-base capacitance. Let's delve into the detailed explanation of why these capacitances play a crucial role and analyze the other options provided.

Correct Option: Option 2 - Collector-Base Junction Capacitance and Emitter-Base Capacitance

A common-emitter amplifier is a widely used configuration in electronic circuits. It provides good voltage gain and is commonly used for amplification purposes. However, at high frequencies, the gain of the amplifier is significantly influenced by certain capacitances inherent to the transistor. The two main capacitances that affect the high-frequency performance of a common-emitter amplifier are:

1. Collector-Base Junction Capacitance (C_cb): This is the capacitance between the collector and the base of the transistor. It is also known as the Miller capacitance because it gets multiplied by the gain of the amplifier in the feedback loop, making its effect more pronounced. This capacitance introduces a feedback path for high-frequency signals, reducing the overall gain of the amplifier at high frequencies.
2. Emitter-Base Capacitance (C_eb): This is the capacitance between the emitter and the base of the transistor. This capacitance affects the input impedance of the amplifier and influences the high-frequency response. It acts as a shunt capacitance at the input, causing the input signal to be partially bypassed to the ground at high frequencies, thus reducing the input signal amplitude and, consequently, the gain.

At high frequencies, these capacitances form low-impedance paths, which shunt the signal away from the intended path, leading to a reduction in gain. The combined effect of these capacitances is to create a low-pass filter, which attenuates the high-frequency components of the signal.

Mathematical Analysis:

The high-frequency response of the common-emitter amplifier can be analyzed using the hybrid-π model of the transistor. In this model, the capacitances C_cb and C_eb are explicitly included. The gain-bandwidth product (GBP) is a key parameter, which remains constant for a given transistor. The gain of the amplifier decreases as the frequency increases, and this relationship can be expressed as:

Gain (A_v) = A_v0 / (1 + jω/ω_T)

where A_v0 is the low-frequency gain, ω is the angular frequency, and ω_T is the transition frequency, which is inversely proportional to the sum of the capacitances.

Concept:

• Common Emitter (CE): Has high output resistance → (c)
• Common Collector (CC): Has low output resistance → (a)
• Common Base (CB): Has very high output resistance → (b)

Final Matching: (i)-(c), (ii)-(a), (iii)-(b)

PNP Transistor

PNP transistor is a three-terminal device i.e. base, emitter, and collector.

The main function of the PNP transistor is amplification.

For amplification, the base and emitter junction must be forward bias, and the collector and emitter junction must be reverse bias.

Operating modes of transistor

Additional Information

NPN Transistor:

• An NPN transistor is a Bipolar Junction Transistor consisting of two PN junctions in such as way that a P-type semiconductor sandwiching by two N-type Semiconductors.

• These back to back PN junction diodes are known as the collector-base junction and base-emitter junction.
• The base-emitter junction is always connected in forward bias in Active mode therefore it has a barrier voltage (VBE) of + 0.7 volts for Si and + 0.3 volts for Ge.
• The collector-base junction is always connected in reverse bias in Active mode.
• In this transistor, electrons are the majority carrier for the input current (IE) and holes are the minority carrier.

Transistors

• A transistor is an electronic device that has 3 terminals.
• The transistor is made up of semiconductor materials such as Silicon and Germanium.
• BJT and FET are examples of transistors.

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Bipolar Junction Transistor (BJT)

• BJT is a semiconductor device that is constructed with 3 doped semiconductor Regions i.e. Base, Collector & Emitter separated by 2 p-n Junctions.
• BJTs are current-driven devices. The current through the two terminals is controlled by a current at the third terminal (base).
• BJT is a bipolar device (current conduction by both types of carriers, i.e. majority and minority electrons and holes) It has a low input impedance.

Junction Field Effect Transistor (JFET)

• JFET is a unipolar voltage-controlled semiconductor device with three terminals: source, drain, and gate.
• In JFET, the current flow is due to the majority of charge carriers. Hence, it is unipolar.

Common base configuration:

Input Terminal Emitter – Base (EB)

Output Terminal Collector – Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

Various transistor configurations and their characteristics are shown in the table

Common base configuration:

Input Terminal Emitter – Base (EB)

Output Terminal Collector – Base (CB)

Since for Amplification Application in BJT

EB Junction → Forward Biased (Low Impedance)

CB Junction → Reversed Biased (High Impedance)

The various characteristics of different configuration are shown 

The correct answer is option 1) 5 mA

Concept:

The common base DC current gain is a ratio of the value of the transistor's collector current to the value of the transistor's emitter current,

i.e.  α= $\frac{I_c}{I_e}$ ---(1)

Also, the common-emitter current gain is the ratio of the value of the transistor's collector current to the value of the transistor's base current in a transistor,

i.e.  β= $\frac{I_c}{I_b}$ ---(2)

Using Equations (1) and (2), we get:

$\beta =\frac{\alpha }{1-\alpha }$

For a transistor, the relation between the collector, base, and emitter current is:

I_E = I_B + I_C

I_C = β I_B

I_E = (β + 1) I_B

Calculation:

With α_dc = 0.98,

β will be:

$\beta =\frac{0.98}{1-0.98}=49$

Now, the emitter current will be:

I_E = (49 + 1) 100 μA

I_E = 5000 μA 

I_E = 5 mA

Bipolar junction transistor

• The Bipolar Junction Transistor (BJT) is a three-terminal device i.e. Base, Emitter, and Collector.
• There are two main types of bipolar junction transistors the NPN and the PNP transistor.
• The emitter is a heavily doped region of the BJT transistor, providing the majority of carriers into the base region.
• The base region is a thin, lightly doped region sandwiched between the emitter and collector.
• The majority of carriers from the emitter pass through the base region and its flow can be externally controlled.

Working configuration of BJT

Concept:

For a transistor, the base current, the emitter current, and the collector current are related as:

IE = IB + IC

where IC = αIE

α = Current gain of the transistor

Calculation:

Given,

IC = 0.95 mA

IB = 0.05 mA

∴ IE = IB + IC = 0.05 + 0.95 = 1 mA

From the above concept,

IC = αIE

$\alpha =\frac{I_C}{I_E}=\frac{0.95}{1}=0.95$

Region of operation of BJT:

In a given circuit, the region of the transistor can be decided through the following procedure.

• Assume that the transistor is in the saturation region.
• Calculate Base current (IB) and collector current (IC) separately.
• Calculate the minimum Base current (IB)min required to operate the transistor in saturation.
• Compare Base current (IB) and (IB)min​

1. IB  ≥   (IB)min, Transistor is in the Saturation region.
2. IB < (IB)min, Transistor is in the Active region
3. IB  < 0, Transistor is in the Cut-off region.

Formula:

In saturation,

Base-Emitter voltage = VBE = 0.8 V

Collector-Emitter voltage VCE = 0.2 V

(IB)min = (IC)sat / β  ----(1)

(IC)sat  = Collector current in Saturation 

β = DC current gain

Calculation:

Assume transistor in Saturation region.

Apply KVL in the outer loop:

5 - I_C(Sat) x 5kΩ - V_CE(Sat) = 0

$I_{C(Sat)}=\frac{4.8}{5k\mathrm{\Omega }}$

I_C(Sat) = 0.96 mA

From equation (1);

$I_{B(Min)}=\frac{0.96mA}{100}$

(IB)min = 9.6 μA

Now, Apply KVL in Emitter Loop; 

5 - 100I_B - 0.7 = 0

I_B = 43 μA

We observe that;

IB  ≥   (IB)min

Hence, BJT is in saturation region.

Concept:

The common base DC current gain is a ratio of the value of the transistor's collector current to the value of the transistor's emitter current, i.e.

$\alpha =\frac{I_c}{I_e}$   ---(1)

Also, the common-emitter current gain is the ratio of the value of the transistor's collector current to the value of the transistor's base current in a transistor, i.e.

$\beta =\frac{I_c}{I_b}$   ---(2)

Using Equations (1) and (2), we get:

$\beta =\frac{\alpha }{1-\alpha }$

For a transistor, the relation between the collector, base, and emitter current is:

IE = IB + IC

IC = β IB

IE = (β + 1) IB

Calculation:

We have,

β = 49

From the above concept,

$\beta =\frac{\alpha }{1-\alpha }$

or $49=\frac{\alpha }{1-\alpha }$

or, 49 - 49α = α

or, 

Hence,50α = 49

α = 49/50 = 0.98

The correct answer is (option 4) i.e. holes flow in the emitter region

Concept:

The arrow on the symbol for bipolar transistors indicates the PN junction between base and emitter and points in the direction in which conventional current travels, i.e. the direction of holes.

NPN:

When the emitter-base is forward biased, holes from the base (p-type) start to flow to the emitter side (n-type) and electrons start to flow from the emitter to the base. The direction, however, represents the direction of the hole flow.

PNP:

When the emitter-base junction is forward biased, holes from the emitter starts to flow to the base. The direction of the arrow also indicates the same.

Concept:

Active mode:

In the active mode, one junction (emitter to base) is forward biased and another junction (collector to base) is reverse biased.

Method to check region of operation:

Step 1 : Assume BJT to be in saturation region (V_CE = 0.2 V), and calculate I_Csat from the circuit.

Step 2: Calculate I_Bmin = ICsat/ β 

Step 3: Calculate I_B from the given circuit.

Step 4: If,  IB > IBmin → Saturation region

                 IB < IBmin → Active region

Analysis:

Assume,

V_BE = 0.7 V

V_CEsat = 0.2 V

ICsat = ( V_CC - V_CEsat)/ R_C

⇒ ICsat = ( 9 - 0.2)/2 K

⇒ ICsat = 4.4 mA

IBmin = ICsat/ β = 4.4/50 = 0.088 mA

I_B = ( 9 - 0.7)/300K = 0.0276 mA

 IB < IBmin → Active region

Additional Information Saturation mode:

In the saturation mode, both the junctions of the transistor (emitter to base and collector to base) are forward biased.

Cutt-off mode:

In the cutoff mode, both the junctions of the transistor (emitter to base and collector to base) are reverse biased.

Inverted mode:

In the inverted mode ( Reverse active mode), one junction (emitter to base) is reverse biased and another junction (collector to base) is forward biased.