Biasing of Transistors MCQ Quiz - Objective Question with Answer for Biasing of Transistors - Download Free PDF

Explanation:

Voltage Divider Bias Circuit in CE Configuration

Problem Statement: For a voltage divider bias circuit in CE configuration using NPN BJT transistor, with given values of V_CC = 10 V, R_C = 2 kΩ, and R_E = 500 Ω, we are tasked to determine the approximate maximum value of the collector current (I_C) for the circuit, assuming a very high DC current gain (β).

Solution:

The maximum collector current (I_C) can be calculated by considering the saturation condition of the transistor, where the transistor acts as a closed switch. In this state, the voltage across the collector-emitter junction (V_CE) is approximately zero.

Step 1: Voltage Distribution in the Circuit

In the saturation condition, the voltage across the collector resistor (R_C) and emitter resistor (R_E) will be:

V_RC + V_RE = V_CC

Where:

• V_RC = Voltage drop across the collector resistor (R_C)
• V_RE = Voltage drop across the emitter resistor (R_E)

Step 2: Current Flow in the Circuit

The current flowing through R_C and R_E is the collector current (I_C), which is equal to the emitter current (I_E) in the saturation condition because the base current (I_B) is negligible for high β values.

Step 3: Maximum Collector Current Calculation

To calculate the maximum collector current (I_C), we use Ohm's Law:

I_C = I_E = V_CC ÷ (R_C + R_E)

Substituting the given values:

V_CC = 10 V, R_C = 2 kΩ = 2000 Ω, R_E = 500 Ω

I_C = 10 ÷ (2000 + 500)

I_C = 10 ÷ 2500

I_C = 0.004 A

I_C = 4 mA

Thus, the maximum collector current for the voltage divider bias circuit is approximately 4 mA.

Important Information

Let’s analyze other options:

Option 1: 5 mA

This value is incorrect because the calculation shows that the maximum collector current (I_C) is 4 mA, based on the circuit parameters. The resistance values and supply voltage do not support a collector current of 5 mA.

Option 3: 2.5 mA

This value is incorrect because it underestimates the current. Using the formula I_C = V_CC ÷ (R_C + R_E), the calculated value is 4 mA, not 2.5 mA.

Option 4: 20 mA

This value is incorrect because it overestimates the current. The resistance values and supply voltage do not support such a high collector current. The calculated maximum value is 4 mA.

Conclusion:

The correct option is Option 2, which gives the maximum collector current (I_C) as approximately 4 mA. This value aligns with the calculations based on the given circuit parameters, considering the saturation condition of the transistor and assuming a very high DC current gain (β).

Explanation:

Significance of the Quiescent Point (Q-point) in a Transistor Amplifier Circuit

Definition: The Quiescent Point (Q-point), also known as the operating point, is a critical parameter in transistor amplifier circuits. It represents the point at which the transistor operates when there is no input signal applied. The Q-point is determined by the DC biasing of the transistor and ensures that the transistor remains in its active region during operation, allowing it to amplify signals effectively.

Working Principle: In a transistor amplifier, the Q-point is established by properly designing the biasing circuit. This involves setting the correct DC voltage and current levels at the transistor terminals (collector, base, and emitter). The goal is to ensure that the transistor operates in its linear region, where the output signal is a faithful amplification of the input signal without distortion. The Q-point is chosen to prevent saturation or cutoff during signal amplification, ensuring the transistor behaves predictably and efficiently.

Importance of Q-point:

• The Q-point ensures that the transistor operates in its active region, where it can amplify signals linearly.
• A properly set Q-point minimizes distortion in the amplified signal.
• The Q-point determines the stability of the amplifier circuit under varying temperature and supply voltage conditions.
• It allows for maximum utilization of the transistor’s capability to amplify signals while avoiding regions of non-linearity (saturation or cutoff).

Correct Option Analysis:

The correct option is:

Option 4: It is the operating point where the transistor operates without an input signal.

This option accurately describes the significance of the Quiescent Point (Q-point). The Q-point is indeed the operating point where the transistor is biased and ready to amplify signals, even in the absence of an input signal. It is established by the DC biasing components in the circuit and ensures the transistor is properly positioned in its active region, avoiding saturation or cutoff.

Additional Information

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

Option 1: It is the point where the transistor is cut off.

This option is incorrect because the Q-point is not set at the cutoff region. The cutoff region is where the transistor does not conduct, and this state is unsuitable for amplification. The Q-point must be set in the active region to ensure proper operation of the amplifier.

Option 2: It represents the minimum current gain of the transistor.

This option is incorrect because the Q-point does not directly represent the current gain of the transistor. The current gain (β) is a property of the transistor itself and is independent of the Q-point. The Q-point is about setting the operating conditions of the transistor, not its intrinsic gain characteristics.

Option 3: It represents the maximum power output of the amplifier.

This option is incorrect. The Q-point does not directly represent the maximum power output of the amplifier. While the position of the Q-point can influence the power output and efficiency of the amplifier, it is not a direct measure of the maximum power output. The Q-point is primarily concerned with ensuring linear operation of the transistor.

Option 5: Not provided in the data row.

The fifth option is not explicitly detailed in the given data row, so it does not apply to this analysis.

Conclusion:

The Quiescent Point (Q-point) is a fundamental concept in transistor amplifier circuits. It represents the operating point where the transistor is biased and ready to amplify signals, even in the absence of an input signal. Setting the Q-point correctly ensures linear operation, minimizes distortion, and provides stability to the amplifier circuit. By understanding the role of the Q-point, engineers can design reliable and efficient amplifier circuits that perform effectively under varying conditions.

Explanation:

Class A Operation in Electronics

Definition: Class A operation is a mode of operation for amplifiers where the transistor or active device is biased such that it conducts during the entire cycle of the input signal. This ensures that the output signal is a faithful reproduction of the input signal without distortion.

Working Principle: In a Class A amplifier, the transistor remains active (i.e., not cut off) throughout the entire input signal cycle. The biasing point, or Q-point, is chosen strategically so that the transistor operates within its active region, allowing continuous conduction of current. This results in high linearity and minimal distortion.

Correct Option Analysis:

The correct option is:

Option 1: Q-point never lies.

This option is correct because the Q-point of a Class A amplifier should be carefully chosen to ensure that the transistor operates within its active region and never enters the cutoff or saturation regions. If the Q-point were to lie outside the active region, the amplifier would either stop conducting (cutoff) or saturate, both of which result in signal distortion. To achieve optimal performance, the biasing resistor is adjusted such that the Q-point stays within the active region of the transistor's operation.

Why the Q-point is critical:

• The Q-point determines the operating conditions of the transistor. In Class A operation, it should be set in the middle of the load line to allow symmetrical signal swing without distortion.
• If the Q-point shifts to the cutoff region, the transistor stops conducting, leading to signal clipping during the negative half-cycle of the input signal.
• If the Q-point shifts to the saturation region, the transistor cannot amplify the signal further, causing distortion during the positive half-cycle of the input signal.

Advantages of Class A Operation:

• High linearity: The output signal is a faithful reproduction of the input signal.
• Minimal distortion: The continuous conduction ensures that the signal remains undistorted.
• Simplicity in design: Class A amplifiers are relatively straightforward to design and implement.

Disadvantages of Class A Operation:

• Low efficiency: Since the transistor conducts throughout the input signal cycle, a significant amount of power is dissipated as heat.
• Heat generation: The continuous conduction results in higher heat dissipation, requiring robust thermal management.
• Limited output power: The efficiency constraints limit the maximum output power that can be achieved.

Applications: Class A amplifiers are commonly used in applications where linearity and signal fidelity are critical, such as audio amplification, instrumentation, and high-precision signal processing.

Additional Information

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

Option 2: Q-point lies in the middle of the load line.

This option is partially correct but does not encompass the full explanation. While it is true that the Q-point for a Class A amplifier is typically set in the middle of the load line to allow symmetrical signal swing, the primary goal is to ensure that the Q-point never lies outside the active region. Setting the Q-point in the middle of the load line is one method to achieve this, but it is not the definitive answer.

Option 3: Q-point lies in the cut-off region.

This option is incorrect because if the Q-point lies in the cutoff region, the transistor would stop conducting during a portion of the signal cycle. This results in signal clipping and severe distortion, defeating the purpose of Class A operation, which aims to maintain continuous conduction and signal fidelity.

Option 4: Q-point lies on the operating region.

While this option might seem correct at first glance, it is too vague to be considered a precise answer. The "operating region" could refer to the active region, the cutoff region, or the saturation region. For Class A operation, the Q-point must specifically be within the active region, not just any operating region.

Conclusion:

In Class A operation, the biasing resistor must be adjusted to ensure that the Q-point remains within the active region of the transistor's operation. This guarantees continuous conduction and prevents signal distortion. While other options provide partial explanations or describe incorrect scenarios, Option 1 correctly emphasizes the importance of keeping the Q-point out of the cutoff or saturation regions, ensuring high signal fidelity and linearity.

Solution: Given \(beta = 100\)) and \(V_{BE} = 0.7 \), V , \(V_{CC} = 15 \), V , \(R_1 = 100 \, k\Omega\) , \(R_2 = 50 \, k\Omega\)), \(R_C = 4.7 \, k\Omega \), and\( R_E = 3.3 \, k\Omega\)

\(V_{BB} = \frac{R_2}{R_1 + R_2}V_{CC} = \frac{50 \times 10^3}{150 \times 10^3} \times 15 = 5 \, V\)


\(V_E = V_{BB} - V_{BE} = 5 - 0.7 = 4.3 \, V\)


\(i_E \approx i_C = \frac{V_E}{R_E} = \frac{4.3}{3.3 \times 10^3} = 1.3 \, mA\)


\(V_C = V_{CC} - I_C R_C = 15 - 1.3 \times 4.7 = 8.9 \, V\)

Thus the correct option is (1): 8.9 V

The correct option is 1

Concept:

Transistor Biasing:

Transistor Biasing is the process of setting a transistor DC operating voltage or current conditions to the correct level so that any AC input signal can be amplified correctly by the transistor

The steady-state operation of a transistor depends a great deal on its base current, collector voltage, and collector current values and therefore, if the transistor is to operate correctly as a linear amplifier, it must be properly biased around its operating point.

Transistor biasing is provided by a biasing circuit and there are five types of biasing in transistor

1. Fixed Base Biasing a Transistor

2. Collector Feedback Biasing a Transistor

3. Dual Feedback Transistor Biasing

4. Transistor Biasing with Emitter Feedback

5. Voltage Divider Transistor Biasing

Common Emitter(CE) Configuration:

In CE configuration input is connected between base and emitter while the output is taken between collector and emitter.

\(β = \frac{α }{{1 - α }} = \frac{{{I_C}}}{{{I_B}}} = \frac{{{I_C}}}{{{I_E} - {I_C}}}\)

I_E = I_B + I_C  

I_C = β I_B + I_CEO 

IC = α I_E + ICBO

I_C = α (I_C + I_B) + ICBO

IC (1 - α ) = α I_B + ICBO

\(\Large{I_C=\frac{\alpha I_B}{1-\alpha}+\frac{I_{CBO}}{1-\alpha}}\)

In CE configuration, when I_B = 0 then I_C = I_CEO

\(\Large{I_{CEO} = \frac{I_{CBO}}{1 - α}} \)

Where, α = Current gain

β = Current Amplification Factor

IE, IB, IC = Emitter, Base and Collector current respectively

ICEO = Collector emitter cutoff current

ICBO = Collector base cutoff current

Calculation: 

Given: α = 0.98, ICBO = 5 μA, IB = 95 μA

\(β = \frac{0.98 }{(1 - 0.98)}= \frac{0.98 }{0.02 }= 49\)

\(I_{CEO} = \frac{5 × 10^{-6}}{(1 - 0.98)} = 250\ \mu A\)

I_C = 49 x (95 × 10^-6) + 250 × 10^-6 = 4905 × 10^-6 A

\(I_E=\frac{(4905 - 5) × 10^{-6}}{0.98}=\frac{4900 × 10^{-6}}{0.98}=5\ \ mA\)

Concept:

• Transistor: A transistor has two PN junctions i.e., it is like two diodes. The junction between base and emitter may be called emitter diode. The junction between base and collector may be called collector diode.

• The transistor can act in one of the three states:
• CUT-OFF: EMITTER DIODE AND COLLECTOR DIODE ARE OFF.
• ACTIVE: EMITTER DIODE IS ON AND COLLECTOR DIODE IS OFF.
• SATURATED: EMITTER DIODE AND COLLECTOR DIODE ARE ON.

Note: Please understand that the question is asking about the regions of a biased transistor. Biasing is a set of DC voltages that we apply at the Base-emitter 'or' emitter-collector terminal. There are three possibilities based on this voltage, i.e. Active Region, Cut-off region, and Saturation region.

Concept:

Transistor operation regions

Either forward or reverse biasing is done to the emitter and collector junctions of the transistor. These biasing methods make the transistor circuit to work in four kinds of regions as Active region, Saturation region, Cutoff region, and the Inverse active region.

The below table shows the different operation regions of the transistor.

NPN and PNP transistors schematic are shown below:

Calculation:

Consider the given transistor and applying the KVL we get:

- 5 + I_E(5k) + 0.7 + I_B(100K) + 5 = 0

We know that I_E = (1 + β)I_B

(1 + β)I_B(5k) + I_B(100) = - 0.7

I_B < 0

The above equation implies that the base current is opposite to the traditional PNP transistor so the emitter junction is reverse biased.

The collector voltage is zero and this junction also reverse bias due to the grounding of the collector terminal.

Conclusion: Both the junctions are Reverse Bias so the transistor is in the cut-off.

Concept:

A general npn transistor is represented as:

Applying KVL from VB to emitter ground, we can write:

VB - IB RB - VBE = 0

Solving the above, we get the base current as:

\(I_B=\frac{V_B-V_{BE}}{R_B}\)

Calculation:

Given

V_B = 10 V

V_BE = 0.7 V

R_B = 100 kΩ

From Concept

\(I_B=\frac{10-0.7}{100k}\)

I_B = 93 μA

PNP transistor: 

It is formed by sandwiching a thin layer of N-type semiconductor between two P-type semiconductor

In a PNP transistor, the current travels from the emitter to the collector. The PNP transistor's letter designates the voltage needed by the emitter, collector, and base of the transistor. . 

The direction of flow of holes is the same as that of conventional current. Hence, the current flows from emitter to collector. 

Concept:

The increase in collector current with an increase in collector-emitter voltage is called as ‘Early Effect’.

The characteristic curve of Ic vs. VCE will no longer be constant and will be as shown:

When we have to consider the small-signal output resistance associated with the collector current, it is given as:

\({r_0} = \frac{{\partial {V_C}}}{{\partial {I_C}}} = \frac{{{V_A}}}{{{I_C}}}\)

V_A = early voltage

I_C = DC collector current

Calculation:

Given, r₀ = 50 kΩ

|V_A| = 50 V

We can write:

\(50k = \frac{{50}}{{{I_C}}}\)

I_C = 1 mA

Voltage divider bias or self-bias provides a highly stable operating point compared to all the other biasing circuits.

It will make the collector current to be independent on temperature, base to emitter voltage and β.

Note:

The goal of transistor biasing is to establish a known quiescent operating point or Q-point for the bipolar transistor to work efficiently and produce an undistorted output signal.

β of a transistor is affected by the temperature and other factors.

This results in a change in the collector current, which shifts the operating point of the transistor.

Fixed bias:

• The above circuit is called a "fixed base bias circuit" because the transistors base current, IB remains constant for given values of VCC.
• With this single resistor type of biasing arrangement the biasing voltages and currents do not remain stable during transistor operation and can vary easily.

Collector Feedback Bias:

This is a beta(β) dependent biasing method which requires two resistors to provide the necessary DC bias for the transistor.

Voltage Divider Bias:

• This configuration provides the highest stability among all.
• Here the common emitter transistor configuration is biased using a voltage divider network to increase stability.

Given:

β = 100

V_C = 5 V

\(I_C = {V_C \over R_C}={5\over3.3\;k}=1.515\;mA \)

\(I_B = {I_C\over β}=0.01515\;mA\)

I_E = (1 + β) I_B = 1.53 mA

- 12 + 1.2 × 10³ × 1.53 × 10^-3 + 0.7 + V_B = 0

V_B = 9.464 V

\(I_X =\dfrac {{12-V_B}}{4.7\;k}=\dfrac{{12-9.464}}{4.7\;k}=0.54\;mA\)

I_Y = I_X + I_B = (0.54 + 0.01515) mA = 0.5547 mA

\(R_2=\dfrac{V_B}{I_Y}=\dfrac{9.464}{0.5547\times10^{-3}}=17.06\;kΩ\)

∴ The value of R₂ = 17.06 kΩ.

Load line in CE amplifier

• When no signal is applied, the transistor voltage and current conditions are as indicated at the quiescent point (Q point) on the dc load line.
• When an ac signal is applied, the transistor voltage and current vary above and below point Q.
• The load line is drawn between collector current I_C and collector voltage V_CE for different values of base current I_B.
• When a value for the maximum possible collector current is considered, that point will be present on the Y-axis, which is nothing but the saturation point.
• When a value for the maximum possible collector-emitter voltage is considered, that point will be present on the X-axis, which is the cutoff point.
• When a line is drawn joining these two points, such a line can be called a Load line.
• This line, when drawn over the output characteristic curve, makes contact at a point called an Operating point or quiescent point, or Q-point.

We have,

V_BE = 0.7 V (As shown in circuit)

V_E = 0 V

V_C = V_CE = 10 V

Since,

Vcc = 10kΩ ∙ I_CC + V_C

or, I_CC = \(\frac{V_{cc}-V_c}{10k}\)

or, \(I_{cc}= \frac{15-10}{10k}=\frac{5}{10k}=0.5\ mA.\)

We know that, For a transistor,

I_E = I_C + I_B

Also, I_C + I_B = I_CC (From circuit)

Hence, I_E = I_CC = 0.5 ≈ 0.51 mA