ETE (Spring – 2023)

(1)

(b) Transistor as an Amplifier (3 Marks):

A transistor can be used as an amplifier by utilizing its ability to control a larger current with a smaller input current. In an amplifier circuit, the transistor is typically connected in a common-emitter configuration. When a small AC signal (input) is applied to the base, it modulates the current flowing from the collector to the emitter. The amplified version of the input signal appears at the output, which is taken across the collector resistor.

Basic Circuit:

• The input signal is connected to the base through a coupling capacitor.
• A biasing circuit is used to set the transistor in the active region (where it can amplify).
• The output is taken across the collector resistor.

This amplification occurs because a small change in base current causes a large change in collector current, which is then reflected as a large voltage change across the load resistor.

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(c) Differences among Common Base, Emitter, and Collector Configurations (4 Marks):

Characteristic	Common Base	Common Emitter	Common Collector
Input Impedance	Low	Medium	High
Output Impedance	High	High	Low
Voltage Gain	High	High	Low
Current Gain	Low	High	High
Phase Shift	180° out of phase	180° out of phase	No phase shift (in-phase)

• Common Base: Used for high-frequency applications, but has low current gain and low input impedance.
• Common Emitter: Most commonly used configuration for amplification, provides high voltage and current gain, but with phase inversion.
• Common Collector: Also called an emitter follower, it has high current gain, no phase shift, and is used for impedance matching.

(2)

(b) Importance of Transistor Biasing (3 Marks):

Transistor biasing is crucial to ensure that the transistor operates in the correct region of its characteristic curve (active region) for amplification. Without proper biasing, the transistor could operate in the cutoff or saturation region, leading to distortion or no amplification. Biasing stabilizes the operating point (Q-point) of the transistor, ensuring consistent performance. It also helps compensate for temperature variations and transistor parameter changes, making the amplifier work reliably under different conditions.

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(c) NPN Transistor Calculations (4 Marks):

Given:

• $V_{CC} = 8V$
• $R_C = 3.5 \, k\Omega$

(i) Maximum Collector Current for Faithful Amplification:

For maximum collector current, the transistor should be at its maximum possible current when $V_{CE}$ is nearly zero. The maximum current can be calculated as:

$$I_C = \frac{V_{CC}}{R_C} = \frac{8V}{3500 \, \Omega} = 2.29 \, mA$$

(ii) Minimum Zero Signal Collector Current:

The minimum collector current, $I_C(min)$, is typically around half of the maximum current to allow the transistor to amplify both positive and negative cycles of the input signal symmetrically. For a maximum current of 2.29 mA, the minimum zero signal current would be approximately:

$$I_C(min) \approx \frac{2.29}{2} = 1.145 \, mA$$

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(a) Base Resistor, Biasing with Collector-Feedback, and Voltage-Divider Biasing (6 Marks):

1. Base Resistor Biasing: In base resistor biasing, a resistor is connected between the base of the transistor and the supply voltage $V_{CC}$. This resistor limits the base current, setting the operating point of the transistor. A common issue with this method is poor stability when the transistor's parameters change with temperature.

   Circuit:

   • The base resistor $R_B$ is connected between $V_{CC}$ and the transistor's base.
   • The emitter is typically grounded or connected to a resistor for stability.

2. Biasing with Collector-Feedback Resistor: In this method, a resistor is placed between the collector and the supply voltage $V_{CC}$. This resistor provides feedback that stabilizes the transistor's operating point by adjusting the base current automatically based on the collector current.

   Circuit:

   • A resistor $R_C$ is connected between the collector and $V_{CC}$.
   • The feedback helps stabilize the Q-point, but the gain is lower compared to other biasing methods.

3. Voltage-Divider Biasing: This is the most stable biasing method. It uses two resistors (usually $R_1$ and $R_2$) to form a voltage divider to provide a stable base voltage. The resistors are connected between $V_{CC}$ and ground, and the junction of the resistors is connected to the base of the transistor.

   Circuit:

   • $R_1$ is connected between $V_{CC}$ and the base of the transistor.
   • $R_2$ is connected between the base and ground.
   • This method provides a stable base voltage, making it ideal for amplifiers.

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(b) Determining Load Resistance and Base Resistor Bias Circuit (4 Marks):

Given:

• $V_{CC} = 20V$
• $V_{CE} = 10V$
• $I_C = 4 \, mA$
• $V_{BE} = 0.6V$
• $\beta = 150$

Step 1: Calculate Load Resistance $R_C$
The voltage drop across the load resistor $R_C$ is the difference between $V_{CC}$ and $V_{CE}$:

$$V_{RC} = V_{CC} - V_{CE} = 20V - 10V = 10V$$

Using Ohm's law, the load resistance $R_C$ is:

$$R_C = \frac{V_{RC}}{I_C} = \frac{10V}{4 \, mA} = 2500 \, \Omega = 2.5 \, k\Omega$$

Step 2: Calculate Base Resistor $R_B$
First, calculate the base current $I_B$ using the transistor's current gain $\beta$:

$$I_B = \frac{I_C}{\beta} = \frac{4 \, mA}{150} = 26.67 \, \mu A$$

Now, the base voltage $V_B$ is:

$$V_B = V_{BE} + V_{CE} = 0.6V + 10V = 10.6V$$

The base resistor $R_B$ is calculated using the supply voltage $V_{CC}$ and the base voltage $V_B$:

$$R_B = \frac{V_{CC} - V_B}{I_B} = \frac{20V - 10.6V}{26.67 \, \mu A} = \frac{9.4V}{26.67 \, \mu A} = 352.5 \, k\Omega$$

So, the base resistor $R_B$ is approximately $352.5 \, k\Omega$.

(3)

(a) Draw and explain the Hybrid equivalent circuit of PNP or NPN Transistor (2 Marks):

The hybrid equivalent circuit of a transistor is a simplified model used to analyze small signals.

For an NPN transistor:

• Base to Emitter: The resistance $r_{\pi}$ represents how much the base resists current flow.
• Current Source: A current source $\beta I_b$ represents the amplified current from the base to the collector (where $\beta$ is the transistor's gain).
• Collector: The resistance $r_o$ represents the resistance in the collector.

In this model:

• The small input signal is applied at the base.
• The transistor amplifies the signal, and the output is measured at the collector.

(4)

(a) Difference between JFET and Bipolar Transistor (4 Marks):

Feature	JFET (Junction Field Effect Transistor)	Bipolar Transistor (BJT)
Type of Current	Works with voltage control (unipolar)	Works with current control (bipolar)
Current Flow	Current flows through a channel controlled by voltage at the gate	Current flows between the collector and emitter controlled by the base current
Device Structure	Has three regions: Source, Gate, Drain	Has three regions: Emitter, Base, Collector
Power Consumption	Lower power consumption	Higher power consumption due to base current

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(b) Working Principle of JFET (4 Marks):

A Junction Field Effect Transistor (JFET) is a voltage-controlled device. The current flowing between the source and drain is controlled by the voltage applied at the gate. When a reverse bias is applied to the gate (relative to the source), it creates an electric field that narrows or shuts off the conducting channel. This controls the flow of current. JFETs are typically used for low noise and high input impedance applications.

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(c) Determining Gate-Source Voltage and Pinch-off Voltage (2 Marks):

Given:

• Drain current $I_D = 6 \, mA$
• $I_{DSS} = 15 \, mA$
• Gate-source cutoff voltage $V_{GS_{off}} = -6V$

(i) Gate-Source Voltage $V_{GS}$:

The equation for the drain current in a JFET is:

$$I_D = I_{DSS} \left( 1 - \frac{V_{GS}}{V_{GS_{off}}} \right)^2$$

Substituting the known values:

$$6 \, mA = 15 \, mA \left( 1 - \frac{V_{GS}}{-6V} \right)^2$$

Solving this equation gives the value of $V_{GS}$.

(ii) Pinch-off Voltage $V_p$:

The pinch-off voltage is equal to $V_{GS_{off}}$, which is given as $-6V$. So, $V_p = -6V$.

(5)

(a) Difference between D-MOSFETs and JFETs (5 Marks):

Feature	D-MOSFET (Depletion MOSFET)	JFET (Junction Field Effect Transistor)
Control Type	Voltage-controlled, gate is isolated from the channel	Voltage-controlled, gate is reverse-biased
Gate Characteristics	Gate voltage can be positive or negative	Gate is always reverse-biased for controlling current
Conducting Channel	Made of semiconductor material between drain and source	Channel is formed naturally with a PN junction
Operation Type	Can operate in both depletion and enhancement modes	Operates only in depletion mode (normally on)
Gate Current	Very low or nearly zero, as gate is insulated by oxide layer	Also very low, but gate is connected to channel via PN junction

Summary:

• D-MOSFETs are more flexible as they can work in both enhancement and depletion modes and have an insulated gate.
• JFETs are simpler and usually operate in depletion mode, with a gate that is directly connected to the channel.

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(a) Transfer Characteristic Curve of D-MOSFET (5 Marks):

The transfer characteristic curve of a D-MOSFET shows the relationship between the gate-source voltage ($V_{GS}$) and the drain current ($I_D$).

Explanation:

• The curve typically has three regions: cutoff, linear (or ohmic), and saturation.
• Cutoff Region: When $V_{GS}$ is below a certain threshold, the MOSFET is off, and no current flows through the channel.
• Linear Region: When $V_{GS}$ is slightly higher than the threshold, the MOSFET starts to conduct, and the current increases linearly with increasing $V_{GS}$.
• Saturation Region: Once $V_{GS}$ exceeds a certain value, the current plateaus, and the MOSFET is fully conducting.

In a D-MOSFET (Depletion mode), the MOSFET is normally on, and when $V_{GS}$ is negative, it depletes the channel, reducing the current. The curve starts at a non-zero current even when $V_{GS} = 0$.