ETE (Spring – 2019)

(1)

(a) Why is a transistor known as a bipolar device? (1 Mark)

A transistor is called a bipolar device because it uses both electrons (negative charge carriers) and holes (positive charge carriers) for conducting current.

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(b) Explain the transistor action in detail. (3 Marks)

A transistor has three regions: emitter, base, and collector. The emitter injects charge carriers (electrons or holes), the base controls their flow, and the collector collects these carriers to create current. The base is very thin and lightly doped, allowing most of the carriers from the emitter to reach the collector.

In NPN transistors, electrons move from the emitter to the collector through the base. In PNP transistors, holes flow from the emitter to the collector. A small current at the base controls a much larger current between the emitter and collector, which is the main transistor action.

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(c) Input and output characteristics of a common collector connection (3 Marks)

Input Characteristics:

The input characteristics show the relationship between the base current ($I_B$) and the base-emitter voltage ($V_{BE}$) for different collector-emitter voltages ($V_{CE}$).

Output Characteristics:

The output characteristics show the relationship between the collector current ($I_C$) and the collector-emitter voltage ($V_{CE}$) for different base currents ($I_B$).

(2)

(a) Show that $\beta = \frac{\alpha}{1 - \alpha}$ (2 Marks)

A transistor's current gain is represented by two terms:

1. $\alpha$: The current gain in common base configuration, defined as:

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

   where $I_C$ is the collector current and $I_E$ is the emitter current.

2. $\beta$: The current gain in common emitter configuration, defined as:

   $$\beta = \frac{I_C}{I_B}$$

   where $I_B$ is the base current.

From the current relationship in a transistor:

$$I_E = I_C + I_B$$

Rearranging for $I_B$:

$$I_B = I_E - I_C$$

Substitute $I_C = \alpha I_E$:

$$I_B = I_E - \alpha I_E = I_E(1 - \alpha)$$

Now, substituting this $I_B$ into $\beta$:

$$\beta = \frac{I_C}{I_B} = \frac{\alpha I_E}{I_E(1 - \alpha)}$$

Simplify:

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

Thus, $\beta = \frac{\alpha}{1 - \alpha}$.

(3)

(a) What is the significance of the arrow in the transistor symbol? (1 Mark)

The arrow in the transistor symbol shows the direction of current flow in the emitter. For NPN transistors, the arrow points outward, and for PNP transistors, the arrow points inward.

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(b) What is faithful amplification? Write the conditions to achieve it. (2 Marks)

Faithful amplification means the transistor amplifies the input signal without distortion. The output signal should be an exact, larger copy of the input signal.

Conditions for faithful amplification:

1. The base-emitter junction must be forward-biased.
2. The collector-base junction must be reverse-biased.
3. The transistor must operate within the active region of its characteristics.

(4)

(a) How will you measure h-parameters of a linear circuit? (3 Marks)

To measure h-parameters (hybrid parameters) of a linear circuit:

1. Identify input and output terminals: Label the input as $V_1$ (voltage) and $I_1$ (current), and the output as $V_2$ (voltage) and $I_2$ (current).
2. Short the output (for $h_{11}$ and $h_{12}$):
   • Apply a known input voltage ($V_1$) and measure the input current ($I_1$) to calculate $h_{11} = \frac{V_1}{I_1}$.
   • Measure the output voltage ($V_2$) to calculate $h_{12} = \frac{V_2}{V_1}$.
3. Open the input (for $h_{21}$ and $h_{22}$):
   • Apply a known output current ($I_2$) and measure the input current ($I_1$) to calculate $h_{21} = \frac{I_1}{I_2}$.
   • Measure the output voltage ($V_2$) to calculate $h_{22} = \frac{V_2}{I_2}$.

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(c) Appraise the voltage divider bias of a JFET (4 Marks)

Voltage divider bias is a stable biasing method for JFETs that keeps the operating point fixed.

Advantages:

1. It provides good thermal stability.
2. It ensures the JFET operates in the active region for faithful amplification.
3. The biasing is less sensitive to variations in $V_{GS}$ or $I_D$.

Working:

1. A resistor network (voltage divider) is connected between the supply voltage ($V_{DD}$) and ground to set the gate voltage ($V_G$).
2. The source resistor ($R_S$) develops a source voltage ($V_S = I_D R_S$), which ensures the required $V_{GS} = V_G - V_S$.
3. The drain resistor ($R_D$) controls the voltage drop and ensures proper drain current ($I_D$).

This setup maintains a stable $V_{GS}$ and $I_D$ regardless of temperature or device variations.

(5)

(a) Distinguish between FET and BJT (3 Marks)

Feature	FET	BJT
Type of Device	Unipolar (uses one carrier type)	Bipolar (uses electrons & holes)
Input Impedance	High	Low
Control Mechanism	Voltage-controlled	Current-controlled
Power Consumption	Low	High
Thermal Stability	Better	Moderate
Speed	Slower	Faster

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(b) Explain the operation of a P-channel JFET (4 Marks)

A P-channel JFET uses holes as the majority carriers.

Operation:

1. Structure: It has a P-type channel with N-type regions forming the gate.
2. Biasing:
   • The gate-source junction is reverse-biased ($V_G < V_S$).
   • The voltage at the drain ($V_D$) is less than the source ($V_S$).
3. Working:
   • When no voltage is applied to the gate ($V_{GS} = 0$), maximum current flows through the channel ($I_{DSS}$).
   • As negative $V_{GS}$ increases, the depletion regions expand, narrowing the channel and reducing current ($I_D$).
   • At a certain $V_{GS}$, the channel closes completely ($V_{GS(off)}$).

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(c) A JFET has a drain current of 5 mA. If $I_{DSS} = 10$ mA and $V_{GS(off)} = -6$ V, find (i) $V_{GS}$ and (ii) $V_P$. (3 Marks)

Using the Shockley's equation:

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

Substitute $I_D = 5$ mA, $I_{DSS} = 10$ mA, $V_{GS(off)} = -6$ V:

$$5 = 10 \left(1 - \frac{V_{GS}}{-6} \right)^2$$ $$0.5 = \left(1 - \frac{V_{GS}}{-6} \right)^2$$

Taking square root:

$$\sqrt{0.5} = 1 - \frac{V_{GS}}{-6}$$ $$0.707 = 1 + \frac{V_{GS}}{6}$$ $$\frac{V_{GS}}{6} = -0.293$$ $$V_{GS} = -1.758 \, \text{V}$$

(i) $V_{GS} = -1.76$ V
(ii) $V_P = V_{GS(off)} = -6$ V

(7)

(a) What is the basic difference between D-MOSFET and E-MOSFET? (2 Marks)

1. D-MOSFET (Depletion MOSFET):

   • Can operate in both depletion (negative gate voltage) and enhancement (positive gate voltage) modes.
   • Conducts current even when $V_{GS} = 0$.

2. E-MOSFET (Enhancement MOSFET):

   • Operates only in enhancement mode (positive gate voltage).
   • Does not conduct current when $V_{GS} = 0$.