CBSE Class 12 Physics Electronic Devices Frequently Asked Questions

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UNIT IX

ELECTRONIC DEVICES

2 MARKS

1. How is a p-type semiconductor formed? Name the majority carriers in it. Draw the energy band diagram of a p-type semiconductor. [2 Marks]

Answer:
1. A p-type semiconductor is formed by doping a trivalent impurity (such as boron, indium, or aluminium) into an intrinsic semiconductor like silicon or germanium.
2. The majority charge carriers in a p-type semiconductor are holes.
3. In the energy band diagram, the acceptor energy level lies just above the valence band.

Teacher's Note:
a) Trivalent impurities create electron vacancies which act as positive mobile holes.
b) The acceptor level accepts electrons from the valence band at room temperature.

2. How is an n-type semiconductor formed? Name the majority carriers in it. Draw the energy band diagram of a n-type semiconductor. [2 Marks]

Answer:
1. An n-type semiconductor is formed by doping a pentavalent impurity (such as phosphorus, arsenic, or antimony) into an intrinsic semiconductor.
2. The majority charge carriers in an n-type semiconductor are free electrons.
3. In the energy band diagram, the donor energy level lies just below the conduction band edge.

Teacher's Note:
a) Pentavalent dopants provide extra free electrons for electrical conduction.
b) Thermal energy easily excites electrons from the donor level into the conduction band.

3. With the help of a diagram, show the biasing of a light emitting diode (LED). Give its two advantages over conventional incandescent lamps. [2 Marks]

Answer:
1. An LED is heavily doped p-n junction diode operated in forward bias, where the p-side is connected to the positive terminal and the n-side to the negative terminal of a battery.
2. Advantage 1: Low operational voltage and low power consumption.
3. Advantage 2: Fast action and no warm-up time required with a long life.

Teacher's Note:
a) Forward biasing injects minority carriers across the junction where they recombine and emit photons.
b) LEDs are widely used due to their durability and high energy efficiency.

4. Draw a circuit diagram to show how a photodiode is biased. Draw its characteristic curves for two different illumination intensities. [2 Marks]

Answer:
1. A photodiode is operated in reverse bias mode, connected across a reverse bias voltage source with a series resistor to measure current.
2. The reverse current increases linearly with an increase in optical illumination intensity for a given reverse voltage.
3. Characteristic curves show distinct parallel current levels for different illumination intensities (\(I_2 \gt I_1\)) under reverse bias.

Teacher's Note:
a) Reverse bias helps in sweeping the photogenerated minority carriers quickly to produce measurable photocurrent.
b) Photodiodes detect optical signals rapidly because the reverse saturation current depends on light intensity.

5. Give the logic symbol for an AND gate. Draw the output wave form for input wave forms for this gate. [2 Marks]

Answer:
1. The AND gate logic symbol has two inputs (A and B) entering a D-shaped body and one output (Y = A · B).
2. The output wave form is HIGH (1) only when both input wave forms A and B are simultaneously HIGH (1), and LOW (0) otherwise.

Teacher's Note:
a) The boolean expression for an AND gate is Y = A AND B.
b) Students must carefully align clock pulses or time intervals when drawing input-output waveforms.

3 MARKS

6. What is rectification? How can a diode valve be used as half wave rectifier and full wave rectifier? [3 Marks]

Answer:
1. Rectification is the process of converting alternating current (AC) into direct current (DC) using rectifying devices like diodes.
2. Half-wave rectifier: Uses a single diode in series with the load; it conducts current only during the positive half-cycle of the AC input, blocking the negative half-cycle.
3. Full-wave rectifier: Uses two diodes with a center-tapped transformer or a bridge arrangement; both half-cycles of the AC input are utilized to pass unidirectional current through the load.

Teacher's Note:
a) Diodes offer low resistance in forward bias and high resistance in reverse bias.
b) Full-wave rectification provides a smoother DC output with a higher ripple frequency than half-wave rectification.

7. Explain how the depletion layer and the barrier potential are formed in a p-n junction diode. [3 Marks]

Answer:
1. When a p-n junction is formed, holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side due to the concentration gradient.
2. This diffusion leaves behind immobile ionized acceptors on the p-side and ionized donors on the n-side near the junction, creating a region devoid of free mobile carriers called the depletion layer.
3. The accumulation of these stationary charges sets up an internal electric field from the n-side to the p-side, which creates a potential barrier opposing further diffusion of charge carriers.

Teacher's Note:
a) Equilibrium is established when the diffusion current equals the drift current across the junction.
b) The height of the potential barrier depends on the doping concentration and temperature.

8. Draw a circuit diagram for use of NPN transistor as an amplifier in common emitter configuration. The input resistance of a transistor is \(1000\,\Omega\). On changing its base current by \(10\,\mu\text{A}\), the collector current increases by \(2\text{ mA}\). If a load resistance of \(5\text{ k}\Omega\) is used in the circuit, calculate (i) the current gain & (ii) voltage gain of the amplifier [3 Marks]

Answer:
1. Given data: Input resistance \(R_{in} = 1000\,\Omega\), change in base current \(\Delta I_b = 10\,\mu\text{A} = 10 \times 10^{-6}\text{ A}\), change in collector current \(\Delta I_c = 2\text{ mA} = 2 \times 10^{-3}\text{ A}\), load resistance \(R_L = 5\text{ k}\Omega = 5000\,\Omega\).
2. (i) Current gain \(\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{2 \times 10^{-3}}{10 \times 10^{-6}} = 200\).
3. (ii) Voltage gain \(A_v = \beta \times \frac{R_L}{R_{in}} = 200 \times \frac{5000}{1000} = 200 \times 5 = 1000\).

Teacher's Note:
a) Always convert microamperes and kilo-ohms to standard SI units before calculating gains.
b) The common emitter configuration provides both high current gain and high voltage gain.
c) The product of current gain and resistance ratio directly yields the voltage gain.

9. The output of an AND gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table. [3 Marks]

Answer:
1. Logic circuit connection: Two inputs A and B are fed into an AND gate. The output of the AND gate (Y' = A · B) is connected to both inputs of a subsequent NAND gate.
2. Since both inputs of the NAND gate are tied together, it functions as a NOT gate, inverting the output of the AND gate to give final output \(Y = \overline{A \cdot B}\), which is equivalent to a NAND gate.
3. Truth table:

Teacher's Note:
a) Connecting both inputs of any universal gate (NAND or NOR) converts it into a NOT gate.
b) The resulting logic combination behaves identically to a standard NAND gate.

10. What is a Zener diode? How it is symbolically represented? With the help of a circuit diagram, explain the use of Zener diode as a voltage stabilizer. [3 Marks]

Answer:
1. A Zener diode is a heavily doped semiconductor diode designed to operate reliably in the reverse breakdown region without getting damaged.
2. It is symbolically represented like a standard p-n junction diode except that the ends of the cathode bar are bent in opposite directions like a letter Z.
3. Voltage stabilizer working: Connected in parallel across the load resistance and reverse biased; when input voltage fluctuates, the Zener diode maintains a nearly constant voltage drop across the load by drawing varying current through a series resistor while operating in the breakdown region.

Teacher's Note:
a) Heavy doping causes a narrow depletion region, resulting in a low and sharp Zener breakdown voltage.
b) The voltage across the Zener diode remains constant even when current through it changes.

11. With the help of a suitable diagram, explain the formation of depletion region in a p-n junction. How does its width change when the junction is: (i) forward biased? & (ii) reverse biased? [3 Marks]

Answer:
1. Depletion region formation: Due to concentration gradients, holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side, leaving behind immobile ionized donors and acceptors near the junction.
2. (i) Forward bias: When the p-type is connected to the positive terminal and n-type to the negative terminal, the applied external field opposes the barrier potential, causing the depletion region width to decrease.
3. (ii) Reverse bias: When the p-type is connected to the negative terminal and n-type to the positive terminal, the external field assists the barrier potential, causing the depletion region width to increase.

Teacher's Note:
a) Forward bias lowers the potential barrier and facilitates easy flow of majority carriers.
b) Reverse bias widens the depletion layer and increases the potential barrier, restricting majority carrier flow.

5 MARKS

12. Explain briefly with the help of a circuit diagram how V-I characteristics of a p-n junction diode are obtained in: (i) forward bias & (ii) reverse bias. [5 Marks]

Answer:
1. Forward bias circuit: The p-type of the diode is connected to the positive terminal and the n-type to the negative terminal of a variable DC power supply through a milliammeter and a protective resistor.
2. Forward characteristics: As forward voltage increases gradually, current remains very small until the barrier potential is crossed (about 0.7 V for Si, 0.2 V for Ge), after which current rises sharply with a small increase in voltage.
3. Reverse bias circuit: The p-type is connected to the negative terminal and the n-type to the positive terminal of a high-voltage variable DC supply through a microammeter.
4. Reverse characteristics: A very small reverse saturation current (of the order of microamperes) flows, which remains almost constant up to a certain critical reverse voltage called the breakdown voltage.
5. Breakdown region: Beyond the breakdown voltage, the reverse current increases suddenly to a large value due to avalanche or zener breakdown mechanisms.

Teacher's Note:
a) Mention the appropriate instruments: milliammeter for forward current and microammeter for reverse current.
b) Clearly state the difference in scales on the graph axes for forward (mA) and reverse (μA) currents.
c) Emphasize the role of the protective resistance in forward biasing to prevent damage from overheating.
d) Explain the distinct physical phenomena governing minority carrier drift in reverse bias versus majority diffusion in forward bias.

13. Explain the function of base region of a transistor. Why this region is made thin and lightly doped? Draw a circuit diagram to study the input and the output characteristics of n-p-n transistor in a common emitter (CE) configuration. Show these characteristics graphically. Explain how current amplification factor of the transistor is calculated using output characteristics. [5 Marks]

Answer:
1. Function of base: The base forms the middle thin section of a transistor; its primary function is to pass majority charge carriers injected from the emitter into the collector.
2. Reason for thin and light doping: It is made thin and lightly doped to minimize recombination of electrons and holes, ensuring that almost all carriers injected by the emitter reach the collector.
3. Circuit diagram and characteristics: In common emitter configuration, emitter is common to input (base-emitter) and output (collector-emitter). Input characteristics plot base current (\(I_B\)) versus base-emitter voltage (\(V_{BE}\)) at constant collector-emitter voltage (\(V_{CE}\)). Output characteristics plot collector current (\(I_C\)) versus collector-emitter voltage (\(V_{CE}\)) at constant base current (\(I_B\)).
4. Graphical representation: Input curves resemble forward diode characteristics shifted right by \(V_{CE}\). Output curves show active, saturation, and cut-off regions where \(I_C\) varies slightly with \(V_{CE}\) for a fixed \(I_B\).
5. Current amplification factor (\(\beta\)): It is calculated from the output characteristics using the formula \(\beta = \frac{\Delta I_C}{\Delta I_B}\) at a constant \(V_{CE}\).

Teacher's Note:
a) Highlight that the base region controls the flow of charge carriers between emitter and collector.
b) Ensure correct polarities are shown in the circuit diagram for an n-p-n transistor in CE mode.
c) Remember that \(\beta\) values are typically high, ranging from 20 to 200 in standard transistors.
d) Verify the active region operating conditions where the emitter-base junction is forward-biased and collector-base junction is reverse-biased.

14. Draw the energy bands of p-type and n-type semiconductors. Explain with a circuit diagram the working of a full wave rectifier. [5 Marks]

Answer:
1. Energy band of n-type: An additional donor energy level lies just below the conduction band, populated by donor electrons ready for thermal excitation.
2. Energy band of p-type: An acceptor energy level lies just above the valence band, creating vacant acceptor states that accept electrons from the valence band.
3. Full wave rectifier circuit: It uses two diodes (\(D_1\) and \(D_2\)) connected to the secondary of a center-tapped transformer, with a load resistor \(R_L\) connected across the common output.
4. Working during positive half cycle: During the positive half cycle of AC input, diode \(D_1\) becomes forward-biased and conducts, while \(D_2\) is reverse-biased and blocks current.
5. Working during negative half cycle: During the negative half cycle, diode \(D_2\) becomes forward-biased and conducts, while \(D_1\) is reverse-biased, ensuring current through \(R_L\) flows in the same direction for both halves.

Teacher's Note:
a) Draw clear energy band diagrams marking the valence band, conduction band, and impurity levels.
b) Emphasize the importance of the center tap on the transformer secondary winding.
c) Mention that the output frequency of a full wave rectifier is double the input AC frequency (2f).
d) Discuss the unidirectional pulsating DC output obtained across the load resistor.

15. Explain with the help of a circuit diagram the use of an n-p-n transistor as an amplifier in common emitter configuration. Draw the input and output wave forms of the signal. Write the expression for its voltage gain. [5 Marks]

Answer:
1. Circuit arrangement: An n-p-n transistor in common emitter configuration has its emitter grounded, base forward-biased by \(V_{BB}\), and collector reverse-biased by \(V_{CC}\) through load resistance \(R_L\).
2. Working principle: A small alternating input signal voltage \(V_i\) is superimposed on the DC base bias, producing a large fluctuation in base current \(\Delta I_B\).
3. Amplification action: Due to transistor action, this causes a much larger fluctuation in collector current \(\Delta I_C\), creating a large amplified voltage drop across \(R_L\).
4. Phase reversal: The input and output waveforms are out of phase by 180 degrees, meaning a positive peak in input results in a negative peak in output.
5. Voltage gain expression: Voltage gain (\(A_v\)) is given by \(A_v = \frac{V_o}{V_i} = \beta \times \frac{R_L}{r_i}\), where \(\beta\) is the current amplification factor and \(r_i\) is the input resistance.

Teacher's Note:
a) Ensure input and output sinusoidal waveforms clearly indicate the 180-degree phase shift.
b) Highlight the distinction between AC load line and DC load line in amplifier analysis.
c) Note that power gain is much higher than voltage gain in a common emitter amplifier configuration.
d) Check the biasing conditions to ensure the transistor operates strictly within the active region.

16. What is an n-p-n transistor? How does it differ from p-n-p transistor? Give their symbols. Explain transistor action. [5 Marks]

Answer:
1. Definition of n-p-n transistor: An n-p-n transistor is a three-terminal semiconductor device formed by sandwiching a thin layer of p-type semiconductor between two n-type semiconductor regions.
2. Differences: In an n-p-n transistor, majority charge carriers are electrons, offering higher mobility and switching speeds compared to a p-n-p transistor where holes are majority carriers.
3. Circuit symbols: In an n-p-n symbol, the arrow on the emitter points outwards, whereas in a p-n-p symbol, the emitter arrow points inwards towards the base.
4. Transistor action (Injection): When the emitter-base junction is forward-biased, a large number of electrons are injected from the n-type emitter into the thin p-type base.
5. Transistor action (Collection): As these electrons cross the base, very few recombine with holes, and the vast majority are swept into the collector due to the strong reverse bias at the collector-base junction, constituting the collector current.

Teacher's Note:
a) Clearly draw circuit symbols with arrowheads pointing in the correct directions for both transistor types.
b) Emphasize that emitter current equals the sum of base current and collector current (\(I_E = I_B + I_C\)).
c) Mention that n-p-n transistors are generally preferred over p-n-p due to higher electron mobility.
d) Explain the biasing requirements: emitter-base forward-biased and collector-base reverse-biased for normal active operation.