NIOS Class 12 Physics Chapter 29 Applications of Semiconductor Devices

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NIOS Class 12 Physics Chapter 29 Applications of Semiconductor Devices

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Applications of Semiconductor Devices

Chapter: 29

Module-VIII: Semiconductor Devices And Communication

INTEXT QUESTIONS 29.1

1. Draw a circuit of full-wave rectifier with a filter capacitor.

Ans: A full-wave rectifier with a filter capacitor is an essential circuit that converts alternating current (AC) to direct current (DC) while reducing the ripple content in the output. The circuit consists of a center-tapped transformer, two p-n junction diodes, a load resistor, and a filter capacitor connected in parallel with the load.

Full-wave rectifier circuit with filter capacitor.

The circuit works on the principle that the two diodes conduct alternately during positive and negative half-cycles of the input AC voltage. During the positive half-cycle of the input, diode D1 conducts while D2 remains reverse-biased. The current flows through the load resistor RL from point B to Y. During the negative half-cycle, diode D2 conducts while D1 is reverse-biased, and current again flows through the load in the same direction from B to Y.

The filter capacitor C is connected parallel to the load resistor RL. This capacitor charges to nearly the peak voltage during the conducting period of the diodes. When the diode current tends to decrease, the capacitor discharges through the load, maintaining current flow and reducing voltage fluctuations. The larger the capacitance value and load resistance, the smaller will be the ripple in the output DC voltage. This filtering action results in a much smoother DC output compared to unfiltered rectification.

2. What will be the output voltage, if you connect a Zener diode in forward bias instead of reverse bias in the regulator circuit of Example 29.1?

Ans: If a Zener diode is connected in forward bias instead of reverse bias in the voltage regulator circuit, it will behave like an ordinary p-n junction diode rather than functioning as a voltage regulator. In this incorrect configuration, the output voltage will not be stabilized at the Zener breakdown voltage.

When a Zener diode is forward-biased, it conducts current just like a regular diode with a forward voltage drop of approximately 0.7V (for silicon diodes). The diode will not exhibit its characteristic voltage regulation property, which only occurs in reverse bias when the voltage exceeds the Zener breakdown voltage. In the forward-biased condition, the diode offers very low resistance to current flow, and most of the input voltage will appear across the series resistance Rs.

Therefore, the output voltage across the load will be approximately (Vi – 0.7V), where Vi is the input voltage and 0.7V is the forward voltage drop across the Zener diode. This output voltage will vary directly with changes in input voltage, completely defeating the purpose of voltage regulation. For Example 29.1, instead of getting a stabilized 6V output, the output would vary between approximately (16.5 – 0.7) = 15.8V to (21 – 0.7) = 20.3V as the input voltage changes, providing no voltage regulation whatsoever.

INTEXT QUESTIONS 29.2

1. For a CE mode amplifier, υ_i is 20 mV and υ_o is one volt. Calculate voltage gain.

Ans: Given: Input voltage, υ_i = 20 mV 

= 20 × 10^-3 V

= 0.02 V

Output voltage, υo = 1 V

The voltage gain (AV) of an amplifier is defined as the ratio of output voltage to input voltage.

Using the formula for voltage gain:

AV = υ_o / υ_i

Substituting the given values:

AV = 1 V / 0.02 V

AV = 50

Therefore, the voltage gain of the CE mode amplifier is 50.

This means that the amplifier increases the input signal voltage by a factor of 50. The CE configuration typically provides high voltage gain, making it suitable for voltage amplification applications.

2. The P_o of an amplifier is 200 times that P_i. Calculate the power gain.

Ans: Given:

P_o = 200 × Pi (where Po is output power and Pi is input power)

The power gain (AP) of an amplifier is defined as the ratio of output power to input power.

Using the formula for power gain:

AP = P_o / P_i

From the given condition:

P_o = 200 × Pi

Substituting this into the power gain formula:

AP = (200 × Pi) / Pi

AP = 200

Therefore, the power gain of the amplifier is 200.

This high power gain indicates that the amplifier is very effective in increasing the power level of the input signal, making it suitable for applications where significant power amplification is required, such as in audio power amplifiers or RF transmitters.

3. For a CE amplifier, R_L = 2000 Ω, r_i = 500 Ω and β = 50. Calculate voltage gain and power gain.

Ans: Given: Load resistance, RL = 2000 Ω

Input resistance, r_i = 500 Ω

Current amplification factor, β = 50

Calculate Voltage Gain (AV)

For a CE amplifier, the voltage gain is given by:

AV = β × RL / ri

Substituting the given values:

AV = 50 × 2000 Ω / 500 Ω

AV = 100000 / 500

AV = 200

Calculate Power Gain (AP)

The power gain is related to current gain and voltage gain:

AP = AI × AV

For a CE amplifier, the current gain AI = β = 50

Therefore:

AP = β × AV

AP = 50 × 200

AP = 10000

Voltage gain, AV = 200

Power gain, AP = 10000

The high voltage gain of 200 means the output voltage is 200 times the input voltage, while the power gain of 10000 indicates that the output power is 10000 times the input power, demonstrating the excellent amplification characteristics of the CE configuration.