All questions of Analog Electronics for Electrical Engineering (EE) Exam

Calculation of Input Impedance Zf:
- The input impedance Zf of a CD MOSFET amplifier with voltage divider bias can be calculated using the formula:
Zf = R1 || R2 + (1 + g_m * (R1 || R2))^-1
- Given that R1 = 1.5 MΩ and R2 = 1 MΩ, we can substitute these values into the formula.

Calculating R1 || R2:
- R1 || R2 = (R1 * R2) / (R1 + R2)
- R1 || R2 = (1.5 MΩ * 1 MΩ) / (1.5 MΩ + 1 MΩ)
- R1 || R2 = 1.5 MΩ * 1 MΩ / 2.5 MΩ
- R1 || R2 = 0.6 MΩ = 600 kΩ

Calculating g_m:
- Given parameters of the MOSFET and biasing conditions, the transconductance g_m is typically provided in the problem statement.

Substitute into the Input Impedance Formula:
- Zf = 600 kΩ + (1 + g_m * 600 kΩ)^-1

Final Calculation:
- Since we are given the options, we can calculate the input impedance for each option and find that the correct answer is option 'B' 600 kΩ.
Therefore, the input impedance Zf of the Common Drain (CD) MOSFET amplifier with voltage divider bias is 600 kΩ.

To find the input resistance of the network, we can use the formula:

Input resistance (Rin) = (Voltage gain / Current gain) * (Feedback factor / Amplifier gain)

Given:
Amplifier gain (Av) = 100
Feedback factor (β) = 5

The voltage gain (Av) can be calculated as the ratio of output voltage (Vo) to input voltage (Vi):

Av = Vo / Vi

Since it is a voltage series feedback network, the output voltage (Vo) and input voltage (Vi) are related by the formula:

Vo = Vi / (1 + β * Av)

Substituting the given values:

Av = 100
β = 5

Vo = Vi / (1 + 5 * 100)
Vo = Vi / 501

Now, we can calculate the output voltage (Vo) using the formula:

Vo = Av * Vi

Vo = 100 * 4V
Vo = 400V

Substituting the calculated values into the equation for Vo:

400V = Vi / 501

Solving for Vi:

Vi = 400V * 501
Vi = 200,400V

Now, we can substitute the calculated values into the formula for input resistance (Rin):

Rin = (Av / Ai) * (β / Av)

Rin = (100 / (2mA / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / (0.002A / 200,400V)) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = (100 / 0.002A) * (5 / 100)
Rin = 1,000 * 0.05
Rin = 50

Therefore, the input resistance of the network is 50 ohms.

Both assertion and reason are individually correct statements. However, the reason for assertion is that due to the applied electric field, and electrostatic force is developed on the electron and the electrons would be accelerated in a direction opposite to the applied electric field and this motion is called directed motion of electron

Understanding Bandwidth in RC-Coupled Amplifiers
The bandwidth of an RC-coupled amplifier is influenced by various factors, primarily the coupling and bypass capacitors, as well as device shunt capacitors. Here’s a detailed breakdown of how these components affect bandwidth:
Low-Frequency Limitations
- Coupling Capacitors:
- These capacitors block DC voltage while allowing AC signals to pass. At low frequencies, their reactance increases, leading to a reduction in gain and consequently limiting bandwidth.
- Bypass Capacitors:
- Used to improve the gain of the amplifier by providing a low impedance path to ground for AC signals. However, their effectiveness diminishes at lower frequencies, contributing to bandwidth limitations in that range.
High-Frequency Limitations
- Device Shunt Capacitors:
- These include parasitic capacitances inherent in the amplifier devices (transistors, etc.). At high frequencies, they can create a feedback path that reduces gain, leading to a decrease in bandwidth.
- Coupling and Bypass Capacitors:
- At high frequencies, the reactance of these capacitors also plays a role in limiting bandwidth. Their behavior can lead to phase shifts and gain reduction, further restricting the upper frequency limit.
Conclusion
The correct answer, option 'C', highlights that both coupling and bypass capacitors limit bandwidth at the low frequency end while device shunt capacitors impose limitations at the high frequency end. Understanding these components’ roles is crucial for designing amplifiers with desired bandwidth characteristics.

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Understanding Voltage and Current Phasors
In electrical engineering, phasors are a convenient way to represent sinusoidal waveforms. The voltage phasor (V) and the current phasor (I) are expressed in complex form, capturing both magnitude and phase information.
What is Apparent Power?
- Apparent power (S) is defined as the product of the voltage phasor and the conjugate of the current phasor.
- Mathematically, it can be expressed as S = V * I* (where I* is the complex conjugate of I).
- Apparent power is measured in volt-amperes (VA) and represents the total power flowing in the circuit, without considering the phase difference between voltage and current.
Why is VI Apparent Power?
- When we consider the product VI, we are essentially calculating the apparent power, which indicates how much power is "apparent" in the circuit.
- This value includes both active (real) power and reactive power, making it a comprehensive measure of power.
Components of Power
- Active Power (P): The actual power consumed in the circuit, measured in watts (W). It is the power that performs work.
- Reactive Power (Q): The power stored and released by inductors and capacitors, measured in reactive volt-amperes (VAR). It does not perform any work but is essential for maintaining the voltage levels.
Conclusion
- Thus, the product VI gives us the apparent power, making option 'A' the correct answer. Understanding this concept is crucial for analyzing AC circuits and managing power effectively in electrical systems.

Introduction:
The common drain amplifier, also known as a source follower, is a type of field-effect transistor (FET) amplifier. It is commonly used to provide a high input impedance and low output impedance. The voltage gain of the common drain amplifier can be determined by analyzing its small-signal equivalent circuit.

Explanation:
The voltage gain of a common drain amplifier is determined by the ratio of the output voltage to the input voltage. In the common drain configuration, the input is applied to the gate terminal, and the output is taken from the drain terminal. The source terminal is connected to a fixed reference voltage.

Operating Principle:
To understand the voltage gain of the common drain amplifier, it is important to understand its operating principle. When a small AC signal is applied to the gate terminal, it modulates the channel current flowing through the transistor. This modulated current flows through the load resistor connected to the drain terminal, producing an output voltage across it.

Factors Affecting Voltage Gain:
The voltage gain of a common drain amplifier is affected by several factors, including the transconductance (gm) of the transistor, the load resistor (RL), and the drain resistor (RD). The transconductance represents the ability of the transistor to convert a small change in the input voltage into a corresponding change in the output current.

Analysis of Small-Signal Equivalent Circuit:
To determine the voltage gain, we can analyze the small-signal equivalent circuit of the common drain amplifier. The small-signal equivalent circuit includes the transistor, the load resistor, and the drain resistor. By applying nodal analysis to this circuit, we can derive an expression for the voltage gain.

Voltage Gain Calculation:
The voltage gain (Av) of the common drain amplifier can be calculated using the following equation:

Av = -gm * RL

Where:
- Av is the voltage gain
- gm is the transconductance of the transistor
- RL is the load resistor

Conclusion:
In conclusion, the voltage gain of a common drain amplifier is always slightly less than 1. This is because the transconductance of the transistor is typically less than 1, and the voltage gain is directly proportional to the transconductance. Therefore, the correct answer is option B) 1.

Calculating Thermal Resistance for 2N338 Transistor:
In this case, we are given the maximum power dissipation (PC,max) of the 2N338 transistor as 125 mW at a free-air temperature of 25°C and a maximum junction temperature of 150°C. To calculate the thermal resistance of the transistor, we can use the formula:

Thermal Resistance (θ) = (Tj - Ta) / PC,max
Where:
- θ is the thermal resistance
- Tj is the junction temperature (150°C)
- Ta is the ambient temperature (25°C)
- PC,max is the maximum power dissipation (125 mW)
Now, substituting the given values into the formula:
θ = (150°C - 25°C) / 125 mW
θ = 125°C / 125 mW
θ = 1.0 °C/mW
Therefore, the thermal resistance of the 2N338 transistor is 1.0 °C/mW. This means that for every 1 mW of power dissipated by the transistor, the junction temperature will increase by 1.0°C above the ambient temperature.

Analogies between FET and BJT

An analogy between Field Effect Transistor (FET) and Bipolar Junction Transistor (BJT) can be drawn in the context of their functioning. Both of these electronic devices are used for amplifying signals, and they have a similar structure in terms of the presence of a source, gate, and drain in FET and collector, base, and emitter in BJT. However, the way they function is different, and the analogy is drawn between the specific parts of the devices.

The drain of FET and collector of BJT

In FET, the drain is the part where the current flows out of the device. Similarly, in BJT, the collector is the part where the current flows out of the device. The analogy between these two parts is drawn based on their functionality. The collector in BJT is responsible for collecting the majority carriers (either electrons or holes) that flow from the emitter and then move towards the base. Similarly, the drain in FET is responsible for collecting the majority carriers that flow from the channel to the drain.

The analogy between the drain of FET and collector of BJT is further strengthened by the fact that both of these parts are operated in the reverse-biased mode. In BJT, the collector-base junction is reverse-biased, while in FET, the gate-source junction is reverse-biased. In both cases, the reverse-biasing results in a depletion region, which helps to control the flow of current through the device.

Conclusion

In conclusion, the analogy between the drain of FET and the collector of BJT is drawn based on their functionality and the way they operate in the device. Both of these parts are responsible for collecting the majority carriers and are operated in the reverse-biased mode. However, it is important to note that this analogy is limited to specific parts of the devices and does not imply that FET and BJT are the same.

Theory.

An amplifier has an open loop voltage gain of "A." The open loop voltage gain, also known as the gain without any feedback applied, is a measure of how much the amplifier amplifies the input voltage. It is typically expressed as a ratio or in decibels.

The open loop voltage gain can be calculated by dividing the output voltage by the input voltage:

Open Loop Voltage Gain (A) = Output Voltage / Input Voltage

For example, if an amplifier has an output voltage of 10V and an input voltage of 1V, the open loop voltage gain would be 10.

It is important to note that the open loop voltage gain is not a fixed value and can vary depending on the frequency of the input signal. Amplifiers often have a frequency response curve that shows how the gain changes with frequency.

Additionally, the open loop voltage gain is usually much higher than the closed loop voltage gain when feedback is applied. Feedback is commonly used to stabilize the amplifier's performance and reduce distortion.

To calculate the charge deposited on the plane when struck by a lightning bolt, we can use the equation Q = I * t, where Q is the charge, I is the current, and t is the time.

Given:
Current (I) = 15,000 A
Time (t) = 100 s

Calculating the Charge:
Using the formula Q = I * t, we can substitute the given values:
Q = 15,000 A * 100 s
Q = 1,500,000 C

Therefore, the charge deposited on the plane when struck by the lightning bolt is 1,500,000 Coulombs or 1.5 C.

In this case, the correct answer is option 'D'.

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Explanation:

When an electric field is applied across a semiconductor, the free electrons in the material experience a force due to the field. This force causes the electrons to accelerate and gain kinetic energy. However, this energy gained by the electrons is not permanently retained. Instead, it is lost as heat through collisions with atoms in the crystal lattice of the semiconductor material.

Collisions with Atoms in the Crystal:
- When the accelerated electrons collide with atoms in the crystal lattice, they transfer some of their kinetic energy to these atoms.
- These collisions cause the atoms to vibrate, increasing their thermal energy and consequently increasing the temperature of the material.
- This process is known as electron-phonon scattering, where phonons represent the lattice vibrations.

Reason for Choosing Option 'B':
- The other options listed (recombination with holes, radiation while being accelerated, and collision with other electrons) do not play a significant role in the loss of energy as heat in a semiconductor under an applied electric field.
- Recombination with holes refers to the combination of a free electron with an empty state (hole) in the valence band, resulting in the release of energy as light or heat. However, this process is not directly related to the loss of energy by accelerated electrons in an electric field.
- Radiation while being accelerated is more relevant in the context of charged particles moving close to the speed of light or in high-energy physics. In the case of electrons in a semiconductor under a typical electric field, the energy lost through radiation is negligible.
- Collisions with other electrons can lead to scattering and resistance in the material but do not directly result in the loss of energy as heat.

Conclusion:
When an electric field is applied across a semiconductor, the kinetic energy gained by the free electrons is lost primarily as heat through collisions with atoms in the crystal lattice. These collisions transfer energy to the lattice, increasing its thermal energy and causing the material to heat up. This phenomenon is an essential aspect of understanding the behavior of semiconductors under electric fields.