Analog Electronics

Step 1: Understand the circuit.
The circuit contains an ideal Op-Amp followed by a diode in series with the output. The diode ensures that the output appears only when the Op-Amp forward-biases it. Since the Op-Amp is ideal, it will produce whatever output is needed to make the diode conduct when possible.

Step 2: Output behaviour for .
When is too small, the Op-Amp must generate internally. However, the diode blocks conduction until the voltage at the diode input exceeds . Thus for , the external output .

Step 3: Output behaviour for .
Once exceeds , the diode forward-biases. The Op-Amp output is now able to appear across the load resistor. Thus , a straight-line graph with slope 1 starting at .

Step 4: Conclusion.
The graph must be a straight line shifted right by , which corresponds to option (B).

Step 1: Determine emitter voltage.
The emitter is connected to a 3V supply through a 1k resistor. With forward diode drop 0.7V, emitter terminal sits at:
.

Step 2: Determine base voltage.
Base is connected to emitter through a 3k resistor to ground. Since is large, base current is negligible.
Thus the base voltage is equal to emitter voltage: .

Step 3: Determine collector current.
Emitter and collector currents are nearly equal because .
.

Step 4: Compute collector voltage.
The collector resistor is 3k to +10V.
Voltage drop = .
Thus collector voltage = .

Step 5: Adjust for actual base-emitter relation.
Accounting for current through resistors more accurately gives .

Step 1: Understand Op-amp behavior.
The Op-Amp keeps the inverting and non-inverting terminals at the same voltage (ideal Op-Amp). Both nodes thus stay at .

Step 2: Compute current through 1k resistor.
The emitter resistor is 1k and bottom node is grounded, so emitter current flows through 1k from +10V:
.

Step 3: Relate emitter and base currents.
For transistor: and .
Thus, .

Step 4: Compute base current.
.

Step 1: Note that .
Thus capacitor charges to peak voltage and discharges very slowly — typical peak detector.

Step 2: Input is 24 Vrms.
Peak voltage = .

Step 3: Ideal diode.
No drop across diode, so capacitor charges to full peak voltage.

Step 4: DC output across .
Since discharge is negligible, .

Step 1: Condition for phase difference.
For an RC circuit, the phase difference between and is when:

Step 2: Substitute given values.

Step 3: Find frequency.

Step 4: Conclusion.
Hence, the input frequency for a phase difference is .

Step 1: Understanding the Concept:
A Zener diode is used for voltage regulation. When reverse biased, it maintains a constant voltage across its terminals, provided the input voltage is high enough to keep it in the "breakdown" region. If the conditions do not support breakdown, the Zener acts like an open circuit. We must check the operating condition for each load resistance.
Step 2: Key Formula or Approach:
1. Determine if the Zener diode is in breakdown. This occurs if the voltage across the load (if the Zener were absent) is greater than . This no-load voltage is found using the voltage divider rule: . 2. If , the Zener is ON, and the load voltage is regulated to . The load current is . 3. If , the Zener is OFF and acts as an open circuit. The load current is then determined by the simple series circuit: .
Step 3: Detailed Explanation:
Given: Source voltage V, Series resistance k , Zener voltage V. Case 1: k - Check the condition for breakdown: - Since , the Zener diode is ON and regulating. - The voltage across the load is V. - The load current is . Case 2: k - Check the condition for breakdown: - Since , the Zener diode is OFF (not in breakdown). - It acts as an open circuit. The circuit is a simple voltage divider with and . - The load current is the total current in this series circuit: Calculate the ratio : Step 4: Final Answer:
The ratio , rounded to two decimal places, is 1.80.

Step 1: Understanding the Concept:
For a BJT to operate in the saturation regime, the base current must be large enough to cause the collector current to reach its maximum possible value, known as the saturation current . The maximum value of the base resistance corresponds to the minimum base current required to just achieve saturation ( ).
Step 2: Key Formula or Approach:
1. Apply Kirchhoff's Voltage Law (KVL) to the collector-emitter loop to find the collector saturation current, . 2. Use the current gain to find the minimum base current required for saturation, . 3. Apply KVL to the base-emitter loop to find the maximum base resistance that allows this minimum saturation current to flow. Step 3: Detailed Explanation:
Given values: V, k , V, V, V, .
1. Calculate : From the collector loop KVL: 2. Calculate : This is the minimum base current to put the transistor in saturation. 3. Calculate the maximum : Using this minimum required base current in the base loop KVL: Step 4: Final Answer:
The maximum value of the base resistance is 20 k .

Step 1: Understanding the Concept:
The given circuit is a bridge circuit, specifically a Maxwell bridge (or a similar AC bridge). We need to find the phase difference between the output voltage across points A and B ( ) and the input voltage . This involves calculating the complex voltages and using the voltage divider rule in the phasor domain.
Step 2: Key Formula or Approach:
We will use complex impedance to analyze the AC circuit.

Impedance of a resistor R:
Impedance of an inductor L:
Impedance of a capacitor C:
The voltage divider rule states that the voltage across an impedance in series with is .
We need to find the phase of . Let be the reference phasor, .
Step 3: Detailed Explanation:
1. Calculate the impedances of the components: Given rad/s.

Resistors: ,
Inductor:
Capacitor:
2. Calculate the voltages at points A and B using the voltage divider rule:

For point A (left branch): is the voltage across the inductor .
For point B (right branch): is the voltage across the capacitor .
3. Calculate the output voltage : Find a common denominator: 4. Determine the phase difference: The relationship is . The complex number corresponds to in polar form, which represents a magnitude of 1 and a phase angle of or radians. This means that the phasor is rotated by with respect to the phasor . Therefore, the output voltage leads the input voltage by . Step 4: Final Answer:
The output voltage is related to the input voltage by . This implies a phase lead of . Thus, option (C) is correct.

Step 1: Understanding the Concept:
The given circuit is an inverting differentiator. An operational amplifier (Op-Amp) with a capacitor in the input path and a resistor in the feedback path acts as a differentiator. The output voltage is proportional to the negative time derivative of the input voltage. We are given a ramp function as input and need to find the corresponding output waveform.
Step 2: Key Formula or Approach:
For an ideal Op-Amp in the inverting configuration, the virtual ground principle applies, meaning the voltage at the inverting input ( ) is equal to the voltage at the non-inverting input ( ). Here, (grounded), so .
The current flowing through the capacitor is . Since , .
The current flowing through the feedback resistor is . Since , .
For an ideal Op-Amp, the input current is zero, so .
Therefore, .
This gives the input-output relationship for a differentiator circuit: \textit{Note: The component labels in the diagram are swapped. The input element should be the capacitor and the feedback element should be the resistor for a standard differentiator. The given circuit is an integrator. Let's solve for the given circuit.} The given circuit is an inverting integrator. The input is through a resistor and the feedback element is a capacitor . For an integrator, the output voltage is: Step 3: Detailed Explanation:
1. Analyze the input signal: The input is a ramp function. This means for some positive constant . The graph shows a straight line with a positive slope starting from the origin. 2. Apply the integrator formula: We need to integrate the input signal. Assume the initial voltage across the capacitor (and hence the initial output voltage) is zero, i.e., . 3. Analyze the output waveform: The output voltage is , where is a positive constant. This equation describes a parabola that opens downwards and starts from the origin (0,0). Let's check the given options:

(A) Shows a function that increases faster than a line (like ), but it is positive. This is incorrect due to the negative sign.
(B) Shows a constant negative output. This would be the output if the input were a constant positive voltage. (Incorrect)
(C) Shows a negative step function. This is incorrect.
(D) Shows a parabola opening downwards, starting from the origin. This matches our derived result .
Step 4: Final Answer:
The circuit is an inverting integrator. The integral of a ramp function ( ) is a parabolic function ( ). The plot that correctly represents a downward-opening parabola starting from the origin is (D).