Q. 1
Electronics Engineering
Difficulty: Easy
(1 Mark)
An ideal operational amplifier (Op-Amp) has which set of theoretical electrical characteristics?
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Step-by-Step Explanation & Concept Rationale
An ideal op-amp draws zero input current (Rin = inf), can drive any load without voltage drop (Rout = 0), has infinite differential gain, infinite bandwidth, infinite CMRR, infinite slew rate, and perfect symmetry (Vout = 0 when V+ = V-).
Q. 2
Electronics Engineering
Difficulty: Easy
(1 Mark)
The 'Virtual Ground' concept in an inverting op-amp configuration is valid only when:
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Step-by-Step Explanation & Concept Rationale
With negative feedback, Vout = Aol * (V+ - V-). Since Aol is enormous (~10^5 to 10^6) and Vout is finite, the differential input voltage (V+ - V-) = Vout / Aol ≈ 0. If the non-inverting terminal V+ is connected to ground (0 V), the inverting terminal V- is held at 0 V ('virtual ground') without being physically wired to ground.
Q. 3
Electronics Engineering
Difficulty: Easy
(1 Mark)
The closed-loop voltage gain of an inverting op-amp amplifier with input resistor R1 and feedback resistor Rf is:
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Step-by-Step Explanation & Concept Rationale
By KCL at the virtual ground node (V- = 0): (Vin - 0) / R1 + (Vout - 0) / Rf = 0. Solving yields Av = Vout / Vin = -Rf / R1.
Q. 4
Electronics Engineering
Difficulty: Easy
(1 Mark)
The closed-loop voltage gain of a non-inverting op-amp amplifier with ground resistor R1 and feedback resistor Rf is:
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Step-by-Step Explanation & Concept Rationale
Voltage at the inverting node tracks input voltage by virtual short: V- = V+ = Vin. From the voltage divider across feedback path: Vin = Vout * [R1 / (R1 + Rf)]. Solving gives Av = Vout / Vin = 1 + (Rf / R1). Gain is strictly >= 1 and non-inverting.
Q. 5
Electronics Engineering
Difficulty: Easy
(1 Mark)
An op-amp Voltage Follower (Buffer) is constructed by setting Rf = 0 and R1 = infinity. Its voltage gain, input impedance, and output impedance are:
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Step-by-Step Explanation & Concept Rationale
For a voltage follower, direct negative feedback forces Vout = Vin (Av = 1). It draws zero current from signal sources (Rin ≈ 10^12 ohms for FET inputs) while driving heavy loads with near-zero output impedance, making it the ideal unity-gain impedance buffer.
Q. 6
Electronics Engineering
Difficulty: Easy
(1 Mark)
Slew Rate (SR) of an operational amplifier is defined as:
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Step-by-Step Explanation & Concept Rationale
Slew Rate SR = (dVout / dt)_max, typically measured in Volts per microsecond (V/μs). It is caused by the finite internal charging current (I_tail) available to charge the internal Miller frequency-compensation capacitor (Cc): SR = I_tail / Cc. Standard 741 has SR ≈ 0.5 V/μs.
Q. 7
Electronics Engineering
Difficulty: Medium
(1 Mark)
What is the Full-Power Bandwidth (fm) of an op-amp with slew rate SR delivering an undistorted sinusoidal output of peak amplitude Vp?
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Step-by-Step Explanation & Concept Rationale
For Vout(t) = Vp * sin(2*pi*f*t), maximum rate of change is (dVout/dt)_max = 2*pi*f*Vp. For distortion-free reproduction without slewing (triangular distortion), this must not exceed SR: 2*pi*fm*Vp <= SR, giving fm = SR / (2 * pi * Vp).
Q. 8
Electronics Engineering
Difficulty: Medium
(1 Mark)
An ideal op-amp Integrator circuit has a resistor R connected to the inverting input and a capacitor C in the feedback loop. Its output voltage vout(t) in response to input vin(t) is:
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Step-by-Step Explanation & Concept Rationale
Current through input resistor is i = vin / R. Because no current enters the op-amp input, this current charges the feedback capacitor: i = -C * (dvout / dt). Equating gives dvout/dt = -(1 / RC) * vin, so vout(t) = -(1 / RC) * integral vin(tau) dtau.
Q. 9
Electronics Engineering
Difficulty: Medium
(1 Mark)
Why must a practical op-amp integrator include a high-value feedback resistor (Rf) in parallel with the integrating capacitor C?
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Step-by-Step Explanation & Concept Rationale
At DC (f = 0), a pure capacitor acts as an open circuit, making the op-amp operate at its massive open-loop gain (~10^5). Any tiny input offset voltage Vio is multiplied by Aol, saturating the output at the power rail. Shunting C with Rf clamps the DC closed-loop gain to -Rf / R1.
Q. 10
Electronics Engineering
Difficulty: Medium
(1 Mark)
An op-amp Differentiator circuit has input capacitor C and feedback resistor R. Its major practical design limitation is:
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Step-by-Step Explanation & Concept Rationale
Differentiator gain magnitude is |Av| = omega * R * C. High-frequency noise components are amplified violently, and the 90-degree phase lead combined with internal op-amp poles can induce Barkhausen oscillation. Practical differentiators add a small series input resistor and feedback capacitor to roll off high-frequency gain.
Q. 11
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Schmitt Trigger circuit uses an op-amp with which type of feedback to provide hysteresis?
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Step-by-Step Explanation & Concept Rationale
A Schmitt Trigger applies positive feedback from output to non-inverting input. This establishes two distinct switching thresholds: Upper Trigger Point (UTP) and Lower Trigger Point (LTP). The difference (UTP - LTP) forms a hysteresis band that prevents noisy, slow-varying signals from causing false multiple output transitions.
Q. 12
Electronics Engineering
Difficulty: Medium
(1 Mark)
An Instrumentation Amplifier (In-Amp) typically uses a 3-op-amp topology. Its primary architectural advantage over a simple 1-op-amp difference amplifier is:
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Step-by-Step Explanation & Concept Rationale
A 3-op-amp In-Amp uses two non-inverting buffer input stages followed by a balanced differential stage. Signal sources see pure non-inverting gate/base inputs (infinite impedance without loading), common-mode signals pass with unity gain while differential signals are amplified by (1 + 2R/Rg), providing massive CMRR (> 100 dB).
Q. 13
Electronics Engineering
Difficulty: Medium
(1 Mark)
A Precision Rectifier (superdiode) solves the major limitation of standard silicon diodes in small-signal rectification by:
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Step-by-Step Explanation & Concept Rationale
Standard silicon diodes cannot rectify signals below their 0.6-0.7 V threshold. Placing the diode in an op-amp feedback loop divides the threshold voltage by the open-loop gain (Aol ≈ 10^5), allowing precise rectification of millivolt signals without threshold distortion.
Q. 14
Electronics Engineering
Difficulty: Easy
(1 Mark)
In a 555 Timer integrated circuit, the internal resistive voltage divider consists of:
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Step-by-Step Explanation & Concept Rationale
The 555 timer gets its famous name from its internal string of three matched 5 kΩ resistors connected in series between VCC and ground. This divider sets precise reference thresholds of (2/3) VCC at the Upper Comparator and (1/3) VCC at the Lower Comparator.
Q. 15
Electronics Engineering
Difficulty: Easy
(1 Mark)
In a 555 Timer configured as a Monostable Multivibrator (one-shot pulse generator), the output pulse duration (T_pulse) is given by:
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Step-by-Step Explanation & Concept Rationale
When triggered by a negative pulse (< 1/3 VCC) at Pin 2, the internal flip-flop sets and the discharge transistor turns off, allowing capacitor C to charge through resistor R toward VCC. When capacitor voltage reaches (2/3) VCC, the upper comparator resets the flip-flop: Vc(t) = VCC * (1 - e^(-T / RC)) = (2/3) VCC, giving T = RC * ln(3) ≈ 1.1 * R * C.
Q. 16
Electronics Engineering
Difficulty: Easy
(1 Mark)
In a 555 Timer connected as an Astable Multivibrator (free-running square wave generator) with resistors RA, RB and timing capacitor C, the charging time (T_high) of the output is:
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Step-by-Step Explanation & Concept Rationale
During the HIGH state, capacitor C charges from (1/3) VCC to (2/3) VCC through both RA and RB in series. The charging time is T_high = ln(2) * (RA + RB) * C ≈ 0.693 * (RA + RB) * C.
Q. 17
Electronics Engineering
Difficulty: Easy
(1 Mark)
In the standard 555 astable multivibrator, the discharging time (T_low) of the output is:
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Step-by-Step Explanation & Concept Rationale
During the LOW state, the internal discharge transistor (Pin 7) is turned on, discharging capacitor C from (2/3) VCC down to (1/3) VCC through resistor RB to ground. The discharge interval is T_low = ln(2) * RB * C ≈ 0.693 * RB * C.
Q. 18
Electronics Engineering
Difficulty: Medium
(1 Mark)
The oscillation frequency f of a standard 555 astable multivibrator is:
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Step-by-Step Explanation & Concept Rationale
Total period T = T_high + T_low = 0.693 * (RA + RB) * C + 0.693 * RB * C = 0.693 * (RA + 2*RB) * C. Therefore, frequency f = 1 / T = 1 / [0.693 * (RA + 2*RB) * C] = 1.44 / [(RA + 2*RB) * C].
Q. 19
Electronics Engineering
Difficulty: Medium
(1 Mark)
The Duty Cycle D of a standard 555 astable circuit is defined as T_high / (T_high + T_low). In the conventional circuit, the duty cycle is always:
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Step-by-Step Explanation & Concept Rationale
Because charging takes place through (RA + RB) and discharging occurs only through RB, T_high is always longer than T_low. Thus, Duty Cycle = (RA + RB) / (RA + 2*RB) is mathematically constrained to be > 50%. Connecting a diode in parallel with RB allows independent charging through RA, enabling duty cycles <= 50%.
Q. 20
Electronics Engineering
Difficulty: Easy
(1 Mark)
What is the function of the Control Voltage pin (Pin 5) on a 555 Timer IC?
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Step-by-Step Explanation & Concept Rationale
Pin 5 accesses the (2/3) VCC reference tap of the upper comparator. Applying an external modulating analog voltage shifts the threshold, producing Pulse Width Modulation (PWM) or Pulse Position Modulation (PPM). When unused, a 0.01 μF capacitor is connected to ground to decouple power rail noise.
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