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Qwestrum Engineering360 · Electrical & Electronics · Analog Electronics

Operational Amplifiers

With negative feedback the ideal op-amp forces its two inputs to the same voltage (virtual short) and draws no input current; from these two rules the gain of any op-amp circuit follows immediately.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.

Key formulas & points

Skim these first — then read the full notes below.

  • Slew rate limits large-signal response
  • Instrument amplifier: high CMRR differential front end
  • Integrator, differentiator, active filters from op-amp

Topic details

Introduction

The two golden rules of an ideal op-amp with negative feedback are: no current flows into the inputs, and the inputs are at the same voltage (virtual short). Applying them gives the inverting gain −R_f/R_i and the non-inverting gain 1 + R_f/R_i.

Scope in B.Tech and GATE syllabus

For the inverting amplifier the inverting input is a virtual ground; the input current R_i flows entirely through R_f, giving the gain directly. These configurations extend to summers, integrators, differentiators and filters.

Key relations & formulas

Formulas (Indian textbook notation)

  • Virtualshort:V+≈V−(negativefeedback,idealop−amp)Virtual short: V_+ \approx V_- (negative feedback, ideal op-amp)

Formulas (Indian textbook notation)

  • Inverting:Vo=−(RfRi)Vi;Non−inverting:Vo=(1+RfRi)ViInverting: V_{o} = -(\frac{R_{f}}{R_{i}}) V_{i}; Non-inverting: V_{o} = (1 + \frac{R_{f}}{R_{i}}) V_{i}
CMRR=AdAcmCMRR = \frac{A_{d}}{A_{cm}}
(differential vs common-mode gain)

Notation and sign conventions

Relation 1 —
Virtualshort:V+≈V−Virtual short: V_+ \approx V_-

Formulas (Indian textbook notation)

  • Virtualshort:V+≈V−(negativefeedback,idealop−amp)Virtual short: V_+ \approx V_- (negative feedback, ideal op-amp)
Write this relation with symbols exactly as in Microelectronic Circuits — Sedra & Smith before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Inverting:Vo=−Inverting: V_{o} = -

Formulas (Indian textbook notation)

  • Inverting:Vo=−(RfRi)Vi;Non−inverting:Vo=(1+RfRi)ViInverting: V_{o} = -(\frac{R_{f}}{R_{i}}) V_{i}; Non-inverting: V_{o} = (1 + \frac{R_{f}}{R_{i}}) V_{i}
Write this relation with symbols exactly as in Microelectronic Circuits — Sedra & Smith before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
CMRR=AdAcmCMRR = \frac{A_{d}}{A_{cm}}
CMRR=AdAcmCMRR = \frac{A_{d}}{A_{cm}}
(differential vs common-mode gain)
Write this relation with symbols exactly as in Microelectronic Circuits — Sedra & Smith before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Common-mode rejection ratio (CMRR = A_d/A_cm) measures how well the op-amp rejects signals common to both inputs; a high CMRR is essential for instrumentation amplifiers sensing small differential signals on a large common-mode background.

Governing relations in practice

Real limitations: slew rate caps how fast the output can change (V/µs), limiting large-signal bandwidth; the gain–bandwidth product limits small-signal gain at high frequency; input offset voltage and bias current cause DC errors.

Design and analysis considerations

An integrator uses a capacitor in the feedback path (V_o = −(1/RC)∫V_i dt); a differentiator swaps R and C. A practical integrator adds a large resistor across C to prevent DC drift.

Assumptions and validity limits

State assumptions explicitly before using any relation for operational amplifiers — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Analog Electronics viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Analog Electronics papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to operational amplifiers.
4. Use equation 1:
Virtualshort:V+≈V−Virtual short: V_+ \approx V_-
.
5. Use equation 2:
Inverting:Vo=−Inverting: V_{o} = -
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Operational Amplifiers appears in signal conditioning and audio. In Indian electrical curricula this topic is tested because it connects theory to amplifiers and op-amp circuits.
GATE and semester exams often combine operational amplifiers with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use operational amplifiers?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Forgetting the non-inverting gain includes the “1 +” term
• Assuming the virtual-short rule without negative feedback present
• Ignoring slew-rate limiting in large-signal high-frequency response
• Omitting the stabilising resistor across the integrator capacitor

Quick revision checklist

Before attempting operational amplifiers problems, confirm you can:
1. Slew rate limits large-signal response
2. Instrument amplifier: high CMRR differential front end
3. Integrator, differentiator, active filters from op-amp
Revise the solved examples in Microelectronic Circuits — Sedra & Smith and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Inverting amplifier gain

Problem

An inverting op-amp amplifier has R_i = 2 kΩ and R_f = 47 kΩ. Find the closed-loop gain and the output for a +100 mV input.

Solution

Gain A_v = −R_f/R_i = −47k/2k = −23.5.
Output V_o = A_v × V_i = −23.5 × 0.1 V.
V_o = −2.35 V.
The output is inverted and 23.5 times larger.

Conceptual check — Operational Amplifiers

Problem

In a Analog Electronics semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of operational amplifiers." What should a complete answer include?

Exams & GATE

Sedra & Smith — design inverting amp for specified gain.

📖 Standard books (India)

  • Microelectronic Circuits — Sedra & Smith

    Read: Syllabus unit

    Analog electronics reference

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