Qwestrum Engineering360 · Electrical & Electronics · Analog Electronics
Feedback Amplifiers
Negative feedback reduces gain to A_f = A/(1+Aβ) but in return improves gain stability, bandwidth, distortion and impedance — all by the same factor (1+Aβ), the amount of feedback.
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.
- Series-shunt (voltage-voltage): stabilises voltage gain
- Shunt-series (current-current): stabilises current gain
- Oscillation when Aβ = −1 at some frequency (Barkhausen)
Topic details
Introduction
The loop gain Aβ (the “amount of feedback”) is the key figure. Closed-loop gain becomes A_f = A/(1+Aβ); for large Aβ it approaches 1/β, set almost entirely by the passive feedback network and nearly independent of the active device.
Scope in B.Tech and GATE syllabus
The same factor (1+Aβ) improves everything: gain sensitivity, bandwidth (gain–bandwidth product is conserved), and distortion all reduce by (1+Aβ); the gain–bandwidth product stays constant.
Key relations & formulas
(negative feedback)
Formulas (Indian textbook notation)
Formulas (Indian textbook notation)
Notation and sign conventions
Relation 1 —
(negative feedback)
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 —
Formulas (Indian textbook notation)
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 —
Formulas (Indian textbook notation)
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
Four topologies set which quantity is stabilised and how impedances change: series-shunt (voltage amplifier) raises input and lowers output impedance; shunt-series (current amplifier) lowers input and raises output impedance; series-series (transconductance) and shunt-shunt (transresistance) complete the set.
Governing relations in practice
Gain desensitivity: a 10% change in open-loop gain becomes a 10%/(1+Aβ) change in closed-loop gain — the practical reason feedback is used.
Design and analysis considerations
If the loop gain reaches −1 (magnitude 1 at −180°) the feedback becomes positive and the amplifier oscillates (Barkhausen condition); phase margin ensures this does not happen.
Assumptions and validity limits
State assumptions explicitly before using any relation for feedback 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 feedback amplifiers.
4. Use equation 1:
5. Use equation 2:
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.
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 feedback amplifiers.
4. Use equation 1:
.
5. Use equation 2:
.
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
Feedback 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 feedback amplifiers with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use feedback amplifiers?" — answer with a lab, mini-project, or plant visit example if possible.
Common mistakes in exams
• Using 1 − Aβ for negative feedback (it is 1 + Aβ)
• Forgetting bandwidth increases by the same factor the gain decreases
• Mixing up which impedance rises/falls for each topology
• Ignoring the sign/phase that turns intended negative feedback into oscillation
• Forgetting bandwidth increases by the same factor the gain decreases
• Mixing up which impedance rises/falls for each topology
• Ignoring the sign/phase that turns intended negative feedback into oscillation
Quick revision checklist
Before attempting feedback amplifiers problems, confirm you can:
1. Series-shunt (voltage-voltage): stabilises voltage gain
2. Shunt-series (current-current): stabilises current gain
3. Oscillation when Aβ = −1 at some frequency (Barkhausen)
2. Shunt-series (current-current): stabilises current gain
3. Oscillation when Aβ = −1 at some frequency (Barkhausen)
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.
Closed-loop gain and desensitivity
Problem
An amplifier has open-loop gain A = 1000 and feedback factor β = 0.09. Find the closed-loop gain and the percentage change in A_f if A drops by 10%.
Solution
Loop gain Aβ = 1000 × 0.09 = 90; 1 + Aβ = 91.
A_f = A/(1+Aβ) = 1000/91 = 10.99.
Desensitivity: dA_f/A_f = (1/(1+Aβ)) dA/A = (1/91)(−10%).
= −0.11%. A 10% drop in A causes only 0.11% change in A_f.
A_f = A/(1+Aβ) = 1000/91 = 10.99.
Desensitivity: dA_f/A_f = (1/(1+Aβ)) dA/A = (1/91)(−10%).
= −0.11%. A 10% drop in A causes only 0.11% change in A_f.
Conceptual check — Feedback 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 feedback amplifiers." What should a complete answer include?
Exams & GATE
Sedra & Smith — identify feedback topology and find A_f.
📖 Standard books (India)
Microelectronic Circuits — Sedra & Smith
Read: Syllabus unit
Analog electronics reference
Explore related topics
See real electrical & electronics careers
After exams and interviews, see how engineers actually built careers — milestones and decisions from people in the field.