--- title: "Gain-bandwidth relationship; GBP is constant" subject: "Advanced Electronics" chapter: 2 key: "semester_4/advanced_electronics:gbp-constant-relationship" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/gbp-constant-relationship site: BCT Engineering --- # Gain-bandwidth relationship; GBP is constant Advanced Electronics · Chapter 2: Operational Amplifier Characterization - Asked in: 2073 Shrawan, 1+6 marks; 2072 Kartik, 4 marks (asked 2 times) - Video: [Op-Amp: Gain Bandwidth Product and Frequency Response](https://www.youtube.com/watch?v=wfkzz1rg-xk) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/gbp-constant-relationship - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2 ## Question Describe relationship of gain and bandwidth of an op amp and prove gain bandwidth product is constant. ## Short answer Open‑loop gain A(f)=A0/(1+jf/fo). With feedback, closed‑loop gain Af0=A0/(1+βA0) and bandwidth fF=fo(1+βA0). Their product Af0·fF = A0·fo = ft, constant. Gain and bandwidth are inversely proportional. ## Answer #### Relationship between gain and bandwidth The open-loop gain of an op-amp is very high (about 2 × 10⁵ for the 741) only at low frequencies. Because of the internal compensation capacitor it falls at −20 dB/decade above a low break frequency fo (about 5 Hz), reaching unity at ft (about 1 MHz). When negative feedback reduces the gain, the bandwidth increases by the same factor. So **gain and bandwidth are inversely proportional**, and their product is constant. ```text Gain (dB) 106 |-------. open loop | \ 40 |---------\---. closed loop, gain 100 | \ : 0 +-----------\-:-----------> f (log) fo fF ft = 1 MHz ``` #### Proof that GBP is constant The open-loop gain of an internally compensated op-amp (e.g. 741) has a single dominant pole at fo: ```text A(f) = A0 / (1 + j f/fo) A0 = dc open-loop gain, fo = open-loop break frequency ``` Apply negative feedback with feedback factor β (for a non-inverting amplifier β = R1/(R1 + Rf)): ```text Af(f) = A(f) / (1 + β A(f)) = [A0/(1 + jf/fo)] / [1 + βA0/(1 + jf/fo)] = A0 / (1 + βA0 + jf/fo) = [A0/(1 + βA0)] / [1 + j f / (fo(1 + βA0))] ``` This is again a single-pole response, with ```text Closed-loop dc gain : Af0 = A0 / (1 + βA0) Closed-loop BW : fF = fo (1 + βA0) Product : Af0 × fF = A0 × fo = constant ``` The factor (1 + βA0) that divides the gain multiplies the bandwidth, so **gain × bandwidth = A0·fo = ft**, the unity-gain bandwidth, which is fixed for a given op-amp. For the 741, A0 ≈ 2 × 10⁵ and fo ≈ 5 Hz, so GBP ≈ 1 MHz: with a gain of 10 the bandwidth is 100 kHz, with a gain of 100 it is 10 kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When negative feedback lowers the closed-loop gain of an op-amp, what happens to the bandwidth? A. It increases by the same factor B. It decreases by the same factor C. It stays unchanged D. It increases only if the gain becomes less than 1 Answer: A. Feedback divides the dc gain by (1 + beta*A0) and multiplies the bandwidth by the same factor. (Key point: Gain-bandwidth relation) 2. What open-loop response of an internally compensated op-amp is assumed in the proof? A. A(f) = A0 at all frequencies B. A(f) = A0/(1 + jf/fo), a single dominant pole C. A(f) = A0 (1 + jf/fo), a single zero D. Two equal poles at fo Answer: B. The compensation capacitor gives one dominant pole at fo, so gain falls at -20 dB/decade above it. (Key point: Single dominant pole) 3. After feedback with factor beta, the closed-loop bandwidth is: A. fF = fo/(1 + beta*A0) B. fF = fo*A0 C. fF = fo(1 + beta*A0) D. fF = fo(1 - beta*A0) Answer: C. Rewriting Af(f) as [A0/(1+beta*A0)] / [1 + jf/(fo(1+beta*A0))] shows the new break frequency is fo(1+beta*A0). (Key point: Closed-loop bandwidth) 4. For the 741 (A0 about 2x10^5, fo about 5 Hz), what is the gain-bandwidth product? A. About 10 kHz B. About 5 Hz C. About 2 x 10^5 Hz D. About 1 MHz, equal to the unity-gain frequency ft Answer: D. A0 x fo = 2x10^5 x 5 Hz = 1 MHz = ft, so gain 10 gives 100 kHz bandwidth. 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