--- title: "Input offset voltage; proof of constant GBP" subject: "Advanced Electronics" chapter: 2 key: "semester_4/advanced_electronics:input-offset-voltage-gbp" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/input-offset-voltage-gbp site: BCT Engineering --- # Input offset voltage; proof of constant GBP Advanced Electronics · Chapter 2: Operational Amplifier Characterization - Asked in: 2071 Shrawan, 2+5 marks - Video: [Op-Amp: Input Offset Voltage Explained](https://www.youtube.com/watch?v=hUu3SqRSYyA) — 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/input-offset-voltage-gbp - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2 ## Question Define input offset voltage. For a given op-amp, prove that gain-bandwidth-product (GBP) is constant. ## Short answer Input offset voltage is the dc voltage required between inputs for zero output. For an internally compensated op-amp, open-loop gain A(f)=A0/(1+jf/fo) yields closed-loop gain Af0=A0/(1+βA0) and bandwidth fF=fo(1+βA0), so Af0·fF = A0·fo is constant. ## Answer #### Input offset voltage **Input offset voltage (Vio)** is the dc differential voltage that must be applied between the inverting and non-inverting inputs of an op-amp to make its output voltage zero. It is caused by mismatch between the input transistors. With feedback it appears at the output as Voo = (1 + Rf/R1)·Vio. For the 741, Vio is 1 mV typical and 6 mV maximum. #### 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. This shows that for an op-amp, gain can be traded for bandwidth, but their product cannot be increased. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With feedback, an input offset voltage Vio appears at the output as: A. Voo = (1 + Rf/R1) x Vio B. Voo = Vio C. Voo = Rf x Vio D. Voo = Vio/(1 + Rf/R1) Answer: A. Vio is amplified by the noise gain. (Key point: Vio effect) 2. What causes the input offset voltage? A. Thermal noise in Rf B. Mismatch between the input transistors C. The supply voltage being too high D. Slew-rate limiting Answer: B. Unequal VBE of the differential pair gives a non-zero output with grounded inputs. (Key point: Cause of Vio) 3. After applying feedback to A(f) = A0/(1 + jf/fo), the response is: A. Two-pole with a peak B. Flat with no break frequency C. Again single-pole, with dc gain A0/(1 + beta*A0) and break frequency fo(1 + beta*A0) D. Single-pole with break frequency fo/(1 + beta*A0) Answer: C. Algebra gives [A0/(1+beta*A0)] / [1 + jf/(fo(1+beta*A0))]. (Key point: Closed-loop response form) 4. What does the proof show about gain and bandwidth? A. Both can be increased together by feedback B. Bandwidth is independent of gain C. The product depends on the feedback factor beta D. Af0 x fF = A0 x fo = ft, so gain can be traded for bandwidth but the product is fixed Answer: D. beta cancels in the product, leaving the op-amp's fixed unity-gain bandwidth. 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