--- title: "Slew rate expressions and constant GBP" subject: "Advanced Electronics" chapter: 2 key: "semester_4/advanced_electronics:slew-rate-gbp-constant" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/slew-rate-gbp-constant site: BCT Engineering --- # Slew rate expressions and constant GBP Advanced Electronics · Chapter 2: Operational Amplifier Characterization - Asked in: 2074 Chaitra, 3+5 marks - 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/slew-rate-gbp-constant - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2 ## Question Explain Slew rate of an Op-Amp with necessary expressions. Prove that the Gain Bandwidth Product of Op-Amp is Constant. ## Short answer Slew rate (SR) is the maximum rate of output change, SR = Imax/C, 0.5 V/µs for 741. For a sine wave, distortion occurs if 2πfVm > SR. Gain-bandwidth product is constant because closed-loop gain Af0 = A0/(1+βA0) and bandwidth fF = fo(1+βA0) multiply to A0fo, the unity-gain frequency. ## Answer #### Slew rate **Slew rate (SR)** is the maximum rate of change of output voltage per unit time that an op-amp can produce, SR = (dVo/dt)max, usually in V/µs (0.5 V/µs for the 741). **Cause:** inside the op-amp, a limited current Imax charges the compensation capacitor C (30 pF in the 741). Since i = C·dv/dt: ```text SR = (dVo/dt)max = Imax / C 741: SR = 15 µA / 30 pF = 0.5 V/µs ``` **Slew rate and sine waves:** for vo = Vm sin(2πft), ```text dvo/dt = 2πf·Vm·cos(2πft) (dvo/dt)max = 2πf·Vm No distortion if 2πf·Vm ≤ SR Full-power bandwidth: fmax = SR / (2π·Vm) Maximum output peak: Vm = SR / (2πf) ``` If the required rate is larger than SR, the output becomes a triangular wave with smaller amplitude (slew-rate distortion). Example: 741 at 10 V peak gives fmax = 0.5 × 10⁶/(2π × 10) ≈ 7.96 kHz. For a step input the output rises as a ramp of slope SR, e.g. a 10 V step takes 20 µs. #### Proof that gain-bandwidth product 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. What limits the slew rate of the 741? A. Limited current Imax charging the internal compensation capacitor, SR = Imax/C B. The load resistance only C. The input offset voltage D. The supply ripple Answer: A. i = C dv/dt, so the maximum dv/dt is Imax/C = 15 uA/30 pF = 0.5 V/us. (Key point: Cause of slew rate) 2. For a sine output of peak Vm, the maximum undistorted frequency is: A. fmax = SR x 2*pi*Vm B. fmax = SR/(2*pi*Vm) C. fmax = SR/Vm D. fmax = 2*pi*Vm/SR Answer: B. The maximum sine slope is 2*pi*f*Vm and must not exceed SR. (Key point: Full-power bandwidth) 3. If the required slope exceeds SR, the output becomes: A. A larger sine wave B. A perfect square wave C. A triangular wave of smaller amplitude D. Unchanged Answer: C. The output can only ramp at SR, so peaks are cut off and shape becomes triangular. (Key point: Result of slew limiting) 4. In the proof, feedback changes the dc gain from A0 to A0/(1 + beta*A0) and the bandwidth from fo to: A. fo/(1 + beta*A0), so the product shrinks B. fo, so the product is unchanged by feedback C. fo*beta*A0 only D. fo(1 + beta*A0), so the product stays A0 x fo Answer: D. Gain divides by (1 + beta*A0) and bandwidth multiplies by it, so Af0 x fF = A0 x fo = ft. 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