--- title: "Voltage follower square-wave output with slew rate" subject: "Advanced Electronics" chapter: 2 key: "semester_4/advanced_electronics:voltage-follower-square-wave" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/voltage-follower-square-wave site: BCT Engineering --- # Voltage follower square-wave output with slew rate Advanced Electronics · Chapter 2: Operational Amplifier Characterization - Asked in: 2069 Asar, 3.5+3.5 marks - Video: [Op-Amp Slew Rate Explained (with Examples)](https://www.youtube.com/watch?v=2DFIr6t1hbc) — 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/voltage-follower-square-wave - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2 ## Question Draw the waveform of output voltage of voltage follower operational amplifier having square wave as input signal with peak-to-peak voltage of 2V and frequencies (i) 100Hz and (ii) 100KHz. Assume that op-amp has slew rate of 0.5V/μs. Label the waveform neatly illustrating timing and amplitude details. ## Short answer At 100 Hz, the 4 µs slew ramp is negligible; output is a ±1 V square wave. At 100 kHz, the output is trapezoidal: ramps linearly between −1 V and +1 V in 4 µs, with 1 µs flat tops, peak-to-peak remains 2 V. ## Answer A voltage follower has gain 1, so ideally the output equals the input square wave: 2 V peak-to-peak, i.e. a step of 2 V at every edge (from −1 V to +1 V and back). The output cannot change faster than SR = 0.5 V/µs, so each edge becomes a ramp. ```text Time to slew through 2 V: t_r = ΔV / SR = 2 V / 0.5 V/µs = 4 µs ``` #### (i) f = 100 Hz ```text T = 1/100 = 10 ms, half-period = 5 ms Ramp time / half period = 4 µs / 5 ms = 0.08 % ``` The 4 µs ramps are negligible, so the output is practically a perfect square wave of ±1 V. ```text +1 V ____________ ____________ | | | | | | -1 V | |__________| 0 5 ms 10 ms t (each edge is a 4 µs ramp, too short to see) ``` #### (ii) f = 100 kHz ```text T = 1/100 kHz = 10 µs, half-period = 5 µs Ramp takes 4 µs, flat top lasts 5 - 4 = 1 µs ``` The output is a **trapezoidal wave**: it rises linearly from −1 V to +1 V in 4 µs, stays at +1 V for 1 µs, falls linearly to −1 V in 4 µs, stays for 1 µs, and repeats. The peak-to-peak value is still 2 V because the 4 µs ramp fits inside the 5 µs half-period. ```text +1 V ___ ___ / \ / \ / \ / \ -1 V __/ \___/ \__ 0 4 5 9 10 14 15 t (µs) rise 0-4 µs, flat 4-5 µs, fall 5-9 µs, flat 9-10 µs, next rise 10-14 µs ... ``` If the frequency were raised above 125 kHz (half-period < 4 µs), the output would become a triangle wave of reduced amplitude. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The time to slew through a 2 V edge at 0.5 V/us is: A. 4 us B. 1 us C. 2 us D. 0.25 us Answer: A. t = dV/SR = 2/0.5 = 4 us. (Key point: Ramp time) 2. At 100 Hz (half-period 5 ms), the output looks like: A. A triangle wave B. Practically a perfect 2 V p-p square wave C. A trapezoid with long ramps D. A 1 V p-p wave Answer: B. 4 us ramps are 0.08% of the half-period. (Key point: 100 Hz output) 3. At 100 kHz (half-period 5 us), the output is: A. A perfect square wave B. A triangle with 1 V p-p C. A trapezoid with 4 us ramps and 1 us flat tops, still 2 V p-p D. A flat dc level Answer: C. The 4 us ramp fits in 5 us, leaving 1 us flat. (Key point: 100 kHz output) 4. The output becomes a reduced-amplitude triangle when the frequency exceeds about: A. 12.5 kHz B. 1 MHz C. 500 kHz D. 125 kHz (half-period < 4 us) Answer: D. Half-period 4 us corresponds to 125 kHz. 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