- 2069 Asar · 3.5+3.5 marks
Voltage follower square-wave output with slew rate
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.
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.
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.
Time to slew through 2 V:
t_r = ΔV / SR = 2 V / 0.5 V/µs = 4 µs
(i) f = 100 Hz
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.
+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
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.
+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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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