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Slew-rate distortion check, 330 kΩ/10 kΩ inverter

Short answer

Output peak 3.3 V at 31.83 kHz requires 0.66 V/µs slew rate, exceeding the 0.5 V/µs limit, so the output distorts. Remedy: reduce Rf to ≤250 kΩ (gain ≤25) or use a faster op-amp.

Define slew rate of an operational amplifier. The inverting op-amp with Rf = 330 KΩ and R1 = 10 KΩ has a slew rate of 0.5 V/μs with input signal equal to 0.1 sin(200000t). Determine whether the output will be distorted due to slew rate limitation, if so find remedy.

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). It is caused by the limited current available to charge the internal compensation capacitor (SR = Imax/C).

Checking for distortion

Given: Rf = 330 kΩ, R1 = 10 kΩ, SR = 0.5 V/µs, vin = 0.1 sin(200000t) V.

|Av| = Rf/R1 = 330/10 = 33
Vm   = 33 × 0.1 = 3.3 V   (output peak)
ω    = 2 × 10⁵ rad/s,  f = ω/2π = 31.83 kHz
Required slope = ω·Vm = 2 × 10⁵ × 3.3
               = 0.66 × 10⁶ V/s = 0.66 V/µs

Since 0.66 V/µs > 0.5 V/µs, the output will be distorted due to slew-rate limiting. (Full-power bandwidth for 3.3 V peak: fmax = SR/(2πVm) = 0.5 × 10⁶/(2π × 3.3) = 24.11 kHz < 31.83 kHz.)

Remedies

  1. Reduce the gain: maximum undistorted output peak Vm = SR/ω = 0.5 × 10⁶/(2 × 10⁵) = 2.5 V, so gain ≤ 2.5/0.1 = 25, i.e. Rf ≤ 25 × 10 kΩ = 250 kΩ.
  2. Use an op-amp with higher slew rate: SR ≥ 0.66 V/µs (e.g. LF351, 13 V/µs).
  3. Reduce the input amplitude: Vin(peak) ≤ 2.5/33 = 75.8 mV.

Answer: distorted (0.66 V/µs required > 0.5 V/µs); remedy Rf ≤ 250 kΩ or a faster op-amp.

✓ Answer checked by Claude Opus 5.5 (high effort)

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