Op-Amp Output Collapses at 20 kHz: Slew Rate or Gain-Bandwidth?
A student builds an inverting amplifier with a 741 op-amp: Rin = 1 kΩ, Rf = 100 kΩ, so the intended gain is −100, powered from ±15 V. The input is a 0.1 V peak sine from a signal generator, and the expected output is 10 V peak. At 1 kHz the output is a clean 10 V peak sine, exactly as designed. At 20 kHz the output collapses to about 3.6 V peak and has visibly straight sides with rounded corners, and the gain measured with the DMM is not 100. There is also a steady DC offset of 0.15 V at the output with no input. The 741 datasheet gives a slew rate of 0.5 V/µs, unity-gain bandwidth 1 MHz, input offset voltage 5 mV maximum, and input bias current 80 nA typical. The student has a TL071 (13 V/µs, 3 MHz) and an LF356 (12 V/µs, 5 MHz) in the drawer.
1. Decide, with numbers, whether the gain-bandwidth product or the slew rate is limiting the output at 20 kHz, and explain how the waveform shape supports your answer.
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Gain-bandwidth check. For a single-pole op-amp the closed-loop bandwidth is fT/|Acl| = 1 MHz/100 = 10 kHz. At 20 kHz: |Acl| = 100/√(1 + (20/10)²) = 100/2.236 ≈ 44.7 → output ≈ 4.47 V peak.
Slew-rate check. The maximum frequency an output of amplitude Vp can follow is f = SR/(2πVp), so Vp(max) = SR/(2πf) = (0.5 × 10⁶)/(2π × 20 000) ≈ 3.98 V peak.
Which one binds? The slew-rate limit (3.98 V) is lower than the small-signal GBW limit (4.47 V), and the measured 3.6 V is just below both. That is consistent with the 0.5 V/µs figure being typical rather than guaranteed. The decisive evidence is the shape: a GBW-limited output is still a clean sine, just smaller, because the amplifier is linear; a slew-limited output has constant-slope straight edges with rounded corners, because the output stage cannot supply the current needed to charge the compensation capacitor any faster. The observed waveform is the slew-limited case. Check the arithmetic on the slope: 2π × 20 kHz × 3.6 V ≈ 0.45 V/µs, close to the quoted 0.5 V/µs.
The 1 kHz result fits: full-power bandwidth at 10 V peak is 0.5 × 10⁶/(2π × 10) ≈ 8 kHz, so 1 kHz is safely inside.
2. The requirement is 10 V peak at 20 kHz with gain −100. State the minimum specifications needed and choose between the TL071 and the LF356.
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Slew-rate requirement. Required SR = 2πfVp = 2π × 20 000 × 10 = 1.26 V/µs. Both candidates clear this easily (13 and 12 V/µs), so slew rate is no longer the issue.
Bandwidth requirement. At 20 kHz the loop must still provide enough open-loop gain. With the single-pole model, |Acl| = 100/√(1 + (f/fT/100)²):
- TL071, fT = 3 MHz: corner at 30 kHz → |Acl| = 100/√(1 + 0.444) ≈ 83 → 8.3 V peak (17 % low).
- LF356, fT = 5 MHz: corner at 50 kHz → |Acl| = 100/√(1 + 0.16) ≈ 93 → 9.3 V peak (7 % low).
For a 1 % gain error at 20 kHz you would need A(20 kHz) ≥ 10 000 (fT ≥ 200 MHz), which is unrealistic here; a 5–10 % amplitude error is the normal price of a 100× stage at 20 kHz.
Choice. The LF356 is the better of the two: it meets the slew requirement with a factor of ten margin and gives about 9.3 V peak with roughly 7 % gain error. If exactly 10 V is mandatory, use a faster device (fT ≥ 10 MHz) or split the gain into two stages of 10 each, which raises the bandwidth by a factor of ten for the same total gain. Do not simply raise the supply rails: the limit is dynamic, not a headroom problem.
3. Explain the 0.15 V DC offset at the output and state two ways to deal with it.
Read the worked answer
Referred to the input: 0.15 V/100 = 1.5 mV at the input, which is well inside the 741's 5 mV maximum input offset voltage and typical for the part. It is not a fault.
Where it comes from. Mismatch in the input differential pair gives a built-in VOS, which the closed-loop gain multiplies like any other DC input. The bias-current term is negligible here: 80 nA flowing through the parallel combination of Rin and Rf (1 kΩ || 100 kΩ ≈ 990 Ω) produces only about 0.08 mV, which is 3 µV at the output - nothing. If Rin had been 100 kΩ instead, the bias-current term would have been 100 nA × 100 kΩ = 10 mV and would have dominated, so the resistor values matter.
Two fixes.
- Null it: connect a 100 kΩ potentiometer between pins 1 and 5 with the wiper to the negative rail (the 741's offset-null pins) and trim for 0 V out. This only works for the 741 and similar; nulling must be done at the temperature of use, since VOS drifts.
- Block it: put a coupling capacitor between the amplifier output and the load. Choose C so that the time constant with the load resistance does not cut into the audio band, e.g. 10 µF into 10 kΩ gives a 1.6 Hz corner.
A third option is to use a low-offset op-amp (precision types are available), which is not justified here since 1.5 mV is within specification.