Op-Amp Gain, Bandwidth and Slew-Rate Limit
A student builds a non-inverting amplifier with an op-amp having a gain–bandwidth product of 1 MHz and a slew rate of 0.2 V/µs. The feedback resistors are 99 kΩ and 1 kΩ, giving a nominal closed-loop gain of 100. The input is a 50 mV peak sine wave at 20 kHz, and the output on the oscilloscope is a triangular wave of about 1.6 V peak instead of the expected sine. At 1 kHz the output is a clean sine. The student proposes doubling the feedback resistor to 199 kΩ to 'restore the gain', and then adding a 100 pF capacitor across it to 'stop the triangle'. The circuit is powered from ±12 V, and the op-amp output is not clipping against the supply rails.
1. Calculate the closed-loop bandwidth, the expected small-signal output at 20 kHz, and the slew-rate-limited amplitude. Explain the triangle.
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Closed-loop bandwidth = 1 MHz / 100 = 10 kHz. At 20 kHz the small-signal gain is 100 / √(1 + (20/10)²) = 44.7, so the expected sine is 50 mV × 44.7 = 2.24 V peak.
That sine would need a slew rate 2πfV = 2π(20,000)(2.24) = 0.281 V/µs, above the op-amp's 0.2 V/µs. The maximum sine amplitude at 20 kHz is SR / (2πf) = 0.2 × 10⁶ / (2π × 20,000) = 1.59 V peak. The op-amp therefore slews, producing a triangle of about 1.6 V peak rather than a sine.
2. Evaluate doubling the feedback resistor to 199 kΩ and adding 100 pF across it.
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Doubling the feedback resistor makes the nominal gain 200 and the bandwidth 1 MHz / 200 = 5 kHz. At 20 kHz the small-signal gain is still limited by the gain–bandwidth product to about 1 MHz / 20 kHz = 50, so the output is not restored to 5 V. The slew-rate limit remains 1.59 V peak, so the triangle remains. At low frequencies the gain is now 200, so a 50 mV input would demand 10 V output and may clip.
The 100 pF capacitor across 199 kΩ adds a pole near 1 / (2π × 199 kΩ × 100 pF) = 8 kHz, reducing phase margin and bandwidth further. It is not a cure.
3. What op-amp specification is needed for a clean 5 V peak at 20 kHz? Would splitting the gain into two stages solve it?
Read the worked answer
For a clean 5 V peak at 20 kHz, required slew rate = 2π(20,000)(5) = 0.628 V/µs. Required gain–bandwidth product is at least 100 × 20,000 = 2 MHz, with margin. Select an op-amp with GBW of several MHz and SR at least 1 V/µs, or reduce the required output amplitude or frequency.
Splitting the gain into two stages raises the small-signal bandwidth, but it does not reduce the slew-rate requirement for the same 5 V output. The output stage still must move 5 V at 20 kHz.