Op-Amp Gain That Looks Right at 1 kHz but Not at 10 kHz
A student builds a non-inverting amplifier with Rf = 99 kΩ and R1 = 1 kΩ, giving a designed closed-loop gain of 100. The op-amp has a gain-bandwidth product of 1 MHz, a slew rate of 0.5 V/µs, and output saturation at ±13 V. The supply is ±15 V. The input is a 0.100 V peak sine wave. At 1 kHz the output is a clean 10.0 V peak sine. At 10 kHz the output amplitude falls to about 7 V peak and the waveform looks triangular. At 20 kHz the output is clearly triangular and even smaller. The student first suspects the breadboard wiring, then the function generator. The lab instructor says the op-amp is simply being asked to do too much. The student wants to know which op-amp limit is responsible at each frequency, and whether lowering the gain to 10 would fix the 10 kHz problem for the same 0.100 V input.
1. Calculate the small-signal closed-loop bandwidth and the full-power bandwidth for a 10 V peak output. Use these two numbers to explain the waveform at 10 kHz.
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The small-signal closed-loop bandwidth is GBW/closed-loop gain = 1 MHz/100 = 10 kHz. At 10 kHz the amplifier is at its −3 dB point, so a 10 V low-frequency output would fall to about 7.07 V peak. That matches the observed 7 V amplitude.
The full-power bandwidth for a 10 V peak sine is set by slew rate: f_FP = SR/(2π V_p) = 0.5 × 10⁶ V/s / (2π × 10 V) ≈ 7.96 kHz. At 10 kHz, the required slew rate is 2π × 10 kHz × 10 V = 0.628 V/µs, which exceeds the 0.5 V/µs limit. The op-amp cannot follow the sine at the zero crossings, so the output becomes triangular.
Both limits are active at 10 kHz: the finite GBW reduces the amplitude, and the slew rate distorts the shape. At 1 kHz, the required slew rate is only 0.063 V/µs and the frequency is well below 10 kHz, so the output is clean.
2. If the gain is reduced to 10 and the input remains 0.100 V peak, determine the new output amplitude, small-signal bandwidth, and full-power bandwidth. Will the 10 kHz output be clean?
Read the worked answer
With a gain of 10, the output amplitude is 0.100 V × 10 = 1.0 V peak. The small-signal closed-loop bandwidth becomes 1 MHz/10 = 100 kHz. The full-power bandwidth for a 1 V peak output is 0.5 × 10⁶/(2π × 1) ≈ 79.6 kHz.
At 10 kHz, the amplifier is a decade below its small-signal bandwidth and well below the 79.6 kHz slew-rate limit. The required slew rate is 2π × 10 kHz × 1 V = 0.0628 V/µs, far below 0.5 V/µs. The output should be a clean 1 V peak sine, assuming the breadboard and load are not introducing other problems. Lowering the gain moves both limits in the right direction because the output amplitude and the closed-loop gain are both reduced.
3. If the gain must stay at 100 and the output must remain 10 V peak, what maximum frequency is possible without slew distortion, and what practical fix would you recommend?
Read the worked answer
With a 10 V peak output, the slew-rate limit gives f_max = SR/(2π V_p) = 0.5 × 10⁶/(2π × 10) ≈ 7.96 kHz. Above that frequency the output will slope-limit and become triangular, regardless of the small-signal gain.
To operate at 10 kHz or higher with gain 100 and 10 V peak, the op-amp needs a higher slew rate, at least 2π × 10 kHz × 10 V ≈ 0.63 V/µs, and enough gain-bandwidth product to keep the closed-loop bandwidth above 10 kHz. A practical fix is to select a faster op-amp with, for example, GBW ≥ 10 MHz and slew rate ≥ 1 V/µs, or to reduce the required output amplitude. Cascading two gain-of-10 stages does not by itself fix the final stage’s slew-rate limit, because the last stage still must swing 10 V peak.