Op Amp Gain Bandwidth Calculator
Free Op amp gain bandwidth Calculator for electronics & circuits. Enter variables to compute results with formulas and detailed steps.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Op Amp Gain Bandwidth Calculator
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Formula: Bandwidth = GBW / Closed-Loop Gain
Worked example — Bandwidth: 90.9 kHz | Rise Time: 3.85 us | Phase at 50kHz: -28.8 deg
Formula
Bandwidth = GBW / Closed-Loop Gain
Where GBW is the gain-bandwidth product of the op amp (a fixed specification), and the closed-loop gain determines how much bandwidth is available. For non-inverting circuits, the noise gain equals the signal gain. For inverting circuits, the noise gain is 1 + |signal gain|.
Worked Examples
Example 1: Inverting Amplifier Bandwidth
Problem:An LM741 op amp (GBW = 1 MHz) is configured as an inverting amplifier with gain = -10 (Rf=10k, Ri=1k). What is the bandwidth?
Solution:Signal gain = -Rf/Ri = -10k/1k = -10 (magnitude 10) Noise gain = 1 + Rf/Ri = 1 + 10 = 11 Bandwidth = GBW / noise gain = 1 MHz / 11 = 90.9 kHz Gain in dB = 20 log10(10) = 20 dB Rise time = 0.35 / 90.9e3 = 3.85 microseconds Phase at 50 kHz = -arctan(50k/90.9k) = -28.8 degrees
Result:Bandwidth: 90.9 kHz | Rise Time: 3.85 us | Phase at 50kHz: -28.8 deg
Example 2: High-Speed Non-Inverting Amplifier
Problem:An OPA637 (GBW = 80 MHz) is set to non-inverting gain of 5. Find bandwidth and gain at 5 MHz.
Solution:Bandwidth = GBW / gain = 80 MHz / 5 = 16 MHz Gain at DC = 5 (14 dB) At 5 MHz: gain = 5 / sqrt(1 + (5/16)^2) = 5 / sqrt(1 + 0.0977) = 5 / sqrt(1.0977) = 5 / 1.0477 = 4.772 (13.57 dB) Gain reduction = 14 - 13.57 = 0.43 dB Phase = -arctan(5/16) = -17.4 degrees
Result:Bandwidth: 16 MHz | Gain at 5MHz: 4.772 (13.57 dB) | Only 0.43 dB rolloff
Frequently Asked Questions
What is gain-bandwidth product (GBW) and why is it constant for op amps?
The gain-bandwidth product is a fundamental specification of operational amplifiers that states the product of closed-loop gain and bandwidth remains approximately constant. If an op amp has a GBW of 1 MHz, it can provide a gain of 10 with a bandwidth of 100 kHz, or a gain of 100 with a bandwidth of 10 kHz. This relationship arises from the single dominant pole in the op amp frequency response, which causes the open-loop gain to roll off at 20 dB per decade (6 dB per octave). The GBW is determined during the IC design process by the internal compensation capacitor. Faster op amps have higher GBW products, with modern high-speed op amps reaching into the GHz range for demanding applications.
How does closed-loop gain affect the available bandwidth of an op amp circuit?
As you increase the closed-loop gain of an op amp circuit, the available bandwidth decreases proportionally according to the GBW relationship: bandwidth = GBW / gain. A unity-gain buffer (gain = 1) has the maximum bandwidth equal to the full GBW. At gain of 10, bandwidth is GBW/10. At gain of 100, bandwidth is GBW/100. This tradeoff is one of the most important considerations in analog circuit design. If you need both high gain and wide bandwidth, you can cascade multiple lower-gain stages. Two stages of gain 10 each (total gain 100) give 10 times more bandwidth than a single stage of gain 100, though at the cost of added noise and complexity.
What is the difference between non-inverting and inverting gain configurations?
In a non-inverting configuration, the closed-loop gain equals 1 + Rf/Ri, where Rf is the feedback resistor and Ri is the input resistor. The output is in phase with the input and the minimum gain is 1 (unity gain follower). In an inverting configuration, the gain equals -Rf/Ri, with the negative sign indicating a 180-degree phase inversion. The noise gain (which determines bandwidth) in the inverting case is (1 + Rf/Ri), which is always one more than the magnitude of the signal gain. This means an inverting amplifier with gain of -1 actually has a noise gain of 2 and therefore half the bandwidth of a non-inverting unity-gain buffer. This distinction between signal gain and noise gain is crucial for correctly predicting bandwidth and stability.
How do I determine if my op amp circuit has adequate phase margin for stability?
Phase margin is the amount of additional phase shift needed to reach -180 degrees (where positive feedback causes oscillation) at the frequency where the loop gain equals unity (0 dB). A minimum of 45 degrees phase margin is generally needed for stable operation, with 60-70 degrees considered ideal for well-damped response. For a single-pole op amp operating within its GBW, the phase margin at the closed-loop bandwidth is inherently 45 degrees. Capacitive loads, stray capacitance at the output, and additional poles from the op amp itself can erode phase margin. If the phase margin drops below 45 degrees, the step response will show ringing and overshoot. Below about 10-15 degrees, the circuit becomes unstable and oscillates.
What is slew rate and how does it limit op amp performance differently from GBW?
Slew rate is the maximum rate at which the op amp output voltage can change, measured in V/microsecond. It limits large-signal performance independently of the GBW. While GBW determines the small-signal frequency response, slew rate limits the maximum output frequency for a given voltage swing. The full-power bandwidth (the maximum frequency at which the op amp can produce a full-amplitude undistorted sine wave) equals slew rate divided by (2 pi times peak voltage). For example, an op amp with 1 V/us slew rate producing a 10V peak signal has a full-power bandwidth of only 15.9 kHz, even if the small-signal bandwidth is much higher. Selecting an op amp requires checking both the small-signal bandwidth and the slew rate to ensure adequate performance.
How does the rise time of an op amp circuit relate to its bandwidth?
Rise time (10% to 90% of final value for a step input) and bandwidth are inversely related by the approximation: rise time = 0.35 / bandwidth. This relationship comes from the single-pole response characteristic of feedback amplifiers. A circuit with 100 kHz bandwidth has a rise time of 3.5 microseconds. A 10 MHz bandwidth circuit has a rise time of 35 nanoseconds. This approximation assumes a single-pole rolloff, which is valid for most properly compensated op amp circuits operating within their GBW. For cascaded stages, the overall rise time is approximately the square root of the sum of squares of individual stage rise times. This rise time calculation is essential for digital interface circuits, pulse amplifiers, and data acquisition systems.
What factors should I consider when selecting an op amp for a specific application?
Op amp selection involves balancing multiple specifications against application requirements. For signal bandwidth, ensure the GBW provides adequate bandwidth at your required gain with at least 5-10 times margin. For output swing, verify the slew rate supports your signal amplitude and frequency. For precision applications, consider input offset voltage, bias current, and their temperature drifts. Noise performance matters for low-level signal amplification and is specified as voltage noise density (nV per root Hz) and current noise density. Supply voltage and current consumption determine compatibility with your power supply. Input impedance type (JFET vs bipolar) affects source loading. Output drive capability must match your load impedance. Finally, package, cost, and availability influence the practical selection.
How do I calculate the actual gain at a specific frequency including rolloff effects?
The actual gain magnitude at any frequency includes the effect of bandwidth rolloff. For a single-pole system, the gain at frequency f is: A(f) = Acl / sqrt(1 + (f/fc)^2), where Acl is the DC closed-loop gain and fc is the closed-loop bandwidth (GBW/Acl). At the bandwidth frequency (f = fc), the gain drops to Acl/sqrt(2), which is -3 dB below the DC gain. At 10 times the bandwidth, the gain drops to about Acl/10 (-20 dB). The phase shift at any frequency is: phase = -arctan(f/fc). At the bandwidth frequency, the phase is -45 degrees. These calculations are essential for determining whether your signal will be accurately amplified at the operating frequency and how much phase distortion will be introduced.
What is noise gain and why does it differ from signal gain in inverting configurations?
Noise gain is the gain that the op amp feedback loop applies to input-referred noise sources and also determines the loop gain and bandwidth. For non-inverting configurations, noise gain equals signal gain (1 + Rf/Ri). For inverting configurations, noise gain equals (1 + Rf/Ri), which is always one more than the magnitude of the signal gain (Rf/Ri). This difference exists because the feedback network attenuates the op amp output differently for signals applied at the non-inverting input (including noise) versus the inverting input. The bandwidth of any op amp circuit is always GBW divided by the noise gain, not the signal gain. This means an inverting amplifier with signal gain of -1 has the same bandwidth as a non-inverting amplifier with gain of +2.
How do cascaded amplifier stages affect overall bandwidth and noise?
When multiple amplifier stages are cascaded, the overall bandwidth is less than the bandwidth of any individual stage. For n identical stages each with bandwidth BW, the overall bandwidth is approximately BW times sqrt(2^(1/n) - 1). For two identical stages, this gives 0.644 times the individual bandwidth. For three stages, it is 0.510 times. This means two 100 kHz stages give an overall bandwidth of about 64.4 kHz. To achieve a target bandwidth with cascaded stages, each stage must have wider bandwidth than the target. For noise, the first stage is most critical because its noise is amplified by all subsequent stages. The total output noise voltage is approximately sqrt(sum of squared individual noise contributions). Using higher gain in the first stage reduces the impact of noise from subsequent stages.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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