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 by Manoj Kumar, Mathematics Educator
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)\nNoise gain = 1 + Rf/Ri = 1 + 10 = 11\nBandwidth = GBW / noise gain = 1 MHz / 11 = 90.9 kHz\nGain in dB = 20 log10(10) = 20 dB\nRise time = 0.35 / 90.9e3 = 3.85 microseconds\nPhase 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\nGain at DC = 5 (14 dB)\nAt 5 MHz: gain = 5 / sqrt(1 + (5/16)^2)\n= 5 / sqrt(1 + 0.0977) = 5 / sqrt(1.0977)\n= 5 / 1.0477 = 4.772 (13.57 dB)\nGain reduction = 14 - 13.57 = 0.43 dB\nPhase = -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 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.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy