Signals & Wireless (RF)
Op-amp gain-bandwidth product explained
An op-amp's gain-bandwidth product (GBW) is a fixed budget: the gain you set times the frequency where that gain starts to fall stays about the same. So the corner frequency ≈ GBW ÷ gain. A 10 MHz op-amp set for ×10 is flat to about 1 MHz; set for ×2, to about 5 MHz. Ask for more gain and you get less bandwidth.
Why gain costs bandwidth
An op-amp's own open-loop gain is enormous at low frequencies, around 100,000, but falls 20 dB per decade from a low frequency upward, like an RC low-pass. Feedback sets a much smaller closed-loop gain, and the amplifier holds that gain only while the op-amp has plenty to spare. Where the falling open-loop gain meets your closed-loop gain, there's nothing left over, and the response rolls off.
The noise gain is the gain for a non-inverting stage, 1 + Rf ÷ Rg, and one more than the gain's size for an inverting stage. At the corner, the gain is down to 70.7%, −3 dB, and the output lags by about 45°.
Live experiment
Spend the gain-bandwidth budget
A non-inverting stage on a 10 MHz op-amp, with a 1 MHz signal. Ask one stage for ×10 and its corner falls to 1 MHz: you get about ×7. Ask for ×2 and it's flat. Then split ×9 into two stages of ×3: about ×8.2, where one ×9 stage gives ×6.7.
“Gain meets bandwidth” is lesson 3 of 6 in Oscillator Outcrop, Signals & Wireless (RF) World 2. Free with an account. Lessons unlock in order, so you’ll do the 9 before it first.
A worked example
A 10 MHz op-amp, a 1 MHz signal, and you need ×9.
- One stage of ×9: corner = 10 MHz ÷ 9 ≈ 1.1 MHz. The signal sits almost at the corner, and you get only about ×6.7.
- Two stages of ×3: each corner = 10 ÷ 3 ≈ 3.3 MHz. Each delivers about ×2.87 at 1 MHz, and together 2.87 × 2.87 ≈ ×8.2.
Gains multiply along a chain, but each stage keeps its own bandwidth. That's why radio receivers and hi-fi preamps spread their gain over several stages.
Splitting the gain for audio
On a 1 MHz op-amp, ×1000 (60 dB) in one stage has its corner at 1 MHz ÷ 1000 = 1 kHz: useless for music. Three ×10 stages also give ×1000, but each keeps a 100 kHz corner. At 20 kHz each is still about ×9.8, so the chain gives about 59.5 dB, nearly flat across the audio band.
GBW isn't the only limit
GBW is a small-signal figure. Big, fast swings also run into the slew rate, the fastest the output can move, and every output stops short of its rails. Datasheets lead with all three. Op-amps for audio typically have a few megahertz of GBW; parts for video and radio reach hundreds of megahertz or more.
Free lesson
Now build it yourself
Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.
“Gain meets bandwidth” is lesson 3 of 6 in Oscillator Outcrop, Signals & Wireless (RF) World 2. Free with an account. Lessons unlock in order, so you’ll do the 9 before it first.
Or start from the very first lesson, no account neededPractice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: Roughly fixed. More gain, less bandwidth.
- Why not “Unlimited”? Every real op-amp has a finite gain-bandwidth, which is why gain falls at high frequencies.
- Why not “Bigger at high gain”? It stays about the same whatever gain you set. That's why more gain means less bandwidth.
- Why not “Always 1”? Gain times bandwidth is a large number in hertz, like 1 MHz or 10 MHz, not 1.
Show the answer
Answer: gets lower. Gain and bandwidth trade off, like a bike's gears. Their product, the gain-bandwidth product, stays about the same, so more gain means a lower corner.
- Why not “stays the same”? Gain times bandwidth stays about the same, so the corner has to move when the gain does.
- Why not “gets higher”? More gain spends more of the fixed gain-bandwidth budget, so the corner comes down.
Show the answer
Answer: 45°. An eighth of a cycle.
- Why not “0°”? Well below the corner the lag is near 0°. At the corner it has reached 45°.
- Why not “90°”? 90° is where the lag heads far above the corner. At the corner itself it's halfway, 45°.
- Why not “180°”? A single RC leg can never lag more than 90°, so 180° isn't possible.
“Gain meets bandwidth” is lesson 3 of 6 in Oscillator Outcrop, Signals & Wireless (RF) World 2. Free with an account. Lessons unlock in order, so you’ll do the 9 before it first.
Questions people ask
What is gain-bandwidth product?
For most op-amps, the closed-loop gain times the frequency where that gain starts to fall stays about constant. It's quoted in hertz: a 10 MHz GBW op-amp set for ×10 is flat to about 1 MHz, and set for ×2, to about 5 MHz.
How do I calculate an op-amp amplifier's bandwidth?
Divide the GBW by the noise gain. For a non-inverting amplifier the noise gain is its gain, 1 + R_f ÷ R_g. For an inverting one it's 1 + R_f ÷ R_in, one more than the size of its gain. A ×100 non-inverting stage on a 1 MHz op-amp has its corner near 10 kHz.
Why use two amplifier stages instead of one?
Gains multiply while each stage keeps its own bandwidth. ×1000 in one stage on a 1 MHz op-amp has a 1 kHz corner; three ×10 stages keep 100 kHz corners each, so the chain stays nearly flat across the audio band.
What is the difference between GBW and unity-gain bandwidth?
For a typical op-amp with a single dominant pole they're nearly the same number: the frequency where its open-loop gain falls to 1. Datasheets may quote either. Op-amps that aren't stable at unity gain are the exception.
Related topics
- Op-amp slew rate explainedThe fastest an op-amp's output can move, in volts per microsecond, and why it bends big, fast waves.
- Non-inverting amplifier explainedAn op-amp, two resistors and gain G = 1 + R_f ÷ R_g: bigger, in step, and clipped at the rails.
- How to read a Bode plotGain in decibels against frequency on a log scale: corners, −3 dB and 20 dB per decade.
- Op-amp inverting amplifier explainedSignal into the − input through a resistor: a bigger copy, upside down, with gain −R_f ÷ R_in.