Learning objective: Compute a non-inverting amplifier gain and approximate its small-signal bandwidth using the op-amp gain-bandwidth product.
Understand the idea
A non-inverting amplifier accepts an input voltage on the op-amp's non-inverting terminal and uses negative feedback on the inverting terminal. Under the usual ideal assumptions, its gain depends on a resistor ratio. A real op-amp also has finite output swing, input limitations and frequency-dependent open-loop gain.
Worked example: given data
- Feedback resistor Rf = 90 kΩ and resistor to ground Rg = 10 kΩ.
- Input is a small 0.20 V DC signal. Assume an op-amp operating within its linear supply limits.
- For bandwidth estimation, suppose its unity-gain bandwidth is 1 MHz and a single-pole approximation applies.
Detailed solution
Use the virtual short assumption
With negative feedback and an ideal op-amp in its linear region, V− ≈ V+ = Vin. Negligible current flows into the op-amp input terminals.
Apply the resistor divider
The feedback network makes V− = VoutRg/(Rf + Rg). Rearranging gives Av = Vout/Vin = 1 + Rf/Rg.
Compute output voltage
Av = 1 + 90/10 = 10 V/V. Thus Vout = 10 × 0.20 = 2.0 V, provided the supply rails and output swing allow this value.
Estimate closed-loop bandwidth
Under the approximate constant gain-bandwidth product model, BW ≈ GBW/Av = 1 MHz/10 = 100 kHz. This is a small-signal estimate, not a complete high-frequency performance specification.
Check for real-world restrictions
The estimate presumes enough supply headroom, appropriate common-mode input range, stable negative feedback and acceptable slew rate. Large-signal, high-frequency waveforms may be limited by slew rate before the small-signal bandwidth is reached.
What the answer means
An op-amp's gain can be precisely set by the feedback resistors only while feedback keeps the device in its linear operating region. Saturation, bandwidth limits and input common-mode range become essential when translating an ideal exam calculation into practical hardware.
Three mistakes to avoid
- Omitting the “1 +” in non-inverting gain.
- Using the ideal gain when the predicted output exceeds the supply limits.
- Treating gain-bandwidth as an exact universal law for every op-amp topology.
Try a related question
Keeping Rg = 10 kΩ, choose Rf for a non-inverting gain of 5 and estimate bandwidth for a 1 MHz GBW.
Show the worked-answer check
Rf = (5 − 1) × 10 kΩ = 40 kΩ. Approximate bandwidth = 1 MHz/5 = 200 kHz under the same single-pole assumption.
How to revise this concept
Study inverting amplifiers, comparator saturation and the effect of a finite input offset voltage. First attempt the example on paper without opening the answer, then compare the intermediate steps. Explain each assumption before memorising a formula.
Content note: This is a newly prepared practice example, not an official GATE previous-year question. It was drafted with AI assistance for this educational website. Students should verify calculations and exam requirements against standard textbooks and official sources; educator review is recommended before publication.