Engineering & Technology · Electronics, Devices & Signal Processing

Operational amplifiers and analogue circuit functions

Two op-amp inputs have almost equal voltage in one calculation. They are not joined by a wire. Feedback changes the output until the circuit approaches that condition, provided the amplifier can operate in the required region and the feedback behaviour is stable.

Electrical Circuits supplies charge/current balance, stored energy, sinusoidal phase, rms and the unloaded RC response. Material Structure & Properties supplies the distinction between a material description and a complete product. This chapter explains bias and feedback before using gain. It introduces spectra through an explicit two-tone sum and sampling through points on known curves. All parameters, code intervals, timing values and noise examples are original teaching models; no physical construction or performance qualification is implied.

Back to the chapter

After reading, you can

Use feedback to derive a simple analogue function while recognising the conditions behind ideal op-amp rules.

Separate the amplifier from its surrounding loop

An operational amplifier responds to the difference between its two input voltages. The ideal voltage-feedback model has very high differential gain, negligible input currents and negligible output resistance. It simplifies a circuit calculation; it does not remove supply connections or imply unlimited available output energy.

With suitable negative feedback and linear operation, the differential input voltage can be approximated as zero. This is a consequence of the operating loop, not an unconditional property of the component. If the output cannot move far enough, an input is outside its permitted range or feedback is unstable, the shortcut may fail.

Follow the divider rather than memorising a rule

In an original non-inverting configuration, the signal reaches the positive input. A feedback resistor Rf connects output to the negative input; Rg connects that input to reference. Negligible input current makes this an unloaded divider, so v−=vout Rg/(Rf+Rg).

If v−≈v+=vin, rearranging gives vout≈(1+Rf/Rg)vin. The ratio sets the ideal closed-loop gain. For an inverting arrangement with the positive input at reference, an input resistor and feedback resistor instead yield vout≈−(Rf/Rin)vin by current balance at the negative-input node. The ideal node is a virtual reference, not an actual connection able to absorb arbitrary current.

Resistors can sum scaled inputs; capacitors can introduce time dependence. Such functions come from charge/current balance and the feedback connection. Naming a circuit “integrator” does not exempt its finite output range or its initial stored state.

Test the requested answer against the amplifier’s limits

The figure adds an explicitly hypothetical ±4 V output limit to an otherwise static gain-five model. This is a teaching clip model, not a particular op amp’s transfer curve. A predicted 5 V output is outside this assumed range even if the resistor algebra is correct.

Real checks also include input common-mode range, output load/current, offset and bias effects, bandwidth, slew and loop stability. Absolute maximum ratings describe stress boundaries, while recommended conditions and performance specifications answer different questions. Meeting a voltage range alone does not demonstrate a complete working analogue function.

Horizontal input −1 to 1 V, 250 units/V; vertical output labelled −5,0,5 V,35 units/V. Orange unrestricted vout=5vin. Blue clipped output max(−4,min(4,5vin)), clipping beyond |vin|=0.8 V. At 1 V input clipped output is 4 V; with 40 kΩ feedback and 10 kΩ reference resistor the negative input is 0.8 V. Model limits are invented, not a product guarantee.

Check the requested output before trusting gain

Take Rg=10 kΩ and Rf=40 kΩ. The ideal non-inverting gain is 1+40/10=5. At vin=0.4 V, predicted output is 2 V. The feedback-divider current is 2 V/50 kΩ=40 μA and the negative input is at 0.4 V, consistent with the assumed ideal loop.

At vin=1 V the unrestricted prediction is 5 V. In our separate clip model the output stops at 4 V and v−=0.8 V. The inputs are then unequal by 0.2 V; imposing equality after clipping would contradict the model.

At output 2 V into a separately assumed 2 kΩ load, load current is 1 mA. The supply must provide the required output energy as well as internal consumption. Input-current idealisation does not supply that load current from the signal source.

Try the reasoning

Are equal op-amp input voltages an unconditional wiring fact?
Show the answer and reasoning
No. They approximate a suitable linear negative-feedback operating condition.
Can doubling both gain-setting resistors preserve the ideal gain?
Show the answer and reasoning
Yes, their ratio remains the same. Real bias, noise, loading and dynamic effects may still change.
Why distinguish maximum stress ratings from operating specifications?
Show the answer and reasoning
Avoiding a stress limit does not establish the required linear behaviour, accuracy, range or stability.

What to carry forward

Derive the feedback relationship, then check whether the required state is feasible under the real constraints.

References

Teaching resources and research supporting this lesson’s explanations and examples.