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Essential Integrated Circuits

The Operational Amplifier

Also known as: LM358, TL072

13 min read

Quick Answer

An operational amplifier has two inputs and one output, and it multiplies the difference between the inputs by an enormous gain. That gain is so large that any real signal drives the output straight to a supply rail, which is why an op-amp is almost never used on its own. Feedback is what makes it useful.

Intuition

It only watches who is ahead

A referee at the finish line is not measuring anything. They are not timing the race, they do not care how fast anyone ran, and if the whole field slows down by half they will still call the same winner. All they do is decide which of two runners is ahead, and then say so as loudly as the occasion demands.

An operational amplifier does that with voltage.

It has two inputs. It compares them, ignores everything they have in common, and drives its output in whichever direction says which one is higher. Lift both inputs by five volts together and nothing about the output changes, because nothing about the comparison changed. That indifference to the common level is not a side effect; it is the definition.

What makes an op-amp strange is how loudly it says so. The gain built into one is not ten or a hundred. It is hundreds of thousands, which means a difference far too small to see on a meter is enough to slam the output as far as the supply lets it go. On its own the part is not an amplifier at all. It is a device with an opinion and no restraint, and the entire craft of using one is about restraining it.

Practitioner

Three elements, and everything else is detail

2.0 MΩ sits between the two inputs, a controlled source produces 200000 times the difference across it, and 75 Ω in series reaches the output

Five nodes and no junction dots anywhere, because the model is a chain rather than a network.

Inside the package there are dozens of transistors: a differential pair at the front, current mirrors setting the bias, a gain stage, an output stage. None of that is what you calculate with. What you calculate with is three elements.

A resistance sits between the two inputs, an invented 2.0 MΩ in this lesson. The voltage across it is the difference, and it is the only input quantity the model contains. A controlled source then produces that difference multiplied by the open-loop gain, an invented 200000. In series with that source sits an output resistance, an invented 75 Ω, and beyond it the output terminal.

Every figure in this lesson is invented in the same way and belongs to no real part. Open-loop gains, input resistances and output currents vary by orders of magnitude across the op-amp family, and the ones that govern a design are on that part's datasheet.

In the other currency the same gain is 106 dB, which is where the decibel figures on a datasheet come from. Either way the number is doing something unusual for an amplifier specification: it is far larger than anybody wants.

Two panels of the same transfer characteristic, clipping at 13.5 V: on a 10 mV scale the slope is 0.68 % of the panel, on a 100 µV scale it is 67.5 %

Same amplifier, same vertical axis, horizontal scales a hundred apart.

Worked example — How much difference the output can stand

The output cannot reach the rails. On a 15 V supply it stops 1.5 V short at each end, so it can reach 13.5 V either way, a total span of 27.0 V.

Working backwards through the gain, the input difference that just puts the output there is 67.5 µV.

Beyond that in either direction the output is against a rail and stays there. The whole range over which this part behaves as an amplifier is 135 µV wide.

Put that beside the output span and it is 0.00050 % of it. The left panel above is drawn on a scale you might plug a signal generator into, and at that scale the sloping region is too narrow to see; the right panel is the same characteristic magnified a hundred times, where it finally looks like a line with a gradient.

Both panels are the same part. Nothing switched between them.

Engineer

More gain means less room

Input window against open-loop gain on log scales, falling one decade per decade: 200000 leaves 135 µV, a gain of 100 would leave 270 mV

One decade of gain costs one decade of usable input range, and the trade never improves.

The window is twice the output's reach divided by the gain, so it shrinks exactly as fast as the gain grows. A part with a gain of 100 would leave 270 mV to work in, which is a perfectly usable amplifier that you could drive from almost anything. A part with a gain of 10000000 leaves 2.7 µV, which is thermal noise and a warm afternoon.

This is the trap that catches people reading datasheets for the first time. Open-loop gain looks like a figure of merit, and manufacturers compete on it, and a bigger one is genuinely better. But it is not better because you get to use it. It is better because of what happens when you throw almost all of it away, which is what the golden rules are about and is the next lesson.

The one job an op-amp does do well open-loop is deciding which input is higher, since that is all the saturated output is telling you. That job has a name and a lesson of its own: comparators.

It genuinely does not care about the common level

Two panels sharing a swept common level: both inputs lie on one trace 60 µV apart, and the output stays flat at 12.0 V throughout

The top panel has two traces on it. At this scale they are one line, and the figure says so rather than pretending otherwise.

Hold the difference between the inputs constant and move both of them up and down together, and the output does not move.

Worked example — A difference small enough to keep the output honest

Set the difference at 60 µV, which is inside the window. The output then sits at 12.0 V, comfortably short of the rail.

Now sweep both inputs together from minus 5.0 V to plus 5.0 V. The output stays at 12.0 V the whole way.

Across that sweep the two input traces are 0.00060 % of the swept range apart, which is why the top panel of the figure shows only one line.

That property is worth more than the gain, and it is the reason op-amps are used to measure things. A sensor bridge sitting on a noisy supply produces a few millivolts of real signal riding on volts of rubbish that both its outputs share, and a differential amplifier can throw the shared part away. How well a real one manages that has a number attached, and instrumentation amplifiers are built around it.

There is a limit on how far the common level may go. The inputs of this invented part accept 12 V either way, which is nearer the rails than the output can reach but not all the way to them. Push past it and the internal stages come out of their working region, and the output does something unhelpful without warning you first.

Professional

What the supply lets it do

Three spans on one voltage scale: supply 15 V, output reaching 13.5 V, inputs accepting 12 V, with 10.0 % of the supply unreachable

All three bands drawn at the same pixels per volt, so the heights are the voltages.

An op-amp needs power, and every voltage it produces has to come from the supply it is given. Nothing it does can go outside those rails, and in practice nothing gets close to them.

Worked example — What the supply buys, and what it does not

On rails of 15 V either side of common, the output reaches 13.5 V either way.

That leaves 1.5 V stranded at the top and the same at the bottom, which is 10.0 % of the whole supply that no signal can ever use.

The part also draws 1.4 mA with no signal and no load at all, so it burns 42 mW sitting still.

Some op-amps do better than this. A rail-to-rail output stage gets within a few tens of millivolts of each rail instead of a volt and a half, which matters enormously on a single 3 V supply and hardly at all on a bench pair. The phrase is on the front page of a datasheet because it is the first thing a designer on a low supply needs to know.

And what the load lets it do

Reachable output against load resistance on a log axis: flat at 13.5 V above 540 Ω, falling below it as the 25 mA limit takes over

Two ceilings on one output. Which one you meet depends on the load.

There is a second ceiling, and it is a current rather than a voltage. This part will supply 25 mA, and no arrangement of the circuit around it will produce more.

Worked example — The smallest load that still reaches the top

At 25 mA the output can only reach 13.5 V into 540 Ω or more. Below that resistance the current limit governs and the output falls short.

A 2.0 kΩ load asks for 6.8 mA at full swing, which is 3.7 times inside the limit.

Driving it that hard puts 91 mW into the load, against the 42 mW the part burns doing nothing.

The output resistance costs something too, though far less than beginners expect.

Worked example — What the output resistance takes

75 Ω in series with a 2.0 kΩ load is a divider, so the gain seen at the load falls from 200000 to 192771.

In decibels that is 105.7 dB instead of 106 dB, a loss of 0.32 dB.

Which is nothing, and it is nothing for a reason: throwing away a third of a decibel out of a hundred and six matters only if you were relying on the exact figure, and nobody sensible is.

What to take from this

The gain is not a specification you use. It is a resource you spend on accuracy, and the spending is done by feedback.

The two inputs are equals in the model. Nothing in the three-element picture distinguishes them except sign, and neither is a reference.

The output has two ceilings, a voltage set by the rails and a current set by the output stage, and a design meets whichever one it reaches first.

Everything in this lesson is static. Move the signal and the picture changes: the gain falls with frequency, the output cannot move faster than a certain rate, and the inputs are not exactly matched. Those are the subject of real op-amp limitations, and none of them is in the model above.

Common mistakes

  • Treating open-loop gain as usable gain — at 200000 the input window is 135 µV wide, which is 0.00050 % of the output's 27.0 V span. Any real signal saturates the part. The gain is spent on accuracy through feedback, not collected at the output.
  • Expecting the output to reach the rails — it stops 1.5 V short at each end here, which is 10.0 % of the supply gone. On a bench pair that is a detail; on a 3 V single supply it is most of the signal, and it is why rail-to-rail parts exist.
  • Referencing one input to ground in the model — nothing in the three-element picture makes either input a reference. What the amplifier responds to is the voltage across the 2.0 MΩ between them, and both inputs may sit anywhere inside the 12 V common-mode range.
  • Forgetting the output current limit — 25 mA into a load below 540 Ω means the output cannot reach 13.5 V no matter what the feedback demands. The circuit will look like a gain error and is a current limit.
  • Worrying about output resistance — 75 Ω into 2.0 kΩ costs 0.32 dB of a 106 dB gain. Feedback removes even that, and the parameter is mostly there to explain why heavy loads are a problem at all.
  • Reading a datasheet gain figure as a promise — it is quoted at DC, into a stated load, at one temperature, and it varies by a factor of several between parts from the same reel. Circuits are designed so that the exact figure does not matter.

Frequently asked questions

Why make the gain so large if it cannot be used?

Because feedback converts spare gain into accuracy. A circuit built around an op-amp sets its behaviour with a ratio of components and relies on the amplifier to be effectively perfect; the larger the open-loop gain, the closer to perfect it is, and the less the exact figure matters. A gain of 200000 is not there to be used. It is there to be thrown away.

What happens if I connect an op-amp with nothing in the feedback path?

The output sits against one rail or the other, decided by whichever input happens to be higher, including by a fraction of a millivolt of internal imbalance. That is a legitimate circuit when the answer you want is which input is higher, and useless when you wanted an amplifier. Comparators are the parts built for the first job.

Does an op-amp need a split supply?

No. It needs a supply, and the model only cares about the total span between the rails and where the signals sit inside it. A single supply works perfectly well provided the inputs and output stay inside their ranges, which usually means arranging a mid-supply reference for signals to sit on rather than using zero volts.

Why does the output stop short of the supply rails?

Because the transistors in the output stage need some voltage across them to keep working, and how much depends on how the stage is built. 1.5 V at each end is typical of the older designs; a rail-to-rail output stage uses a different topology and gets within tens of millivolts. Neither reaches the rail exactly, because a device with zero volts across it is a short circuit rather than a transistor.

Is the input resistance really megohms?

For a bipolar-input part it is that order, and for an input stage built on field-effect transistors it is far higher and the number stops meaning much. In either case the resistance is rarely the thing that limits a circuit; the small current the inputs actually draw usually matters more, and that is a separate parameter.

Knowledge check

An op-amp has an open-loop gain of 200000 and an output that reaches 13.5 V either way. Over what range of input difference is it an amplifier? (Show answer)
Working back through the gain, 13.5 V corresponds to 67.5 µV, so the output leaves the rail only between plus and minus that: a window 135 µV wide. That is 0.00050 % of the 27.0 V the output covers.
Both inputs are swept together from -5.0 V to +5.0 V while the difference between them is held at 60 µV. What does the output do? (Show answer)
Nothing. It sits at 12.0 V for the whole sweep, because the amplifier responds only to the difference and that never changed. The 60 µV difference is 0.00060 % of the swept range, which is why both input traces draw as one line.
Why can this part not reach 13.5 V into a 200 Ω load? (Show answer)
Because the output current limit of 25 mA governs below 540 Ω. Into 200 Ω the limit allows 5.0 V and no more, whatever the feedback demands. Above 540 Ω the rail is the ceiling instead.
On ±15 V rails, how much of the supply can no signal ever use, and why? (Show answer)
The output stops 1.5 V short of each rail, so 3 V of the 30 V span is unreachable: 10.0 % of the supply. The output stage's transistors need voltage across them to keep working.
How much gain does 75 Ω of output resistance cost when driving 2.0 kΩ? (Show answer)
The two form a divider, so 200000 becomes 192771 at the load. In decibels that is 105.7 dB against 106 dB, a loss of 0.32 dB, which is negligible and is why the parameter rarely decides anything.