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Operational amplifiers

An amplifier with a gain of a hundred thousand is useless on its own, since a fraction of a millivolt drives its output hard against the supply rail. Hand some of the output back to the input and the same chip becomes whatever amplifier a pair of resistors describes.

Year 13AQA 3.13.2.1, 3.13.3.2, 3.13.4.1, 3.13.4.2, 3.13.4.4

Builds on Potential dividers and Discrete semiconductor devices.

IN THIS TOPIC

State the properties of the ideal op-amp and use the open-loop relation.
Explain comparator action and saturation, say which input carries the signal, and say what a comparator is for.
Say what negative feedback sacrifices and what it gains.
Use the inverting gain, and explain the virtual earth that produces it.
Use the non-inverting gain, and use the gain-bandwidth product to find a bandwidth.

COMMON MISCONCEPTION

An op-amp with an open-loop gain of 100 000 multiplies any input by 100 000.

The ideal amplifier, and the rails that own it

An operational amplifier amplifies the difference between its two inputs, the non-inverting input $V_{+}$ and the inverting input $V_{-}$ . At A-level it is treated as ideal, and the ideal has four properties worth learning as a list. Its open-loop gain $A_{OL}$ is infinite. Its input resistance is infinite, so neither input draws current. Its output resistance is zero, so the output does not sag under load. Its bandwidth is infinite. A representative general-purpose device manages about 10⁵, a few megohms and tens of ohms; the numbers vary from chip to chip, and the data sheet is the authority for any particular part.

V_{out} = A_{OL}(V_{+} - V_{-})

ON THE AQA DATA SHEET

Take that equation seriously for a moment. With $A_{OL}$ = 10⁵ and an output that can reach 13 V before the supply rails stop it, a difference of 13/10⁵ = 130 μV already drives the output the whole way. Anything larger asks for a voltage beyond the rails, and the amplifier cannot deliver it. The output sticks a volt or so inside the rail and stays there, saturated. So much for multiplying any input by a hundred thousand.

Saturation put to work: the comparator

Being saturated is useful if the question you are asking is which input is larger. Wire the signal to the non-inverting input $V_{+}$ and the fixed reference to the inverting input $V_{-}$ . Then $V_{+} - V_{-}$ is positive whenever the signal is above the reference, so the output sits at the positive rail, and it is negative whenever the signal is below the reference, so the output sits at the negative rail. Swap the two connections, reference on $V_{+}$ and signal on $V_{-}$ , and both statements reverse. The circuit is a comparator, and it turns a smoothly varying voltage into a two-level decision. Which input carries the signal is part of the answer, never an afterthought.

FIG. 1A sine wave on the non-inverting input crossing a fixed reference held on the inverting input, and the comparator's output beneath it: pinned at +13 V while the signal is above the reference and −13 V while it is below, switching exactly at the crossings.

The switch-over is sharp because the whole swing from one rail to the other takes only 26/10⁵ = 260 μV of input difference. A thermostat is a comparator with a thermistor divider on one input and a preset divider on the other. A light-operated alarm is the same circuit with an LDR, and a frost warning is the same circuit again with the inputs swapped, which is the polarity reversal above.

Giving the gain away on purpose

Feed a fraction of the output back to the inverting input and the amplifier acquires a conscience. Should the output start to rise too far, the fed-back share raises $V_{-}$ , which cuts the difference the amplifier is working on and pulls the output back down. It settles wherever the two inputs are almost equal, and with so much open-loop gain, almost equal means within microvolts.

The gain then belongs to the feedback network rather than to the chip, and that is the trade-off the rest of the subject is built on. Give up most of the gain available and what you get in return is a gain set by resistors you chose, a far wider bandwidth, less distortion, and behaviour that survives a change of chip.

The inverting amplifier and its virtual earth

The inverting amplifier earths the non-inverting input and feeds the signal through $R_{in}$ to the inverting input, with $R_{f}$ bridging that input to the output. Feedback holds the two inputs within microvolts of each other, and one of them is at 0 V, so the other is held at 0 V as well without being connected to earth at all. That node is the virtual earth.

FIG. 20.20 V across a 10 kΩ resistor drives 20 μA into the virtual earth. No current enters the op-amp, so the same 20 μA continues through the 47 kΩ feedback resistor and the output sits at −0.94 V.

Two facts finish the derivation. The current arriving through $R_{in}$ is $V_{in}/R_{in}$ , since one end of that resistor is at $V_{in}$ and the other is at 0 V. None of it enters the op-amp, so all of it runs on through $R_{f}$ , and the output must sit at minus that current times $R_{f}$ . Divide one by the other and the chip has vanished from the answer:

\frac{V_{out}}{V_{in}} = -\frac{R_{f}}{R_{in}}

ON THE AQA DATA SHEET

WORKED EXAMPLE

Following the current round

In the figure $V_{in}$ = 0.20 V, $R_{in}$ = 10 kΩ and $R_{f}$ = 47 kΩ. Find the current into the virtual earth and the output voltage, then check against the gain equation.

The input resistor has the whole 0.20 V across it, so I = 0.20/(10 × 10³) = 20 μA.

That current continues through the 47 kΩ resistor from a node at 0 V, so the output is −20 × 10⁻⁶ × 47 × 10³ = −0.94 V.

The gain equation agrees, −(47/10) × 0.20 = −0.94 V. Learn the current route anyway, because the derivation marks are written for it.

The non-inverting amplifier

Move the signal to the non-inverting input and return the feedback through a divider of $R_{f}$ and $R_{1}$ hung across the output. The divider offers the inverting input the fraction $R_{1}/(R_{1} + R_{f})$ of whatever the output is doing, and feedback drives the output until that fraction matches $V_{in}$ . Rearranging gives a gain of

\frac{V_{out}}{V_{in}} = 1 + \frac{R_{f}}{R_{1}}

ON THE AQA DATA SHEET

FIG. 339 kΩ over 10 kΩ gives a gain of 4.9, so 0.30 V in becomes 1.47 V out. The divider returns exactly 0.30 V to the inverting input, and that is what holds the output where it is.

Two differences from the inverting circuit matter in an exam. The output is in phase with the input, and the gain can never fall below one, since the 1 is stuck at the front. The signal also arrives straight at the op-amp's own input terminal, so the circuit draws almost nothing from whatever drives it. A high-resistance sensor is far better read this way.

GUIDED PRACTICE

Designing to a number

An instrument needs a non-inverting amplifier of gain 11, built with 1.0 kΩ as $R_{1}$ . Find $R_{f}$ , and state the gain the same two resistors would give in the inverting arrangement.

Show the working

1 + $R_{f}$ /1.0 = 11, so $R_{f}$ = 10 kΩ.

In the inverting circuit the same pair gives −10/1.0 = −10.

The non-inverting gain is always one greater, and it keeps the phase. Quoting 10 for a non-inverting design is the standard slip, and it costs the mark.

What the bargain costs: gain-bandwidth product

The bandwidth a real op-amp has is small. For the same representative compensated device, the open-loop gain of 10⁵ only holds up to about 10 Hz, and above that the gain falls in proportion to frequency until it reaches one at around 1 MHz. Multiply the gain by the frequency at which it runs out and the answer is the same everywhere along that slope:

\text{gain} \times \text{bandwidth} = \text{constant}

NOT ON THE AQA DATA SHEET: LEARN IT

FIG. 4Gain against frequency for a chip with a gain-bandwidth product of 1 MHz. Closed-loop gains of 1000, 100 and 10 are drawn as straight-line approximations, running flat until they meet the falling open-loop line at 1 kHz, 10 kHz and 100 kHz, the corner at which the true gain has already fallen to 0.71 of its flat value.

The trade in negative feedback is therefore a quantitative one. Ask a 1 MHz chip for a closed-loop gain of 1000 and its corner lands at 1 kHz; settle for a gain of 10 and the same chip corners at 100 kHz. Every factor of ten of gain surrendered gives a factor of ten of bandwidth.

Be careful with the word flat. The corner quoted as a bandwidth is where the gain has already fallen to $1/\sqrt{2}$ , about 0.71, of its low-frequency value, the same 0.71 that fixed a filter's cut-off earlier in this unit, which is why it is called the 3 dB point. A bandwidth of 1 kHz therefore means flat to within that 0.71 up to 1 kHz, not untouched up to it. Cascade two identical stages and each contributes its own 0.71 at the shared corner, leaving the pair at 0.71 × 0.71 = 0.50 there, so a chain is always narrower than any one of its stages.

INDEPENDENT PRACTICE

Enough bandwidth for audio

An op-amp has a gain-bandwidth product of 1.0 MHz, and an audio stage must reach 20 kHz. Find the gain that puts one stage's corner at 20 kHz. Two such stages are then cascaded: give the overall gain, and say where the pair has fallen to 0.71. Then find the gain per stage that would put the pair's own 0.71 point at 20 kHz.

Show the working

Gain = 1.0 × 10⁶/(20 × 10³) = 50, so one stage of gain 50 has already dropped to 0.71 of its flat value at 20 kHz.

Two of them give an overall gain of 50 × 50 = 2500, but each is at 0.71 there, so the pair is at 0.71 × 0.71 = 0.50 of flat at 20 kHz, not flat at all. The pair reaches 0.71 far earlier, where $1 + (f/f_{c})^{2} = \sqrt{2}$ , that is at 0.644 of the stage corner: 20 × 0.644 = 12.9 kHz.

For the pair to hold 0.71 out to 20 kHz each stage needs its corner at 20/0.644 = 31.1 kHz, so a gain of 1.0 × 10⁶/(31.1 × 10³) = 32. Two stages of 32 give 1024 overall and reach 0.71 at 20.1 kHz.

Splitting a large gain between stages still gains bandwidth, since one stage of gain 1024 would corner at 1.0 × 10⁶/1024 = 977 Hz. The lesson is to test the tolerance on the whole chain rather than on one stage.

ASSESSMENT FOCUS

The four ideal properties are a list, so write them as one. Infinite open-loop gain, infinite input resistance, zero output resistance, infinite bandwidth.
Comparator answers need the reference named, the input it sits on stated, and both rails given. With the signal on $V_{+}$ and the reference on $V_{-}$ the output is at the positive rail above the reference and at the negative rail below it; swapping the inputs reverses both, and either way the change takes only a few hundred microvolts.
The virtual earth carries the derivation marks. The inverting input is held at 0 V, the input current is $V_{in}/R_{in}$ , no current enters the op-amp, so all of it flows on through $R_{f}$ .
Mind the signs. Inverting gain is negative and the output is in antiphase; non-inverting gain is positive and can never drop below 1.
Hold every calculated output against the supply rails before writing it down. If it exceeds them, say the amplifier saturates and give the output as the rail voltage.
Bandwidth questions are one division. Gain-bandwidth product over the closed-loop gain, and the two always move in opposite directions. That answer is the frequency at which the gain has fallen to 0.71, so a cascade of stages reaches 0.71 sooner than any single stage in it.

CHECK YOURSELF

An op-amp with an open-loop gain of 1.0 × 10⁵, a gain-bandwidth product of 1.0 MHz and an output that saturates at ±13 V is wired as an inverting amplifier with $R_{in}$ = 5.0 kΩ and $R_{f}$ = 100 kΩ. Find the closed-loop gain and the bandwidth, then the output for an input of 0.10 V and for an input of 1.0 V.

Show a hint

Gain from the resistor ratio, bandwidth from the product, and hold each answer against the rails.

Show the answer

Gain = − $R_{f}/R_{in}$ = −100/5.0 = −20, so the output is inverted and twenty times larger.

Bandwidth = 1.0 × 10⁶/20 = 50 kHz.

For 0.10 V in, the output is −20 × 0.10 = −2.0 V, comfortably inside the rails.

For 1.0 V in, the equation asks for −20 V. The output cannot pass the rail, so the amplifier saturates at about −13 V and the waveform is clipped flat.

With no feedback the op-amp only compares: a few hundred microvolts of difference send the output to a rail.

Negative feedback hands the gain to the resistors, and gain times bandwidth stays constant.

Inverting gain is minus Rf over Rin about a virtual earth; non-inverting gain is 1 + Rf over R1.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

17 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the operational amplifiers questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Rearranging equations is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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State the properties of the ideal op-amp and use the open-loop relation.
Explain comparator action and saturation, say which input carries the signal, and say what a comparator is for.
Say what negative feedback sacrifices and what it gains.
Use the inverting gain, and explain the virtual earth that produces it.
Use the non-inverting gain, and use the gain-bandwidth product to find a bandwidth.

Open the full revision checklist to track your progress across the whole unit.

Operational amplifiers

The ideal amplifier, and the rails that own it

Saturation put to work: the comparator

Giving the gain away on purpose

The inverting amplifier and its virtual earth

The non-inverting amplifier

What the bargain costs: gain-bandwidth product

Question workbook

Question workbook

Answer book