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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 slams its output 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

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, and say what a comparator is for.
Say what negative feedback gives away and what it buys.
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.

WHAT YOU PROBABLY THINK

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. Real chips manage 10⁵, a few megohms and tens of ohms.

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

ON YOUR 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. Feed the signal to one input and a fixed reference to the other. The output then sits at the positive rail whenever the signal is above the reference and at the negative rail whenever it is below. The circuit is a comparator, and it turns a smoothly varying voltage into a two-level decision.

FIG. 1A sine wave crossing a fixed reference level, and the comparator's output beneath it: pinned at +13 V above the reference and −13 V below, switching exactly at the crossings.

The switch-over is sharp because the whole transition takes only that 130 μV. 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.

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 bargain the rest of the subject is built on. Surrender most of the gain available and you get back 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 YOUR 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 YOUR 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.

YOUR TURN

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. Its 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 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 run flat until they meet the falling open-loop line at 1 kHz, 10 kHz and 100 kHz.

The trade in negative feedback is therefore a quantitative one. Ask a 1 MHz chip for a closed-loop gain of 1000 and the amplifier is flat only to 1 kHz. Settle for a gain of 10 and the same chip runs to 100 kHz. Every factor of ten of gain surrendered buys a factor of ten of bandwidth.

TRY IT UNSEEN

Enough bandwidth for audio

An op-amp has a gain-bandwidth product of 1.0 MHz, and an audio stage must stay flat to 20 kHz. Find the largest gain one stage can give, then the overall gain if two such stages of gain 50 are cascaded.

Show the working

Largest gain = 1.0 × 10⁶/(20 × 10³) = 50.

Two stages of gain 50 give an overall gain of 50 × 50 = 2500, and each stage still reaches its own 20 kHz.

One stage asked for 2500 would have run out at 1.0 × 10⁶/2500 = 400 Hz, useless for audio. Splitting a large gain between stages is how the bandwidth is kept.

THE EXAM BIT

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 and both rails stated. Above the reference the output sits at the positive rail, below it at the negative, and the change takes only 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.

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

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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, and say what a comparator is for.
Say what negative feedback gives away and what it buys.
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