Notes · ElectronicsUK · A-Levels

Operational amplifiers

The operational amplifier is the workhorse of analogue design. This chapter develops the ideal op-amp rules, the comparator, and the inverting, non-inverting and summing configurations, deriving each gain from the virtual-earth analysis. It closes with the real-world limits of bandwidth, gain-bandwidth product and slew rate that decide whether a design actually works.

5 sections~19 min reading time3 competenciesLevel Standard 3 · Advanced 2

T·0222 / 16

Exam profile

AO1 · State the ideal op-amp rules and describe the comparator, inverting, non-inverting and summing configurations.AO2 · Calculate closed-loop gain and output voltage for each configuration using the virtual-earth analysis.AO3 · Evaluate the effect of finite gain-bandwidth product and slew rate on a proposed amplifier and choose components to meet a specification.

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basic level

At AS the focus is the comparator and the inverting and non-inverting amplifiers, with gain and output-voltage calculations and the idea of the virtual earth.

higher level

The full A-Level adds the summing amplifier, and the quantitative treatment of gain-bandwidth product and slew rate as limits on a real design.

Depth

Text

Contents · 5 sections

Operational amplifiers

The ideal op-amp and the comparator#

StandardWJEC/Eduqas A Level Electronics — Unit 1 Analogue Electronics (op-amp basics)

Comparator transfer characteristic

Fig. 1Open-loop, the output snaps between the two saturation voltages (here about

\pm 13\,\text{V}

) as the input crosses the reference — a step-like transfer characteristic.

An operational amplifier is a very high-gain differential voltage amplifier: it amplifies the difference between its non-inverting (

+

) and inverting (

-

) inputs. Its ideal properties are worth committing to memory because every configuration is analysed from them: infinite open-loop gain, infinite input impedance (so no current flows into either input), zero output impedance, and infinite bandwidth. Real devices only approximate these, but the idealisations make the analysis tractable and are accurate enough for design.

Two 'golden rules' follow from the ideal model when negative feedback is present. Rule one: no current flows into either input, because the input impedance is infinite. Rule two: the op-amp adjusts its output until the two inputs are at the same voltage, because the gain is so large that even a microvolt of difference would saturate the output. These two rules — no input current, and equal input voltages — are all you need to derive the gain of every feedback configuration in this chapter.

Without feedback the op-amp is used open-loop as a comparator. Because the open-loop gain is enormous, the tiniest difference between the inputs drives the output hard to one supply rail or the other. If the

+

input is above the

-

input the output sits near the positive rail; if it is below, the output sits near the negative rail. The transfer characteristic is therefore essentially a step: the comparator answers the yes/no question 'is this input above that reference?' and is the natural bridge from an analogue signal to a digital decision.

The comparator is the input-processing element of countless control systems: a light-level, temperature or voltage signal from a potential-divider sensor is compared against a reference, and the digital output then drives a transistor switch or a logic system. Adding a little positive feedback turns the comparator into a Schmitt trigger with hysteresis, which you meet in the timing chapter; here the key idea is simply that an op-amp with no feedback is a decision-maker, not a linear amplifier.

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

Open-loop relation

With huge open-loop gain the output saturates at a rail for any non-zero input difference.

Worked example

Comparator output states

An op-amp comparator on $\pm 15\,\text{V}$ rails has a $2.0\,\text{V}$ reference on the inverting input and saturates about $1.5\,\text{V}$ short of each rail. Find the output when the non-inverting input is $1.0\,\text{V}$ and when it is $3.0\,\text{V}$ .

Input at 1.0 V
The $+$ input ( $1.0\,\text{V}$ ) is below the $-$ input ( $2.0\,\text{V}$ ), so the difference is negative and the output is driven to the negative rail: about $-13.5\,\text{V}$ .
Input at 3.0 V
Now the $+$ input ( $3.0\,\text{V}$ ) is above the reference, the difference is positive, and the output is driven to the positive rail: about $+13.5\,\text{V}$ .

Result: The output is about $-13.5\,\text{V}$ for a $1.0\,\text{V}$ input and about $+13.5\,\text{V}$ for a $3.0\,\text{V}$ input — the comparator has turned an analogue level into a two-state digital decision.

Typical mistakes

Assuming current flows into an op-amp input — the ideal input impedance is infinite, so no input current flows.
Expecting a comparator output to sit somewhere between the rails; open-loop it is driven fully to one rail or the other.

Active revision

An op-amp comparator runs from $\pm 15\,\text{V}$ rails with the reference on its $-$ input set to $2.0\,\text{V}$ . State the approximate output voltage when the $+$ input is at $1.0\,\text{V}$ and when it is at $3.0\,\text{V}$ .

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The inverting amplifier and the virtual earth#

StandardWJEC/Eduqas A Level Electronics — Unit 1 Analogue Electronics (inverting amplifier)

Inverting amplifier with virtual earth

Fig. 2The

-

input is held at

0\,\text{V}

(virtual earth). The input current

V_{in}/R_{in}

flows on through

R_f

, forcing

V_{out} = -R_f\,V_{in}/R_{in}

The inverting amplifier applies the input through a resistor

R_{in}

to the inverting (

-

) input, with a feedback resistor

R_f

from the output back to the same point, and the non-inverting (

+

) input grounded. Because the

+

input is at

0\,\text{V}

and the golden rules force the two inputs to the same voltage, the

-

input also sits at

0\,\text{V}

— even though it is not physically connected to ground. This is the virtual earth, and it is the key to the whole analysis.

Analysing the currents at the virtual earth gives the gain directly. The input current is

V_{in}/R_{in}

because the left end of

R_{in}

is at

V_{in}

and the right end is at the virtual earth (

0\,\text{V}

). No current enters the op-amp (golden rule one), so all of this current must flow on through the feedback resistor

R_f

. The output must therefore sit at whatever voltage makes that current flow through

R_f

, namely

V_{out} = -R_f \times (V_{in}/R_{in})

. Rearranging gives the closed-loop gain

A_V = -R_f/R_{in}

The minus sign is the phase inversion: a positive input produces a negative output and vice versa. The magnitude of the gain is set purely by the ratio of two resistors, not by the op-amp itself, which is exactly why negative feedback is so powerful — it trades away the op-amp's huge but ill-defined open-loop gain for a precise, stable, predictable closed-loop gain that you can design to any value simply by choosing the resistor ratio.

The virtual earth also fixes the input impedance of the stage at

R_{in}

, because the source 'sees'

R_{in}

terminated in a virtual earth. This matters when a stage must not load its source. It also makes the inverting configuration the natural basis of the summing amplifier and of many filter and instrumentation circuits, because several signals can be fed into the same virtual earth without interfering with one another — each contributes its own current independently.

In design questions the task is usually to pick

R_f

and

R_{in}

to give a required gain while keeping the resistors in a sensible range (typically kilohms to hundreds of kilohms, so that bias currents are negligible but noise stays low). Because only the ratio matters, there is freedom to scale both resistors together; the examiner rewards a justified choice, for example choosing

R_{in}

to set the input impedance and then

R_f

to set the gain.

A_V = -\dfrac{R_f}{R_{in}}

Inverting gain

Closed-loop gain set by the resistor ratio; the minus sign is the phase inversion.

V_{out} = -\dfrac{R_f}{R_{in}}\,V_{in}

Output voltage

Follows from the virtual-earth current balance.

Worked example

Designing an inverting amplifier

Design an inverting amplifier with a gain of $-20$ and an input impedance of $5.0\,\text{k}\Omega$ . Then find the output when the input is $+150\,\text{mV}$ .

Input resistor
The input impedance of an inverting amplifier equals $R_{in}$ , so $R_{in} = 5.0\,\text{k}\Omega$ .
Feedback resistor
$A_V = -R_f/R_{in}$ , so $R_f = |A_V| \times R_{in} = 20 \times 5.0\,\text{k}\Omega = 100\,\text{k}\Omega$ .
$R_f = |A_V|\,R_{in} = 20 \times 5.0\,\text{k}\Omega = 100\,\text{k}\Omega$
Output voltage
$V_{out} = -\dfrac{R_f}{R_{in}} V_{in} = -20 \times 0.150 = -3.0\,\text{V}$ . The input is inverted and amplified twentyfold.

Result: $R_{in}=5.0\,\text{k}\Omega$ , $R_f=100\,\text{k}\Omega$ ; an input of $+150\,\text{mV}$ gives $V_{out}=-3.0\,\text{V}$ .

Typical mistakes

Dropping the minus sign, or forgetting that the inverting amplifier inverts the signal.
Thinking the $-$ input is connected to ground — it is a virtual earth held at $0\,\text{V}$ by feedback, not a physical ground.

Active revision

Design an inverting amplifier with a voltage gain of $-15$ and an input impedance of $10\,\text{k}\Omega$ . State both resistor values and the output for an input of $-0.20\,\text{V}$ .

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The non-inverting amplifier and the voltage follower#

StandardWJEC/Eduqas A Level Electronics — Unit 1 Analogue Electronics (non-inverting amplifier)

Non-inverting amplifier

Fig. 3The signal drives the

+

input directly (high input impedance). The feedback divider

R_f

R_{in}

sets

A_V = 1 + R_f/R_{in}

In the non-inverting amplifier the signal is applied directly to the non-inverting (

+

) input, and the feedback network —

R_f

from the output to the

-

input and

R_{in}

from the

-

input to ground — sets the gain. The golden rules again do the work: no current enters the inputs, and the op-amp holds the

-

input equal to the

+

input, which is now at

V_{in}

. The feedback network is therefore a potential divider from

V_{out}

to ground, tapped at

V_{in}

Setting the divider output equal to

V_{in}

gives

V_{in} = V_{out}\,R_{in}/(R_{in}+R_f)

, which rearranges to the closed-loop gain

A_V = 1 + R_f/R_{in}

. Notice two things: the gain is positive, so there is no phase inversion, and it can never be less than one, because the '

1

' is always present. The non-inverting amplifier is the natural choice whenever a signal must be amplified without inversion or when a very high input impedance is needed.

That high input impedance is the configuration's great strength. Because the signal drives the

+

input directly, and no current flows into it, the source is barely loaded at all — the input impedance is essentially that of the op-amp itself, which is enormous. This is the opposite of the inverting amplifier, whose input impedance is only

R_{in}

. When a sensor or a high-impedance source must not be loaded, the non-inverting configuration is preferred.

The special case

R_f = 0

(or

R_{in}

removed) gives a gain of exactly one: the voltage follower, or buffer. It seems pointless to build an amplifier with unity gain, but the buffer is one of the most useful blocks in all of electronics. Its huge input impedance means it takes almost no current from the source, while its near-zero output impedance means it can drive a heavy load; it therefore isolates a delicate signal source from a demanding load — impedance matching without changing the voltage.

Choosing between the inverting and non-inverting configurations is a genuine design decision. The inverting amplifier can attenuate (gain magnitude below one) and easily sums signals, but has a modest input impedance and inverts. The non-inverting amplifier preserves phase and offers very high input impedance but cannot attenuate. Matching the configuration to the demands of the surrounding system — input impedance, sign, gain range — is exactly the AO3 reasoning the examiner is looking for.

A_V = 1 + \dfrac{R_f}{R_{in}}

Non-inverting gain

From the feedback potential divider; always at least one, no phase inversion.

A_V = 1 \ \text{(voltage follower)}

Buffer

With Rf = 0 the stage has unity gain, very high input impedance and low output impedance.

Worked example

Non-inverting gain and the buffer

A non-inverting amplifier has $R_f=90\,\text{k}\Omega$ and $R_{in}=10\,\text{k}\Omega$ . Find the gain and the output for a $250\,\text{mV}$ input. Then state what happens if $R_f$ is replaced by a direct wire.

Gain
$A_V = 1 + R_f/R_{in} = 1 + 90/10 = 1 + 9 = 10$ .
$A_V = 1 + \dfrac{90\,\text{k}\Omega}{10\,\text{k}\Omega} = 10$
Output
$V_{out} = A_V V_{in} = 10 \times 0.250 = 2.5\,\text{V}$ , in phase with the input.
Follower case
Replacing $R_f$ with a wire (and removing $R_{in}$ ) makes $A_V = 1$ : a voltage follower that copies the input voltage but presents a very high input impedance and a low output impedance.

Result: $A_V = 10$ gives $V_{out}=2.5\,\text{V}$ ; shorting the feedback turns the stage into a unity-gain buffer.

Typical mistakes

Forgetting the ' $+1$ ' and quoting the gain as $R_f/R_{in}$ .
Claiming a non-inverting amplifier can have a gain below one — its minimum gain is exactly one (the follower).

Active revision

A non-inverting amplifier uses $R_f = 47\,\text{k}\Omega$ and $R_{in}=10\,\text{k}\Omega$ . Calculate its gain and its output for a $+0.40\,\text{V}$ input, and state one advantage over the inverting configuration.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The summing amplifier#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (summing amplifier)

Three-input summing amplifier

Fig. 4Each input drives its own current into the virtual earth; the feedback resistor sums them, giving

V_{out} = -R_f\sum V_i/R_i

The summing amplifier extends the inverting configuration by feeding several inputs, each through its own resistor, into the same virtual earth. Because the virtual earth holds that node at

0\,\text{V}

and no current enters the op-amp, each input drives its own independent current into the node: input one contributes

V_1/R_1

, input two contributes

V_2/R_2

, and so on. The virtual earth is what keeps them from interfering — each source sees

0\,\text{V}

at its resistor's far end regardless of what the other sources are doing.

All of these currents must flow on through the feedback resistor

R_f

, so the output is

V_{out} = -R_f(V_1/R_1 + V_2/R_2 + V_3/R_3 + \dots)

. When all the input resistors are equal to a common value

R

, this simplifies to

V_{out} = -(R_f/R)(V_1 + V_2 + V_3 + \dots)

: the output is the inverted, scaled sum of the inputs. The summing amplifier is therefore an analogue adder — it performs arithmetic on voltages.

By choosing the input resistors independently you can give each input a different weight. If

R_1 = R

R_2 = R/2

and

R_3 = R/4

, then successive inputs are weighted 1, 2 and 4 — the binary weights. This is precisely how a binary-weighted digital-to-analogue converter works, as you will see in the signal-conversion chapter: each bit of a binary number is fed to a summing amplifier through a resistor scaled by its place value, and the output is the analogue equivalent of the number.

The summing amplifier is also the basis of audio mixing, where several signal sources are combined into one output, and of offset and level-shifting circuits, where a fixed reference is added to a signal. In every case the design proceeds from the same virtual-earth current balance: work out each input current, add them, and multiply by

R_f

with a minus sign. The independence of the inputs — a direct consequence of the virtual earth — is what makes the block so useful.

A subtle but examinable point is that the inputs do not load one another precisely because the summing junction is a virtual earth. If the op-amp's feedback failed, the node would no longer be held at

0\,\text{V}

and the inputs would interact. This is a good illustration of why negative feedback is doing the heavy lifting throughout op-amp design: it creates the well-defined conditions (here, a rock-steady

0\,\text{V}

node) that make the simple current analysis valid.

V_{out} = -R_f\left(\dfrac{V_1}{R_1} + \dfrac{V_2}{R_2} + \dfrac{V_3}{R_3}\right)

Summing amplifier

Each input drives its own current into the virtual earth; all flow through Rf.

V_{out} = -\dfrac{R_f}{R}\,(V_1 + V_2 + V_3) \quad (R_1=R_2=R_3=R)

Equal-resistor case

The output is the inverted, scaled sum of the inputs.

Worked example

Output of a weighted summing amplifier

A summing amplifier has $R_f=20\,\text{k}\Omega$ . Input A is $0.5\,\text{V}$ through $10\,\text{k}\Omega$ , input B is $1.0\,\text{V}$ through $20\,\text{k}\Omega$ and input C is $0.8\,\text{V}$ through $40\,\text{k}\Omega$ . Find the output.

Input currents
$I_A = 0.5/10\,\text{k} = 50\,\mu\text{A}$ ; $I_B = 1.0/20\,\text{k} = 50\,\mu\text{A}$ ; $I_C = 0.8/40\,\text{k} = 20\,\mu\text{A}$ .
Total feedback current
$I = 50 + 50 + 20 = 120\,\mu\text{A}$ , all flowing through $R_f$ .
Output
$V_{out} = -I R_f = -(120\,\mu\text{A})(20\,\text{k}\Omega) = -2.4\,\text{V}$ .
$V_{out} = -R_f\left(\dfrac{0.5}{10\text{k}} + \dfrac{1.0}{20\text{k}} + \dfrac{0.8}{40\text{k}}\right) = -2.4\,\text{V}$

Result: The output is $-2.4\,\text{V}$ : the inverted, weighted sum of the three inputs.

Typical mistakes

Averaging the inputs instead of summing them — the output is a weighted sum, not a mean.
Forgetting that each input current is $V/R$ referenced to the virtual earth, and mishandling unequal input resistors.

Active revision

A summing amplifier has $R_f = 10\,\text{k}\Omega$ and inputs $V_1=1.0\,\text{V}$ through $10\,\text{k}\Omega$ , $V_2=2.0\,\text{V}$ through $10\,\text{k}\Omega$ and $V_3=0.5\,\text{V}$ through $5.0\,\text{k}\Omega$ . Calculate the output voltage.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Bandwidth, gain-bandwidth product and slew rate#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (op-amp limits)

Open-loop gain roll-off (Bode plot)

Fig. 5The open-loop gain falls at

-20\,\text{dB}

/decade; it reaches unity (0 dB) at the transition frequency, here

1\,\text{MHz}

— the gain-bandwidth product.

A real op-amp does not have infinite bandwidth: its open-loop gain is enormous at DC but falls off with frequency, dropping at a steady

-20\,\text{dB}

per decade above a low break frequency until it reaches unity (0 dB). The frequency at which the open-loop gain has fallen to one is called the transition frequency, and the product of gain and bandwidth is constant along the roll-off. This constant is the gain-bandwidth product (GBP), a single figure that captures the speed of the device.

The practical consequence is a direct trade-off between gain and bandwidth in any feedback amplifier: the closed-loop gain multiplied by the closed-loop bandwidth equals the GBP. Double the gain and you halve the usable bandwidth. So an op-amp with a

1\,\text{MHz}

GBP configured for a gain of 100 can only amplify signals up to about

10\,\text{kHz}

before its response starts to fall; the same op-amp at a gain of 10 reaches

100\,\text{kHz}

. Recognising and calculating this trade-off is a core A2 skill.

Slew rate is a separate, large-signal limit: it is the maximum rate at which the output voltage can change, quoted in volts per microsecond. However fast the input demands the output to move, the output cannot rise or fall faster than the slew rate, so a large, fast signal is distorted into a triangular ramp. For a sinusoid of peak voltage

V_p

and frequency

f

, the steepest part of the wave has slope

2\pi f V_p

; setting this equal to the slew rate gives the full-power bandwidth, the highest frequency at which the full output swing is undistorted.

Bandwidth and slew rate are different limits and must not be confused. Bandwidth is a small-signal limit set by the GBP and the gain; slew rate is a large-signal limit set by how fast the internal capacitances can be charged. A design can be perfectly within its bandwidth yet still slew-limited if the wanted output is both large and fast. The examiner expects you to test a proposed amplifier against both limits before declaring it fit for purpose.

The final large-signal reality is clipping: the output cannot exceed the supply rails, so any demand for a bigger swing is flattened at the rail, distorting the waveform. Together, finite gain-bandwidth product, slew rate and clipping convert the ideal op-amp of the earlier sections into a real device with real limits — and evaluating a design against all three is precisely the analyse-and-evaluate reasoning the A-Level rewards.

A_{CL}\times f_{BW} = \text{GBP}

Gain-bandwidth product

Closed-loop gain times bandwidth is constant for a given op-amp.

f_{max} = \dfrac{\text{SR}}{2\pi V_p}

Full-power bandwidth

Highest frequency at which a peak-Vp sinusoid stays within the slew rate.

Worked example

Testing an amplifier against both limits

An op-amp has a gain-bandwidth product of $2.0\,\text{MHz}$ and a slew rate of $0.5\,\text{V}/\mu\text{s}$ . It is configured for a gain of 40 and must deliver a $4.0\,\text{V}$ peak sinusoid. Find the small-signal bandwidth and the full-power bandwidth, and state which limits the design.

Small-signal bandwidth
$f_{BW} = \text{GBP}/A_{CL} = 2.0\,\text{MHz}/40 = 50\,\text{kHz}$ .
Full-power bandwidth
$f_{max} = \text{SR}/(2\pi V_p) = (0.5\times 10^{6}\,\text{V/s})/(2\pi \times 4.0) = 500000/25.13 = 19.9\,\text{kHz}$ .
$f_{max} = \dfrac{0.5\times 10^{6}}{2\pi \times 4.0} \approx 19.9\,\text{kHz}$
Which limits first
The full-power bandwidth ( $\approx 20\,\text{kHz}$ ) is lower than the small-signal bandwidth ( $50\,\text{kHz}$ ), so at a $4\,\text{V}$ peak swing the slew rate limits the design first.

Result: Small-signal bandwidth $50\,\text{kHz}$ ; full-power bandwidth $\approx 20\,\text{kHz}$ . The slew rate is the binding limit for a $4\,\text{V}$ peak output.

Typical mistakes

Confusing slew-rate (large-signal) distortion with gain-bandwidth (small-signal) roll-off.
Forgetting the factor $2\pi$ when relating slew rate to the maximum slope of a sinusoid.

Active revision

An op-amp has a gain-bandwidth product of $1.5\,\text{MHz}$ and a slew rate of $0.6\,\text{V}/\mu\text{s}$ . It is used at a gain of 30 to produce a $5\,\text{V}$ peak sinusoid. Find the small-signal bandwidth and the full-power bandwidth, and state which limit applies first.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

References & sources

Sources

WJEC / Eduqas

WJEC/Eduqas GCE Electronics specification

Operational amplifiers

Key points

Comparator output states

Key points

Designing an inverting amplifier

Key points

Non-inverting gain and the buffer

Key points

Output of a weighted summing amplifier

Key points

Testing an amplifier against both limits

Sources