Physics › Electronics › Summing and difference amplifiers

Summing and difference amplifiers

Hang extra input resistors on an inverting amplifier and it becomes a mixer, adding voltages with whatever weights you choose. Rearrange the same op-amp and it subtracts instead, so a few millivolts from a sensor can be read cleanly through a room full of interference.

Year 13AQA 3.13.2

Builds on Operational amplifiers.

IN THIS TOPIC

Use the summing amplifier equation, choosing a separate gain for each input through its own resistor.
Explain how weighted resistors turn a binary number into an analogue voltage.
Use the difference amplifier equation, and explain how it removes interference shared by both inputs.
Spot when saturation or resistor tolerance makes the ideal equations fail.

WHAT YOU PROBABLY THINK

A summing amplifier simply adds up its input voltages.

Every input meets a virtual earth

The inverting amplifier of the last lesson had one input resistor feeding the op-amp's inverting input, with negative feedback holding that input within microvolts of 0 V. The point is called a virtual earth, and it is the hinge of everything here. Because the node sits at 0 V, the current arriving through an input resistor is just the input voltage over that resistance, and because the op-amp inputs draw no appreciable current, every bit of that current must leave through the feedback resistor.

Now hang several input resistors on the same node, one per signal. Each input drives its own current $V/R$ into the virtual earth without disturbing the others, the currents add at the node, and the whole total is forced through $R_{f}$ . The output swings to whatever voltage makes that happen, so it carries an inverted, weighted copy of every input at once:

V_{out} = -R_{f}(\frac{V_{1}}{R_{1}} + \frac{V_{2}}{R_{2}} + \frac{V_{3}}{R_{3}} + ...)

ON YOUR DATA SHEET

FIG. 1Three inputs, three resistors, one summing point. Each input sends the current V over R into the virtual earth, the currents add, and all of it drains through the feedback resistor to the output.

Read the equation as a set of independent gains. Input 1 is amplified by $R_{f}/R_{1}$ , input 2 by $R_{f}/R_{2}$ , and so on, each channel deaf to its neighbours. That is the opening lie dealt with. The circuit does not add voltages; it adds weighted voltages, inverts the total, and the weights are yours to set by choosing resistors.

WORKED EXAMPLE

Two microphones, one wire

A mixer uses $R_{f}$ = 100 kΩ. A singer's microphone feeds $R_{1}$ = 100 kΩ and a quieter guitar feeds $R_{2}$ = 50 kΩ. At one instant $V_{1}$ = 0.20 V and $V_{2}$ = −0.15 V. Find the output.

The gains first: $R_{f}/R_{1}$ = 1 for the voice, $R_{f}/R_{2}$ = 2 for the guitar, boosting the weaker source.

$V_{out}$ = −(1 × 0.20 + 2 × (−0.15)) = −(0.20 − 0.30) = +0.10 V.

Note the double negative. The guitar's negative swing, amplified and inverted along with everything else, left the output positive. Sign discipline is most of the marks in these questions.

Weighted resistors: from binary to analogue

Choose the weights in powers of two and the mixer becomes a digital-to-analogue converter, a DAC. A 4-bit binary number arrives as four logic levels, each 0 V or 5.0 V. Give the most significant bit the strongest channel and halve the gain at each step down, say $R_{f}$ = 10 kΩ with input resistors of 10, 20, 40 and 80 kΩ. The gains run 1, $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{8}$ , matching the place values 8, 4, 2, 1.

One least significant bit is then worth 5.0 × 10/80 = 0.625 V of output. Feed in the binary count 0000, 0001, 0010 and upward, and the output descends an even staircase, one 0.625 V step per count, sixteen levels in all. Every digital audio player ends in some refinement of this idea.

FIG. 2A weighted-resistor DAC walking through the sixteen 4-bit codes. Each count adds exactly one least-significant-bit step, so the analogue output is a staircase in strict proportion to the binary number.

YOUR TURN

The staircase, step by step

Using the DAC above, with logic 1 = 5.0 V, find the magnitude of the output for the input 1011, and check it against the step size.

Show the working

Only the bits at logic 1 contribute. MSB: 5.0 × 1 = 5.0 V. Third bit: 5.0 × $\frac{1}{4}$ = 1.25 V. LSB: 5.0 × $\frac{1}{8}$ = 0.625 V.

Total magnitude 5.0 + 1.25 + 0.625 = 6.875 V, at the output as −6.875 V.

The check writes itself. 1011 is eleven, and 11 × 0.625 V = 6.875 V. If your weighted sum disagrees with count times step, a resistor weight went astray.

The difference amplifier

Swap the geometry and the op-amp subtracts. A difference amplifier feeds one signal towards each input through matched resistor pairs, and its output amplifies only the gap between them:

V_{out} = (V_{+} - V_{-})\frac{R_{f}}{R_{1}}

ON YOUR DATA SHEET

Where subtraction earns its keep is small sensors. A strain gauge or thermistor usually sits in a bridge, a pair of potential dividers across one supply. In balance both divider midpoints sit at the same voltage. When the sensor shifts, the two midpoints separate by millivolts, and that tiny difference rides on top of a large steady level that carries no information. The difference amplifier ignores the shared level and amplifies the gap alone.

The same subtraction is the standard cure for interference. Mains wiring induces the same 50 Hz hum on both leads of a sensor cable, since the leads run side by side. Identical on both inputs, the hum is common-mode, and $V_{+} - V_{-}$ wipes it out before the gain is applied. The wanted signal, applied as a genuine difference between the leads, survives at full gain.

WORKED EXAMPLE

Reading a strain gauge through the hum

A bridge produces $V_{+}$ = 2.5124 V and $V_{-}$ = 2.5100 V, both carrying an identical 0.30 V mains hum. The difference amplifier has $R_{f}/R_{1}$ = 1000. Find the output.

$V_{out}$ = (2.5124 − 2.5100) × 1000 = 2.4 × 10⁻³ × 1000 = 2.4 V.

The 2.51 V standing level and the 0.30 V of hum appear in both terms of the subtraction, so neither reaches the output. Amplifying $V_{+}$ alone by 1000 would have demanded an absurd 2500 V and buried the signal in 300 V of amplified hum.

Where the ideal story stops

The equations promise any output you like; the hardware does not. An op-amp runs from supply rails, typically ±15 V, and its output cannot pass them. Ask the summing equation for −20 V and the real output slams into saturation a volt or so inside the negative rail and sits there, clipping the waveform flat. Every design should end with the same check, worst-case output against the rails.

The other limit hides in the resistors. Every gain in this lesson is a ratio of real resistors, and real resistors come with a tolerance, commonly 5% or 1%. Two 5% resistors can leave a gain almost 10% adrift. A DAC feels this hardest, because its staircase is only trustworthy if each weight is accurate to under half a step. For an 8-bit converter that means resistor ratios good to about 0.2%, and needing eight precision values spanning a 128 to 1 range is exactly what drives real converters away from the weighted-resistor design.

TRY IT UNSEEN

Pushed past the rails

A summing amplifier has gains of 2 and 4 on its two inputs and runs from ±15 V rails, saturating at ±13 V. Inputs are $V_{1}$ = 2.0 V and $V_{2}$ = 3.0 V. What does the output actually do? Keeping $V_{1}$ = 2.0 V, find the largest $V_{2}$ the amplifier can handle linearly.

Show the working

The ideal output is −(2 × 2.0 + 4 × 3.0) = −16 V. That lies beyond the −13 V limit, so the real output saturates at −13 V and the waveform is clipped.

For linear working the magnitude must stay within 13 V, so 4 + 4 $V_{2}$ ≤ 13, giving $V_{2}$ ≤ 2.25 V.

The exam habit to build: compute the ideal answer, then hold it against the rails before writing it down.

THE EXAM BIT

Work out each input's own gain $R_{f}/R_{in}$ before touching the voltages. The classic error is adding the voltages first and applying one gain to the lot.
The output is inverted. State the sign, then the size. A missing minus sign in a summing amplifier answer costs a mark every time it appears.
Difference amplifier answers subtract first and multiply after. Quote the order: $(V_{+} - V_{-})$ , then times $R_{f}/R_{1}$ .
Common-mode rejection is a one-sentence explanation. Interference induced identically on both inputs appears in both terms of the subtraction, so it cancels, while the wanted signal is a genuine difference and is amplified.
Whenever supplies are given, check the ideal output against the rails. If it exceeds them, say the amplifier saturates and give the output as (near) the rail voltage.
Derivation marks come from the virtual earth. Say the inverting input is held at 0 V, the input currents are $V/R$ , they add at the node, and all of the total flows through $R_{f}$ .

CHECK YOURSELF

A summing amplifier has $R_{f}$ = 120 kΩ, with $V_{1}$ = 1.5 V through $R_{1}$ = 60 kΩ and $V_{2}$ = −0.8 V through $R_{2}$ = 30 kΩ. The supplies are ±15 V and the op-amp saturates at ±13 V. Find the output, and then the output if $V_{2}$ falls to −5.0 V.

Show a hint

Each input has its own gain of Rf over its own resistor. Mind both signs, then mind the rails.

Show the answer

The gains are 120/60 = 2 and 120/30 = 4.

$V_{out}$ = −(2 × 1.5 + 4 × (−0.8)) = −(3.0 − 3.2) = +0.2 V. The larger-gain channel wins the sign.

With $V_{2}$ = −5.0 V the ideal output is −(3.0 − 20) = +17 V, beyond the +13 V limit, so the amplifier saturates at about +13 V and the output no longer follows the equation.

Each input rides its own gain, Rf over its own resistor, into a virtual earth; the output is minus the weighted sum.

The difference amplifier gives Rf over R1 times (V+ − V−), so interference shared by both inputs subtracts to nothing.

No output passes the supply rails; past them the amplifier saturates and clips.

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

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

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.

Use the summing amplifier equation, choosing a separate gain for each input through its own resistor.
Explain how weighted resistors turn a binary number into an analogue voltage.
Use the difference amplifier equation, and explain how it removes interference shared by both inputs.
Spot when saturation or resistor tolerance makes the ideal equations fail.

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