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How do op-amp circuits amplify and combine analogue signals, and how do we predict their output?

The operational amplifier as a high-gain difference amplifier, and the inverting, non-inverting, summing and difference configurations with their closed-loop gain relationships and saturation.

An SQA Advanced Higher Engineering Science answer on operational amplifier circuits, covering the ideal op-amp, open-loop gain and saturation, negative feedback, and the inverting, non-inverting, summing and difference amplifier configurations with their closed-loop gain relationships.

Generated by Claude Opus 4.816 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. The ideal op-amp and open-loop gain
  3. Negative feedback and the virtual earth
  4. The four configurations
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to treat the operational amplifier as a near-ideal, very high-gain difference amplifier, to explain why negative feedback tames that gain into a stable, predictable value, and to calculate the output of the four standard configurations: inverting, non-inverting, summing and difference amplifiers. You must also recognise saturation, where the output can rise no further than the supply rails.

The ideal op-amp and open-loop gain

The ideal op-amp assumed at Advanced Higher has three properties that make the analysis simple: infinite open-loop gain, infinite input resistance (so no current flows into either input), and zero output resistance. Because the open-loop gain is so large, even a few microvolts of difference between the inputs would demand an impossibly large output, so in practice the output saturates at a value a little below the positive or negative supply rail.

Negative feedback and the virtual earth

Used open-loop, the op-amp is only a comparator. Feeding a fraction of the output back to the inverting input creates negative feedback, which stabilises the gain at a value set by external resistors and makes it almost independent of the device's own (variable) open-loop gain. The two golden rules for an ideal op-amp with negative feedback follow directly:

  • No current enters the inputs (infinite input resistance).
  • The two inputs sit at almost the same voltage (V+VV^+ \approx V^-), because any difference, multiplied by the huge gain, would saturate the output.

When the non-inverting input is earthed, the second rule forces the inverting input to almost 0 V0\ \text{V}. It is not wired to earth, yet it behaves as if it were: this is the virtual earth, and it is the key to analysing the inverting and summing circuits.

The four configurations

The inverting amplifier applies the input through R1R_1 to the virtual-earth node and feeds back through RfR_f; the gain is the resistor ratio, with a minus sign because the output is inverted. The non-inverting amplifier applies the input directly to V+V^+, giving a gain of 1+Rf/R11 + R_f/R_1 that is always greater than one and does not invert. The summing amplifier is an inverting amplifier with several input resistors meeting at the virtual earth, so the output is the weighted, inverted sum of the inputs, which is how analogue signals are mixed. The difference amplifier amplifies the difference between two inputs while rejecting any signal common to both, which is why it is used to read small sensor signals against noise.

Examples in context

A microphone preamplifier uses a non-inverting stage to lift a few millivolts to a usable level without inverting the waveform. An audio mixing desk sums several channels with a summing amplifier, each input resistor setting that channel's level. An instrumentation front end reading a strain-gauge bridge uses a difference amplifier so that the small bridge imbalance is amplified while mains hum picked up equally on both leads is rejected. In every case the resistor values, not the chip, fix the gain, which is why the design is reliable across devices.

Try this

Q1. State the relationship for the closed-loop gain of an inverting amplifier. [1 mark]

  • Cue. Av=RfR1A_v = -\dfrac{R_f}{R_1}.

Q2. A non-inverting amplifier has R1=2 kΩR_1 = 2\ \text{k}\Omega and Rf=18 kΩR_f = 18\ \text{k}\Omega. Find its gain. [2 marks]

  • Cue. Av=1+182=10A_v = 1 + \dfrac{18}{2} = 10.

Q3. Explain why the inverting input of an op-amp with negative feedback is called a virtual earth. [2 marks]

  • Cue. The huge open-loop gain forces the two inputs to almost the same potential; with V+V^+ earthed, VV^- sits at almost 0 V0\ \text{V} without a physical connection to earth.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style4 marksAn inverting amplifier has an input resistor of 10 kΩ10\ \text{k}\Omega and a feedback resistor of 150 kΩ150\ \text{k}\Omega. Calculate the closed-loop voltage gain and the output voltage when the input is +0.20 V+0.20\ \text{V}.
Show worked answer →

For an inverting amplifier the closed-loop gain is set by the resistor ratio.

Relationship: Av=RfR1A_v = -\dfrac{R_f}{R_1}.

Substitution: Av=15010=15A_v = -\dfrac{150}{10} = -15.

Output voltage: Vout=AvVin=15×0.20=3.0 VV_{out} = A_v V_{in} = -15 \times 0.20 = -3.0\ \text{V}.

Markers reward the gain relationship with the minus sign, the value of 15-15, and the output of 3.0 V-3.0\ \text{V}, the minus sign showing the inversion. A common slip is dropping the sign and reporting +3.0 V+3.0\ \text{V}.

SQA AH style5 marksA summing amplifier has two inputs of 0.5 V0.5\ \text{V} and 1.5 V1.5\ \text{V} applied through input resistors of 20 kΩ20\ \text{k}\Omega each, with a feedback resistor of 40 kΩ40\ \text{k}\Omega. Calculate the output voltage and explain why the inverting input is called a virtual earth.
Show worked answer →

The summing amplifier output is the weighted sum of the inputs, inverted.

Relationship: Vout=(RfR1V1+RfR2V2)V_{out} = -\left(\dfrac{R_f}{R_1}V_1 + \dfrac{R_f}{R_2}V_2\right).

Substitution: Vout=(4020(0.5)+4020(1.5))=(1.0+3.0)=4.0 VV_{out} = -\left(\dfrac{40}{20}(0.5) + \dfrac{40}{20}(1.5)\right) = -(1.0 + 3.0) = -4.0\ \text{V}.

Virtual earth: because the open-loop gain is enormous, the difference between the two inputs is almost zero, and with the non-inverting input earthed the inverting input sits at almost 0 V0\ \text{V} without being physically connected to earth.

Markers reward the summing relationship, the value of 4.0 V-4.0\ \text{V}, and a correct statement that the huge gain forces the inputs to almost the same potential, fixing the inverting input near zero.

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