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Summing and difference amplifiers questions
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.
17 original questions · 52 marks · the summing and difference amplifiers notes · Electronics
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Explain why a summing amplifier does not simply add its input voltages.
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Each input drives its own current V/R into the virtual earth, so each input is amplified by its own gain Rf/Rin before it joins the total (1). The output is therefore minus the weighted sum, with the weights set by the input resistors (1). Adding the voltages first and applying one gain to the lot is the standard error, and it only ever gives the right answer when every input resistor happens to be equal.State the two facts about the summing point of an inverting amplifier that carry the derivation marks.
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The inverting input is held at 0 V by the feedback, a virtual earth, so the current through each input resistor is simply that input's voltage divided by its own resistance (1). The op-amp's input resistance is taken as infinite, so no current enters the chip and the whole total leaves through Rf (1). Omitting the second statement is the standard error, and without it the derivation does not close.Write down the equation for the output of a difference amplifier, and state the order in which the two operations must be carried out.
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Vout = (V+ − V−) × Rf/R1 (1). Subtract first, then multiply by the gain (1). Amplifying each input separately and subtracting afterwards is the standard error: with inputs of a few volts and a gain of a thousand it demands outputs of kilovolts, and the real chip would simply saturate.Explain how a difference amplifier removes interference picked up on the leads of a sensor.
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The two leads run side by side, so mains wiring induces the same interference on both; it is common-mode (1). Identical on both inputs, it appears in both terms of V+ − V− and cancels before any gain is applied, while the wanted signal, a genuine difference between the leads, survives at full gain (1). Saying the amplifier 'filters out' the hum is the standard error: nothing is filtered, it is subtracted.State what happens to the output of a summing amplifier when the equation asks for a voltage larger than the supply rails, and state the check a designer should always make.
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The output cannot pass the rails, so it saturates a volt or so inside the rail and stays there, clipping the waveform flat (1). Every design should end by computing the worst-case output and holding it against the rails (1). Writing down the ideal figure, such as −20 V from a ±15 V supply, is the standard error, and it is the one examiners mark hardest.State what limits the accuracy of a weighted-resistor digital-to-analogue converter, and state how accurate each resistor weight must be.
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Every gain is a ratio of real resistors, and real resistors carry a tolerance, commonly 5% or 1%, so two 5% resistors can leave a weight almost 10% adrift (1). Each weight must be accurate to better than half a step of the staircase, or the output stops rising in equal steps with the code (1). Assuming the ideal staircase holds however coarse the resistors are is the standard error, and it is what drives real converters away from this design.A summing amplifier has Rf = 100 kΩ. Its three inputs are 0.20 V through 100 kΩ, 0.30 V through 50 kΩ and 0.10 V through 20 kΩ. Calculate the current from each input and the output voltage.
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Each input sees a virtual earth, so the currents are 0.20/100 kΩ = 2 μA, 0.30/50 kΩ = 6 μA and 0.10/20 kΩ = 5 μA (1). They add at the node to 13 μA, all of which leaves through Rf (1). Vout = −13 × 10−6 × 100 × 103 = −1.30 V (1). Dropping the minus sign is the standard error here, and it costs a mark every time it appears.A summing amplifier has Rf = 47 kΩ, with V1 = 0.50 V through R1 = 47 kΩ and V2 = −0.25 V through R2 = 4.7 kΩ. Calculate the output voltage.
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Gains first: 47/47 = 1 and 47/4.7 = 10 (1). Vout = −(1 × 0.50 + 10 × (−0.25)) = −(0.50 − 2.50) (1) = +2.0 V (1). Watch the double negative: the negative input, amplified ten times and then inverted along with everything else, leaves the output positive. Working the voltages before the gains, or losing one of the two minus signs, are the standard errors.A 4-bit weighted-resistor DAC uses Rf = 10 kΩ with input resistors of 10, 20, 40 and 80 kΩ from the most significant bit downward, and logic 1 = 5.0 V. Calculate the step size and the magnitude of the output for the input 0110.
A difference amplifier with Rf/R1 = 50 has V+ = 3.4120 V and V− = 3.3800 V. Calculate the output voltage.
Two leads from a sensor arrive at a difference amplifier of gain 200. The wanted signal is a difference of 12 mV between them, and each lead also carries an identical 0.25 V of mains hum. Calculate the output, and state the contribution the hum makes to it.
A summing amplifier gives gains of 3 and 5 to its two inputs and saturates at ±13 V. Determine the output when V1 = 1.0 V and V2 = 2.4 V, and the largest V2 the amplifier can handle linearly with V1 held at 1.0 V.
A 4-bit weighted-resistor DAC built on a summing amplifier has a step size of 0.625 V. Calculate the magnitude of its full-scale output and state the number of distinct output levels.
An 8-bit weighted-resistor DAC uses Rf = 10 kΩ, logic 1 = 5.0 V, and 10 kΩ as the input resistor of the most significant bit, each successive bit's resistor being double the one before. Calculate the resistance of the least significant bit's resistor, the step size and the magnitude of the full-scale output, then determine the accuracy each resistor weight needs if the staircase is to stay correct to within half a step, and comment on the practicality of the design.
An audio mixer is built round a summing amplifier with Rf = 100 kΩ and must give voltage gains of 1, 2 and 5 to its three channels. Calculate the three input resistors, and calculate the output at an instant when the channel inputs are 0.10 V, 0.15 V and 0.02 V.
A strain-gauge bridge sits balanced with both midpoints at 2.5000 V. Under load one midpoint rises to 2.5085 V while the other stays put. A difference amplifier built with R1 = 1.0 kΩ is to give an output of 4.0 V under that load. Calculate the difference voltage, the gain needed and Rf, and state what would happen if the raised midpoint were fed to an ordinary inverting amplifier of the same gain instead.
A sensor delivers 1.00 V at the bottom of its range and 1.50 V at the top. A summing amplifier with Rf = 100 kΩ is to convert this to 0 V at the bottom of the range and −5.0 V at the top, using the sensor through R1 and a steady −1.00 V reference through R2. Calculate both input resistors, and verify the output at each end of the range.
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