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Operational amplifiers questions
An amplifier with a gain of a hundred thousand is useless on its own, since a fraction of a millivolt drives its output hard 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.
17 original questions · 51 marks · the operational amplifiers notes · Electronics
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State the four properties of an ideal operational amplifier.
Mark scheme
All four of the following, one mark per two correct (2): infinite open-loop gain; infinite input resistance, so neither input draws current; zero output resistance, so the output does not sag under load; and infinite bandwidth. Real chips manage about 105, a few megohms and tens of ohms. Writing 'very high gain' alone is the standard error: the examiner is marking a list of four, so give all four.Write down the open-loop relation for an operational amplifier, and explain what is meant by saturation.
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Vout = AOL(V+ − V−) (1). Saturation is what happens when that equation asks for an output beyond the supply rails: the output cannot pass them, so it sticks a volt or so inside the rail and stays there, clipping the waveform flat (1). Forgetting that the amplifier works on the difference between the inputs, and applying the gain to one input alone, is the standard error.State what negative feedback gives away in an operational amplifier circuit, and give two things it buys.
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It gives away most of the available gain (1). In return, any two of: the closed-loop gain is set by the feedback resistors rather than by the chip; the bandwidth is far wider; distortion falls; the circuit's behaviour survives a change of chip (1). Saying feedback 'increases the gain' is the standard error; it reduces it on purpose.Explain what is meant by the virtual earth in an inverting amplifier.
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Negative feedback holds the two inputs within microvolts of each other, and the non-inverting input is connected to 0 V, so the inverting input is held at 0 V as well (1) without being connected to earth at all, which is why it is called virtual (1). Saying the inverting input 'is earthed' is the standard error: it carries the input and feedback currents, which a real earth connection would swallow.State two differences between an inverting and a non-inverting amplifier, other than the equations for their gains.
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The inverting amplifier's output is in antiphase with its input, while the non-inverting amplifier's output is in phase (1). The non-inverting gain can never fall below 1, and the signal arrives straight at the op-amp's own input terminal, so the circuit draws almost nothing from the source and suits a high-resistance sensor (1). Forgetting the minus sign on the inverting gain is the standard error this question is built to catch.State the relationship between the closed-loop gain of an amplifier and its bandwidth, and state what happens to the bandwidth when the gain is increased by a factor of ten.
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Gain × bandwidth = constant, the gain-bandwidth product of the chip (1). Ten times the gain therefore means a tenth of the bandwidth (1). Expecting the two to rise together is the standard error; they always move in opposite directions.An operational amplifier has an open-loop gain of 1.0 × 105 and its output saturates at ±13 V. Calculate the difference between its input voltages that is just enough to drive the output to the positive rail, and state what happens for any larger difference.
Mark scheme
V+ − V− = Vout/AOL = 13/(1.0 × 105) (1) = 1.3 × 10−4 V, that is 130 μV (1). Any larger difference asks for an output beyond the rail, which the amplifier cannot deliver, so it saturates at about +13 V and stays there (1). Multiplying by the open-loop gain instead of dividing is the standard error, and it gives a wholly impossible 1.3 × 106 V.An inverting amplifier uses Rin = 4.7 kΩ and Rf = 82 kΩ. Calculate its closed-loop gain and its output for an input of 0.25 V.
Mark scheme
Gain = −Rf/Rin = −82/4.7 (1) = −17.4 (1). Vout = −17.4 × 0.25 = −4.36 V (1), comfortably inside the rails. The minus sign is worth a mark on its own: an inverting amplifier answer written as +4.4 V has thrown it away, and it is the single most common lost mark in this topic.A non-inverting amplifier uses R1 = 2.2 kΩ with Rf = 33 kΩ. Calculate its closed-loop gain and its output for an input of 0.60 V.
An inverting amplifier has Rin = 22 kΩ and Rf = 56 kΩ, and its input is 0.44 V. Calculate the current into the inverting input node and hence the output voltage, working from the currents rather than from the gain equation.
An operational amplifier has a gain-bandwidth product of 1.0 MHz. Calculate the bandwidth available at a closed-loop gain of 200, and the bandwidth at a closed-loop gain of 25.
An inverting amplifier with Rin = 6.0 kΩ and Rf = 120 kΩ runs from supplies that let its output saturate at ±13 V. Determine the output for an input of 0.90 V, and the largest input the amplifier can handle linearly.
An operational amplifier with an open-loop gain of 1.0 × 105 and rails at ±13 V is used with no feedback as a comparator. Its non-inverting input is at 2.62 V and its inverting input is held at a reference of 2.50 V. State the output voltage, and justify it.
A sensor delivers a 15 mV peak signal that must be amplified to 3.0 V peak and must have a bandwidth of at least 25 kHz. The available op-amps have a gain-bandwidth product of 1.0 MHz and saturate at ±13 V. Calculate the overall gain needed, show that a single stage cannot provide it at that bandwidth, and design a two-stage inverting chain that can, checking the bandwidth of each stage and the output against the rails.
An inverting amplifier is to have a gain of −25 and an input resistance of 4.7 kΩ. Calculate the feedback resistance required, state the gain obtained if the nearest preferred value of 120 kΩ is fitted instead, and calculate the percentage error in the gain.
A comparator runs from ±13 V rails and senses light from a 12 V sensing supply. Its inverting input is fed from the junction of two equal 10 kΩ resistors connected across that 12 V supply. Its non-inverting input is fed from the junction of a light-dependent resistor connected to +12 V and a 4.7 kΩ resistor connected to 0 V. Calculate the reference voltage, calculate the LDR resistance at which the output changes state, and state which way the output moves as the light level falls.
An operational amplifier has an open-loop gain of 1.0 × 105 and a gain-bandwidth product of 1.0 MHz. Calculate the frequency up to which the open-loop gain holds up, and explain why a designer who wants a flat response to 100 kHz cannot have a closed-loop gain of 1000.
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