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The operational amplifier — universally abbreviated op-amp — is the single most flexible analogue building block in electronics. A monolithic integrated circuit containing typically 20–30 transistors and a few precision resistors, it presents to the user a deceptively simple specification: two high-impedance inputs (inverting and non-inverting), one low-impedance output, and an enormous open-loop voltage gain. Wrap a few external resistors around it in the right topology and you can make an inverting amplifier with any gain you choose, a non-inverting amplifier, a unity-gain buffer, a summing junction that adds several signals weighted by resistors, a differential amplifier that rejects common-mode noise, an integrator, a differentiator, a precision rectifier, a logarithmic amplifier, or a comparator. The classical 741 op-amp, introduced in 1968 and still in production today, has an open-loop gain of around 10⁵ and remains the textbook example. Modern op-amps achieve open-loop gains of 10⁶ to 10⁷ with picoamp input bias currents and sub-millivolt input offsets. This lesson develops the ideal op-amp model, the principle of negative feedback, and the three canonical configurations — inverting, non-inverting and summing — together with the closely related voltage follower.
Spec mapping (AQA 7408 §3.13.3 — Analogue signal processing, option E): This lesson covers the ideal operational amplifier model (infinite open-loop gain, infinite input impedance, zero output impedance); the inverting amplifier configuration (G_v = −R_f/R_in); the non-inverting amplifier (G_v = 1 + R_f/R_in); the summing amplifier; the voltage follower (buffer) with unity gain; the principle of negative feedback and the resulting "virtual short" / "virtual ground" at the input; the trade-off between closed-loop gain and bandwidth (gain–bandwidth product). (Refer to the official AQA specification document for exact wording.)
Synoptic links: Op-amps build directly on the junction transistor picture from Lesson 0 — every op-amp is internally a multi-stage transistor amplifier with negative feedback applied externally to set the closed-loop behaviour. They feed forward into active filters (Lesson 3), where the combination of an op-amp and a few RC components produces a sharper roll-off than a passive RC filter alone can achieve. The summing amplifier is the analogue heart of a DAC's resistor-ladder network — connecting back to digital signals (Lesson 1) and forward to transducers (Lesson 6), where op-amps condition the millivolt-level signals from thermocouples, strain gauges and photodiodes up to volt-scale signals suitable for ADC input.
An op-amp has five external pins: two power supplies (typically ±V_S, e.g. ±15 V or single 5 V), two signal inputs labelled V+ (non-inverting) and V− (inverting), and one signal output V_out. The op-amp is defined by:
V_out = A_OL (V+ − V−)
where A_OL is the open-loop voltage gain. In the idealisation we use for circuit analysis:
Ideal op-amp assumptions:
- Infinite open-loop gain: A_OL → ∞
- Infinite input impedance: no current flows into V+ or V−
- Zero output impedance: V_out can drive any load without sagging
- Infinite bandwidth and zero offset: the op-amp responds instantly with no DC error
These idealisations are excellent approximations for low-frequency, moderate-gain circuits using a modern op-amp. The 741 has A_OL ≈ 2 × 10⁵, input impedance ≈ 2 MΩ, output impedance ≈ 75 Ω — close enough to ideal for textbook purposes.
For circuit analysis of any op-amp configured with negative feedback (output fed back to the inverting input), two simple rules follow directly from the ideal assumptions:
Rule 1 (virtual short): V+ = V−. Because A_OL is enormous, any tiny difference between the input terminals would saturate the output. For the output to sit at a finite value, V+ and V− must be essentially equal.
Rule 2 (no input current): I+ = I− = 0. Because the input impedance is infinite, no current flows into either input pin.
Apply these two rules to any standard op-amp configuration and the closed-loop gain falls out by a few lines of algebra. This is the only analysis technique you need at A-Level — every standard topology yields its gain from Rules 1 and 2 plus Kirchhoff's current law at the inverting node.
A bare op-amp running open-loop has an enormous gain, but that gain is poorly controlled (it varies by a factor of 2–3 between samples and with temperature) and the output saturates against the supply rails for any input difference larger than a few microvolts. To build a useful amplifier we deliberately sacrifice most of the open-loop gain in exchange for predictability — by feeding part of the output back to the inverting input with a known resistor ratio. The remaining "closed-loop" gain depends only on the ratio of two external resistors, which can be made precisely as 1% or 0.1% components.
If the open-loop gain is A and a fraction β of the output is fed back to the inverting input (subtracting from V+), the closed-loop gain G is:
G = A / (1 + Aβ)
For Aβ ≫ 1, G ≈ 1/β — the closed-loop gain depends only on the feedback ratio, not on the open-loop gain. With A = 10⁵ and a feedback fraction β = 0.01 (giving a target gain of 100), the closed-loop gain comes out at 99.9 instead of 100 — accurate to one part in a thousand even though A varied by 30%.
The benefits of negative feedback:
The cost is that some open-loop gain must be deliberately discarded.
The single most common op-amp topology:
R_f
┌────/\/\────┐
│ │
V_in ──/\/\──┬──[−]│
R_in │ │op-amp───→ V_out
│ [+]│
│ │
GND GND
The signal enters through R_in, the non-inverting input is grounded, and a feedback resistor R_f bridges output to inverting input.
Apply Rule 1: V− = V+ = 0 (since V+ is grounded). The inverting input is at virtual ground — held at 0 V by the feedback even though it is not directly connected to ground.
Apply Rule 2 (no current into the op-amp) plus KCL at the inverting node:
I_in = I_f (V_in − 0)/R_in = (0 − V_out)/R_f
Rearranging:
G_v = V_out / V_in = −R_f / R_in
The negative sign is the "inverting" part — the output is 180° out of phase with the input. A 1 V positive input produces an R_f/R_in volts negative output.
Question: Design an inverting amplifier with a gain of −20 using a 741 op-amp powered from ±15 V. Choose suitable resistor values.
Solution: G_v = −R_f/R_in = −20. Choose R_f = 100 kΩ, R_in = 5 kΩ. (Equally valid: R_f = 20 kΩ, R_in = 1 kΩ.) The choice within the constraint R_f/R_in = 20 is governed by two practical considerations:
A typical sensible choice is R_in in the range 1 kΩ to 10 kΩ. The 5 kΩ / 100 kΩ pair is fine.
The output range: the 741 swings to within about 2 V of either rail, so V_out can sit between −13 V and +13 V approximately. Therefore V_in must remain within about ±13/20 = ±0.65 V to avoid output saturation.
R_f
┌────/\/\────┐
│ │
├────[−] │
│ │op-amp───→ V_out
R₁ │
│ [+]
GND │
│
V_in
The signal enters at the non-inverting input. The output is divided by R_1 and R_f and fed back to the inverting input.
Apply Rule 1: V− = V+ = V_in. The inverting input therefore sits at V_in (not at ground). Apply KCL at the inverting node, with no current into the op-amp:
(V_in − 0)/R_1 = (V_out − V_in)/R_f
Rearranging:
V_out (1/R_f) = V_in (1/R_1 + 1/R_f) V_out / V_in = R_f/R_1 + 1
G_v = 1 + R_f / R_1
The gain is always positive (no inversion) and always ≥ 1 (you cannot attenuate with this topology). It has very high input impedance — essentially the bare op-amp input impedance, often hundreds of megohms.
Question: A microphone delivers a 5 mV peak signal with source impedance 600 Ω. Design a non-inverting amplifier with a gain of +50.
Solution: G_v = 1 + R_f/R_1 = 50, so R_f/R_1 = 49. Choose R_1 = 1 kΩ, R_f = 49 kΩ — or, using the nearest standard E12 value, R_f = 47 kΩ giving G_v = 48.
The non-inverting topology is preferred here because its very high input impedance (≫ 600 Ω microphone source) avoids loading the microphone. The output is 5 × 48 = 240 mV peak — comfortable for a 741 powered from ±15 V.
A special case of the non-inverting amplifier in which R_f = 0 and R_1 = ∞ (i.e. R_f is a short and R_1 is removed):
V_in ──[+]
│op-amp──┬─→ V_out
[−]──────┘
The output is wired directly to the inverting input. Apply Rule 1: V− = V+ = V_in; therefore V_out = V_in.
G_v = 1 (unity gain, in phase)
The follower has unity voltage gain but enormous current gain — it presents a very high input impedance and a very low output impedance, isolating the source from the load. Use it whenever a high-impedance source (e.g. a pH probe, a piezoelectric transducer, a long resistor divider) needs to drive a low-impedance load (a wire-wound potentiometer, an analogue panel meter, the input of an ADC) without being loaded down.
Question: A 100 kΩ potentiometer is used as a voltage divider to set a reference voltage. The reference must drive an ADC with 10 kΩ input impedance. Without a buffer, by how much is the reference voltage altered? Explain how a voltage follower removes this error.
Solution: Without buffer: the ADC's 10 kΩ load in parallel with one half of the 50 kΩ potentiometer arm gives R_eff = (50 × 10)/(50 + 10) = 8.33 kΩ. The divider ratio is no longer the intended ½ but 8.33 / (50 + 8.33) = 0.143 — the reference is loaded down by a factor of 3.5. With a voltage follower between the potentiometer wiper and the ADC, the follower presents 10⁸ Ω to the potentiometer (negligible loading) and ≈ 0 Ω to the ADC. The ADC sees the un-loaded potentiometer voltage exactly.
Extend the inverting topology to multiple inputs, each through its own resistor:
V_1 ──/\/\──┐
R_1 │ R_f
V_2 ──/\/\──┼──/\/\──┐
R_2 │ │
V_3 ──/\/\──┼───[−] │
R_3 │ │op-amp───→ V_out
[+]
│
GND
The inverting input is at virtual ground (V− = V+ = 0). KCL at this node, with no current into the op-amp:
V_1/R_1 + V_2/R_2 + V_3/R_3 = −V_out / R_f
Therefore:
V_out = −R_f (V_1/R_1 + V_2/R_2 + V_3/R_3)
If all input resistors are equal to R_in, this simplifies to V_out = −(R_f/R_in)(V_1 + V_2 + V_3) — a weighted sum that adds the inputs. The summing amplifier is the analogue version of an addition operation. It is used in audio mixing desks (each fader sets a gain, the summer combines the channels) and in the resistor-ladder DAC (each bit of the digital word contributes a binary-weighted voltage to the analogue output).
Question: A four-channel audio mixer uses a summing amplifier with R_f = 10 kΩ and each input resistor set to 10 kΩ. Two microphones supply 20 mV and 30 mV; an instrument supplies 50 mV; a backing track supplies 100 mV. Calculate V_out.
Solution: With R_f = R_in, V_out = −(V_1 + V_2 + V_3 + V_4) = −(0.020 + 0.030 + 0.050 + 0.100) = −0.200 V = −200 mV. The negative sign indicates inversion; in audio it is inaudible (the human ear cannot detect absolute phase) and is removed by a second inverting stage with unity gain if positive output is needed.
The open-loop gain of a real op-amp is not constant with frequency. The 741 has A_OL ≈ 2 × 10⁵ at DC, falls off at −20 dB/decade above an internal "compensation pole" at about 10 Hz, and reaches unity gain at about 1 MHz. This product:
Gain–Bandwidth Product (GBP): A × f_3dB = constant
is fixed for a given op-amp. The 741 has GBP ≈ 1 MHz; a high-speed op-amp like the AD8056 has GBP ≈ 300 MHz.
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