Op-Amps: the analogue building block you'll use forever
Gain of 100 000, and the first thing every design does is throw most of it away.
An operational amplifier has such enormous open-loop gain that it's useless on its own — a microvolt of input slams the output against a supply rail. One insight makes it the most useful analogue part ever made. Wrap it in negative feedback. The circuit's behaviour then stops depending on the op-amp's messy internal characteristics, and starts depending only on a ratio of two resistors. Precision from imprecise parts.
The ideal op-amp, and the two golden rules
Ideal characteristics:
Open-loop gain A → ∞
Input impedance Z_in → ∞ (no input current)
Output impedance Z_out → 0
Bandwidth → ∞
CMRR → ∞
Offset voltage = 0
From these follow the two rules that solve every circuit:
RULE 1 — VIRTUAL SHORT: V₊ = V₋
(if A is infinite, any output needs zero input
difference, so the two inputs sit at the same voltage)
RULE 2 — NO INPUT CURRENT: I₊ = I₋ = 0
(infinite input impedance)
Those two rules do all the work. Analysing an op-amp circuit means writing KCL at the inverting node, setting V₋ equal to V₊, and assuming no current enters the input pins. Every configuration below is derived from just that — you don't need to memorise the gain formulas if you can apply the rules.
The standard configurations
INVERTING: A_v = −R_f/R₁
R_in = R₁ (the virtual ground makes it exactly R₁)
NON-INVERTING: A_v = 1 + R_f/R₁
R_in = very high (input goes straight to the + pin)
Note: gain can never be below 1
VOLTAGE FOLLOWER: A_v = 1 (R_f = 0, R₁ = ∞)
Pure buffer — huge R_in, tiny R_out
SUMMING (inverting):
V_out = −R_f(V₁/R₁ + V₂/R₂ + V₃/R₃)
DIFFERENCE (with matched resistors):
V_out = (R_f/R₁)(V₂ − V₁)
INTEGRATOR (C in feedback):
V_out = −(1/R₁C)∫V_in dt
DIFFERENTIATOR (C at input):
V_out = −R_f C dV_in/dt
Worked numerical 1 — inverting amplifier, derived from the rules
An inverting amplifier has R₁ = 2.2 kΩ and R_f = 47 kΩ, with V_in = 0.15 V. Find the gain, output voltage and input impedance — deriving rather than quoting.
Apply the two rules at the inverting node:
V₊ = 0 (grounded), so by RULE 1, V₋ = 0 ← "virtual ground"
By RULE 2 no current enters the − pin, so all the current
through R₁ must continue through R_f:
I₁ = I_f
(V_in − 0)/R₁ = (0 − V_out)/R_f
V_in/R₁ = −V_out/R_f
A_v = V_out/V_in = −R_f/R₁ ∎ (derived, not memorised)
Numerically:
A_v = −47 000/2200 = −21.36
V_out = −21.36 × 0.15 = −3.20 V
Input impedance:
Since the − pin is a virtual ground, the source sees just
R₁ to ground: R_in = 2.2 kΩ
That relatively low R_in is the inverting configuration's
weakness — it loads the source.
Worked numerical 2 — non-inverting and follower
Design a non-inverting amplifier with a gain of 11. Then find what happens if R_f is shorted.
A_v = 1 + R_f/R₁ = 11
R_f/R₁ = 10
Choose R₁ = 1 kΩ → R_f = 10 kΩ
With V_in = 0.4 V: V_out = 11 × 0.4 = 4.4 V (in phase)
If R_f is shorted (R_f = 0):
A_v = 1 + 0/1000 = 1
→ becomes a VOLTAGE FOLLOWER
Why bother with a gain of 1? Because of impedance:
R_in ≈ 10¹² Ω (draws essentially nothing from the source)
R_out ≈ 0.1 Ω (drives heavy loads without sagging)
A follower is a buffer — it lets a high-impedance sensor
drive a low-impedance load without the signal collapsing.
Worked numerical 3 — summing amplifier (a 3-bit DAC)
A summing amplifier has R_f = 10 kΩ with inputs through R₁ = 10 kΩ, R₂ = 20 kΩ, R₃ = 40 kΩ. Find V_out when all three inputs are at 4 V, and explain what the circuit does.
V_out = −R_f(V₁/R₁ + V₂/R₂ + V₃/R₃)
= −10k(4/10k + 4/20k + 4/40k)
= −10k × (0.4m + 0.2m + 0.1m)
= −10 000 × 0.7 × 10⁻³
= −7.0 V
Look at the weightings: the three inputs contribute
in the ratio 4 : 2 : 1 — binary weights.
Feeding logic levels (0 V or 4 V) into the three inputs:
000 → 0.0 V 100 → −4.0 V
001 → −0.5 V 101 → −4.5 V
010 → −1.0 V 110 → −5.0 V
011 → −1.5 V 111 → −7.0 V
This is a binary-weighted DIGITAL-TO-ANALOGUE CONVERTER.
Each doubling of resistance halves the contribution, so
the analogue output is proportional to the binary input.
Worked numerical 4 — real-world limits
An op-amp has a gain-bandwidth product of 1 MHz and a slew rate of 0.5 V/µs. Find the bandwidth at a gain of 100, and the largest 10 kHz sine it can output undistorted.
Gain–bandwidth product is constant:
GBW = A_v × BW = 1 MHz
At A_v = 100:
BW = 1 MHz/100 = 10 kHz
→ beyond 10 kHz the gain falls off. Wanting gain AND
bandwidth means cascading stages or a faster op-amp.
Slew-rate limit — the maximum rate the output can change:
For V_out = V_p sin(2πft), max slope = 2πf·V_p
Setting that equal to the slew rate:
V_p(max) = SR/(2πf)
= 0.5 V/µs / (2π × 10 000)
= 0.5 × 10⁶/(62 832)
= 7.96 V
So at 10 kHz the output can swing at most ±7.96 V before
the waveform's peaks get flattened into triangles —
regardless of the supply rails. At 100 kHz that falls to
just 0.8 V.
Full-power bandwidth (for a 10 V peak output):
f_max = SR/(2πV_p) = 0.5 × 10⁶/(2π × 10) = 7.96 kHz
💡 Exam angle: the most reliable marks come from deriving A_v = −R_f/R₁ using the virtual-ground and no-input-current rules — state both rules explicitly, then write KCL. Memorise the gain formulas for inverting, non-inverting, follower and summing. The trap to watch: a non-inverting amplifier's gain is 1 + R_f/R₁, not R_f/R₁, and so it can never be less than unity. Slew rate and gain-bandwidth appear in "why doesn't it work at high frequency?" questions.
Syllabus points
Inverting & non-inverting gain (numerical)
Op-amp applications
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