University Physics — Year 1 · Bachelor Year 1

10The Operational Amplifier

A strain gauge glued to a bridge girder changes its resistance by a thousandth when a truck passes; the Wheatstone bridge around it (Chapter 6) turns that into two millivolts. Two millivolts drive nothing. Between the gauge and the alarm that will close the bridge to traffic sits a chip the size of a fingernail, sold for a few cents, which multiplies the signal by a thousand, filters it, compares it with a threshold and lights a lamp — four jobs, one component, four different wirings. This chapter introduces the operational amplifier through its ideal model, derives the handful of circuits that every instrument is built from, and shows what happens when the feedback is removed.

10.1 The component and its ideal model

Definition 10.1 (Operational amplifier)

An operational amplifier is an integrated circuit with two inputs — the non-inverting input at potential $V_+$ and the inverting input at $V_-$ — and one output at potential $V_s$ , powered by two supplies $\pm V_{\mathrm{cc}}$ (typically $\pm15\,\mathrm{V}$ ) that are usually left off the diagrams. It amplifies the difference $\epsilon = V_+ - V_-$ :

V_s = \mu\,\epsilon \quad\text{as long as } \abs{V_s} < V_{\mathrm{sat}},

with an open-loop gain $\mu$ of order $10^5$ and a saturation voltage $V_{\mathrm{sat}}$ slightly below $V_{\mathrm{cc}}$ . When $\mu\abs\epsilon$ would exceed $V_{\mathrm{sat}}$ , the output sticks at $+V_{\mathrm{sat}}$ or $-V_{\mathrm{sat}}$ : the saturated regime.

Definition 10.2 (Ideal operational amplifier)

The ideal model assumes: (i) no current enters either input, $i_+ = i_- = 0$ (infinite input impedance); (ii) the output imposes its voltage whatever the current drawn (zero output impedance); (iii) infinite gain, $\mu \to \infty$ . In the linear regime ( $\abs{V_s} < V_{\mathrm{sat}}$ ) the third assumption forces

\epsilon = V_+ - V_- = 0 :

the two inputs sit at the same potential (a “virtual short circuit”) although no current flows between them.

The operational amplifier: symbol with its two inputs, and its transfer characteristic V_s( ) — a steep linear segment of slope 105 (drawn far too shallow), then saturation at ± V_ sat. In the ideal model the linear segment is vertical: = 0. — The operational amplifier: symbol with its two inputs, and its transfer characteristic $V_s(\epsilon)$ — a steep linear segment of slope $\mu \sim 10^5$ (drawn far too shallow), then saturation at $\pm V_{\mathrm{sat}}$ . In the ideal model the linear segment is vertical: $\epsilon = 0$ .

Proposition 10.3 (Negative feedback and the linear regime)

If the output is connected back to the inverting input through a network (negative feedback), the circuit has a stable linear operating point at $\epsilon = 0$ and the ideal-amplifier rules apply. If the output is fed back to the non-inverting input (positive feedback), or not fed back at all, the amplifier saturates: $V_s = \pm V_{\mathrm{sat}}$ according to the sign of $\epsilon$ .

Partial proof. With negative feedback, suppose $V_s$ rises a little above its equilibrium value; through the feedback network $V_-$ rises, $\epsilon$ falls, and the amplifier pulls $V_s$ back down: the perturbation is corrected. With positive feedback the same perturbation raises $V_+$ , increases $\epsilon$ and drives $V_s$ further away until it saturates. The full argument (a first-order model of the amplifier, $\mu(j\omega) = \mu_0/(1 + j\omega/\omega_0)$ , and the stability of the resulting differential equation) is the subject of the Year 2 volume; the rule above is what one uses. ∎

Remark 10.4 (Real amplifiers)

Three departures from the ideal matter in practice. The open-loop gain falls with frequency as $\mu_0/(1 + jf/f_0)$ with $f_0 \sim 10\,\mathrm{Hz}$ : the product $\mu_0 f_0 = f_T \sim 1\,\mathrm{MHz}$ (the gain–bandwidth product) is the bandwidth available to a follower, and a circuit of closed-loop gain $G$ has bandwidth $f_T/G$ . The output cannot change faster than the slew rate, $\sim1\,\mathrm{V}/\text{µ}\mathrm{s}$ . And a small input offset voltage ( $\sim1\,\mathrm{mV}$ ) is amplified like a signal — fatal for millivolt inputs unless trimmed.

10.2 The basic linear circuits

Method 10.5 (Analyzing an ideal op-amp circuit in the linear regime)

Check that the feedback goes to the inverting input.
Write $i_+ = i_- = 0$ : the input nodes obey Kirchhoff’s node law with no current into the amplifier (dividers and Millman’s theorem apply).
Write $V_+ = V_-$ and solve for $V_s$ .
Verify afterwards that $\abs{V_s} < V_{\mathrm{sat}}$ ; if not, the amplifier is saturated and the result is $\pm V_{\mathrm{sat}}$ .

Proposition 10.6 (Follower, non-inverting and inverting amplifiers)

Follower: output wired to $V_-$ , input on $V_+$ : $V_s = V_e$ ; infinite input impedance, zero output impedance.
Non-inverting amplifier: output to $V_-$ through a divider $R_2$ (feedback) and $R_1$ (to ground), input on $V_+$ :

$V_s = \left(1 + \frac{R_2}{R_1}\right)V_e .$
Inverting amplifier: $V_+$ grounded, input through $R_1$ to $V_-$ , feedback $R_2$ from output to $V_-$ :

$V_s = -\frac{R_2}{R_1}\,V_e ,$

with $V_-$ held at ground potential (a virtual ground) and an input impedance $R_1$ .

Proof. Follower: $V_- = V_s$ and $V_+ = V_e$ ; $\epsilon = 0$ gives $V_s = V_e$ . Non-inverting: no current into $V_-$ , so the divider gives $V_- = V_sR_1/(R_1 + R_2)$ ; equate to $V_+ = V_e$ . Inverting: $V_- = V_+ = 0$ ; node law at $V_-$ with $i_- = 0$ : $(V_e - 0)/R_1 + (V_s - 0)/R_2 = 0$ . The source delivers $V_e/R_1$ : input impedance $R_1$ . ∎

The three basic amplifiers. In each the output feeds the inverting input, the amplifier works in its linear regime, and the gain is set by resistor ratios alone — not by the amplifier’s own enormous and ill-defined . — The three basic amplifiers. In each the output feeds the inverting input, the amplifier works in its linear regime, and the gain is set by resistor ratios alone — not by the amplifier’s own enormous and ill-defined $\mu$ .

Example 10.7 (Why a follower)

A sensor of Thévenin resistance $10\,\mathrm{k}\Omega$ feeding a $1\,\mathrm{k}\Omega$ load delivers only $1/11$ of its open-circuit voltage (Remark 6.14). A follower between them draws no current from the sensor and drives the load from a zero-impedance output: the full voltage arrives. The follower is an impedance adapter — and it makes the cascade rule of Proposition 9.14 exact.

Proposition 10.8 (Summing and difference amplifiers)

Inputs $V_1, V_2$ through $R_1, R_2$ to the inverting node, feedback $R_f$ , $V_+$ grounded:

V_s = -R_f\left(\frac{V_1}{R_1} + \frac{V_2}{R_2}\right) \quad (\text{summing amplifier}).

With $V_1$ through $R_1$ to $V_-$ (feedback $R_2$ ) and $V_2$ through $R_1$ to $V_+$ (with $R_2$ from $V_+$ to ground):

V_s = \frac{R_2}{R_1}\,(V_2 - V_1) \quad (\text{difference amplifier}).

Proof. Summing: node law at the virtual ground, $V_1/R_1 + V_2/R_2 + V_s/R_f = 0$ . Difference: $V_+ = V_2R_2/(R_1 + R_2)$ (divider); node law at $V_-$ : $(V_1 - V_-)/R_1 + (V_s - V_-)/R_2 = 0$ , so $V_s = V_-(1 + R_2/R_1) - V_1R_2/R_1$ ; set $V_- = V_+$ and simplify. ∎

10.3 Integrator and differentiator

Proposition 10.9 (Integrator)

Input through $R$ to $V_-$ , a capacitor $C$ as feedback, $V_+$ grounded:

V_s(t) = -\frac{1}{RC}\int_0^t V_e(t')\,\dd t' + V_s(0), \qquad \underline H = -\frac{1}{jRC\omega} .

The gain falls $20\,\mathrm{dB}$ per decade at every frequency and the phase is $+90^\circ$ : an exact integrator — until the smallest offset or bias, integrated forever, drives the output into saturation.

Proof. Virtual ground: the current $V_e/R$ flows into the capacitor, whose voltage (from $V_-$ to the output) is $-V_s$ ; hence $C\,\dd(-V_s)/\dd t = V_e/R$ , and in complex notation $\underline H = -1/(jRC\omega)$ . ∎

Remark 10.10 (Taming the integrator)

A resistor $R'$ in parallel with $C$ gives

\underline H = -\frac{R'/R}{1 + jR'C\omega},

a first-order low-pass of DC gain $-R'/R$ and cutoff $1/R'C$ , which integrates for $\omega \gg 1/R'C$ but cannot drift to saturation. Likewise the differentiator (capacitor at the input, resistor as feedback: $V_s = -RC\,\dd V_e/\dd t$ , $\underline H = -jRC\omega$ ) amplifies high-frequency noise without limit unless a small series resistor caps its gain.

The integrator and its Bode diagram (blue, -20\, dB/ decade throughout, infinite gain at DC); with a resistor across C the gain is capped at low frequency (red): a low-pass that integrates above its cutoff. — The integrator and its Bode diagram (blue, $-20\,\mathrm{dB}/\mathrm{decade}$ throughout, infinite gain at DC); with a resistor across $C$ the gain is capped at low frequency (red): a low-pass that integrates above its cutoff.

Example 10.11 (Triangle from square)

$R = 10\,\mathrm{k}\Omega$ , $C = 100\,\mathrm{nF}$ , $1/RC = 1000\,\mathrm{s}^{-1}$ . A $1\,\mathrm{kHz}$ square wave of $\pm1\,\mathrm{V}$ gives a triangle of slope $\mp1000\,\mathrm{V}/\mathrm{s}$ and amplitude $ET/4RC = 0.25\,\mathrm{V}$ — without the $1/\omega$ attenuation of a passive RC, and with a zero-impedance output to feed the next stage.

10.4 Active filters

Proposition 10.12 (First-order active low-pass)

The inverting amplifier with a capacitor $C$ across its feedback resistor $R_2$ has

\underline H = -\frac{R_2/R_1}{1 + jR_2C\omega}:

a low-pass of DC gain $-R_2/R_1$ (which can exceed $1$ ) and cutoff $\omega_c = 1/R_2C$ , with zero output impedance — stages cascade by simple multiplication.

Proof. Inverting amplifier with the feedback impedance $R_2 \parallel 1/jC\omega = R_2/(1 + jR_2C\omega)$ in place of $R_2$ . ∎

10.5 Without negative feedback: comparators

Proposition 10.13 (Comparator)

With no feedback, the ideal amplifier is saturated: $V_s = +V_{\mathrm{sat}}$ if $V_+ > V_-$ , $-V_{\mathrm{sat}}$ if $V_+ < V_-$ . With a reference $V_{\mathrm{ref}}$ on one input and a signal on the other, the output tells whether the signal exceeds the reference — a one-bit converter.

Proof. $\mu\epsilon$ exceeds $V_{\mathrm{sat}}$ for any $\abs\epsilon > V_{\mathrm{sat}}/\mu \sim 0.1\,\mathrm{mV}$ . ∎

Proposition 10.14 (Schmitt trigger)

Feed a fraction $\beta = R_1/(R_1 + R_2)$ of the output back to $V_+$ (divider $R_1$ to ground, $R_2$ from the output) and apply the signal $V_e$ to $V_-$ . The output switches from $+V_{\mathrm{sat}}$ to $-V_{\mathrm{sat}}$ when $V_e$ rises past $+\beta V_{\mathrm{sat}}$ , and back when $V_e$ falls below $-\beta V_{\mathrm{sat}}$ : two thresholds, a hysteresis of width $2\beta V_{\mathrm{sat}}$ , immune to noise smaller than that width.

Proof. Positive feedback: saturated. If $V_s = +V_{\mathrm{sat}}$ then $V_+ = \beta V_{\mathrm{sat}}$ , and $\epsilon = \beta V_{\mathrm{sat}} - V_e$ stays positive until $V_e$ exceeds $\beta V_{\mathrm{sat}}$ ; then the output flips, $V_+$ drops to $-\beta V_{\mathrm{sat}}$ , and the new state holds until $V_e$ falls below that. ∎

The Schmitt trigger: positive feedback through R_1, R_2 gives two thresholds ± V_ sat. The output flips high-to-low when V_e rises past the upper threshold and low-to-high when it falls below the lower one (arrows): a hysteresis cycle. — The Schmitt trigger: positive feedback through $R_1$ , $R_2$ gives two thresholds $\pm\beta V_{\mathrm{sat}}$ . The output flips high-to-low when $V_e$ rises past the upper threshold and low-to-high when it falls below the lower one (arrows): a hysteresis cycle.

Example 10.15 (The relaxation oscillator)

Connect the Schmitt trigger’s output back to its own inverting input through $R$ , with $C$ from $V_-$ to ground. The capacitor charges toward $\pm V_{\mathrm{sat}}$ and flips the trigger each time it reaches $\pm\beta V_{\mathrm{sat}}$ : the output is a square wave, the capacitor voltage a chain of exponential arcs, and the period is

T = 2RC\ln\frac{1 + \beta}{1 - \beta}

(Exercise 10.12): $2.2\,RC$ for $\beta = \tfrac12$ . No input, one capacitor, and a clock — a system that oscillates because it cannot settle, a theme that returns with the feedback oscillators of the Year 2 volume.

10.6 Exercises

Exercise 10.1 ★

A non-inverting amplifier has $R_1 = 1.0\,\mathrm{k}\Omega$ , $R_2 = 9.0\,\mathrm{k}\Omega$ , $V_{\mathrm{sat}} = 15\,\mathrm{V}$ . Gain? Output for $V_e = 0.50\,\mathrm{V}$ ? Largest input before saturation?

Solution

Solution of Exercise 10.1.

$G = 1 + 9 = 10$ ; $V_s = 5.0\,\mathrm{V}$ ; saturation at $V_e = 15/10 = 1.5\,\mathrm{V}$ .

Exercise 10.2 ★

An inverting amplifier has $R_1 = 10\,\mathrm{k}\Omega$ , $R_2 = 100\,\mathrm{k}\Omega$ . Gain, input impedance, current drawn from a $1.0\,\mathrm{V}$ source, output.

Solution

Solution of Exercise 10.2.

$G = -10$ ; input impedance $R_1 = 10\,\mathrm{k}\Omega$ ; $i = 1.0/10^4 = 0.10\,\mathrm{mA}$ ; $V_s = -10\,\mathrm{V}$ .

Exercise 10.3 ★

A sensor of internal resistance $10\,\mathrm{k}\Omega$ and open-circuit voltage $1.0\,\mathrm{V}$ must drive a $1.0\,\mathrm{k}\Omega$ load. Voltage across the load without and with a follower between them?

Solution

Solution of Exercise 10.3.

Without: divider $1/(1 + 10) = 0.091$ , $91\,\mathrm{mV}$ . With the follower: $1.0\,\mathrm{V}$ — the follower draws nothing from the sensor and supplies the $1\,\mathrm{mA}$ the load needs.

Exercise 10.4 ★

A summing amplifier has $R_f = 10\,\mathrm{k}\Omega$ and inputs through $10\,\mathrm{k}\Omega$ and $20\,\mathrm{k}\Omega$ . Write $V_s$ ; compute it for $V_1 = 1.0\,\mathrm{V}$ , $V_2 = 2.0\,\mathrm{V}$ .

Solution

Solution of Exercise 10.4.

$V_s = -R_f(V_1/R_1 + V_2/R_2) = -(V_1 + V_2/2) = -(1.0 + 1.0) = -2.0\,\mathrm{V}$ .

Exercise 10.5 ★★

Derive the difference amplifier’s $V_s = (R_2/R_1)(V_2 - V_1)$ . With $R_1 = 10\,\mathrm{k}\Omega$ , $R_2 = 100\,\mathrm{k}\Omega$ , $V_1 = 1.00\,\mathrm{V}$ , $V_2 = 1.05\,\mathrm{V}$ : output? What happened to the $1\,\mathrm{V}$ common to both inputs?

Solution

Solution of Exercise 10.5.

$V_+ = V_2R_2/(R_1 + R_2)$ ; at $V_-$ : $(V_1 - V_-)/R_1 = (V_- - V_s)/R_2$ , so $V_s = V_-(1 + R_2/R_1) - V_1R_2/R_1$ ; with $V_- = V_+$ : $V_s = V_2R_2/R_1 - V_1R_2/R_1$ . Numbers: $10 \times 0.05 = 0.50\,\mathrm{V}$ ; the common $1\,\mathrm{V}$ contributes nothing — common-mode rejection.

Exercise 10.6 ★★

Integrator with $R = 10\,\mathrm{k}\Omega$ , $C = 100\,\mathrm{nF}$ , output initially zero. A $1.0\,\mathrm{V}$ step is applied: slope of the output and time to saturation ( $15\,\mathrm{V}$ ). A $1.0\,\mathrm{kHz}$ square wave of $\pm1.0\,\mathrm{V}$ : shape and amplitude of the output.

Solution

Solution of Exercise 10.6.

$V_s = -(1/RC)\int V_e$ : slope $-1000 \times 1.0 = -1000\,\mathrm{V}/\mathrm{s}$ ; saturation after $15/1000 = 15\,\mathrm{ms}$ . Square wave: triangle of amplitude $ET/4RC = 1.0 \times 10^{-3}/(4 \times 10^{-3}) = 0.25\,\mathrm{V}$ (plus whatever constant the initial charge left).

Exercise 10.7 ★★

A $1.0\,\mathrm{M}\Omega$ resistor is added across the capacitor of the previous integrator. Give the transfer function, the DC gain, the cutoff frequency, and the gain at $1.0\,\mathrm{kHz}$ ; in which range does the circuit still integrate?

Solution

Solution of Exercise 10.7.

$\underline H = -(R'/R)/(1 + jR'C\omega)$ : DC gain $-100$ ; cutoff $1/(2\pi R'C) = 1.6\,\mathrm{Hz}$ ; at $1\,\mathrm{kHz}$ , $R'C\omega = 628$ , $G = 100/628 = 0.16$ — the integrator value $1/RC\omega$ . It integrates for $f \gg 1.6\,\mathrm{Hz}$ .

Exercise 10.8 ★★

Design an inverting active low-pass with a gain of $20\,\mathrm{dB}$ in the passband and a cutoff at $500\,\mathrm{Hz}$ , using $R_1 = 1.0\,\mathrm{k}\Omega$ .

Solution

Solution of Exercise 10.8.

$R_2/R_1 = 10$ : $R_2 = 10\,\mathrm{k}\Omega$ ; $C = 1/(2\pi R_2f_c) = 1/(2\pi \times 10^4 \times 500) = 32\,\mathrm{nF}$ .

Exercise 10.9 ★★

A real amplifier has $\mu_0 = 10^5$ and $f_T = 1.0\,\mathrm{MHz}$ . For a non-inverting amplifier designed for a gain of $100$ : actual DC gain (keep $\mu_0$ finite in the analysis), and bandwidth. Bandwidth of a follower?

Solution

Solution of Exercise 10.9.

With finite $\mu$ : $V_s = \mu(V_e - \beta V_s)$ , $\beta = 1/100$ , so $G = \mu/(1 + \mu\beta) = 10^5/(1 + 10^3) = 99.9$ . Bandwidth $f_T/G = 10\,\mathrm{kHz}$ ; follower: $1\,\mathrm{MHz}$ .

Exercise 10.10 ★★★

A comparator with $V_{\mathrm{ref}} = 2.0\,\mathrm{V}$ on $V_-$ watches a slowly rising signal carrying $50\,\mathrm{mV}$ of noise. Describe the output as the signal crosses $2.0\,\mathrm{V}$ . Remedy?

Solution

Solution of Exercise 10.10.

Each noise excursion across $2.0\,\mathrm{V}$ flips the output: a burst of rapid switching (“chatter”) around the crossing. Remedy: hysteresis wider than the noise, i.e. a Schmitt trigger.

Exercise 10.11 ★★★

Schmitt trigger with $R_1 = 1.0\,\mathrm{k}\Omega$ , $R_2 = 9.0\,\mathrm{k}\Omega$ , $V_{\mathrm{sat}} = 15\,\mathrm{V}$ : derive and compute the two thresholds. Sketch $V_s$ for a triangular $V_e$ of amplitude $3\,\mathrm{V}$ .

Solution

Solution of Exercise 10.11.

$V_+ = V_sR_1/(R_1 + R_2) = V_s/10$ : thresholds $\pm1.5\,\mathrm{V}$ . The output is $+15\,\mathrm{V}$ while $V_e$ rises until $1.5\,\mathrm{V}$ , flips to $-15\,\mathrm{V}$ , stays there until $V_e$ falls below $-1.5\,\mathrm{V}$ , flips back: a square wave of the triangle’s frequency, its edges at the threshold crossings.

Exercise 10.12 ★★★

Relaxation oscillator with $R_1 = R_2$ ( $\beta = \tfrac12$ ). Derive the period $T = 2RC\ln\frac{1 + \beta}{1 - \beta}$ from the charging of $C$ through $R$ between the two thresholds; choose $R$ for $1.0\,\mathrm{kHz}$ with $C = 100\,\mathrm{nF}$ ; sketch $V_s(t)$ and $V_C(t)$ .

Solution

Solution of Exercise 10.12.

With $V_s = +V_{\mathrm{sat}}$ , $V_C$ charges from $-\beta V_{\mathrm{sat}}$ toward $+V_{\mathrm{sat}}$ : $V_C = V_{\mathrm{sat}} - (1 + \beta)V_{\mathrm{sat}} \eu^{-t/RC}$ , reaching $+\beta V_{\mathrm{sat}}$ when $\eu^{-t/RC} = (1 - \beta)/(1 + \beta)$ : half-period $RC\ln\frac{1 + \beta}{1 - \beta}$ , hence $T$ . $\beta = \tfrac12$ : $T = 2RC\ln 3 = 2.2RC$ ; $RC = 1/(2.2 \times 10^3) = 0.455\,\mathrm{ms}$ , $R = 4.6\,\mathrm{k}\Omega$ . $V_s$ : square wave $\pm15\,\mathrm{V}$ ; $V_C$ : exponential arcs between $\pm7.5\,\mathrm{V}$ .

10.7 Problem: An instrumentation chain for a strain gauge

Problem 10.1

Weekend problem — from two millivolts on a bridge girder to a red lamp: followers, a difference amplifier, an active filter and a Schmitt trigger, and the two imperfections that decide whether the chain works

A strain gauge of resistance $R(1 + \epsilon)$ , $R = 1.00\,\mathrm{k}\Omega$ , $0 \leq \epsilon \leq 2.0 \times 10^{-3}$ , forms a Wheatstone bridge with three fixed $1.00\,\mathrm{k}\Omega$ resistors, fed by $E = 5.00\,\mathrm{V}$ . The bridge output $u = V_B - V_A$ (between the two midpoints) is $u \approx E\epsilon/4$ (the sign is chosen by wiring). Amplifiers: ideal unless stated; $V_{\mathrm{sat}} = 15\,\mathrm{V}$ ; where needed, $f_T = 1.0\,\mathrm{MHz}$ , slew rate $0.5\,\mathrm{V}/\text{µ}\mathrm{s}$ , input offset $1\,\mathrm{mV}$ .

Part I — The bridge and its loading.

Recall why $u \approx E\epsilon/4$ for small $\epsilon$ .
Compute $u$ at full scale ( $\epsilon = 2.0 \times 10^{-3}$ ).
Give the potential of each midpoint (relative to the bridge’s negative terminal) at zero strain and at full scale. What is their common value called?
Each midpoint is the output of a divider of two $1\,\mathrm{k}\Omega$ resistors: what is the Thévenin resistance seen between the two midpoints?
An amplifier whose input resistance is $10\,\mathrm{k}\Omega$ is connected across the bridge output: what fraction of $u$ does it receive?
Why does a follower on each midpoint cure this? State the two properties used.
What are the followers’ output voltages (in terms of $V_A$ , $V_B$ ), and what current do they draw from the bridge?

Part II — Amplifying the difference.

The followers feed a difference amplifier ( $R_1$ at both inputs, $R_2$ as feedback and to ground). Derive $V_s = (R_2/R_1)(V_B - V_A)$ .
Choose $R_1 = 10\,\mathrm{k}\Omega$ and $R_2$ for a gain of $100$ .
A second, non-inverting stage multiplies by $10$ : give its resistors, the total gain, and the output at full scale.
Both midpoints sit near $E/2 = 2.5\,\mathrm{V}$ (the common-mode voltage). What does an exactly matched difference amplifier do with it? If one resistor ratio is off by $1\%$ , estimate the spurious input-referred voltage and compare with the signal.
The first amplifier has a $1\,\mathrm{mV}$ input offset. What output error does it produce after the total gain? Conclude.
With $f_T = 1\,\mathrm{MHz}$ , what is the bandwidth of each stage? Is it sufficient for a signal below $5\,\mathrm{Hz}$ ?
The output must swing by $2.5\,\mathrm{V}$ when a truck arrives in about $10\,\mathrm{ms}$ : does the slew rate limit?

Part III — Filtering. The amplified signal carries a $50\,\mathrm{Hz}$ hum picked up from the mains.

Design an inverting active low-pass with gain $-1$ and cutoff $10\,\mathrm{Hz}$ , using $R_1 = R_2 = 16\,\mathrm{k}\Omega$ : find $C$ .
Compute its attenuation at $50\,\mathrm{Hz}$ in decibels.
Compute the phase shift at $5\,\mathrm{Hz}$ and the corresponding delay. Acceptable for an alarm?
Why is an active filter preferable here to a passive RC (two reasons)?
A second identical stage is cascaded: attenuation at $50\,\mathrm{Hz}$ ? Why is the product rule exact here?

Part IV — Threshold and alarm. The alarm must trip when the strain exceeds $1.6 \times 10^{-3}$ .

To what output voltage of the chain does this correspond? A comparator with this reference: output states.
The filtered signal still carries $50\,\mathrm{mV}$ of noise near the threshold. What does the plain comparator do?
A Schmitt trigger with the reference on one input and feedback $\beta$ : show that the thresholds are $V_{\mathrm{ref}} \pm \beta V_{\mathrm{sat}}$ ; choose $\beta$ for $\pm0.10\,\mathrm{V}$ and give $R_1$ , $R_2$ .
State the two thresholds in volts and in strain.
The output drives an LED ( $2.0\,\mathrm{V}$ , $13\,\mathrm{mA}$ ) through a resistor: compute it. What protects the LED when the output is at $-V_{\mathrm{sat}}$ ?
Summarize the chain, stage by stage, and name the two non-idealities that set its real limits.

Solution

Solution of Problem 10.1.

1. Two dividers: $u = E[\tfrac12 - (1 + \epsilon)/(2 + \epsilon)] \approx -E\epsilon/4$ (Proposition 6.22); wiring fixes the sign.

2. $u = 5.00 \times 2.0\times10^{-3}/4 = 2.5\,\mathrm{mV}$ .

3. $V_A = E/2 = 2.500\,\mathrm{V}$ always; $V_B = E(1 + \epsilon)/(2 + \epsilon) \approx 2.500 + 0.0025 = 2.5025\,\mathrm{V}$ at full scale. The $2.5\,\mathrm{V}$ shared by both is the common-mode voltage.

4. Each midpoint: $1 \parallel 1 = 500\,\Omega$ ; between the two midpoints, $1.0\,\mathrm{k}\Omega$ .

5. $10/(10 + 1) = 0.91$ : a $9\%$ loss.

6. Infinite input impedance (no current from the bridge, so the midpoints keep their open-circuit potentials) and zero output impedance (the next stage can draw what it needs).

7. $V_A$ and $V_B$ exactly; zero current.

8. As in Proposition 10.8: $V_+ = V_BR_2/(R_1 + R_2)$ , node law at $V_-$ , $V_- = V_+$ .

9. $R_2 = 1.0\,\mathrm{M}\Omega$ .

10. $1 + R_2/R_1 = 10$ : e.g. $1\,\mathrm{k}\Omega$ and $9\,\mathrm{k}\Omega$ . Total gain $1000$ ; full scale $2.5 \times 10^{-3} \times 1000 = 2.5\,\mathrm{V}$ .

11. Matched: the common $2.5\,\mathrm{V}$ cancels exactly (only $V_B - V_A$ survives). A $1\%$ mismatch passes about $1\%$ of it: an input-referred error of order $25\,\mathrm{mV}$ , ten times the full-scale signal — the four resistors must be matched to better than $0.01\%$ (or the common-mode voltage removed first).

12. $1\ \mathrm{mV} \times 1000 = 1\,\mathrm{V}$ , forty percent of full scale: unusable without offset trimming, a low-offset (chopper) amplifier, or a zero calibration with no load on the girder.

13. Stage 1: $f_T/100 = 10\,\mathrm{kHz}$ ; stage 2: $100\,\mathrm{kHz}$ ; far above $5\,\mathrm{Hz}$ .

14. Required $2.5/10^{-2} = 250\,\mathrm{V}/\mathrm{s} = 0.25\,\mathrm{mV}/\text{µ}\mathrm{s}$ , two thousand times below the slew rate: no.

15. $C = 1/(2\pi R_2f_c) = 1/(2\pi \times 1.6\times10^4 \times 10) = 1.0\,\text{µ}\mathrm{F}$ .

16. $x = 5$ : $G = 1/\sqrt{26}$ , $-14\,\mathrm{dB}$ .

17. $\varphi = -\arctan 0.5 = -27^\circ$ ; delay $\varphi/\omega = 15\,\mathrm{ms}$ — nothing against a truck’s passage.

18. No loading of the following stage (zero output impedance) and gain available; also no attenuation of the wanted signal below cutoff.

19. $-28\,\mathrm{dB}$ (gains multiply, decibels add), exact because the first stage’s output impedance is zero: the second stage does not load it.

20. $\epsilon = 1.6\times10^{-3} \to u = 2.0\,\mathrm{mV} \to 2.0\,\mathrm{V}$ ; output $+V_{\mathrm{sat}}$ when the signal exceeds $2.0\,\mathrm{V}$ , $-V_{\mathrm{sat}}$ below (or the reverse, by wiring).

21. It chatters: the lamp flickers as noise crosses the reference.

22. With $V_{\mathrm{ref}}$ through $R_2$ and the output through $R_1$ … in the standard form, the non-inverting input sits at $V_{\mathrm{ref}}(1 - \beta) + \beta V_s$ ; taking the reference shifted accordingly, the switching levels are $V_{\mathrm{ref}} \pm \beta V_{\mathrm{sat}}$ . $\beta V_{\mathrm{sat}} = 0.10\,\mathrm{V}$ : $\beta = 1/150$ , e.g. $R_1 = 1.0\,\mathrm{k}\Omega$ , $R_2 = 149\,\mathrm{k}\Omega$ .

23. $2.1\,\mathrm{V}$ (trip) and $1.9\,\mathrm{V}$ (release): strains $1.68 \times 10^{-3}$ and $1.52 \times 10^{-3}$ .

24. $R = (15 - 2.0)/0.013 = 1.0\,\mathrm{k}\Omega$ . At $-V_{\mathrm{sat}}$ the LED is reverse-biased by $15\,\mathrm{V}$ , beyond its rating: an ordinary diode in antiparallel (or in series) protects it.

25. Bridge ( $2.5\,\mathrm{mV}$ full scale) $\to$ two followers (no loading) $\to$ difference amplifier $\times100$ $\to$ non-inverting $\times10$ $\to$ active low-pass $10\,\mathrm{Hz}$ $\to$ Schmitt trigger at $2.0\,\mathrm{V} \pm 0.1\,\mathrm{V}$ $\to$ LED. The real limits: the input offset voltage and the matching of the difference amplifier’s resistors, both comparable to the millivolt signal.

Terms defined in this chapter

See all 393 terms in the glossary