Notes · ElectronicsUK · A-Levels

Instrumentation systems

Instrumentation systems measure a physical quantity and turn it into a usable electrical signal. This chapter develops resistive sensors and the potential divider, the Wheatstone bridge and its balance condition, the difference and instrumentation amplifiers that amplify a small differential signal while rejecting interference, and position sensing with encoded discs and the Gray code.

4 sections~15 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·0888 / 16

Exam profile

AO1 · Describe resistive sensors, the potential divider, the Wheatstone bridge, the difference and instrumentation amplifiers, and encoded discs.AO2 · Calculate potential-divider outputs, bridge outputs and difference-amplifier gain, allowing for loading.AO3 · Design a complete measurement sub-system and evaluate its sensitivity and noise rejection.

Operators:describecalculatedeterminedesignanalyseevaluate

basic level

At AS the focus is resistive sensors and the potential divider as a sensing sub-system, including the loading effect.

higher level

The full A-Level develops the Wheatstone bridge, the difference and instrumentation amplifiers with common-mode rejection, and encoded-disc position sensing.

Depth

Text

Contents · 4 sections

Instrumentation systems

Sensors and the potential divider#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (sensors)

Potential divider sensing circuit

Fig. 1Two resistors share the supply; the output across

R_2

V_{in}R_2/(R_1+R_2)

. Replacing one arm with a sensor turns a resistance change into a voltage.

Many sensors work by changing their resistance in response to a physical quantity. A thermistor (of the common negative-temperature-coefficient type) has a resistance that falls sharply as its temperature rises; a light-dependent resistor (LDR) has a resistance that falls as the light on it increases; a strain gauge has a resistance that changes slightly as it is stretched or compressed. Each turns a physical change into a resistance change — but a resistance is not directly useful, so it must be converted into a voltage.

The potential divider is the standard sub-system that does this conversion. Two resistances in series across a supply share the supply voltage in proportion to their values, and the voltage across one of them is

V_{out} = V_{in} \times R_2/(R_1 + R_2)

. If one of the two is a sensor, its resistance change moves the output voltage, so the divider turns the sensor's resistance into a signal voltage that the rest of the system can use. Whether the output rises or falls with the measured quantity depends on which arm carries the sensor.

The choice of the fixed resistor sets the sensitivity and the operating point. Making the fixed resistor roughly equal to the sensor's resistance at the middle of the measured range places the output near half the supply, where a given resistance change produces the largest voltage change — the steepest, most sensitive part of the divider's response. Choosing this fixed value to match the range of interest is the key design decision, and a favourite exam calculation.

A crucial practical caution is the loading effect. If the divider output is connected to a following stage that draws current, that stage appears in parallel with the lower resistor, reducing the effective resistance and pulling the output voltage below its unloaded value. To avoid this, the divider must feed a high-impedance input — most simply a voltage-follower buffer (from the op-amp chapter), whose enormous input impedance draws negligible current and so does not disturb the divider. Recognising and preventing loading is a mark of a careful designer.

The sensor-and-divider combination is the input sub-system of countless control and measurement systems: a temperature controller, a light-operated switch, a strain-measuring load cell. Its output feeds a comparator (for a switching decision) or an amplifier and ADC (for measurement). Understanding that the divider converts resistance to voltage, that the fixed resistor sets the sensitivity, and that the output must be buffered against loading, equips you to design the front end of almost any instrumentation system.

V_{out} = V_{in}\,\dfrac{R_2}{R_1 + R_2}

Potential divider

The output is the supply times the fraction of the total resistance in the output arm.

Thermistor resistance falling with temperature

Fig. 2A negative-temperature-coefficient thermistor: its resistance falls sharply as temperature rises (marker at

25\,^{\circ}\text{C}

Worked example

Thermistor divider output at two temperatures

A thermistor is $10\,\text{k}\Omega$ at $25\,^{\circ}\text{C}$ and $4\,\text{k}\Omega$ at $50\,^{\circ}\text{C}$ . It is the upper arm of a divider with a $10\,\text{k}\Omega$ lower resistor across a $5\,\text{V}$ supply, output taken across the fixed resistor. Find the output at each temperature.

At 25 C
$V_{out} = 5 \times R_2/(R_{th}+R_2) = 5 \times 10/(10+10) = 2.50\,\text{V}$ .
At 50 C
The thermistor has fallen to $4\,\text{k}\Omega$ : $V_{out} = 5 \times 10/(4+10) = 5 \times 0.714 = 3.57\,\text{V}$ .
$V_{out} = 5 \times \dfrac{10}{4 + 10} = 3.57\,\text{V}$
Interpretation
As the temperature rises the thermistor resistance falls, so a larger fraction of the supply appears across the fixed resistor and the output rises from $2.50\,\text{V}$ to $3.57\,\text{V}$ .

Result: The output rises from $2.50\,\text{V}$ at $25\,^{\circ}\text{C}$ to $3.57\,\text{V}$ at $50\,^{\circ}\text{C}$ — a temperature-dependent signal voltage.

Typical mistakes

Using the wrong resistor in $R_2/(R_1+R_2)$ — the output is across the lower arm.
Ignoring loading, so a following low-impedance stage pulls the divider output below its calculated value.

Active revision

A thermistor of resistance $10\,\text{k}\Omega$ at $25\,^{\circ}\text{C}$ (falling to $4\,\text{k}\Omega$ at $50\,^{\circ}\text{C}$ ) is placed in series with a $10\,\text{k}\Omega$ fixed resistor across a $5\,\text{V}$ supply, with the output across the fixed resistor. Find the output voltage at each temperature.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The Wheatstone bridge#

StandardWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (bridge)

Wheatstone bridge

Fig. 3Two dividers (

R_1

R_3

and

R_2

R_4

) across one supply; the meter reads the difference between their midpoints, zero at balance.

The Wheatstone bridge is two potential dividers connected across the same supply, with the measurement taken between their two output points. It is the standard circuit for detecting a small change in a sensor resistance, because it compares the sensor against a reference rather than reading it directly. When the two dividers give the same output voltage the bridge is said to be balanced, and the voltage between the two midpoints is zero.

The balance condition is a simple ratio: the bridge is balanced when

R_1/R_2 = R_3/R_4

, where

R_1

and

R_2

form one divider and

R_3

and

R_4

the other. At balance the output is zero regardless of the supply voltage, which is a great advantage — a zero reading is easy to detect precisely and does not drift with the supply. A bridge is often set up to be balanced at the reference condition, so that any output at all signals a change.

When the sensor resistance changes, the bridge goes out of balance and produces an output voltage equal to the difference between the two divider outputs:

V_{out} = V_{in}\left(\dfrac{R_2}{R_1+R_2} - \dfrac{R_4}{R_3+R_4}\right)

. For a small change this output is small but proportional to the change, making the bridge a sensitive, linear detector of tiny resistance variations — exactly what is needed for a strain gauge, whose resistance changes by only a fraction of a percent under load.

The bridge's great strength is that it measures a difference, so effects that change both arms equally cancel out. If temperature affects both the sensor and a matching dummy gauge in the same way, the bridge stays balanced and rejects the temperature drift, responding only to the strain that affects the active gauge. This common-mode rejection is why the bridge, followed by a difference amplifier, is the heart of precision measurement — it separates the wanted signal from unwanted disturbances that affect the whole circuit.

The bridge output is typically only millivolts, so it is fed to a difference amplifier (next section) that amplifies the difference between the two midpoints while ignoring their common level. The bridge-plus-amplifier combination is a classic instrumentation sub-system: the bridge converts a resistance change into a small differential voltage and rejects common disturbances, and the amplifier raises that voltage to a useful level. Designing and analysing this front end is a core A2 skill.

\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4} \ \text{(balance)}

Balance condition

The bridge output is zero when the two divider ratios are equal.

V_{out} = V_{in}\left(\dfrac{R_2}{R_1+R_2} - \dfrac{R_4}{R_3+R_4}\right)

Out-of-balance output

The difference between the two divider outputs.

Worked example

Out-of-balance output of a strain-gauge bridge

A Wheatstone bridge has three fixed $120\,\Omega$ resistors ( $R_1 = R_2 = R_3 = 120\,\Omega$ ) and an active strain gauge $R_4$ that rises from $120\,\Omega$ to $120.6\,\Omega$ under load, with a $5.0\,\text{V}$ supply. Find the output.

Left divider
The left midpoint is $V_{in}R_3/(R_1+R_3) = 5.0 \times 120/240 = 2.500\,\text{V}$ .
Right divider under load
The right midpoint is $V_{in}R_4/(R_2+R_4) = 5.0 \times 120.6/(120+120.6) = 5.0 \times 120.6/240.6 = 2.5062\,\text{V}$ .
Output
$V_{out} = 2.5062 - 2.500 = 0.0062\,\text{V} = 6.2\,\text{mV}$ .
$V_{out} = 5.0\left(\dfrac{120.6}{240.6} - \dfrac{120}{240}\right) \approx 6.2\,\text{mV}$

Result: The bridge produces about $6.2\,\text{mV}$ out of balance — a small but definite signal, ready to be amplified by a difference amplifier.

Typical mistakes

Misremembering the balance ratio — the products of opposite arms are equal at balance.
Forgetting that the useful output is the difference between the two divider outputs, not one divider alone.

Active revision

A Wheatstone bridge has three fixed $120\,\Omega$ resistors and a strain gauge whose resistance rises from $120\,\Omega$ to $120.6\,\Omega$ under load, with a $5.0\,\text{V}$ supply. Find the out-of-balance output voltage.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

The difference and instrumentation amplifier#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (instrumentation amp)

Difference amplifier

Fig. 4One op-amp with matched resistors gives

V_{out} = (R_f/R_{in})(V_2 - V_1)

, amplifying the difference while rejecting what is common to both inputs.

A bridge produces a small voltage between two points, neither of which is at ground, so what must be amplified is the difference between two signals rather than a single voltage. The difference (differential) amplifier does exactly this. Built from a single op-amp with four matched resistors, it produces an output proportional to the difference of its two inputs,

V_{out} = (R_f/R_{in})(V_2 - V_1)

, amplifying the wanted differential signal by the resistor ratio.

Its value lies in what it ignores. Any voltage common to both inputs — mains hum picked up equally on both wires, or a common DC level — is a common-mode signal, and the ideal difference amplifier rejects it completely, responding only to the difference. The measure of how well it does this is the common-mode rejection ratio (CMRR): the ratio of the differential gain to the common-mode gain, usually quoted in decibels. A high CMRR means interference that affects both inputs equally is strongly suppressed.

The simple one-op-amp difference amplifier has a drawback: its input impedance is set by the resistors and is relatively low, so it loads the bridge and its performance depends on the resistors being precisely matched. The instrumentation amplifier solves both problems by adding two non-inverting buffer amplifiers, one in front of each input. These present a very high input impedance so they do not load the source, and they provide gain, before the difference amplifier does the subtraction.

A particularly elegant feature of the standard three-op-amp instrumentation amplifier is that its overall gain can be set by a single resistor, without needing to keep several resistors matched as that resistor is changed. This makes the gain easy to adjust — often with one external component — while preserving the high input impedance and high common-mode rejection. These properties are exactly what precision measurement demands, which is why the instrumentation amplifier is the standard front end for sensors, bridges and biomedical signals.

The bridge, the difference amplifier and the instrumentation amplifier form a coherent measurement chain: the bridge converts a tiny resistance change into a small differential voltage and rejects effects common to both arms; the amplifier raises that difference to a useful level while continuing to reject common-mode interference; and the result feeds a display or an ADC. Understanding how each stage contributes to sensitivity and to noise rejection — and being able to calculate the gain and reason about the CMRR — is the culmination of the instrumentation chapter.

V_{out} = \dfrac{R_f}{R_{in}}\,(V_2 - V_1)

Difference amplifier

Amplifies the difference of the two inputs by the resistor ratio.

\text{CMRR} = 20\log_{10}\!\left(\dfrac{A_{diff}}{A_{common}}\right)\,\text{dB}

Common-mode rejection ratio

How strongly a signal common to both inputs is suppressed relative to the difference.

Worked example

Amplifying a bridge output

A difference amplifier with $R_f = 100\,\text{k}\Omega$ and $R_{in} = 1.0\,\text{k}\Omega$ amplifies the $6.2\,\text{mV}$ differential output of a strain-gauge bridge. Find the output, and explain the fate of a $50\,\text{Hz}$ hum of $0.5\,\text{V}$ present equally on both inputs.

Differential gain
$A_{diff} = R_f/R_{in} = 100\,\text{k}\Omega/1.0\,\text{k}\Omega = 100$ .
Output
$V_{out} = A_{diff}(V_2 - V_1) = 100 \times 6.2\,\text{mV} = 0.62\,\text{V}$ .
$V_{out} = 100 \times 6.2\,\text{mV} = 0.62\,\text{V}$
The common hum
The $0.5\,\text{V}$ hum is identical on both inputs, so its difference is zero. The amplifier responds only to the difference, so the hum is rejected — with a high CMRR it contributes almost nothing to the output.

Result: The output is $0.62\,\text{V}$ ; the common-mode hum is rejected because the amplifier amplifies only the $6.2\,\text{mV}$ difference, not the level common to both inputs.

Typical mistakes

Treating the difference amplifier as amplifying each input separately rather than their difference.
Forgetting that a high CMRR is what lets the amplifier reject interference common to both inputs.

Active revision

A difference amplifier with $R_f = 100\,\text{k}\Omega$ and $R_{in} = 1.0\,\text{k}\Omega$ amplifies the $6.2\,\text{mV}$ output of a bridge. Find the output voltage and state why a common $50\,\text{Hz}$ hum on both inputs is not amplified.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

Position sensing and encoded discs#

AdvancedWJEC/Eduqas A Level Electronics — Unit 3 Analogue Design (position sensing)

Binary and Gray code for positions 0 to 7

Fig. 5The 3-bit Gray code: exactly one bit changes between each row and the next (and between 7 and 0), so an encoded disc never misreads badly at a boundary.

Measuring angular position digitally is done with an encoded disc: a disc marked with a pattern of transparent and opaque (or reflective and non-reflective) sectors, read by a row of optical sensors. Each sensor produces a 1 or 0 depending on whether it sees light through its track, so the row of sensors reads off a binary number that identifies the disc's angular position directly. This is an absolute encoder — every position has its own unique code, so the position is known immediately at switch-on, without needing to count from a reference.

A naive design would mark the tracks in ordinary binary counting order, but this causes a serious problem at the boundaries between positions. Where several bits change at once — for example from

0111

1000

— the sensors cannot switch at exactly the same instant, so during the transition the reading passes through spurious intermediate codes. A momentary misread of the position could be far from either true value, which in a control system could be dangerous.

The Gray code solves this elegantly. It is a binary code arranged so that only one bit changes between any two adjacent values. Because just one track changes at each boundary, there is no instant when several sensors are switching together, and the worst a boundary misread can do is give one of the two neighbouring positions — never a wildly wrong value. Encoded discs are therefore marked in Gray code, and the reading is converted to ordinary binary in software or logic if arithmetic is needed.

Converting between binary and Gray code is a standard skill. To turn binary into Gray, the most significant bit is copied, and each following Gray bit is the exclusive-OR of adjacent binary bits. The reverse conversion rebuilds the binary from the Gray code by a running exclusive-OR. These conversions are neat applications of the XOR gate from the logic chapter and show how the digital tools of the course serve a real measurement problem.

Position and motion sensing pull the whole instrumentation chapter together: a physical quantity (angle) is turned into a clean digital signal by optical sensors and an encoded disc, the Gray code guarantees the reading is robust at every boundary, and the result feeds a digital system for display or control. The same ideas — an incremental encoder counting equally spaced marks, or a linear encoder for straight-line position — extend the principle, but the absolute Gray-coded disc is the clearest example of turning a mechanical position into a trustworthy number.

G_i = B_i \oplus B_{i+1} \ (\text{binary to Gray, MSB copied})

Gray code conversion

Each Gray bit is the exclusive-OR of adjacent binary bits.

Worked example

Converting binary to Gray code

Convert the binary number $1011$ (decimal 11) to Gray code, and confirm it differs by exactly one bit from the Gray code of its neighbour $1010$ (decimal 10).

Copy the MSB
For $1011$ , the first Gray bit equals the first binary bit: $G_3 = 1$ .
XOR adjacent bits
$G_2 = B_3 \oplus B_2 = 1 \oplus 0 = 1$ ; $G_1 = B_2 \oplus B_1 = 0 \oplus 1 = 1$ ; $G_0 = B_1 \oplus B_0 = 1 \oplus 1 = 0$ . So $1011 \to 1110$ .
Check the neighbour
By the same method $1010 \to 1111$ . The two Gray codes $1110$ and $1111$ differ in only the last bit, confirming the single-bit-change property.
$1011_2 \to 1110_{Gray}, \quad 1010_2 \to 1111_{Gray}$

Result: $1011$ in binary is $1110$ in Gray code, and it differs from its neighbour by exactly one bit — the property that makes Gray-coded discs robust.

Typical mistakes

Marking the disc in binary counting order, so multiple bits change at a boundary and cause misreads.
Muddling the binary-to-Gray conversion — copy the MSB, then XOR adjacent binary bits.

Active revision

Write the 3-bit Gray code sequence for positions 0 to 7 and state how many bits change between each adjacent position and its neighbour.

Active recall

Recall the key points — then reveal.

Sources: WJEC/Eduqas GCE Electronics specification (WJEC / Eduqas)

References & sources

Sources

WJEC / Eduqas

WJEC/Eduqas GCE Electronics specification

Instrumentation systems

4 sections~15 min reading time3 competenciesLevel Standard 2 · Advanced 2

Instrumentation systems

Key points

Thermistor divider output at two temperatures

Key points

Out-of-balance output of a strain-gauge bridge

Key points

Amplifying a bridge output

Key points

Converting binary to Gray code

Sources

Instrumentation systems

Key points

Thermistor divider output at two temperatures

Key points

Out-of-balance output of a strain-gauge bridge

Key points

Amplifying a bridge output

Key points

Converting binary to Gray code

Sources