Signal conditioning circuits for sensors

Signal conditioning chain for sensors: bridges and converters, instrumentation amplifiers and CMRR, filters and anti-aliasing, isolation, ADC resolution and 4–20 mA loops, with worked numericals.

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Why it matters

A sensor's raw output — a few millivolts from a bridge, a picocoulomb charge, a resistance change of 0.1 % — is rarely usable directly. Signal conditioning turns it into a clean, scaled, isolated signal that an ADC, controller or 4–20 mA loop can use. Most real measurement errors (noise, ground loops, aliasing, offset) are created or removed here.

Key ideas

A typical chain is: excitation → sensor → bridge or converter → amplifier → filter → isolation → ADC or transmitter, with linearisation and compensation either in hardware or in software.

Converting the sensor's quantity to a voltage.

  • Resistive sensors: Wheatstone bridge (strain gauges, RTDs) or a constant-current source (RTDs, 4-wire). Ratiometric measurement — using the same reference for excitation and the ADC — cancels supply drift.
  • Reactive sensors (LVDT, capacitive): AC excitation, AC bridge, then phase-sensitive demodulation and low-pass filtering.
  • Charge sources (piezoelectric): charge amplifier, Vo = −Q/Cf.
  • Current sources (photodiodes): transimpedance amplifier, Vo = −I·Rf.
  • Thermocouples: cold-junction compensation plus high-gain amplification.

Amplifiers.

  • Inverting gain −Rf/Rin; non-inverting gain 1 + Rf/Rin (very high input impedance — good for buffering a sensor).
  • A bridge output is a small differential signal sitting on a large common-mode voltage (about half the excitation). An instrumentation amplifier (three op-amps: two input buffers with a gain-setting resistor RG, then a difference amplifier) offers very high input impedance, gain set by one resistor and high CMRR. Gain G = (1 + 2R/RG)·(R3/R2).
  • CMRR = differential gain / common-mode gain, usually in dB. The common-mode error at the output is G·Vcm/CMRR (referred to input: Vcm/CMRR).
  • Other error sources: input offset voltage and its drift, bias currents through source resistance, gain error, noise.

Filtering.

  • Low-pass filters remove high-frequency noise; a first-order RC has cut-off fc = 1/(2πRC) and rolls off at only 20 dB/decade — it attenuates, it does not "remove". High-pass filters remove DC offsets and drift; band-pass filters select a carrier; notch filters reject 50 Hz mains pickup.
  • Anti-aliasing: before sampling at fs, frequencies above fs/2 (the Nyquist frequency) must be attenuated, otherwise they fold back as false low-frequency signals.

Isolation and protection. Isolation amplifiers (transformer, optical or capacitive barriers) break ground loops, tolerate large common-mode voltages and protect people and equipment. Add input protection (clamp diodes, series resistors) and shielding with single-point grounding; use twisted pairs for differential signals.

Analog-to-digital conversion. An n-bit ADC with full-scale range FSR has LSB = FSR/2ⁿ; quantisation error is ±½ LSB. Choose gain so the sensor's full-scale output nearly fills the ADC range, otherwise resolution is wasted.

Transmission: the 4–20 mA loop. A transmitter converts the measurement into a current from 4 mA (0 % of span) to 20 mA (100 %). Current is unaffected by wire resistance, the 4 mA "live zero" distinguishes a broken wire (0 mA) from a zero reading, and the loop can power the transmitter. A 250 Ω receiver resistor converts it to 1–5 V.

Linearisation: analog (feedback networks, parallel resistors for thermistors) or digital (look-up tables, polynomials such as the Callendar–Van Dusen or thermocouple polynomials).

Formulas

G = (1 + 2R/RG)·(R3/R2) (three-op-amp instrumentation amplifier) R = input-stage feedback resistors (Ω); RG = gain resistor (Ω); R3/R2 = difference-stage ratio.

CMRR(dB) = 20·log10(Ad/Acm); V_error,out = Ad·Vcm/CMRR Ad = differential gain; Acm = common-mode gain; Vcm = common-mode voltage (V); CMRR as a ratio.

fc = 1/(2π·R·C), |H| = 1/√(1 + (f/fc)²) (first-order RC low-pass)

fs > 2·f_max (sampling theorem)

LSB = FSR / 2ⁿ FSR = ADC full-scale range (V); n = number of bits.

I = 4 + 16·(x − x_min)/(x_max − x_min) (mA, 4–20 mA transmitter)

Worked examples

Example 1 — load cell to ADC. A load cell with rated output 2 mV/V is excited at 5 V. Its output feeds a three-op-amp instrumentation amplifier (R = 25 kΩ, difference stage unity gain) and a 12-bit ADC with a 0–5 V range. Find the required gain, RG, the LSB, and the LSB referred to the load-cell output.

  1. Full-scale output: 2 mV/V × 5 V = 10 mV.
  2. Gain to fill the ADC: G = 5 V/10 mV = 500.
  3. 1 + 2R/RG = 500 → RG = 2 × 25 000/499 = 100.2 Ω.
  4. LSB = 5/2¹² = 5/4096 = 1.221 mV; referred to input 1.221 mV/500 = 2.44 µV.

Answer: G = 500, RG ≈ 100 Ω, LSB = 1.22 mV (2.44 µV at the sensor).

Example 2 — CMRR and the 4–20 mA loop (GATE level). (a) The amplifier in Example 1 has CMRR = 100 dB. The bridge's common-mode voltage is 2.5 V. Find the output error and express it as a percentage of the 10 mV full-scale input. (b) A temperature transmitter ranged 0–200 °C sends 12.8 mA. Find the temperature and the voltage across a 250 Ω receiver.

  1. CMRR = 10^(100/20) = 10⁵.
  2. Output error = Ad·Vcm/CMRR = 500 × 2.5/10⁵ = 12.5 mV; referred to input 2.5/10⁵ = 25 µV.
  3. As % of full scale: 25 µV/10 mV × 100 = 0.25 %.
  4. Temperature: T = (12.8 − 4)/16 × 200 = 110 °C.
  5. Receiver voltage: 12.8 mA × 250 Ω = 3.2 V.

Answer: (a) 12.5 mV at the output, 0.25 % FS; (b) 110 °C, 3.2 V.

Example 3 — a filter that only attenuates. An RC low-pass with fc = 50 Hz receives 60 Hz interference. |H| = 1/√(1 + 1.2²) = 0.640 (−3.9 dB): only 36 % of the interference is removed. A notch filter, a higher-order filter or a lower fc is needed.

Common mistakes

  • Using a single-ended amplifier on a bridge and amplifying the common-mode voltage.
  • Believing a filter "completely removes" noise just above its cut-off frequency.
  • Sampling without an anti-aliasing filter.
  • Choosing a gain that uses only a fraction of the ADC range, wasting resolution.
  • Forgetting the 4 mA offset: 12 mA is 50 % of span, not 60 %.
  • Grounding a shield at both ends and creating a ground loop.

For GATE IN

NATs: instrumentation-amplifier gain and RG, CMRR in dB and the resulting error, RC filter cut-off and attenuation, ADC resolution and LSB, 4–20 mA scaling, bridge outputs. MCQs: purpose of isolation, anti-aliasing, charge vs transimpedance amplifiers, ratiometric measurement. Practise expressing errors referred to input and as % of full scale.

Quick check

  1. Instrumentation amplifier with R = 25 kΩ, RG = 1 kΩ, unity difference stage: gain?
  2. What is the LSB of a 10-bit ADC with a 0–10 V range?
  3. A 4–20 mA transmitter ranged 0–10 bar reads 8 mA. Pressure?
  4. CMRR of 80 dB as a ratio? Answers: 1. 51; 2. 9.77 mV; 3. 2.5 bar; 4. 10 000.

Interview questions

All Sensors and Transducers interview questions

Try answering each one aloud before you open it.

  1. 1.What is signal conditioning in the context of sensors and transducers?Concept

    Signal conditioning refers to the process of manipulating a sensor's output signal to make it suitable for further processing. This can include amplification, filtering, converting, and isolating the signal to ensure it is compatible with the input requirements of data acquisition systems or controllers.

  2. 2.Explain the role of an amplifier in a signal conditioning circuit.Concept

    An amplifier in a signal conditioning circuit increases the amplitude of a sensor's output signal. This is crucial when the sensor's signal is too weak to be processed by subsequent stages. Amplifiers ensure that the signal is strong enough to be accurately digitized by an analog-to-digital converter (ADC) or processed by other electronic components.

  3. 3.Why is filtering important in signal conditioning circuits?Application

    Filtering is important in signal conditioning circuits to remove unwanted noise and interference from the sensor's output signal. By using filters, such as low-pass, high-pass, or band-pass filters, the circuit can isolate the desired frequency range and improve the accuracy and reliability of the data being processed.

  4. 4.What happens if a signal conditioning circuit lacks proper isolation?Application

    If a signal conditioning circuit lacks proper isolation, it can lead to ground loops and interference from other electrical systems, which can distort the sensor's output signal. Isolation helps to prevent these issues by electrically separating different parts of the circuit, ensuring that the signal remains clean and accurate.

  5. 5.Explain the purpose of a Wheatstone bridge in signal conditioning.Concept

    A bridge converts a small resistance change into a differential voltage centred on zero, so the large constant part of the sensor's resistance is cancelled and only the change is amplified. With suitable arm placement it also cancels temperature effects (dummy gauge or 3-wire RTD connection) and doubles or quadruples the output with multiple active elements. Its output rides on a common-mode voltage of about half the excitation, so it is read with an instrumentation amplifier with high CMRR.

  6. 6.Why is an analog-to-digital converter (ADC) used in signal conditioning circuits?Application

    An ADC is used in signal conditioning circuits to convert the analog signal from a sensor into a digital format. This conversion is necessary for digital systems, such as microcontrollers or computers, to process and analyze the data. The ADC ensures that the analog signal is accurately represented in a digital form.

  7. 7.What is the effect of using a low-pass filter in a signal conditioning circuit?Application

    Using a low-pass filter in a signal conditioning circuit allows low-frequency signals to pass through while attenuating high-frequency noise. This is beneficial when the desired signal is at a lower frequency and high-frequency noise or interference needs to be minimized to improve signal clarity and accuracy.

  8. 8.Calculate the output voltage of a Wheatstone bridge if R1 = 100 Ω, R2 = 100 Ω, R3 = 100 Ω, and R4 = 105 Ω, with a supply voltage of 10 V.Numerical

    The output voltage (V_out) of a Wheatstone bridge can be calculated using the formula: V_out = (V_s * (R4/(R3+R4) - R2/(R1+R2))). Substituting the given values: V_out = 10 * (105/(100+105) - 100/(100+100)) = 10 * (0.5122 - 0.5) = 0.122 V.

  9. 9.What is the purpose of using a differential amplifier in signal conditioning?Concept

    A differential amplifier is used in signal conditioning to amplify the difference between two input signals while rejecting any signals that are common to both inputs. This is particularly useful in reducing noise and interference that may affect both inputs equally, thereby enhancing the accuracy of the measurement.

  10. 10.If a sensor outputs a signal of 2 mV and the ADC requires a minimum input of 1 V, what gain should the amplifier provide in the signal conditioning circuit?Numerical

    To determine the required gain, divide the ADC's minimum input voltage by the sensor's output voltage. Gain = 1 V / 2 mV = 1000. Therefore, the amplifier should provide a gain of 1000 to ensure the sensor's signal is compatible with the ADC's input requirements.

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