Electrical Pressure Measurements, Strain Gauges & Transducers
Introduction to Electrical Measurement Systems
Electrical measurement systems form the backbone of modern instrumentation and control systems. They convert physical parameters like pressure, strain, temperature, and displacement into electrical signals that can be processed, displayed, and used for control purposes.
Electrical measurements provide numerous advantages over mechanical methods, including higher accuracy, remote measurement capability, easier data processing, and compatibility with digital systems.
Basic Components of an Electrical Measurement System
1. Sensor/Transducer
Converts the physical parameter into an electrical signal. Examples include strain gauges, thermocouples, and pressure sensors.
2. Signal Conditioning
Modifies the electrical signal for better compatibility with measurement systems. Includes amplification, filtering, and linearization.
3. Signal Processing
Manipulates the signal to extract useful information. May involve A/D conversion, data analysis, and computation.
4. Output/Display
Presents the measurement results in a useful format, such as digital displays, recording devices, or control signals.
Key Principles of Electrical Measurements
- Accuracy: The closeness of a measurement to the true value
- Precision: The repeatability of measurements
- Resolution: The smallest change that can be detected
- Sensitivity: The ratio of output change to input change
- Range: The span between minimum and maximum measurable values
- Linearity: The consistency of sensitivity across the measurement range
Strain Gauges
Strain gauges are sensors that measure strain (deformation) on an object. When a strain gauge is attached to an object and the object deforms, the strain gauge also deforms, causing its electrical resistance to change. This resistance change can be measured and related to the strain.
Working Principle
Strain gauges operate on the principle of piezoresistance: the electrical resistance of a conductor changes when it is subjected to mechanical strain.
Types of Strain Gauges
- Metal Foil Gauges: Most common, made of thin metal foil patterns
- Semiconductor Gauges: Higher sensitivity but affected by temperature
- Wire Gauges: Older technology, made of fine wire
- Optical Fiber Gauges: Based on fiber optic technology for harsh environments
The gauge factor (GF) relates the relative change in resistance to the strain:
GF = (ΔR/R) / ε
Where:
- GF = gauge factor
- ΔR = change in resistance
- R = original resistance
- ε = strain (ΔL/L)
Wheatstone Bridge Configuration
Strain gauges are typically connected in a Wheatstone bridge circuit to measure small resistance changes accurately.
Wheatstone Bridge Output Voltage:
Vout = Vin × [(R3/(R3+R4)) - (R2/(R1+R2))]
For a balanced bridge (R1/R2 = R4/R3), Vout = 0
Bridge Configurations
| Configuration | Description | Sensitivity |
|---|---|---|
| Quarter Bridge | One active gauge in the bridge | Low (ΔR/R) |
| Half Bridge | Two active gauges in the bridge | Medium (2 × ΔR/R) |
| Full Bridge | All four gauges active | High (4 × ΔR/R) |
Temperature compensation is crucial for accurate strain measurements. This can be achieved by using dummy gauges, self-temperature-compensated gauges, or software correction.
Transducers
A transducer is a device that converts one form of energy to another. In the context of electrical measurement systems, transducers typically convert physical quantities (like pressure, temperature, or displacement) into electrical signals.
Classification of Transducers
By Energy Conversion
- Active Transducers: Generate electrical output without external power (e.g., thermocouples, piezoelectric sensors)
- Passive Transducers: Require external power to operate (e.g., strain gauges, RTDs)
By Parameter Measured
- Pressure Transducers
- Temperature Transducers
- Force/Strain Transducers
- Displacement Transducers
- Flow Transducers
- Level Transducers
Common Transducer Technologies
These transducers operate based on resistance change. Examples include:
- Strain Gauges: Change resistance with applied strain
- RTDs (Resistance Temperature Detectors): Change resistance with temperature
- Potentiometers: Variable resistors used for displacement measurement
For RTDs: R = R₀[1 + α(T - T₀)]
Where:
- R = resistance at temperature T
- R₀ = resistance at reference temperature T₀
- α = temperature coefficient of resistance
These transducers work on the principle of capacitance change. The capacitance can change due to:
- Change in distance between plates
- Change in overlapping area of plates
- Change in dielectric constant
Applications include pressure sensors, displacement sensors, and humidity sensors.
C = ε₀εᵣA/d
Where:
- C = capacitance
- ε₀ = permittivity of free space
- εᵣ = relative permittivity of dielectric
- A = area of plates
- d = distance between plates
These transducers operate based on inductance change or induced EMF. Examples include:
- LVDT (Linear Variable Differential Transformer): For precise linear displacement measurement
- RVDT (Rotary Variable Differential Transformer): For angular displacement
- Proximity Sensors: Detect metal objects without contact
These transducers generate an electrical charge when subjected to mechanical stress. They are used for:
- Dynamic pressure measurement
- Vibration measurement
- Acceleration measurement
- Force measurement
Advantages include high sensitivity, wide frequency response, and self-generating operation (no power supply needed).
Piezoelectric transducers are suitable only for dynamic measurements as the charge leaks over time, making them unsuitable for static measurements.
Transducer Characteristics
| Characteristic | Description |
|---|---|
| Sensitivity | Output change per unit input change |
| Range | Minimum to maximum measurable values |
| Resolution | Smallest detectable change in input |
| Linearity | Consistency of sensitivity across range |
| Hysteresis | Difference in output for same input during increasing vs. decreasing values |
| Response Time | Time required to reach a percentage of final output after step input |
| Frequency Response | Range of frequencies over which transducer operates within specifications |
Electrical Pressure Measurements
Pressure is the force exerted per unit area and is a fundamental parameter in many industrial processes. Electrical pressure measurements convert pressure into electrical signals using various transduction methods.
Pressure Measurement Basics
Pressure Types
- Absolute Pressure: Measured relative to perfect vacuum (0 pressure)
- Gauge Pressure: Measured relative to atmospheric pressure
- Differential Pressure: Difference between two pressure points
- Vacuum Pressure: Pressure below atmospheric pressure
Pressure Units
| Unit | Symbol | Relation to Pa |
|---|---|---|
| Pascal | Pa | 1 Pa = 1 N/m² |
| Bar | bar | 1 bar = 10⁵ Pa |
| Pounds per Square Inch | psi | 1 psi ≈ 6895 Pa |
| Atmosphere | atm | 1 atm ≈ 101325 Pa |
| Torr (mmHg) | Torr | 1 Torr ≈ 133.3 Pa |
Electrical Pressure Sensors
These sensors use strain gauges attached to a diaphragm. When pressure is applied, the diaphragm deforms, causing the strain gauges to change resistance.
Typically arranged in a Wheatstone bridge configuration for temperature compensation and higher sensitivity.
These sensors use a flexible diaphragm as one plate of a capacitor. As pressure changes, the diaphragm deforms, changing the distance between plates and thus the capacitance.
- Advantages: High sensitivity, good stability, low hysteresis
- Disadvantages: Complex circuitry, affected by stray capacitance
C = ε₀εᵣA/d
As pressure increases, d decreases, and C increases
These sensors use piezoelectric materials that generate electrical charge when deformed by pressure.
- Best suited for dynamic pressure measurements
- High frequency response (up to hundreds of kHz)
- Not suitable for static pressure due to charge leakage
Applications include blast pressure measurement, engine combustion analysis, and hydraulic pulsation monitoring.
Microelectromechanical systems (MEMS) pressure sensors are miniature devices that integrate mechanical elements and electronics on a silicon chip.
- Small size and low cost
- Good accuracy and repeatability
- Low power consumption
- Easy integration with electronic circuits
Applications include consumer electronics (smartphones, wearables), automotive (tire pressure monitoring), and medical devices.
Signal Conditioning for Pressure Sensors
Amplification
Most pressure sensors produce small output signals (mV range) that need amplification to be compatible with measurement systems.
Instrumentation amplifiers are commonly used due to their high CMRR and input impedance.
Linearization
Many pressure sensors have non-linear response. Linearization can be achieved through:
- Analog circuits
- Digital look-up tables
- Mathematical functions
Temperature Compensation
Pressure sensors are affected by temperature. Compensation methods include:
- Bridge circuit with temperature-sensitive elements
- Software correction using temperature measurement
- Use of self-compensating sensors
Filtering
Filtering removes unwanted noise from pressure signals:
- Low-pass filters for high-frequency noise
- Digital filtering for specific frequency bands
- Averaging for random noise reduction
Electrical Measurement Calculators
Strain Gauge Calculator
Results
Strain (ε): microstrain
Change in Resistance (ΔR): Ω
Relative Change in Resistance (ΔR/R):
Wheatstone Bridge Calculator
Results
Output Voltage (Vout): V
Bridge Balance:
Pressure Conversion Calculator
Results
Converted Value:
Sensitivity & Error Calculator
Results
Sensitivity:
Percentage of Full Scale: %
Examples and Practice Problems
Example 1: Strain Gauge Calculation
Problem: A strain gauge with an initial resistance of 120 Ω and a gauge factor of 2.1 experiences a strain of 500 microstrain (μɛ). Calculate the new resistance of the gauge.
Solution:
Given:
- Initial resistance, R = 120 Ω
- Gauge factor, GF = 2.1
- Strain, ε = 500 μɛ = 500 × 10⁻⁶
Using the gauge factor formula: GF = (ΔR/R) / ε
Rearranging: ΔR/R = GF × ε
ΔR/R = 2.1 × (500 × 10⁻⁶) = 1.05 × 10⁻³
ΔR = R × 1.05 × 10⁻³ = 120 × 1.05 × 10⁻³ = 0.126 Ω
New resistance = R + ΔR = 120 + 0.126 = 120.126 Ω
Answer: The new resistance of the strain gauge is 120.126 Ω.
Example 2: Wheatstone Bridge Output
Problem: A Wheatstone bridge has the following resistances: R₁ = 120 Ω, R₂ = 120 Ω, R₃ = 120 Ω, and R₄ = 120.5 Ω. If the excitation voltage is 10 V, calculate the output voltage.
Solution:
Given:
- R₁ = 120 Ω
- R₂ = 120 Ω
- R₃ = 120 Ω
- R₄ = 120.5 Ω
- Excitation voltage, Vₑₓ = 10 V
Using the Wheatstone bridge output voltage formula:
Vₒᵤₜ = Vₑₓ × [(R₃/(R₃+R₄)) - (R₂/(R₁+R₂))]
Vₒᵤₜ = 10 × [(120/(120+120.5)) - (120/(120+120))]
Vₒᵤₜ = 10 × [0.4988 - 0.5]
Vₒᵤₜ = 10 × [-0.0012]
Vₒᵤₜ = -0.012 V = -12 mV
Answer: The output voltage of the Wheatstone bridge is -12 mV.
Example 3: Pressure Transducer Calibration
Problem: A pressure transducer has the following calibration data:
| Pressure (kPa) | Output (mV) |
|---|---|
| 0 | 0 |
| 100 | 20 |
| 200 | 40 |
| 300 | 60 |
| 400 | 80 |
| 500 | 100 |
a) Determine the sensitivity of the transducer.
b) Calculate the pressure when the output is 55 mV.
Solution:
a) Sensitivity calculation:
Sensitivity = Change in output / Change in input
From the data, when pressure changes from 0 to 500 kPa, output changes from 0 to 100 mV.
Sensitivity = (100 - 0) mV / (500 - 0) kPa = 0.2 mV/kPa
b) Pressure calculation when output is 55 mV:
Using the linear relationship: Pressure = Output / Sensitivity
Pressure = 55 mV / 0.2 mV/kPa = 275 kPa
Answer:
a) The sensitivity of the transducer is 0.2 mV/kPa.
b) The pressure when the output is 55 mV is 275 kPa.
Example 4: Temperature Effect on Strain Gauge
Problem: A strain gauge with a gauge factor of 2.0 and a temperature coefficient of resistance of 3.9 × 10⁻³/°C experiences a temperature change of 20°C. If no mechanical strain is applied, calculate the apparent strain due to temperature change.
Solution:
Given:
- Gauge factor, GF = 2.0
- Temperature coefficient of resistance, α = 3.9 × 10⁻³/°C
- Temperature change, ΔT = 20°C
The change in resistance due to temperature is given by:
ΔR/R = α × ΔT
ΔR/R = 3.9 × 10⁻³/°C × 20°C = 0.078
The apparent strain can be calculated from the gauge factor formula:
GF = (ΔR/R) / ε
Rearranging: ε = (ΔR/R) / GF
ε = 0.078 / 2.0 = 0.039 = 39,000 μɛ
Answer: The apparent strain due to temperature change is 39,000 μɛ (microstrain).
Note: This example illustrates the importance of temperature compensation in strain gauge measurements.
Practice Problems
Problem: A quarter-bridge strain gauge circuit has three fixed resistors of 120 Ω each. The active strain gauge has an initial resistance of 120 Ω and experiences a tensile strain of 1000 μɛ. If the gauge factor is 2.0 and the excitation voltage is 5 V, calculate the output voltage.
Answer: 2.5 mV
Problem: A pressure transducer has a range of 0-100 bar with an output of 4-20 mA. The accuracy is specified as ±0.25% of full scale. Calculate the maximum error in pressure measurement and the maximum error in output current.
Answer: ±0.25 bar, ±0.04 mA
Problem: An LVDT has a sensitivity of 40 mV/mm and a linear range of ±25 mm. The output voltage is 500 mV. Determine the displacement of the core from the null position.
Answer: 12.5 mm
