Test Ideas: Linearize measurements from bridge circuits
Linearizing eases the conversion from temperature to voltage prior to digitizing.
By Camilo Quintáns-Graña and Jorge Marcos-Acevedo, University of Vigo, Spain -- Test & Measurement World, 3/1/2010 12:00:00 AM
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Bridge circuits have long been popular for conditioning signals from resistive sensors. These circuits are sensitive to small changes in resistance, and they provide a differential output from a single current or voltage source. But the sensors you connect to a passive bridge with one measuring branch don’t produce linear outputs.
Temperature sensors such as RTDs produce small resistance changes as a function of temperature. You can linearize a bridge circuit’s output by adding external linearizing circuits. But adding op amps to linearize the output means you’ll need a bipolar power supply. The circuit in Figure 1 represents an active bridge providing a linear voltage output using a unipolar power supply.
 Figure 1. Amplifier U2 produces a single-ended voltage output from the voltage across the bridge circuit. |
The circuit uses the popular Pt100 RTD, which has a resistance of 100 Ω at 0°C. Its temperature coefficient of 0.00385 Ω/Ω/ºC produces a 38.5-Ω increase in resistance from 0°C to 100°C. Thus, the resistance is 138.5 Ω at 100°C.
In Figure 1, resistors R3 and R4 convert the output from the 3-V power supply into two 1-mA constant currents, one in each branch of the bridge. At 0°C, the bridge is balanced and, thus, V1 – V2 = 0 V. The equation below describes the circuit’s output:
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Figure 2 shows a simulation of the circuit’s output from 0°C to 50°C. If you use a passive bridge that produces a nonlinear resistance, then you still need linearizing circuits prior to digitizing the circuit’s analog output or you need to linearize the output in software after digitizing.
Resistive sensors require excitation current, which this circuit provides. A Pt100 RTD requires 1 mA of excitation current to get its specified performance. As the bridge voltage is 3 V, you get 1 mA through the RTD, for which the circuit consumes 0.1 mW in the sensor. That low power dissipation minimizes self-heating, which can affect measurement accuracy.
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this circuit is indicated for resistive sensor, if you want to adapt a sensor with output in source current mode, perhaps you need other kind of conditioning circuit. If you provide me more information about your sensor and application, I could help you better.
thank for your question
Camilo
Camilo Quintans - 2010-7-11 05:29:45 EST -
How could I modify this circuit for an anemometer application for a current of 75 mA through each arm of the bridge? Thank in anticipation of your response
Bob Bhatia - 2010-24-7 00:02:10 EDT
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