11.1 Use Case

A multitude of electronic applications requires the measurement and control of physical quantities, such as temperature, humidity, light, air pressure, and so on. The physical quantities are converted into electrical quantities with the help of transducers. The output of a transducer is an electrical signal representing the measured variable. Usually, the transducer’s output signal levels are very low and need to be amplified before they are sampled in the data acquisition process. The electric signals are further processed and monitored to offer adequate actions based on their changes. For a general block diagram of a simple measurement and control system see Figure 11-2. The amplifiers used for amplifying electrical signals relating to a physical quantity come under the commonly used term of instrumentation amplifiers. Hence the input to an instrumentation amplifier is the output signal from a transducer.

Transducers can be classified in different ways. However, an important factor to consider is whether a power source is required or not for their operation. Active transducers do not need an external power source in their operation. They are self-generating the output signal and operate under the energy conversion principle (i.e., photovoltaic, thermoelectric, electromagnetic, piezoelectric, etc.). On the other hand, passive transducers need an external source of energy for their operation. They produce the output signal in the form of variation in an electrical parameter such as resistance, capacitance or inductance.

The Wheatstone bridge (or resistance bridge) circuit can be used to interface various resistive passive transducers to instrumentation amplifiers (Figure 11-3).

In this circuit, R_A, R_B and R_C are known and given. R_T represents the resistance of the transducer and varies depending on the physical quantity that changes over time. The values for the resistors are chosen for a specific point, which allows for the bridge to be balanced.

When balanced, $\frac{R_C}{R_B}=\frac{R_T}{R_A}=1$, the output voltage of the Wheatstone bridge (and the differential input to the instrumentation amplifier) is zero, thus the output of the amplifier is zero.

When there is a change in the physical quantity being measured, the voltage V2 will no longer be equal to V1. The resistance of the transducer device changes from R_T to R_T ± ΔR. This produces a differential voltage input for the instrumentation amplifier, and the output of the amplifier will no longer be zero.

$V_1=\frac{R_B}{R_B+R_C}\times V_{DD}$

$V_2=\frac{R_A}{R_A+R_T\pm \mathrm{\Delta }R}\times V_{DD}$

Assuming the resistance are chosen to be of same value, i.e.: R_A = R_B = R_C = R_T = R, the differential voltage input becomes:

$V_2-V_1=\frac{R}{2R}\times V_{DD}-\frac{R}{2R\pm \mathrm{\Delta }R}\times V_{DD}$

$V_2-V_1=\frac{\pm \mathrm{\Delta }R}{2(2R\pm \mathrm{\Delta }R)}\times V_{DD}$

If the change in resistance ΔR, is much smaller than 2R (ΔR << 2R) the equation can be simplified to:

$V_2-V_1=\frac{\pm \mathrm{\Delta }R}{4R}\times V_{DD}$

It follows that the output from the instrumentation amplifier can be expressed as:

$V_{OUT}=Gain\times \frac{\pm \mathrm{\Delta }R}{4R}\times V_{DD}$

An important aspect, which needs to be considered is that the differential voltage, V₂-V₁, has to be positive at all times. This is a consequence of the op amp rail supply voltage of 0V for the negative rail and +V_DD for the positive rail. Hence care must be taken when positioning the transducer in the Wheatstone bridge configuration. The transducer position depends on the negative or positive variation of the resistance when excited.