Bridge Circuits: Voltage Divider & AC Bridge Circuits | Sensor & Industrial Instrumentation - Electronics and Communication Engineering (ECE) PDF Download

Bridge Circuits

Resistor temperature compensation can be achieved in voltage divider circuits or bridge circuits. However, when trying to measure small changes in resistance, as required in strain gauges, the signal resolution in bridge circuits is much higher than with voltage dividers. Bridge circuits are used to convert small changes in impedance into voltages. These voltages are referenced to zero, so that the signals can be amplified to give high sensitivity to impedance changes.Bridge Circuits: Voltage Divider & AC Bridge Circuits | Sensor & Industrial Instrumentation - Electronics and Communication Engineering (ECE)

 Voltage Dividers
The two resistive elements R1 and R2 in a strain gauge sensor can be connected in series to form a voltage divider. The elements are driven from a supply, Vs. Since the temperature coefficient of resistance is the same in both elements, the voltage at the junction of the elements, VR, is independent of temperature, and is given by:
Bridge Circuits: Voltage Divider & AC Bridge Circuits | Sensor & Industrial Instrumentation - Electronics and Communication Engineering (ECE)

Bridge Circuits: Voltage Divider & AC Bridge Circuits | Sensor & Industrial Instrumentation - Electronics and Communication Engineering (ECE)

➢ Ac Bridge Circuits
The basic concept of dc bridges can also be extended to ac bridges, the resistive elements are replaced with impedances, as shown in Figure 3.13(a). The bridge supply is now an ac voltage. This type of set up can be used to measure small changes in capacitance as required in the capacitive pressure sensor.
The differential voltage δV across S is then given by:
Bridge Circuits: Voltage Divider & AC Bridge Circuits | Sensor & Industrial Instrumentation - Electronics and Communication Engineering (ECE)where E is the ac supply electromotive force (EMF).
When the bridge is balanced, V = 0, and (3.20) reduces to:
Z2Z3 = Z1Z4.........(3.21)
Or
R2(R3 + j/ωC1) C1) = R1(R4 + J/ωC2)
Because the real and imaginary parts must be independently equal
R3 x R2 = R1 x R4.......(3.22)
and
R2 x C2 = R1 x C1.......(3.23)

Figure 3.13 (a) An ac bridge using block impedances, and (b) a bridge with R and C componentsFigure 3.13 (a) An ac bridge using block impedances, and (b) a bridge with R and C components

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