B RIDGE C IRCUIT T ECHNIQUES

B RIDGE C IRCUIT T ECHNIQUES

Bridge circuits use resistive elements to sense measurement changes. Depending on how the circuit is configured, the bridge will change the voltage or current of its output in proportion to changes in its resistive measurement element. This resistance change generally creates a bridge imbalance . Under normal (balanced) operation of a bridge circuit, the current that passes through one section of a current-sensitive bridge is the same as that in the other section (or in a voltage-sensitive bridge, the voltage differential between the two sections is zero). An imbalance occurs when the resistance of one element changes, thus creating a current or voltage offset that is proportional to the resistance change. The bridge circuit utilizes this offset measurement to determine the value of the measured variable. Figure 13-3 shows a bridge circuit.

Output

Figure 13-3. Simple bridge circuit.

Voltage-Sensitive Bridge. A voltage-sensitive bridge senses a voltage differential at the output of the bridge that is proportional to the resistance change in the bridge. Figure 13-4 illustrates a voltage-sensitive bridge, where

D is the detector device and R D is its resistance. The value of R D for a voltage- sensitive bridge is very high. This amount of resistance could be provided by the input impedance of an amplifier module of a PLC. The following example illustrates the relationship between the resistors in a voltage-sensitive bridge

circuit. Note that a change in the resistance of R 4 (the measuring element) creates the bridge imbalance; the other resistors have fixed, known values.

E X AM PLE 1 3 -3

For the voltage-sensitive circuit shown in Figure 13-4, find (a) the equation that describes the voltage differential measurement

between point A and point B and (b) the bridge resistance ratio when

the voltage differential is 0 (balanced state).

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24 VDC

Figure 13-4. Voltage-sensitive bridge circuit.

S OLU T I ON

(a) Assuming that R D = ∞ (i.e., it is very large) and the excitation voltage impedance ( R i ) equals 0, the voltages at points V A and V B are:

 R 1 + R 3 

 R 2 + R 4 

The voltage differential between points A and B is:

(b) When the differential voltage ( ∆ V ) is 0, V A equals V B , so:

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Therefore, the bridge resistance ratio is: R 1 = R 3

where R 4 is the measuring resistance element.

Current-Sensitive Bridge.

A current-sensitive bridge creates a current flow change through the output of the bridge, that is, between point A and point B

(refer to Figure 13-4). The current flow is the result of a bridge imbalance created by resistance changes in the measuring element. The other resistors in the bridge have known, fixed values.

When current changes are being measured, the detecting device D has a very low resistance R D , allowing current to flow from point A to B through the detector. Typical devices that have very low impedance include galvanom- eters and low-input impedance current amplifier interfaces (PLC modules).

The following equation describes the current that flows through a current- sensitive bridge’s detector as a result of a bridge imbalance:

VR 4

+ R + R 2 + R 4 B R D 1 + R ++ R 3 R 4 B

The term R 4B is the resistance value when the bridge is balanced. The following example illustrates how this equation is used to obtain a current proportional to the change in resistance.

E X AM PLE 1 3 -4

A bridge circuit uses a thermistor with a nominal resistance of 10 Ω to measure small changes in temperature (see Figure 13-5). An amplifier input module, which has an input impedance of 300 Ω , measures small changes in current. What is the current if a change in temperature results in a 10% change in resistance?

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The resistance of the thermistor ( R 4 ) changes 10% due to temperature change, which translates into an R 4 value of 11 Ω (10 Ω +1 Ω ). The term ∆ R 4 is the absolute value of the difference between R 4 B and the new value of R 4 due to the measurement change. Therefore, the difference

in thermistor resistance is calculated as:

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Figure 13-5. Bridge circuit.

= 10 K Ω − 11 K Ω = 1 K Ω

The current measurement is defined by (values of R in K Ω ):

VR ∆

= 24 ( . )( 18 2 20 675 . ) = . 0 06378 mA