INSTRUMENTATION AND CONTROL(MCT 3224)CONCEPT OF SIGNAL CONDITIONING, DIVIDER CIRCUIT AND BRIDGE CIRCUITNadzril SulaimanDepartment of Mechatronics Engineering,International Islamic University Malaysia

ContentConcept of signal conditioningDivider circuitWheatstone BridgeBridge sensitivityBridge resolutionCurrent balance bridgeWheatstone bridge for potential measurement

Objectives of topic Explain the principles of signal conditioning Identify different elements used for signal conditioning Identify different types of Wheatstone bridge circuit Derive equations related to Wheatstone bridge Apply formulas related to Wheatstone bridge to make necessary calculations

Principles of signal conditioningConvert signal to a form suitable for interface with other elements

Signal level and bias changesThe method to adjust the level (magnitude) and bias (zero value) of voltage signal

Signal level and bias changesAmplificationSensor output is made largerGain greater than unity

AttenuationSensor output is made smallerGain is less than unity

LinearizationOften, the characteristic of a sensor is non-linearSpecial circuits were devised to linearize signalsModern approach is to use computer software to linearize

ConversionsSignal transmissionConverting resistance or voltage to current (4- to 20mA)Voltage-to-current and current-to-voltage converter is essential

Digital InterfaceThe use of computers in process control requires the conversion of analog to digital signal and vice versa- ADC and DAC is needed

FilteringEliminating unwanted signal from the loop using filters

Concept of loadingLoading of one circuit by another circuit produces uncertainties in the amplitude of the voltage of the overall system.Example, when a sensor is connected to a circuit (hence, becoming a load), this will introduce additional voltage across the sensor that affects the amplitude of the total circuit voltage.

Rx

RL

Vx

Sensor

Vy

Concept of loadingConsider:

Open circuit = NO Load connected to the circuitVoltage is VxLoading occur when something is connected across the output.Causing output voltage of the circuit to drop to some valueVy < Vx

Rx

RL

Vx

Sensor

Vy

Concept of loadingThe output voltage is calculated using voltage division as

Output voltage is reduced by the voltage dropTo reduce the uncertainty,

Rx

RL

Vx

Sensor

Vy

Concept of loadingExample 1:An amplifier outputs a voltage that is 10 times the voltage on its input terminals. It has an input resistance of 10 k. A sensor outputs a voltage proportional to temperature with a transfer function of 20 mV/ C. The sensor has an output resistance of 5.0 k. If the temperature is 50C, find the amplifier output; (a) without considering the effect of loading (b) by considering the effect of loading

Signal conditioning circuitElements used for signal conditioning:

1) Divider circuit 2) Wheatstone Bridge 3) Filters 4) Amplifiers

Divider circuitUseful to convert resistance into voltage

Divider circuit

Divider CircuitIt is important to consider the following issuesThe variation of VD with either R1 or R2 is non-linear (ie. if resistance varies linearly with measured variable, divider voltage [VD ] will not vary linearly)The effective output impedance of the divider is the parallel combination of R1 and R2.The current flows to both R1 and R2. So, the power rating of both resistors should be considered.

Divider circuitExample 2:The divider shown has R1 = 10.0 k and Vs = 5.00 V. Suppose R2 is a sensor whose resistance varies from 4.00k to 12.0 k as some dynamic variables varies over a range. Then find (a) the minimum and maximum of VD (b) the range of output impedance (c) the range of power dissipated by R2

Bridge circuitConvert resistance variation into voltage variation

Bridge circuitAdvantages:1) Accurate and reliable2) Widely used in industry3) Higher accuracy (of 0.1%) than ordinary ohmmeter for resistance measurement.

Main application is to determine an unknown resistor.Vary one of the remaining resistors until the current through the null detector is zero.This indicates that the bridge is in balanced condition.

Bridge circuit

Bridge circuit- Voltage across galvanometer offset voltage OR voltage at balance condition OR voltage at null condition - is given by:

Bridge circuitExample 3:

A Wheatstone bridge has a ratio arm of 1/100 (R2/ R1). At first balance, R3 is adjusted to 1000.3. The value of Rx is then changed by the temperature change. The new value of R3 to achieve balance condition again is 1021.1 . Find the change of Rx before and after the temperature change.

Sensitivity of GalvanometerA galvanometer is used to detect unbalance condition Wheatstone bridge

Sensitivity of GalvanometerGalvanometer sensitivity is governed by 1) current sensitivity (current per deflection) 2) internal resistance

Sensitivity of Galvanometer

Sensitivity of GalvanometerProblem 4 (2.10):A Wheatstone bridge has R1 = 250, R3 = 500, R4 = 340, and V = 1.5V. The detector is a galvanometer with RG = 150.

Find the value of R2 that will null the bridge.Find the offset current that will result if R2 = 190.

Bridge resolutionThe resolution of the bridge circuit is a function of the resolution of the detector used to determine the bridge offset.

If a detector can measure a change of 100V, this sets a limit on the minimum measurable resistance change in a bridge using this detector.

Bridge resolutionExample 5 (2.8) : A sensor with a nominal resistance of 50 is used in a bridge with R1=R2= 100, V=10V, and R3=100 potentiometer. It is necessary to resolve 0.1 changes of the sensor resistance.

At what value of R3 will the bridge null?What voltage resolution must the null detector possess?

Current balance bridgeSometimes, obtaining a null by varying resistance is not efficientThus, the bridge is nulled by changing current, IIf R4 >> R5 or (R2 + R4) >> R5 thenV = Va - Vb

Current balance bridgeExample 6 (2.11) :A current balance bridge, has resistance of R1 = R2 = 1k, R4 = 590, R5 = 10, and V = 10V.

Find the value of R3 that nulls the bridge with no current.Find the value of R3 that balances the bridge with a current of 0.25mA.

Bridge for potential measurementA bridge can also measure small potential at a very high impedance, using either conventional or current bridge.The potential to be measured is connected in series with the detector.If current balance bridge is used

Bridge for potential measurementExample 7 (2.12) :

A potential measurement bridge has V = 10V,R1 = R2 = R3 = 10k. Find the unknown potential if the bridge nulls with R4 = 9.73k.

Bridge characteristicNull voltage is non-linear for large changes in resistanceNull voltage is nearly linear for small changes in resistance

Bridge ends

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