Page 349 - Mechatronics with Experiments
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October 9, 2014 8:1
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                        JWST499-Cetinkunt
            JWST499-c06
                                                                                        SENSORS  335  254mm×178mm
                             The voltage measurement error percentage due to the electrical loading in this case is
                                                    (1000∕2001) − (1∕2)
                                                e =                   ⋅ 100 = 0.0499%           (6.10)
                                                v
                                                          (1∕2)
                                                                                    3
                             For cases where the measurement device input impedance is about 10 times the resistance
                             of the equivalent two port device whose voltage is measured, the voltage measurement error
                             introduced due to the electrical loading effect is negligible.
                                  Let us consider a poor case of loading error condition: R = R = R . Then the
                                                                                       2
                                                                                           m
                                                                                  1
                             mesasured voltage,
                                                              R R ∕(R + R )
                                                               1 m
                                                                         m
                                                                     1
                                                     V = V i                                    (6.11)
                                                      o
                                                           R R ∕(R + R ) + R 2
                                                             1 m
                                                                   1
                                                                       m
                                                               2
                                                              R ∕(2R )
                                                                    1
                                                       = V     1                                (6.12)
                                                           i  2
                                                           R ∕(2R ) + R 1
                                                                 1
                                                             1
                                                              R ∕2
                                                               1
                                                       = V i                                    (6.13)
                                                           R ∕2 + R 1
                                                             1
                                                             1∕2
                                                       = V                                      (6.14)
                                                           i
                                                           1∕2 + 1
                                                             1
                                                       = V ⋅                                    (6.15)
                                                           i
                                                             3
                             The voltage measurement error due to electrical loading error in this case V is and the
                                                                                            0
                             ideal voltage is V ∗
                                           o
                                                               1000
                                                         ∗
                                                        V = V                                   (6.16)
                                                         o    i
                                                               2000
                                                               1
                                                        V = V                                   (6.17)
                                                         0    i
                                                               3
                                                             (       )
                                                              1000  1
                                                                  −
                                                              2000  3
                                                        e =            ⋅ 100                    (6.18)
                                                         v
                                                                1∕2
                                                             1
                                                           =  ⋅ 100                             (6.19)
                                                             3
                                                           = 33.33                              (6.20)
                                  Therefore, in order to minimize the effect of electrical loading errors due to the circuits
                             used for the measurement, the measurement device should have a large input resistance
                             (input impedance). The larger the input impedance, the smaller the electrical loading error
                             in the voltage measurement.
                      6.3 WHEATSTONE BRIDGE CIRCUIT
                             A Wheatstone bridge circuit is used to convert the change in resistance into voltage output.
                             It is a standard circuit used as part of sensor signal conditioners (Figure 6.8). A Wheatstone
                             bridge has
                                1. a power supply voltage, V , and
                                                       i
                                2. four resistances arranged in a bridge circuit, R , R , R , R .
                                                                          2
                                                                       1
                                                                                4
                                                                             3
                             One of the resistance branches is the resistance of the sensor. The sensor resistance changes
                             as a function of the measured variable, i.e. RTD resistance as a function of temperature or
                             strain-gauge resistance as a function of strain.
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