E1C06 09/14/2010
11:55:5 Page 223
the bridge becomes
E o þ dE o ¼ E i
R
0
1
R
0
1 þ R 2
À
R 3
R 3 þ R 4
¼ E i
R
0
1 R 4 À R 3 R 2
R
0
1 þ R 2
À
Á R 3 þ R 4
ð
Þ
ð6:15Þ
In many designs the bridge resistances are initially equal. Setting R 1 ¼ R 2 ¼ R 3 ¼ R 4 ¼ R allows
Equation 6.15 to be reduced to
dE o
E i
¼
dR=R
4 þ 2 dR=R
ð
Þ
ð6:16Þ
In contrast to the null method of operation of a Wheatstone bridge, the deflection bridge
requires a meter capable of accurately indicating the output voltage, as well as a stable and known
input voltage. But the bridge output should follow any resistance changes over any frequency input,
up to the frequency limit of the detection device! So a deflection mode is often used to measure timevarying signals.
If the high-impedance voltage measuring device in Figure 6.14 is replaced with a relatively lowimpedance current measuring device and the bridge is operated in an unbalanced condition, a
current-sensitive bridge circuit results. Consider Kirchhoff’s laws applied to the Wheatstone bridge
circuit for a current-sensing device with resistance R g . The input voltage is equal to the voltage drop
in each arm of the bridge,
E i ¼ I 1 R 1 þ I 2 R 2
ð6:17Þ
but I 2 ¼ I 1 À I g , which gives
E i ¼ I 1 R 1 þ R 2
ð
ÞÀI g R 2
ð6:18Þ
If we consider the voltage drops in the path through R 1 , R g , and R 3 , the total voltage drop must
be zero:
I 1 R 1 þ I g R g À I 3 R 3 ¼ 0
ð6:19Þ
For the circuit formed by R g , R 4 , and R 2 ,
I g R g þ I 4 R 4 À I 2 R 2 ¼ 0
ð6:20Þ
or with I 2 ¼ I 1 À I g and I 4 ¼ I 3 þ I g ,
I g R g þ I 3 þ I g
À
Á R 4 À I 1 À I g
À
Á R 2 ¼ 0
ð6:21Þ
Equations 6.18 to 6.21 form a set of three simultaneous equations in the three unknowns I 1 , I g , and I 3 .
Solving these three equations for I g yields
I g ¼
E i R 1 R 4 À R 2 R 3
ð
Þ
R 1 R 4 R 2 þ R 3
ð
ÞþR 2 R 3 R 1 þ R 4
ð
ÞþR g R 1 þ R 2
ð
ÞR 3 þ R 4
ð
Þ
ð6:22Þ
and E o ¼ I g R g . Then, the change in resistance of R 1 can be found in terms of the bridge deflection
voltage, E o , by
dR
R 1
¼
R 3 =R 1
ð
ÞE o =E i þ R 2 = R 2 þ R 4
ð
Þ
½
1 À E o =E i À R 2 = R 2 þ R 4
ð
Þ
À 1
ð6:23Þ
6.4 Analog Devices: Resistance Measurements 223
11:55:5 Page 223
the bridge becomes
E o þ dE o ¼ E i
R
0
1
R
0
1 þ R 2
À
R 3
R 3 þ R 4
¼ E i
R
0
1 R 4 À R 3 R 2
R
0
1 þ R 2
À
Á R 3 þ R 4
ð
Þ
ð6:15Þ
In many designs the bridge resistances are initially equal. Setting R 1 ¼ R 2 ¼ R 3 ¼ R 4 ¼ R allows
Equation 6.15 to be reduced to
dE o
E i
¼
dR=R
4 þ 2 dR=R
ð
Þ
ð6:16Þ
In contrast to the null method of operation of a Wheatstone bridge, the deflection bridge
requires a meter capable of accurately indicating the output voltage, as well as a stable and known
input voltage. But the bridge output should follow any resistance changes over any frequency input,
up to the frequency limit of the detection device! So a deflection mode is often used to measure timevarying signals.
If the high-impedance voltage measuring device in Figure 6.14 is replaced with a relatively lowimpedance current measuring device and the bridge is operated in an unbalanced condition, a
current-sensitive bridge circuit results. Consider Kirchhoff’s laws applied to the Wheatstone bridge
circuit for a current-sensing device with resistance R g . The input voltage is equal to the voltage drop
in each arm of the bridge,
E i ¼ I 1 R 1 þ I 2 R 2
ð6:17Þ
but I 2 ¼ I 1 À I g , which gives
E i ¼ I 1 R 1 þ R 2
ð
ÞÀI g R 2
ð6:18Þ
If we consider the voltage drops in the path through R 1 , R g , and R 3 , the total voltage drop must
be zero:
I 1 R 1 þ I g R g À I 3 R 3 ¼ 0
ð6:19Þ
For the circuit formed by R g , R 4 , and R 2 ,
I g R g þ I 4 R 4 À I 2 R 2 ¼ 0
ð6:20Þ
or with I 2 ¼ I 1 À I g and I 4 ¼ I 3 þ I g ,
I g R g þ I 3 þ I g
À
Á R 4 À I 1 À I g
À
Á R 2 ¼ 0
ð6:21Þ
Equations 6.18 to 6.21 form a set of three simultaneous equations in the three unknowns I 1 , I g , and I 3 .
Solving these three equations for I g yields
I g ¼
E i R 1 R 4 À R 2 R 3
ð
Þ
R 1 R 4 R 2 þ R 3
ð
ÞþR 2 R 3 R 1 þ R 4
ð
ÞþR g R 1 þ R 2
ð
ÞR 3 þ R 4
ð
Þ
ð6:22Þ
and E o ¼ I g R g . Then, the change in resistance of R 1 can be found in terms of the bridge deflection
voltage, E o , by
dR
R 1
¼
R 3 =R 1
ð
ÞE o =E i þ R 2 = R 2 þ R 4
ð
Þ
½
1 À E o =E i À R 2 = R 2 þ R 4
ð
Þ
À 1
ð6:23Þ
6.4 Analog Devices: Resistance Measurements 223