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04-1 The Wheatstone Bridge *
* 04-1
Raison d'^etre: convert tiny changes in resistance to voltage.
Shown with an instrumentation amplifier.
Like an ideal op-amp but with finite gain.
Gain of instrumentation amplifier denoted by A.
vo = A(v+ vi).
The Wheatstone bridge consists of four arms.
RB RD
vo = A ______________ ______________ vE .
RA + RB RC + RD
04-1 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-2 *
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Transducer can be placed in one, two, or four arms.
Typical function: Ht(x) = R(1 + xk), xk o 1
where R is the nominal resistance of the transducer and k is a con-
stant.
For simplicity write function as: Ht(x) = R + Rs,
where R is independent of the process-variable value and Rs is dependent
on the process-variable value.
Typically, R AE Rs.
Usually, need to convert Rs to a voltage.
04-2 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-3 *
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Complementary Pairs
Frequently, transducer pairs can have complementary responses.
If so, there are two (usually identical) transducers: : :
: : :positioned so they react oppositely to the process variable: : :
: : :so that when their responses are subtracted: : :
: : :their response to the process variable add: : :
: : :and unwanted quantities cancel out.
For example, consider:
Ht1 (x) = R(1 + xk) and Ht2 (x) = R(1 xk).
Sum: Ht1 (x) + Ht2 (x) = R. (Not helpful.)
Difference: Ht1 (x) Ht2 (x) = 2xk. (Much better.)
04-3 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-4 *
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One-Transducer Configuration
Arm B: Ht(x) = R + Rs = R(1 + xk).
Other Arms: Resistor of value R.
Rs Rs xk
vo = A _________________ vE ss A _____vE = A ____vE .
2(2R + Rs) 4R 4
04-4 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-5 *
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Two-Transducer Configuration
Arm A: Ht2 (x) = R Rs = R(1 xk).
Arm B: Ht1 (x) = R + Rs = R(1 + xk).
Other Arms: Resistor of value R.
Rs xk
vo = A _____vE = A ____vE .
2R 2
As one might expect, twice as sensitive.
04-5 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-6 *
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Four-Transducer Configuration
Arms A and D: Ht2 (x) = R Rs = R(1 xk).
Arms B and C: Ht1 (x) = R + Rs = R(1 + xk).
Rs
vo = A ____vE = AxkvE .
R
04-6 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-7 Wheatstone Bridge Transfer Functions *
* 04-7
Goal
Let Rt = R Rs = R(1 xk) be the transducer response(s).
Assume bridge designed properly.
Need to find two functions:
Hc(Rt) = : : : and Hc(Rs) = : : :
Both functions are equivalent.
Choose whichever is more convenient.
Four-Transducer Configuration
Hc(Rs) = vo = A Rs__RvE .
Let Rt = R + Rs. Then Rs = Rt R.
Hc(Rt) = A Rt__R_____RvE = A Rt__R 1 vE .
Two-Transducer Configuration
Hc(Rs) = vo = A__2Rs__RvE .
Hc(Rt) = A__2Rt__R_____RvE = A__2 Rt__R 1 vE .
One-Transducer Configuration
Hc(Rs) = vo = A__4Rs__RvE .
Hc(Rt) = A__4Rt__R_____RvE = A__4 Rt__R 1 vE .
04-7 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-8 Wheatstone Bridge Sample Problem *
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Design a system with output vo = H (x), where process variable x is
strain and, x 2 [0; 105 ], and H (x) = 106 x V.
Strain will be covered in more detail later.
For now, all we need to know is that strain is dimensionless.
Strain is measured by a strain gauge.
Strain gauges frequently used in complementary pairs.
Use strain gauges with response:
Ht(- ) = R(1 + - Gf ),
where - denotes strain and
constant Gf = 2.
(Gf called gauge factor, a dimensionless quantity.)
Position the two strain gauges to obtain response:
Ht(x) = R(1 + xGf ) and Ht0(x) = R(1 xGf ).
04-8 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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04-9 *
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Derivation of Conditioning Circuit Needed
A Wheatstone bridge is the obvious choice because transducer response is
in form R Rs.
Nevertheless, conditioning-circuit derivation will be presented.
H(x) = Hc(Ht(x))
(Analysis performed as though there were one transducer.)
y_ 1
y = Ht(x) = Rt = R(1 + xGf ). Then x = _R______G.
f
y_ 1 _y_ 1
Then Hc(y) = H R_______G = 106 R_______ V.
f Gf
Response for two-transducer configuration: Hc(Rt) = A__2 Rt__R 1 vE .
6
Choose A and vE so that A__2vE = 10___GV is satisfied.
f
For example, vE = 10 V and A = 105 .
04-9 EE 4770 Lecture Transparency. Formatted 13:06, 22 January 1999 from lsli0*
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| David M. Koppelman - koppel@ee.lsu.edu | Modified 22 Jan 1999 13:08 (19:08 UTC) |