Bridge Method
EIE 240 Electrical and Electronic Measurement
Class 7, March 13, 2015
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Bridge Method
• Diode bridge is an arrangement of four or
more diodes for AC/DC full-wave rectifier.
• Component bridge methods are used for
measurement of resistance, capacitance,
inductance, etc.
• The network will be balanced when the
detector reading becomes zero.
Component
Being
Measured
Bridge
Network
Detector
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1
Wheatstone Bridge
• Wheatstone bridge was invented by
Samuel Hunter Christie in 1833 and
improved and popularized by Sir Charles
Wheatstone in 1843.
B
R1
• DC supply, Vs
R2
• Output voltage, Vo I1
Vo
A
C
I2
R3
Sir Charles Wheatstone (1802 – 1875)
D
R4
+ Vs –
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Wheatstone Bridge (Cont’d)
• When Vo = 0 V, the potential at B must
equal to the potential at D
I1R1 = I2R3
I1R2 = I2R4
Hence
I1R1 = I2R3 = (I1R2/R4) R3
R1/R2 = R3/R4
• The balance condition is independent of Vs
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2
Wheatstone Bridge (Cont’d)
• R1 is the input resistance to be measured
by comparing to accurately known
resistors (standard)
(standard).
• R2 and R4 are known-fixed resistances.
• R3 can be adjusted to give the zero
potential difference condition.
A
R1
B
Adjust R3
Vo = 0 V G
B
D
Wheatstone Bridge
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Wheatstone Bridge (Cont’d)
• Change in R1, change R3
• The precision is about 1 to 1 M.
M
• The accuracy is mainly up to the known
resistors and the sensitivity of the null
detector ( 0.1 to 0.2%).
• Error comes from changes
g in resistances
of the bridge arms by changes in
temperatures or thermoelectric EMF in
contacts.
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3
Sensitivity of the Bridge
• If no galvanometer at the output,
VAB = Vs R1/(R1+R2)
VAD = Vs R3/(R3+R4)
Thus, Vo = VAB – VAD
Vo = Vs ( R1/(R1+R2) – R3/(R3+R4) )
• The relationship between Vo and R1 is
non-linear
• Vo = 0 V when R1/R2 = R3/R4
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Sensitivity of the Bridge (Cont’d)
• Changing R1 to R1+R1 gives a change of
Vo to Vo+Vo
Vo+Vo=Vs((
((R1+R1))/((R
(( 1+R1))+R2) – R3/(R
( 3+R4))
Then (Vo+Vo)–Vo = Vs
R1+R1 – R3
R1+R1+R2 R3+R4
–Vs R1 – R3
R1+R2 R3+R4
Vo = Vs R1+R1 – R1
R1+R1+R2 R1+R2 8
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Sensitivity of the Bridge (Cont’d)
• If small changes R1 << R1 then the
g can be
sensitivityy of Wheatstone bridge
computed from,
Vo  R1 Vs/(R1+R2)
Vo/R1  Vs/(R1+R2)
• Higher R1 to be measured, lower sensitivity
• Amplifier can be used to amplify Vo
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Unbalanced Bridge
• If there is a galvanometer, Rg, between the
p terminals, the current Ig can be
two output
determined by Thévenin equivalent circuit.
Vth = Vo = Vs ( R1/(R1+R2) – R3/(R3+R4) )
• Sh
Shortt voltage
lt
source, then
th Thévenin
Thé
i
resistance is R1//R2 + R3//R4
Rth = R1R2/(R1+R2) + R3R4/(R3+R4)
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5
Unbalanced Bridge (Cont’d)
A
B
R2
R1
C
B
R3
R4
G

D
Rth Ig
Rg
Vth
Vs
For unbalanced bridge,
Ig = Vth / (Rth + Rg)
and
Vg = IgRg
= VthRg/(Rth+Rg)
If balanced or Vg = 0 V
like no movement
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Slightly Unbalanced Bridge
• If R2 = R3 = R4 = R and R1 = R+R
• VAB = Vs (R+R)/(R+R+R)
= Vs (R+R)/(2R+R)
B
• VAD = Vs R/(R+R) = Vs/2
R+R
• Vo = VAB – VAD
I1
= Vs (R+R) – 1
Vo
A
(2R+R) 2
I2
= VsR / (4R+2R)
R
• If R < 5% of R
D
Vo  VsR / 4R
+ Vs –
R
C
R
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Slightly Unbalanced Bridge (Cont’d)
• For Thévenin’s equivalent circuit,
Vth = Vo
= VsR / 4R
Rth = (R+R)//R + R//R
= R(R+R)/(2R+R) + R/2
 R/2 + R/2 = R
Ig = Vth / (Rth + Rg)
Vg = IgRg
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Kelvin Double Bridge
• A modification of Wheatstone bridge for low
resistance measurement (R1 < 1)
p
wire resistances affect
• Because non-perfect
the measurement.
r
R1
r3
R2
r4
Vo
R3
R4
Vs
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7
• Using four-terminal resistors (two for voltage
supply and 2 for current supply)
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• The yoke r is connected to R1 and R2
• The relationship between r3, r4, R3 and R4
R3/R4 = r3/r4
• Using the delta-star transformation, the
equivalent circuit
ra
R1
rb
R2
rc
ra = r3r / (r
( 3+r4+r))
rb = r4r / (r3+r4+r)
R3
R4
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8
Note
• -Y Transformation
Ra = R1R2 / (R1+R2+R3)
Rb = R2R3 / (R1+R2+R3)
Rc = R1R3 / (R1+R2+R3)
C
C
R1
A
Rc
R3
R2
Ra
B
Rb
A
B
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• The balance condition is the same as
Wheatstone bridge (Null Vo = 0 V)
(R1+ra) / (R2+rb) = R3 / R4
R1 = R3((R2+rb))/R4 – ra
= R3R2/R4 + R3rb/R4 – ra
= R3R2/R4 + R3r4r/(r3+r4+r)R4
– r3r/(r3+r4+r)
= R3R2/R4
+ r4r(R3/R4 – r3/r4) / (r3+r4+r)
R1 = R3R2/R4
Therefore R1/R2 = R3/R4 = r3/r4
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9
High Resistance Bridge
• For very high resistance, e.g. 1,000 M,
there is leakage currents over the surface
p
of the insulated post.
• Using three-terminal
resistors (parallel with 2
R1
R2
leakage resistances)
Rs2
• Rs1>>R3 and Rs1//R3 to
Rs1
avoid
id th
the lleakage
k
effect
ff t
R3
R4
• Rs2 may affect the
detector sensitivity
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Bridge Compensation
• The resistance of long leads will be affected
b changes
by
h
iin ttemperatures
t
• To avoid this, 3 leads are required to
connect to the coils
• They are all the same length and resistance
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10
Bridge Compensation (Cont’d)
• Any changes in lead resistance will affect
all 3 leads equally and occur in 2 arms of
b id and
bridge
d will
ill cancell out.
t
3
R1
R2
Vo
1
2
R4
R3
Vs
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Bridge Controlled Circuits
• The bridge can be used as an error detector
in a control circuit, using the potential
difference at the output of the bridge that is
sensitive to any physical parameters.
• Passive circuit elements such as strain
gauges, temperature sensitive resistors
((thermistors)) or p
photo resistors are used as
one arm of Wheatstone bridge.
• A change in the elements (pressure, heat or
light) causes the bridge to be unbalanced.
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References
• Hotek Technologies, Inc webpage :
http://www hotektech com/
http://www.hotektech.com/
• Yokogawa webpage:
http://tmi.yokogawa.com/us/
• MAGNET LAB – Wheatstone Bridge webpage:
http://www.magnet.fsu.edu/education/tutorials/jav
a/wheatstonebridge/index.html
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