bridgeppt.docx
TRANSCRIPT
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2102311 Electrical Measurement a
Instruments (Part II)
Bridge Circuits (DC and AC)
Electronic Instruments (Analog &
Digital) Signal Generators
Frequency and Time IntervalMeasuremen Introduction to Transducers
Textbook:
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Resistor TypesImportance parameters
Value Tolerance
Power rating Temperature c
Type Values () Power rating Tolerance (%) Temperature
(W) coefficient
(ppm/C)
Wire wound 10m~3k 3~1k 1~10 30~300(power)
Wire wound10m~1M 0.1~1 0.005~1 3~30
(precision)
Carbon film 1~1M 0.1~3 2~10 100~200
Metal film 100m~1M 0.1~3 0.5~5 10~200
Metal film10m~100k 0.1~1 0.05~5 0.4~10
(precision)
Metal oxide film 100m~100k 1~10 2~10 200~500
Data: Transistor technology (10/2000)
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Resistor Values Color codes
Alphanumeric4 ban
Color Digit Multiplier Tolerance Temperature
(%) coefficient
(ppm/C)
Silver - 10-2 10 K - -Gold 10
-1 5 J
Black 0 10
0
- - 250 K
Brown 1 101 1 F 100 H
Red 2 102 2 G 50 G
Orange 3 103
-- 15 D
Yellow 4 104 - 25 F
Green 5 105 0.5 D 20 E
Blue 6 106 0.25 C 10 C
Violet 7 107 0.1 B 5 B
Gray 8 108
-- 1 A
White 9 109 -
- -- 20 M
:
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Most sig. fig. of value
Least siof value
Ex.
Green
Bl
R=
Alph
R, K, M, G, and
x100, x10
3, x
Ex. 6M8 =
5904
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Resistor Values Commonly available resistance
R= x%x
Tolerance
Nominal value
Ex. 1 k10% 900-1100
For 10% resistor10, 12, 15, 18,
10 12 15R
RE
10n
where E= 6, 12, 24, 96
for 20, 10, 5, 1% tolerance
n= 0, 1, 2, 3,
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For 10% resistorE= 12 n = 0;R =
1.00000 n = 1;R =1.21152 n = 2;
R =1.46779 n = 3;R =1.77827
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Resistance Measurement Techniques
Bridge circuit
Voltmeter-ammeter
Substitution
Ohmmeter
Voltmeter-ammeter
V V
A A
R
Substitution
A A
Supply
Unknow
Rx Supplyresistance
R
Decade resis
box substitu
place of the
unknown
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Voltmeter-ammeter method
Pro and con:
Simple and theoretical oriented
Requires two meter and calculationsSubject to error: Voltage drop in ammeter
Current in voltmeter (Fig.
+ VA-I
A
+
A
+ + + II
VS V V Rx x VS V V
- - - - -
Fig. (a)Fig. (b)
MeasuredR: R = =V
x+
V
A =R +V
A MeasuredRx:R
meas= V =x meas I I x I I
ifVx>>VA ifIx>>IVR
meas
Rx
Rmeas
Rx
Therefore this circuit is suitable for measure Therefore this circuit is su
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large resistance small resistance
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Ohmmeter
Voltmeter-ammeter method is rarely used in practical ap
(mostly used in Laboratory)Ohmmeter uses only one meter by keeping one paramet
Example: series ohmmeter
Resistance to
be measuredStandard
resistance
Rx
R1Battery
VS Rm
Rx=VIs R1Rm
Basic series ohmmeter
1
45 k 5
Meter Infinity 2 5
resistance
Meter0
Ohmme
Basic series ohmmeter consisting of a PMMC and a series-connected standard
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the ohmmeter terminals are shorted (Rx = 0) meter full scale defection occurs.
ARx= R1+ Rm, and at zero defection the terminals are open-circuited.
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Bridge Circuit
Bridge Circuit is a null method, operates on the principle
comparison. That is a known (standard) value is
adjusted equal to the unknown value.
Bridge Circuit
DC Bridge AC Bridge
(Resistance)
Inductance Capacitance
Wheatstone Bridge Maxwell Bridge Schering BridgeKelvin Bridge Hay BridgeMegaohm Bridge Owen Bridge
Etc.
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Wheatstone Bridge and Balance Co
Suitable for moderate resistance values: 1 to 10 M
A
R1 R2
I1 I2
V D
I4I3
R3
R4
C
Balance condition:
No potential difference
a galvanometer (there is
n the galvanometer)
BUnder this condition
I1 R1=
And also VDC= VB
I3 R3=
whereI1,I2,I3, andI4 are cur
arms respectively, sinceI1 =I
R1 =R2
or Rx= R4
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Example
1 1
12 V
1 1
(a) Equal resistance
1 10
12 V
2 20
(c) Proportional resistance
1
12 V
2
(b)Proportion
1
12 V
2
(d) 2-Volt unb
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Measurement Errors
1. Limiting error of the known resistors R = (RR)Rx 3 3
Using 1st order approximation:R = R
R R
A
x 3 1 1
R1 R2 2. Insufficient sensitivi
3. Changes in resistan
V D Barms due to the heat
temperatures
4. Thermal emf or cont
R3
bridge circuit
Rx 5. Error due to the lead
C 3, 4 and 5 play the impor
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measurement of low val
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Example In the Wheatstone bridge circuit, R3 is a decade resistance wi
accuracy 0.2% and R1 and R2 = 500 0.1%. If the value of R3 at the
520.4 , determine the possible minimum and maximum value of R X
SOLUTIONApply the error equation R =RR2 1R1 R2
x 3R1 R1 R2 R
x=520.4 500
0.1
0.1
0.2
= 520.4( 1 0.004)= 520.41500 100 100 100
Therefore the possible values of R3 are 518.32 to 522.48
ExampleA Wheatstone bridge has a ratio arm of 1/100 (R2/R1). At first b
adjusted to 1000.3 . The value of Rx is then changed by the
temperature value of R3 to achieve the balance condition again is 1002.1
. Find the c the temperature change.
SOLUTIONAt first balance: R old= R 2 = 1000.3 1 =10.003x 3R 100
1
After the temperature change: R new= R 2= 1002.1 1 =10.021R 100x 3
1
Therefore, the change of Rx due to the temperature change is 0.018
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Sensitivity of Galvanometer
A galvanometer is use to detect an unbalance cond
Wheatstone bridge. Its sensitivity is governed by: Current
(currents per unit defect ion) and in ternal resis tance.
consider a bridge circuit under a small unbalance condition, and
apply analysis to solve the current through galvanometer
Thvenin Equivalent Circuit
Thvenin Voltage (VTH)
I1
AVCD= VAC VAD= I1 R1
I2
VR R
where = andI2
=
S
1 21
R1 +R3 RC G D
R3 R4 ThereforeVTH = VCD= V
R1
R1+B
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Sensitivity of Galvanometer (contin
Thvenin Resistance (RTH)
R1A R2
C
R3
D RTH = R1// R3+ R2// R4
R4
B
Completed Circuit
RTH C V
TH VIg=I =RTH+Rg
TH
Gg
RTH+ R VTHD
whereIg= the galvanometer curren
Rg= the galvanometer resista
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Example 1 Figure below show the schematic diagram of a Wheatstone
the bridge elements. The battery voltage is 5 V and its internal resistanc
galvanometer has a current sensitivity of 10 mm/A and an internal resis
Calculate the deflection of the galvanometer caused by the 5- unbalan
SOLUTION The bridge circuit is in the small unbalance condition since th
resistance in arm BC is 2,005 .
A
100 R1 R2
1000
5 V D G C
200 R3 R4
2005
B
(a)
100 A 1000
C
200 2005
D
B
(b)
RTH= 734 C
Ig=3.34A
THG Rg= 1002.77 mV
D
(c)
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Thvenin Voltage (VTH)
VTH
=VAD
100VAC = 5 V
100 + 200
2.77 mV
Thvenin Resistance
(RTH)
RTH =100 // 200+
1000 // 2005=
The galvanometer current
Ig
=
VTH = 2.77 mV = 3.
RTH+ Rg 734 +100
Galvanometer deflection
d=3.32A10
mm
A=33.2mm
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Example 2 The galvanometer in the previous example is replaced by on
resistance of 500 and a current sensitivity of 1mm/A. Assuming thatcan be observed on the galvanometer scale, determine if this new
galva of detecting the 5- unbalance in arm BC
SOLUTION Since the bridge constants have not been changed, the
equ is again represented by a Thvenin voltage of 2.77 mV and a
Thvenin 734 . The new galvanometer is now connected to the output
terminals galvanometer current.
Ig
=
VTH
= 2.77 mV = 2.24A
RTH+ Rg 734 +500
The galvanometer deflection therefore equals 2.24 A x 1 mm/A = 2.2
indicating that this galvanometer produces a deflection that can be eas
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Example 3 If all resistances in the Example 1 increase by 10 times, and
galvanometer in the Example 2. Assuming that a deflection of 1 mmcan galvanometer scale, determine if this new setting can be detected
(the 5 arm BC)
SOLUTION
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Application of Wheatstone Bridge
Murray/Varrley Loop Short Circuit Fault (Loop Te
Loop test can be carried out for the location of either a
groun circuit fault.
R3 X
short1
circuit fault
R4X2
groundMurray Loop T fault
Let R = R1+R
2
R3At balance condition:4
R3Assume: earthis a R1 =R R2=
good conductor R3+R4
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The value ofR1 andR2 are used to calculat
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Murray/Varrley Loop Short Circuit Fault (Loop Te
Examples of commonly used cables (Approx. R at 20oC)
Wire dia. In mm Ohms per km. Meter per ohm
0.32 218.0 4.59
0.40 136.0 7.35
0.50 84.0 11.90
0.63 54.5 18.35
0.90 27.2 36.76
Remark The resistance of copper increases 0.4% for 1oC rise i
Let R = R1+R2and define Ratio=R4/R5 R4
R4 R1At balance condition: Ratio= =RR +R R5
5 2 3
1=Ratio
R +R3 R2=R - RatioR3
R3
Ratio+1 Ratio +1 Varley Loop
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Example Murray loop test is used to locate ground fault in a telephone
s resistance, R =R1+R2 is measured by Wheatstone bridge, and its
value i conditions for Murray loop test are as follows:
R3= 1000and R4= 500Find the location of the fault in meter, if the length per Ohm is 36.67 m.
Power orSOLUTION
R3
communication cableX
1
R1
R1 R3 = 300R2
=R 1R4
X2
3 4
Short
R4circuitR2 =R = 300
1fault 3 4
Murray Loop Test
Therefore, the location from the measurement point is 10036.67 m/
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Application of Wheatstone Bridge
Unbalance bridge
A
R R
Consider a bridge circuit whic
resistors, R in three arms, and the
l resistance ofR +R. ifR/R
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tempera
ture,
strain
etc.
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Example Circuit in Figure (a) below consists of a resistorRvwhich issentemperature change. The plot ofR VS Temp. is also shown in Figure(b). temperature at which the bridge is balance and (b) The output
signal at 60oC.
5 k 5 k6
5
6 V)
4
3(k
4.5 k2R
1
Rv Output0
5 k
0 20 40 60 80 100
signalTemp (
oC)
(a)
(b)
R3R2= 5 k5 kSOLUTION (a) at bridge balance, we have R =v R1 5 k
The value ofR = 5 k corresponding to the temperature of 80 C in thev
(b) at temperature of 60
o
C,Rv is read as 4.5 k, thus R =5 - 4.5 = 0.5use Thvenin equivalent circuit to solve the above problem.
V = V R = 6 V 0.5 k = 0.15 VTH 4R 45 k
It should be noted that R =0.5 k in the problem does not satisfy the
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Low resistance Bridge:Rx< 1
Effect of co nnect ing lead
R2 R3
V Gm
n
R1 Rx
The effects of the connecting lead
a terminals are prominent when the
v to a few Ohms
Ry= the resistance of the conneR
y Rx
At point m:Ry is added to the unknown
high and indication ofRx
At point n:Ry is added toR3, therefore thwill be lower than it should
At pointp: Rx+Rnp=(R3+Rmp)R1R2
rearrange Rx= R3R
1+ Rmp
R1 Rnp
R2 R2
WhereRmp andRnp are
the lead resistance from mtop and n top,respectively.
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The effect of the connecti canceled out, if
the sum ozero.
R
RmpR1
Rnp=0 o2
Rx= R3
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Kelvin Double Bridge: 1 to 0.00001
Four-Terminal ResistorCurrent Current
terminals terminals
Voltage Voltage
terminals terminals
Four-terminal resistor
and potential terminal
defined as that betwe
terminals, so that con
current terminals do
Four-Terminal Resistor and Kelvin Double Bridge
R2R
3
Rb
GR
a
R1 r2
r3 r4
Rx
r1
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r1causes no effect
on the balance
condit
The effects of r2
and r3 could be
minimi r2 and Ra
>> r3.The main errorcomes from r4,
even tho is verysmall.
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Kelvin Double Bridge: 1 to 0.00001
l
R3R
IRb
m
V k Gp
Ran
R1 x
o
2 ratio arms: R1-R2 and Ra-
the connecting lead betwe
The balance conditions: Vlk=V
R
Vlk =
R2 V
y R1+ R2here V= IR =I[R +R +
lo 3 x
RyVlm = I 3 +
R +R +R
a b
Eq. (1) = (2) and rearrange: R =RR
+b y R R
1 1 a
2x
3R2 Ra+ Rb+ Ry b
If we set R1/R2=Ra/Rb, the second term of the right hand side will be zerreduce to the well known relation. In summary, The resistance of the yok
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on the measurement, if the two sets of ratio arms have equal resistance
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High Resistance Measurement
Guard ring technique:Volume resistance,RV
Surface leakage resistanc
High
voltage
supply
A
V
Is Iv High
voltage
supply
AIs
V
Is
(a) Circuit that measures insulation
volume resistance in parallel with surface
leakage resistance
Rmeas
=
Rs
//
Rv
=Is
V+Iv
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(b) Use of guard ring to m resistance
Rmeas
=
Rv
=
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High Resistance Measurement
Example The Insulation of a metal-sheath electrical cable is tested usin
and a microammeter. A current of 5 A is measured when the
compone without guard wire. When the circuit is connect with guard
wire, the curr Calculate (a) the volume resistance of the cable
insulation and (b) the s resistance
SOLUTION
(a ) Volume resistance:
IV=1.5A
R =V =10000 V = 6.7 109
V IV 1.5 A
(b ) Surface leakage resistance:
IV+ IS= 5A IS= 5AIV= 3.5A
R = V =10000 V = 2.9109
S
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MegaOhm Bridge
Just as low-resistance measurements are affected by series lead im
resistance measurements are affected by shunt-leakage resistance.
RA RB RA
E G E
G
R2
R1
RC Rx
the guard terminal is connect to abridge corner such that the leakage
resistances are placed across bridge
arm with low resistances
1// RC RC sinceR1>> RC
R2// R R sinceR2>
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Rx RARC
RB
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Capacitor
Capacitance the ability of a dielectric to store electrical
c unit voltage
conductorArea,A
A 0rC =Dielectric,Typical valur dthickness, d
Dielectric Construction Capacitance Brea
Air Meshed plates 10-400 pF 100 (0.0
Ceramic Tubular 0.5-1600 pF 500
Disk 1pF to 1 F
Electrolytic Aluminum 1-6800 F 1
Tantalum 0.047 to 330 FMic
a Stacked sheets 10-5000 pF 500
Paper Rolled foil 0.001-1 F 20
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Plastic film Foil or Metallized 100 pF to 100 F 5
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Inductor
Inductance the ability of a conductor to produce induced
when the current varies.Nturns
L =
o
rN A
A l
l o= 410-7
H/m
rrelative permeability of core material AirNi ferrite: r> 200
Mn ferrite: r> 2,000
L R
e
Cd Iro
Equivalent circuit of an RF coil Distributed capacitance Cd
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between turns
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Quality Factor of Inductor and Capa
Equivalent circuit of capacitance
Cp
Rp
Parallel equivalent circuit
Equivalent circuit of Inductance
Ls Rs
Series equivalent circuit
R =R
2+X
2
X =R
2+X
2
s s s s
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Cs Rs
Series equivalent circuit
Lp
Rp
Parallel equivalent circuit
R =R X
2
p
s R + 2p p
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Quality Factor of Inductor and Capa
Quality factor of a coil: the ratio of reactance toresistance dependent and circuit configuration)
Inductance series circuit: Q = s =Ls
RsRs
Inductance parallel circuit: Q = p = p
L X
p
Typical Q~
Dissipation factor of a capacitor: the ratio of reactance tore (frequency dependent and circuit configuration)
Capacitance parallel circuit: D =X
=1
Typical D~C R R
Capacitance series circuit: D =Rs =CsRs
Xs
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Inductor and Capacitor
LS =
R2
LP
VP
R + L P P
RS LSR = 2L
2P R I
R + L S PP P
Q =LS
V
ILSRS
IRS
2 2 2 V
C =1 + P P
CS P2 C2R2
P P I RS LS1
RS = RP1 + C R IRSP P
D =CSRSI/C
S
V
I V
L
P
R
P
V/
R
P
V
/
L
P
I
I
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V
CP
RP
I
VCP
V/RP
L
R
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AC Bridge: Balance Condition
Ball four arms are considered
Z2Z (frequency dependent compo1 The detector is an ac respon
1 2
V C
headphone, ac meter
A D Source: an ac voltage at des
Z3 ZZ1, Z2, Z3and Z4are the impedanc
At balance point: EBA = EBCorI1 ZD
VI1 = and I2=
General Form of the ac Bridge Z1 + Z3
Complex Form: 1 4 = Z2 Z3
Polar Form:Magnitude balance: Z1 Z 4
Z1 Z 4(1+4) =Z 2 Z3(2+3) Phase balance:
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1+
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Example The impedance of the basic ac bridge are given as follows:
Z1=10080o(inductive impedance) Z3= 40030o(inductiveZ2=250(pure resistance) Z4= unknown
Determine the constants of the unknown arm.
SOLUTION The first condition for bridge balance requires that
Z4=Z
2
Z3 =
250
400=1,000
Z1100
The second condition for bridge balance requires that the sum of the
ph opposite arms be equal, therefore
4=2+31=0+3080= 50o
Hence the unknown impedance Z4 can be written in polar form as
Z4=1,00050oIndicating that we are dealing with a capacitive element, possibly
cons series combination of at resistor and a capacitor.
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Example an ac bridge is in balance with the following constants: arm
in series withL = 15.9 mHR; arm BC,R = 300 in series with C= 0.2
unknown; arm DA, = 450 . The oscillator frequency is 1 kHz. Find th
arm CD. BSOLUTION
Z Z21
Z1=R+jL=200+j100V 1 2
A D C Z2=R+1/jC=300j
Z3=R=450
Z3 Z4Z4=unknown
D
The general equation for bridge balance states that Z 1 Z 4 = Z 2 Z3
Z4 =
2 3 =450 (200 +j100) =j150Z1 (300 j600)
This result indicates that Z4 is a pure inductance with an inductive
reac at at frequency of 1kHz. Since the inductive reactance XL =
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2fL, we s obtainL = 23.9 mH
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Comparison Bridge: Capacitance
R1Measure an unknown induct
R2capacitance by comparing wi
inductance or capacitance.
Vs D At balance point: Z 1 Z x
C3
Rx
where
Z=R ; Z
2
=R ; and Z
R3 Cx Unknown1 1 2
1capacitance
R =RDiagram of Capacitance
R +1 x j Cx 2
Comparison Bridge=R
2
R3Separation of the real and imaginary terms yields: R and
x
1
Frequency independentTo satisfy both balance conditions, the bridge must contain
two elements in its configuration.
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Comparison Bridge: Inductance
R1
Measure an unknown induct
2 capacitance by comparing wi
inductance or capacitance.
Vs D At balance point: Z 1 Z x
L3 Lx
whereZ
1=R1
;
Z2
=R2
; and
R3 xUnknown
inductance
R1( Rx+jLx)=R2Diagram of InductanceComparison Bridge
Separation of the real and imaginary terms yields: R=2
3 and
x R1
Frequency independent
To satisfy both balance conditions, the bridge must contain
two elements in its configuration.
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Maxwell Bridge
R1Measure an unknown induct
R2 a known capacitance
V
C1
At balance point: Zx =D
R3
L
x where Z2
=R2; Z 3=R3; an
RxUnknown
inductance
Diagram of Maxwell Bridge Zx =Rx+jLx=R
Separation of the real and imaginary terms yields: R = 2 3 anx
R1
Frequency independent
Suitable for Medium Q coil (1-10), impractical for high Q coil:
si large.
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Hay Bridge
R1Similar to Maxwell bridge: but
R2C1At balance point: Z1 Zx
V D where Z =R j ; Z =RC
Lx
1 1 2
1
R3Rx
UnknownR + 1 +inductance R
Diagram of Hay Bridge1 C1 x
R R+ x
R R+ x jRx1 x
C1which expands to +jL R =R R1 x
C1 C1x 12 3
Rx =C
1
Solve the above equations simultaneously
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Hay Bridge: continues
R2 R3C1 C R R R1 1 2 3Rx= and x 1+2C12R121+ 2 C2R2
1 1
Lx Z R1
tanL =
XL
=
L
C
x
R x
tanC
=XC = 1
C1
L 1Rx Z tanL= tanCorQ=C1
Phasor diagram of arm 4 and 1
L = R2 R3C
Thus,Lx can be rewritten as 1+(1/ Qx
For high Q coil (> 10), the term (1/Q)2
can be neglected Lx R
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Schering Bridge
C1 R
2
Used extensively for the measure
R1
and the quality of capacitor in ter
At balance point: Zx
V Dwhere Z =R; Z = 1
Cx
2 2 3 C
C33
jRxUnknown
R =Rcapacitance
Diagram of Schering Bridgex Cx 2 Cx
which expands to R j =R
2
C1 jR2
x Cx C C R3 3 1
Separation of the real and imaginary terms yields: RxC1
a2 C3
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Schering Bridge: continues
Dissipation factor of a seriesRCcircuit:D =
Rx =R C
xx
x
Dissipation factor tells us about the quality of a capacitor, how
clos phase angle of the capacitor is to the ideal value of 90o
For Schering Bridge: D =Rx Cx=R1C1
For Schering Bridge,R1 is a fixed value, the dial ofC1 can be
calibra at one particular frequency
Wien Bridge
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Wien Bridge
Unknown R1
R2
Measure frequency of the voltageFreq.
RC in one arm and parallel RC in 1
At balance point: Z 2 =
Vs D Z =R+ 1 ; Z =R;Y = 1
R3
R4
1 1 C1 2 2 3 R3
C3R = R R 1 +
Diagram of Wien Bridge 2 1 1 4 R3
R2 RR1 R4 jR4
R4
C3
=
which expands to R = +jC R R + R RR C R C
4
2 3 1 4
3 1 3 1C3R1=
Rearrange Eq. (2) gives =1 In most, Wien Bridge,R
2C C R R1 3 1 3
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Wagner Ground Connection
COne way to control stray ca
C6 Shielding the arms, reduceC5
R1 R2
capacitances but cannot eli
w completely.
A 1 D B Stray across arm
C1 2 Cannot eliminate
Cw
C3 Rx Wagner ground connection
R3 x effects of stray capacitanceD
C4
Simultaneous balance of boWagner ground point 1 and 2 at the ground
and C2 to ground, C4 and C5
Diagram of Wagner ground detector circuit)
The capacitance across thecannot be eliminated by Wa
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Capacitor Values
Ceramic Capacitor
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Capacitor Values
Film Capacitor
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Capacitor Values
Chip Capacitor
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Capacitor Values
Tantalum Capacitor
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Capacitor Values
Chip Capacitor