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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