bridgeppt.docx

7/27/2019 BridgePPT.docx

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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)


3/74

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

:


4/74

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


5/74

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


8/74

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


10/74

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


11/74

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


31/74

tempera

ture,

strain

etc.


32/74

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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34/74

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


36/74

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


39/74

on the measurement, if the two sets of ratio arms have equal resistance


40/74

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


41/74

(b) Use of guard ring to m resistance

Rmeas

=

Rv

=


42/74

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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44/74

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>


45/74

Rx RARC

RB


46/74

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


47/74

Plastic film Foil or Metallized 100 pF to 100 F 5


48/74

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


49/74

between turns


50/74

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


51/74

Cs Rs

Series equivalent circuit

Lp

Rp

Parallel equivalent circuit

R =R X

2

p

s R + 2p p


52/74

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


53/74

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


54/74

V

CP

RP

I

VCP

V/RP

L

R


55/74

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:


56/74

1+


57/74

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.


58/74

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 =


59/74

2fL, we s obtainL = 23.9 mH


60/74

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.


61/74

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.


62/74

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.


63/74

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


64/74

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


65/74

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


66/74

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


67/74

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


68/74


69/74

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


70/74

Capacitor Values

Ceramic Capacitor


71/74

Capacitor Values

Film Capacitor


72/74

Capacitor Values

Chip Capacitor


73/74

Capacitor Values

Tantalum Capacitor


74/74

Capacitor Values

Chip Capacitor

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