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2102311 Electrical Measurement and Instruments (Part II) ¾ Bridge Circuits (DC and AC) ¾ Electronic Instruments (Analog & Digital) ¾ Signal Generators ¾ Frequency and Time Interval Measurements ¾ Introduction to Transducers อาภรณ ธีรมงคลรัศมี ตึกไฟฟา 6 ชัน้ หอง 306 Textbook: -A.D. Helfrick, and W.D. Cooper, “Modern Electronic Instrumentation and Measurement Techniques” Prentice Hall, 1994. - D.A. Bell, “Electronic Instrumentation and Measurements”, 2nd ed., Prentice Hell, 1994.

Resistor Resistor Types Types

Importance parameters ™Value ™Tolerance ™Power rating ™Temperature coefficient

Type

Values (Ω)

Power rating (W)

Tolerance (%)

Temperature coefficient (ppm/°C)

Wire wound (power)

10m~3k

3~1k

±1~±10

±30~±300

Wire wound (precision)

10m~1M

0.1~1

±0.005~±1

±3~±30

Carbon film

1~1M

0.1~3

±2~±10

±100~±200

Metal film

100m~1M

0.1~3

±0.5~±5

±10~±200

Metal film (precision)

10m~100k

0.1~1

±0.05~±5

±0.4~±10

Metal oxide film

100m~100k

1~10

±2~±10

±200~±500

Data: Transistor technology (10/2000)

picture

™Color codes ™Alphanumeric

Resistor Resistor Values Values Color

Silver Gold

Digit

-

Multiplier

Tolerance (%)

Temperature coefficient (ppm/°C)

±10

K

10-1

±5

J

-

-

±250

K

10-2

-

-

4 band color codes

Most sig. fig. of value Tolerance Least sig. fig. Multiplier of value

Ex.

Black

0

100

Brown

1

101

±1

F

±100

H

Red

2

102

±2

G

±50

G

Orange

3

103

-

±15

D

Yellow

4

104

-

±25

F

R = 560 Ω ± 2%

Green

5

105

±0.5

D

±20

E

Blue

6

106

±0.25

C

±10

C

Alphanumeric

Violet

7

107

±0.1

B

±5

B

Gray

8

108

-

±1

A

White

9

109

-

-

Data: Transistor technology (10/2000)

-

±20

M

Green Blue

Brown

Red

R, K, M, G, and T = x100, x103 , x106 , x109 , and x1012 Ex. 6M8 = 6.8 x 106 Ω 59Ρ04 = 59.04 Ω

Resistor Values

Commonly available resistance for a fixed resistor

R = x ± %∆x Tolerance Nominal value Ex. 1 kΩ ± 10%

≡

900-1100 Ω

For 10% resistor 10, 12, 15, 18, … 10

12

15

R

E

R ≈ √10n where E = 6, 12, 24, 96 for 20, 10, 5, 1% tolerance n = 0, 1, 2, 3, …

For 10% resistor E = 12 n = 0; R = 1.00000… n = 1; R = 1.21152… n = 2; R = 1.46779… n = 3; R = 1.77827…

Resistance Resistance Measurement Measurement Techniques Techniques Bridge circuit Voltmeter-ammeter Substitution Ohmmeter Voltmeter-ammeter V

V

A

A R

R

Substitution A Supply

Unknow resistance

A Rx

Supply

Decade resistance box substituted in place of the unknown

Voltmeter-ammeter method Pro and con: •Simple and theoretical oriented •Requires two meter and calculations •Subject to error: Voltage drop in ammeter (Fig. (a)) Current in voltmeter (Fig. (b)) + +

VA A

I

A

+

IV

+

+

Vx

VS

V V

-

-

-

Ix

I VS -

V V -

Rx

Fig. (b)

Fig. (a)

V V + VA V Measured Rx: Rmeas = = x = Rx + A I I I if Vx>>VA Rmeas ≈ Rx Therefore this circuit is suitable for measure large resistance

Rx

Measured Rx: Rmeas = if Ix>>IV

Rx V V = = I I x + IV 1 + IV / I x

Rmeas ≈ Rx

Therefore this circuit is suitable for measure small resistance

Ohmmeter •Voltmeter-ammeter method is rarely used in practical applications (mostly used in Laboratory) •Ohmmeter uses only one meter by keeping one parameter constant Example: series ohmmeter

Battery

k 45

R1 Rm

A 0µ

Basic series ohmmeter

75

10

Vs − R1 − Rm I

25

5k

50

0

VS

Meter Infinity resistance

15k

0

Rx

Rx =

Nonlinear scale

Standard resistance

∞

Resistance to be measured

Meter

Ohmmeter scale

Basic series ohmmeter consisting of a PMMC and a series-connected standard resistor (R1). When the ohmmeter terminals are shorted (Rx = 0) meter full scale defection occurs. At half scale defection Rx = R1 + Rm, and at zero defection the terminals are open-circuited.

Bridge Circuit Bridge Circuit is a null method, operates on the principle of comparison. That is a known (standard) value is adjusted until it is equal to the unknown value.

Bridge Circuit AC Bridge

DC Bridge (Resistance)

Wheatstone Bridge Kelvin Bridge Megaohm Bridge

Inductance

Capacitance

Maxwell Bridge Hay Bridge Owen Bridge Etc.

Schering Bridge

Frequency Wien Bridge

Wheatstone Bridge and Balance Condition Suitable for moderate resistance values: 1 Ω to 10 MΩ

A

Balance condition:

R2

R1 I1 V

No potential difference across the galvanometer (there is no current through the galvanometer)

I2

D

B

I4

I3 R3

R4

C

Under this condition: VAD = VAB

I1R1 = I 2 R2 And also VDC = VBC

I3 R3 = I 4 R4 where I1, I2, I3, and I4 are current in resistance arms respectively, since I1 = I3 and I2 = I4

R1 R2 or = R3 R4

R2 Rx = R4 = R3 R1

Example 1Ω

1Ω

1Ω

1Ω

12 V

12 V 1Ω

2Ω

1Ω

(a) Equal resistance

1Ω

2Ω

(b) Proportional resistance 1Ω

10 Ω

10 Ω

12 V

12 V 2Ω

20 Ω

(c) Proportional resistance

2Ω

10 Ω

(d) 2-Volt unbalance

Measurement Errors 1. Limiting error of the known resistors Using 1st order approximation:

A R1 V

R2

R2  ∆R1 ∆R2 ∆R3  Rx = R3 1 ± ± ±  R1  R1 R2 R3  2. Insufficient sensitivity of Detector

B

D

 R ± ∆R2  Rx = ( R3 ± ∆R3 )  2  R ± ∆ R 1   1

3. Changes in resistance of the bridge arms due to the heating effect (I2R) or temperatures 4. Thermal emf or contact potential in the bridge circuit

R3

C

Rx

5. Error due to the lead connection 3, 4 and 5 play the important role in the measurement of low value resistance

Example In the Wheatstone bridge circuit, R3 is a decade resistance with a specified in accuracy ±0.2% and R1 and R2 = 500 Ω ± 0.1%. If the value of R3 at the null position is 520.4 Ω, determine the possible minimum and maximum value of RX SOLUTION Apply the error equation

Rx =

R2  ∆R1 ∆R2 ∆R3  Rx = R3 1 ± ± ±  R1  R1 R2 R3 

520.4 × 500  0.1 0.1 0.2  ± ± 1 ±  = 520.4( 1 ± 0.004) = 520.4 ± 0.4% 500  100 100 100 

Therefore the possible values of R3 are 518.32 to 522.48 Ω Example A Wheatstone bridge has a ratio arm of 1/100 (R2/R1). At first balance, R3 is adjusted to 1000.3 Ω. The value of Rx is then changed by the temperature change, the new value of R3 to achieve the balance condition again is 1002.1 Ω. Find the change of Rx due to the temperature change.

R2 1 = × = 10.003 Ω 1000.3 SOLUTION At first balance: R1 100 R2 1 = 1002.1× = 10.021 Ω After the temperature change: Rx new = R3 R1 100 Rxold = R3

Therefore, the change of Rx due to the temperature change is 0.018 Ω

Sensitivity of Galvanometer A galvanometer is use to detect an unbalance condition in Wheatstone bridge. Its sensitivity is governed by: Current sensitivity (currents per unit defection) and internal resistance. consider a bridge circuit under a small unbalance condition, and apply circuit analysis to solve the current through galvanometer

Thévenin Equivalent Circuit Thévenin Voltage (VTH) I1 VS

A

VCD = VAC − VAD = I1 R1 − I 2 R2

I2 R1

C

R2 D

G R3

where I1 =

R4 B

Therefore

V V = and I 2 R1 + R3 R2 + R4

 R1 R2  − VTH = VCD = V   R + R R + R 3 2 4   1

Sensitivity of Galvanometer (continued) Thévenin Resistance (RTH) R1 C

R2

A

R3

D

R4

RTH = R1 // R3 + R2 // R4

B

Completed Circuit RTH

C Ig= G

VTH

VTH RTH+Rg

Ig =

VTH RTH + Rg

D

where Ig = the galvanometer current Rg = the galvanometer resistance

Example 1 Figure below show the schematic diagram of a Wheatstone bridge with values of the bridge elements. The battery voltage is 5 V and its internal resistance negligible. The galvanometer has a current sensitivity of 10 mm/µA and an internal resistance of 100 Ω. Calculate the deflection of the galvanometer caused by the 5-Ω unbalance in arm BC SOLUTION The bridge circuit is in the small unbalance condition since the value of resistance in arm BC is 2,005 Ω.

Thévenin Voltage (VTH)

A 1000 Ω

100 Ω R1 5V

G

D R3

1000  100  VTH = VAD − VAC = 5 V ×  −  100 + 200 1000 + 2005   ≈ 2.77 mV

R2 C R4 2005 Ω

200 Ω

Thévenin Resistance (RTH)

B (a) 100 Ω C

RTH = 100 // 200 + 1000 // 2005 = 734 Ω

1000 Ω

A

200 Ω

D

2005 Ω B

Ig =

(b) RTH= 734 Ω

C Ig=3.34 µA

VTH 2.77 mV

G D (c)

The galvanometer current

Rg= 100 Ω

VTH 2.77 mV = = 3.32 µ A RTH + Rg 734 Ω + 100 Ω

Galvanometer deflection d = 3.32 µ A ×

10 mm = 33.2 mm µA

Example 2 The galvanometer in the previous example is replaced by one with an internal resistance of 500 Ω and a current sensitivity of 1mm/µA. Assuming that a deflection of 1 mm can be observed on the galvanometer scale, determine if this new galvanometer is capable of detecting the 5-Ω unbalance in arm BC SOLUTION Since the bridge constants have not been changed, the equivalent circuit is again represented by a Thévenin voltage of 2.77 mV and a Thévenin resistance of 734 Ω. The new galvanometer is now connected to the output terminals, resulting a galvanometer current. Ig =

VTH 2.77 mV = = 2.24 µ A RTH + Rg 734 Ω + 500 Ω

The galvanometer deflection therefore equals 2.24 µA x 1 mm/µA = 2.24 mm, indicating that this galvanometer produces a deflection that can be easily observed.

Example 3 If all resistances in the Example 1 increase by 10 times, and we use the galvanometer in the Example 2. Assuming that a deflection of 1 mm can be observed on the galvanometer scale, determine if this new setting can be detected (the 50-Ω unbalance in arm BC) SOLUTION

Application of Wheatstone Bridge Murray/Varrley Loop Short Circuit Fault (Loop Test) •Loop test can be carried out for the location of either a ground or a short Power or circuit fault. R3

communication cable

X1

R1

short circuit fault

R2 R4

ground fault

Short circuit fault

Murray Loop Test Let R = R1+R2

Assume: earth is a good conductor

X2

At balance condition:

R3 R1 = R4 R2

 R3  R1 = R   R R +  3 4

 R4  R2 = R   R R +  3 4

The value of R1 and R2 are used to calculate back into distance.

Murray/Varrley Loop Short Circuit Fault (Loop Test) 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 in Temp. Let R = R1+R2 and define Ratio = R4/R5

R R1 At balance condition: Ratio = 4 = R5 R2 + R3

Ratio R1 = R + R3 Ratio + 1

R - RatioR3 R2 = Ratio + 1

X1

R4

R1 R2

R5

X2

R3

Varley Loop Test

Short circuit fault

Example Murray loop test is used to locate ground fault in a telephone system. The total resistance, R = R1+ R2 is measured by Wheatstone bridge, and its value is 300 Ω. The conditions for Murray loop test are as follows: R3 = 1000 Ω and R4 = 500 Ω Find the location of the fault in meter, if the length per Ohm is 36.67 m.

R3

Power or communication cable X1

 R3  1000 R1 = R  = 200 Ω  = 300 × 1000 + 500  R3 + R4 

R1 R2 R4

X2

Murray Loop Test

SOLUTION

Short circuit fault

 R4  500 R2 = R  300 = × = 100 Ω  1000 + 500  R3 + R4 

Therefore, the location from the measurement point is 100 Ω× 36.67 m/Ω = 3667 m

Application of Wheatstone Bridge Unbalance bridge Consider a bridge circuit which have identical resistors, R in three arms, and the last arm has the resistance of R +∆R. if ∆R/R r2 and Ra >> r3. • The main error comes from r4, even though this value is very small.

Rb G

Ra R1

r3 r2 r1

Rx

r4

Kelvin Double Bridge: 1 to 0.00001 Ω 2 ratio arms: R1-R2 and Ra-Rb the connecting lead between m and n: yoke

l

R2

R3

I

m

Rb V

k

G

p

Vlk =

Ry

Ra

n

R1 o

Eq. (1) = (2) and rearrange:

The balance conditions: Vlk = Vlmp or Vok = Vonp

Rx

R2 V R1 + R2

(1)

here V = IRlo = I [ R3 + Rx + ( Ra + Rb ) // Ry ]

Vlmp

  Ry Rb  = I  R3 + Ra + Rb + Ry  

Rb Ry R Rx = R3 1 + R2 Ra + Rb + Ry

 R1 Ra   −   R2 Rb 

(2)

Rx = R3

R1 R2

If we set R1/R2 = Ra/Rb, the second term of the right hand side will be zero, the relation reduce to the well known relation. In summary, The resistance of the yoke has no effect on the measurement, if the two sets of ratio arms have equal resistance ratios.

High Resistance Measurement Guard ring technique:

Volume resistance, RV Surface leakage resistance, Rs Guard ring

µA High voltage supply

Is

Iv

V

(a) Circuit that measures insulation volume resistance in parallel with surface leakage resistance

Rmeas = Rs // Rv =

V I s + Iv

Is High voltage supply

µA

Iv

V

Material under test

Is

(b) Use of guard ring to measure only volume resistance

Rmeas = Rv =

V Iv

High Resistance Measurement Example The Insulation of a metal-sheath electrical cable is tested using 10,000 V supply and a microammeter. A current of 5 µA is measured when the components are connected without guard wire. When the circuit is connect with guard wire, the current is 1.5 µA. Calculate (a) the volume resistance of the cable insulation and (b) the surface leakage resistance SOLUTION

(a ) Volume resistance:

IV =1.5 µA RV =

V 10000 V = = 6.7 ×109 Ω IV 1.5 µA

(b ) Surface leakage resistance:

IV+ IS = 5 µA RS =

IS = 5 µA – IV = 3.5 µA

V 10000 V = = 2.9 × 109 Ω IS 3.5 µA

MegaOhm Bridge Just as low-resistance measurements are affected by series lead impedance, highresistance measurements are affected by shunt-leakage resistance.

RA E

E

G

RC the guard terminal is connect to a bridge corner such that the leakage resistances are placed across bridge arm with low resistances

R1 // RC ≈ RC R2 // Rg ≈ Rg

RA

RB

since R1 >> RC since R2 >> Rg

RB R2

Rx

G R1

RC Rx ≈ RA RB

RC

Capacitor Capacitance – the ability of a dielectric to store electrical charge per unit voltage Area, A

conductor

Dielectric, εr thickness, d

A ε 0ε r C = d

Typical values pF, nF or µF

Dielectric

Construction

Capacitance

Breakdown,V

Air

Meshed plates

10-400 pF

100 (0.02-in air gap)

Ceramic

Tubular

0.5-1600 pF

500-20,000

Disk

1pF to 1 µF

Aluminum

1-6800 µF

10-450

Tantalum

6-50 500-20,000

Electrolytic Mica

Stacked sheets

0.047 to 330 µF 10-5000 pF

Paper

Rolled foil

0.001-1 µF

200-1,600

Plastic film

Foil or Metallized

100 pF to 100 µF

50-600

Inductor Inductance – the ability of a conductor to produce induced voltage when the current varies. N turns

L =

A

µo µ r N

2

A

l

µo = 4π×10-7 H/m

l

µr – relative permeability of core material Ni ferrite: µr > 200 Mn ferrite: µr > 2,000 L

Air core inductor

Re

Cd Equivalent circuit of an RF coil

Iron core inductor Distributed capacitance Cd between turns

Quality Factor of Inductor and Capacitor Equivalent circuit of capacitance Cp

Cs

Rs

Rp Parallel equivalent circuit

Series equivalent circuit

Equivalent circuit of Inductance Ls

Lp

Rs

Rp Series equivalent circuit

Rs2 + X s2 Rp = Rs

Rs2 + X s2 Xp = Xs

Parallel equivalent circuit

Rs =

R p X p2 R +X 2 p

2 p

Xs =

X p R p2 R p2 + X p2

Quality Factor of Inductor and Capacitor Quality factor of a coil: the ratio of reactance to resistance (frequency dependent and circuit configuration) Inductance series circuit:

Q=

Inductance parallel circuit: Q =

X s ω Ls = Rs Rs Rp Xp

=

Typical Q ~ 5 – 1000

Rp

ω Lp

Dissipation factor of a capacitor: the ratio of reactance to resistance (frequency dependent and circuit configuration) Capacitance parallel circuit:

D=

Capacitance series circuit:

D=

Xp Rp

=

1 ω C p Rp

Rs = ω Cs Rs Xs

Typical D ~ 10-4 – 0.1

Inductor and Capacitor RP2 LS = 2 ⋅ LP RP + ω 2 L2P

RS

ω 2 L2P I RS = 2 ⋅ R P RP + ω 2 L2P Q=

I

V

V

RS2 + ω 2 L2S LP = ⋅ LS ω 2 L2S

LP

LS

RS2 + ω 2 L2S RP = ⋅ RS RS2

RP V

ω LS

IωLS

RS

V/ωLP

V/RP

θ

Q=

RP ω LP

θ I

IRS

CS = RS =

1+ ω C R ⋅ CP ω 2CP2 RP2 2

2 P

1 ⋅ RP 1 + ω CP2 RP2

I

RS

V CP

LS

RP

2

D = ω CS RS

I

V

2 P

IRS I/ωCS

I VωCP δ

δ V

CP =

1 ⋅ CS 1 + ω 2CS2 RS2

1 + ω 2CS2 RS2 ⋅ RS RP = ω 2CS2 RS2 D=

V/RP

1 ω CP RP

AC Bridge: Balance Condition B Z2

Z1 I1

V

I2

C

D

A

all four arms are considered as impedance (frequency dependent components) The detector is an ac responding device: headphone, ac meter Source: an ac voltage at desired frequency

Z3

Z4

Z1, Z2, Z3 and Z4 are the impedance of bridge arms At balance point:

D

I1 =

General Form of the ac Bridge Complex Form: Polar Form: Z1Z4 ( ∠θ1 + ∠θ 4 ) =Z2 Z3 ( ∠θ 2 + ∠θ 3 )

EBA = EBC or I1 Z1 = I 2 Z 2 V V and I 2 = Z1 + Z 3 Z2 + Z4 Z1 Z 4 = Z 2 Z 3

Magnitude balance: Phase balance:

Z1Z4 =Z2 Z3

∠θ1 + ∠θ 4 =∠θ 2 + ∠θ 3

Example The impedance of the basic ac bridge are given as follows: Z1 = 100 Ω ∠80o (inductive impedance)

Z3 = 400 ∠30o Ω (inductive impedance)

Z 2 = 250 Ω (pure resistance)

Z 4 = unknown

Determine the constants of the unknown arm. SOLUTION The first condition for bridge balance requires that

Z4 =

Z 2 Z 3 250 × 400 = = 1, 000 Ω Z1 100

The second condition for bridge balance requires that the sum of the phase angles of opposite arms be equal, therefore

∠θ 4 =∠θ 2 + ∠θ 3 − ∠θ1 = 0 + 30 − 80 = −50o Hence the unknown impedance Z4 can be written in polar form as

Z 4 = 1, 000 Ω ∠ − 50o Indicating that we are dealing with a capacitive element, possibly consisting of a series combination of at resistor and a capacitor.

Example an ac bridge is in balance with the following constants: arm AB, R = 200 Ω in series with L = 15.9 mH R; arm BC, R = 300 Ω in series with C = 0.265 µF; arm CD, unknown; arm DA, = 450 Ω. The oscillator frequency is 1 kHz. Find the constants of arm CD. B

Z1 I1

V

Z1 = R + jω L = 200 + j100 Ω

I2

C

D

A

SOLUTION

Z2

Z3

Z4

Z 2 = R + 1/ jω C = 300 − j 600 Ω Z3 = R = 450 Ω Z 4 = unknown

D The general equation for bridge balance states that

Z4 =

Z1 Z 4 = Z 2 Z 3

Z 2 Z 3 450 × (200 + j100) = = j150 Ω (300 − j 600) Z1

This result indicates that Z4 is a pure inductance with an inductive reactance of 150 Ω at at frequency of 1kHz. Since the inductive reactance XL = 2πfL, we solve for L and obtain L = 23.9 mH

Comparison Bridge: Capacitance R1 Vs

Measure an unknown inductance or capacitance by comparing with it with a known inductance or capacitance.

R2

At balance point:

D C3

Rx R3

Cx

Z1 Z x = Z 2 Z 3

1 1 =R1 ; Z 2 = R2 ; and Z 3 = R3 + jω C3   1  1  R1  Rx +  = R2  R3 +  j ω C j ω C x  3   

where Z

Unknown capacitance

Diagram of Capacitance Comparison Bridge Separation of the real and imaginary terms yields:

R2 R3 Rx = R1

and

C x = C3

Frequency independent To satisfy both balance conditions, the bridge must contain two variable elements in its configuration.

R1 R2

Comparison Bridge: Inductance R1 Vs

Measure an unknown inductance or capacitance by comparing with it with a known inductance or capacitance.

R2

At balance point:

D

L3

Lx R3

Rx

where

Z1 Z x = Z 2 Z 3

Z1 =R1 ; Z 2 = R2 ; and Z3 = R3 + jω L3

Unknown inductance

Diagram of Inductance Comparison Bridge Separation of the real and imaginary terms yields:

R1 ( Rx + jω Lx ) = R2 ( RS + jω LS ) R2 R3 Rx = R1

and

Lx = L3

Frequency independent To satisfy both balance conditions, the bridge must contain two variable elements in its configuration.

R2 R1

Maxwell Bridge R1

Measure an unknown inductance in terms of a known capacitance

R2 C1

V

R3

Z x = Z 2 Z3 Y1

At balance point:

D Lx Rx

where

Unknown inductance

Diagram of Maxwell Bridge

1 Z 2 = R2 ; Z3 = R3 ; and Y1 = + jω C1 R1 1  Z x = Rx + jω Lx = R2 R3  + jω C1   R1 

Separation of the real and imaginary terms yields:

Rx =

R2 R3 R1

and

Lx = R2 R3C1

Frequency independent Suitable for Medium Q coil (1-10), impractical for high Q coil: since R1 will be very large.

Hay Bridge Similar to Maxwell bridge: but R1 series with C1

R1 R2

C1 V

At balance point:

D

where

Lx R3

Z1 = R1 −

Z1Z x = Z 2 Z3

j ; Z 2 = R2 ; and Z3 = R3 ω C1

 1  + R  1  ( Rx + jω Lx ) = R2 R3 jω C1   Diagram of Hay Bridge L (1) R1 Rx + x = R2 R3 C1 Lx jRx − + jω Lx R1 = R2 R3 which expands to R1 Rx + C1 ω C1 Rx (2) = ω Lx R1 ω C1 Rx

Unknown inductance

Solve the above equations simultaneously

Hay Bridge: continues

ωLx

θL

ω 2C12 R1 R2 R3 Rx = 1 + ω 2C12 R12

and

Z

R1

Rx

θC

1 ωC1

Z

Lx =

R2 R3C1 1 + ω 2C12 R12

X L ω Lx tan θ L = = =Q R Rx X 1 tan θ C = C = R ω C1 R1 tan θ L = tan θ C or Q =

Phasor diagram of arm 4 and 1 Thus, Lx can be rewritten as For high Q coil (> 10), the term (1/Q)2 can be neglected

1 ω C1 R1

R2 R3C1 Lx = 1 + (1/ Q 2 )

Lx ≈ R2 R3C1

Schering Bridge C1

Used extensively for the measurement of capacitance and the quality of capacitor in term of D

R2 R1

V

Z x = Z 2 Z3 Y1 where Z = R ; Z = 1 ; and Y = 1 + jω C 2 2 3 1 jω C3 R1 At balance point:

D

C3

Cx Rx

Unknown capacitance

Rx −

 − j  1  j j ω C = R2  +  1 ω Cx ω C R x  1  

Diagram of Schering Bridge which expands to

j R2C1 jR2 Rx − = − C3 ω C3 R1 ω Cx

Separation of the real and imaginary terms yields:

Rx = R2

C1 C3

and

C x = C3

R1 R2

Schering Bridge: continues Dissipation factor of a series RC circuit:

D=

Rx = ω Rx C x Xx

Dissipation factor tells us about the quality of a capacitor, how close the phase angle of the capacitor is to the ideal value of 90o For Schering Bridge:

D = ω Rx Cx = ω R1C1

For Schering Bridge, R1 is a fixed value, the dial of C1 can be calibrated directly in D at one particular frequency

Wien Bridge R1

Unknown Freq.

R2

C1

Measure frequency of the voltage source using series RC in one arm and parallel RC in the adjoining arm

Z 2 = Z1Z 4 Y3 1 1 Z1 = R1 + ; Z 2 = R2 ; Y3 = + jωC3 ; and Z 4 = R4 jωC1 R3 At balance point:

Vs

D R3 R4

C3

Diagram of Wien Bridge which expands to R2 =

  j   1 R2 =  R1 − R j ω C +  4 3 ω C R 1    3 

RC R1 R4 jR4 + jω C3 R1 R4 − + 4 3 R3 C1 ω C1 R3

Rearrange Eq. (2) gives

f =

1 2π C1C3 R1 R3

R2 R1 C3 = + R4 R3 C1

ω C3 R1 =

(1)

1 ω C1 R3

(2)

In most, Wien Bridge, R1 = R3 and C1 = C3 (1)

R2 = 2 R4

(2)

f =

1 2π RC

Wagner Ground Connection C C5

R1

Rw A Cw

C6

R2

1 2

C1

D

B

C2

Rx

C3 R3

Wagner ground

Cx D

C4

Diagram of Wagner ground

One way to control stray capacitances is by Shielding the arms, reduce the effect of stray capacitances but cannot eliminate them completely. Stray across arm Cannot eliminate Wagner ground connection eliminates some effects of stray capacitances in a bridge circuit Simultaneous balance of both bridge makes the point 1 and 2 at the ground potential. (short C1 and C2 to ground, C4 and C5 are eliminated from detector circuit) The capacitance across the bridge arms e.g. C6 cannot be eliminated by Wagner ground.

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