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 ธีรมงคลรศมั ี ตึกไฟฟา 6 ชนั้ 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. ResistorResistor TypesTypes Importance parameters Value Tolerance Power rating Temperature coefficient

Type Values (Ω) Power rating Tolerance (%) Temperature picture (W) coefficient (ppm/°C)

Wire wound 10m~3k (power) 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) Resistor Values Color codes Resistor Values Alphanumeric 4 band color codes

Color Digit Multiplier Tolerance Temperature (%) coefficient (ppm/°C) Most sig. fig. of value Silver 10-2 ±10 K Tolerance - - - Least sig. fig. Multiplier Gold 10-1 J of value ±5 Ex. Black 0 100 - - ±250 K Brown 1 101 ±1 F ±100 H Red 2 102 ±2 G ±50 G Green

3 Red Orange 3 10 - ±15 D Blue Brown - Yellow 4 104 - ±25 F R = 560 Ω ±2% Green 5 5 D E 10 ±0.5 ±20 Alphanumeric Blue 6 106 ±0.25 C ±10 C Violet 7 107 ±0.1 B ±5 B R, K, M, G, and T = 0 3 6 9 12 Gray 8 108 - ±1 A x10 , x10 , x10 , x10 , and x10 - White 9 109 - Ex. 6M8 = 6.8 x 106 Ω - - - ±20 M 59Ρ04 = 59.04 Ω Data: Transistor technology (10/2000) 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 n R ≈√10 For 10% resistor E = 12 n = 0; R = 1.00000… where E = 6, 12, 24, 96 n = 1; R = 1.21152… for 20, 10, 5, 1% tolerance n = 2; R = 1.46779… n = 0, 1, 2, 3, … n = 3; R = 1.77827… ResistanceResistance MeasurementMeasurement TechniquesTechniques Bridge circuit Voltmeter-ammeter Substitution Ohmmeter

Voltmeter-ammeter

V V

A A R R Substitution A A Decade resistance Unknow Supply Supply box substituted in resistance Rx 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)) I + VA - A A + + + + IV Ix I V V V V R VS V Rx Vx S x

- - - - - Fig. (b) Fig. (a) VV R VVVV+ R == = x Measured R : xA A Measured Rx: meas x RRmeas == =+x I II++1/ II I II x VVx if I >>I RR≈ if Vx>>VA RRmeas ≈ x x V meas x

Therefore this circuit is suitable for measure Therefore this circuit is suitable for measure large resistance 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 Resistance to be measured Nonlinear scale Standard resistance 15k R k x 45 5k R 50 Battery 1 Meter Infinity 25 75 Rm resistance 0 VS ∞ 1 0 0 0 V µ RRR=−−s Meter A x I 1 m

Basic series ohmmeter 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

DC Bridge AC Bridge (Resistance) Inductance Capacitance Frequency

Wheatstone Bridge Maxwell Bridge Schering Bridge Wien Bridge Kelvin Bridge Hay Bridge Megaohm Bridge Owen Bridge Etc. Wheatstone Bridge and Balance Condition

Suitable for moderate resistance values: 1 Ω to 10 MΩ A Balance condition: No potential difference across the R1 R2 galvanometer (there is no current through the galvanometer)

I1 I2 Under this condition: V = V V D B AD AB

I IR11= IR 2 2 I3 4 And also V = V R DC BC 3 R4 IR33= IR 44

where I1, I2, I3,and I4 are current in resistance arms respectively, since I = I and I = I C 1 3 2 4 RR R 12or 2 = RRRx ==43 RR34 R1 Example

1 Ω 1 Ω 1 Ω 1 Ω

12 V 12 V

1 Ω 1 Ω 2 Ω 2 Ω

(a) Equal resistance (b) Proportional resistance

1 Ω 10 Ω 1 Ω 10 Ω

12 V 12 V

2 Ω 20 Ω 2 Ω 10 Ω

RR22±∆ 1. Limiting error of the known resistors RRRx =±∆()33 RR11±∆

Using 1st order approximation: RRR212∆∆∆R3 RRx =±±±3 1 A RRRR1123

R 2. Insufficient sensitivity of Detector 1 R2 3. Changes in resistance of the bridge arms due to the heating effect (I2R) or V D B temperatures

4. Thermal emf or contact potential in the bridge circuit

R3 Rx 5. Error due to the lead connection C 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 RRR∆∆∆R SOLUTION Apply the error equation 2123 RRx =±±±3 1 RRRR1123 520.4× 500 0.1 0.1 0.2 Rx =±±±=±=±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 SOLUTION At first balance: RRxold= 3 =×=Ω 1000.3 10.003 R1 100 R2 1 After the temperature change: RRxnew= 3 =×=Ω 1002.1 10.021 R1 100

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) A VVVIRIRCD=−= AC AD 11 − 2 2 I1 I 2 VV where II==and R1 R2 12 R13++RRR 24 VS CDG

R3 R4 Therefore RR12 VVVTH== CD  − R13++RRR 24 B Sensitivity of Galvanometer (continued)

Thévenin Resistance (RTH)

R1 A R2 R =+RRRR// // CDTH 1324 R3 R4

Completed Circuit

R TH C V I = TH g R +R VTH TH g I = g R + R VTH G TH g 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.

A Thévenin Voltage (VTH)

1000 Ω 100 Ω  100 1000  R1 R2 VTH = VAD −VAC = 5 V ×  −  5 V D G C 100 + 200 1000 + 2005  R R 3 4 ≈ 2.77 mV 200 Ω 2005 Ω

B Thévenin Resistance (RTH) (a)

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

C D 200 Ω 2005 Ω The galvanometer current B V 2.77 mV (b) TH I g == =3.32 µA RTH= 734 Ω C RRTH+Ω+Ω g 734 100

Ig=3.34 µA V TH G Galvanometer deflection 2.77 mV Rg= 100 Ω D 10 mm d =×=3.32 µA 33.2 mm (c) A µ 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.

VTH 2.77 mV I g == =2.24 µA RRTH+Ω+Ω g 734 500 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 circuit fault. Power or communication cable R 3 X1

short R1 circuit fault R2

R X2 4 Short circuit fault ground Murray Loop Test fault Let R = R +R 1 2 R R At balance condition: 31= RR42   Assume: earth is a R3 R4 RR1 = RR2 =  good conductor RR34+ RR34+

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 = R +R and define Ratio = R /R 1 2 4 5 X R 1 4 R RR 1 At balance condition: Ratio ==41 R2 RRR+ R5 523 X 2 Short R circuit Ratio RR- Ratio 3 fault RRR=+R = 3 13Ratio+ 1 2 Ratio+ 1 Varley Loop Test 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.

Power or communication cable SOLUTION R 3 X1 R 1 R 1000 RR= 3 =×300 = 200 Ω R 1  2 RR34++1000 500 X R4 2 Short  circuit R4 500 RR2 = =×300 =Ω 100 fault RR34++1000 500 Murray Loop Test

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

Unbalance bridge Consider a bridge circuit which have identical A resistors, R in three arms, and the last arm has the resistance of R +∆R. if ∆R/R << 1 R R

Thévenin Voltage (VTH) V CDG ∆R VVVTH=≈ CD R R+∆R 4R

B Thévenin Resistance (RTH) Small unbalance occur by the external RTH ≈ R environment R = R TH C This kind of bridge circuit can be found in sensor ∆R V =V G applications, where the resistance in one arm is TH 4R sensitive to a physical quantity such as pressure, D temperature, strain etc. Example Circuit in Figure (a) below consists of a resistor Rv which is sensitive to the temperature change. The plot of R VS Temp. is also shown in Figure (b). Find (a) the temperature at which the bridge is balance and (b) The output signal at Temperature of 60oC. 6 5 kΩ 5 kΩ 5 4 6 V Ω)

(k 3 v 4.5 kΩ R 2 1 0 Rv Output 0 20 40 60 80 100 120 5 kΩ signal Temp (oC)

(b) (a) RR32× 5 kΩ×Ω 5 k SOLUTION (a) at bridge balance, we have Rv = ==Ω5 k R1 5 kΩ o The value of Rv = 5 kΩ corresponding to the temperature of 80 C in the given plot. o (b) at temperature of 60 C, Rv is read as 4.5 kΩ, thus ∆R = 5 - 4.5 = 0.5 kΩ. We will use Thévenin equivalent circuit to solve the above problem. ∆R 0.5 kΩ VV==×6 V = 0.15 V TH 445R ×Ω k It should be noted that ∆R = 0.5 kΩ in the problem does not satisfy the assumption ∆R/R

<< 1, the exact calculation gives VTH = 0.158 V. However, the above calculation still gives an acceptable solution. Low resistance Bridge: Rx < 1 Ω

Effect of connecting lead The effects of the connecting lead and the connecting

terminals are prominent when the value of Rx decreases R2 R3 to a few Ohms m p Ry = the resistance of the connecting lead from R3 to V G Ry n Rx

R1 At point m: Ry is added to the unknown Rx, resulting in too Rx high and indication of Rx At point n: Ry is added to R3, therefore the measurement of Rx will be lower than it should be.

The effect of the connecting lead will be R1 nd rd At point p: RRxnp+=+() RR3 mp canceled out, if the sum of 2 and 3 term is R2 zero. R11Rnp R RRmp−= np 0 or = RR11 R RR rearrange Rx =+RR3 mp − R np 22mp RR22 Where R and R are the lead resistance R1 mp np RRx = 3 from m to p and n to p, respectively. R2 Kelvin Double Bridge: 1 to 0.00001 Ω

Four-Terminal Resistor Current Current terminals terminals Four-terminal resistors have current terminals and potential terminals. The resistance is defined as that between the potential terminals, so that contact voltage drops at the current terminals do not introduce errors. Voltage Voltage terminals terminals Four-Terminal Resistor and Kelvin Double Bridge

R 3 • r1 causes no effect on the balance condition. R2 • The effects of r2 and r3 could be minimized, if R1 >> r2 and Ra >> r3. Rb G • The main error comes from r4, even though this value is very small. Ra

r3 r4 R1 r2 Rx

r1 Kelvin Double Bridge: 1 to 0.00001 Ω

l 2 ratio arms: R1-R2 and Ra-Rb R the connecting lead between m and n: yoke R2 3 I m The balance conditions: Vlk = Vlmp or Vok = Vonp Rb p R2 V k VV= (1) G Ry lk R12+ R R a here n VIRIRR==lo[()//]3 +++ x RR a b R y R 1 R x Ry o VIRlmp=+3 R b (2) RRRaby++ R RR11RRby Ra 1 Eq. (1) = (2) and rearrange: RR=+ − RRx = 3 x 3  R R22RRRRRaby++ b 2

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 µA Is Is High Iv Iv V High voltage voltage V Material supply supply under test Is

(a) Circuit that measures insulation volume (b) Use of guard ring to measure only volume resistance in parallel with surface leakage resistance resistance V V RRRmeas== s// v RRmeas= v = Is + Iv 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

V 10000 V 9 RV = ==×Ω6.7 10 IV 1.5 µA

(b ) Surface leakage resistance: I = 5 µA–I = 3.5 µA IV+ IS = 5 µA S V

V 10000 V 9 RS = ==×Ω2.9 10 IS 3.5 µA MegaOhm Bridge

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

R RA RB A RB

E G E G R2

R1 Rx R RC C

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

RC R1 // RRCC≈ since R1 >> RC RR≈ xAR R // RR≈ B 2 g g since R2 >> Rg Capacitor

Capacitance – the ability of a dielectric to store electrical charge per unit voltage Area, A conductor ε A 0ε r Dielectric, εr C = Typical values pF, nF or µF thickness, d d

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 Electrolytic Aluminum 1-6800 µF 10-450 Tantalum 0.047 to 330 µF 6-50 Mica Stacked sheets 10-5000 pF 500-20,000 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 µ µ N 2 A L = o r A l

-7 l µo = 4π×10 H/m

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

L Re

Cd Iron core inductor Equivalent circuit of an RF coil Distributed capacitance Cd between turns Quality Factor of Inductor and Capacitor

Equivalent circuit of capacitance

Cp C s Rs

Parallel equivalent circuit Series equivalent circuit Equivalent circuit of Inductance Lp L s Rs

Series equivalent circuit Parallel equivalent circuit

22 22 2 2 RXss+ RXss+ RXpp XRpp Rp = X p = Rs = 22 X s = 22 Rs X s RXpp+ RXpp+ Quality Factor of Inductor and Capacitor

Quality factor of a coil: the ratio of reactance to resistance (frequency dependent and circuit configuration) XLω Inductance series circuit: Q ==ss Typical Q ~ 5 – 1000 RRss RR Inductance parallel circuit: Q ==pp XLppω

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

X p 1 Capacitance parallel circuit: D == Typical D ~ 10-4 – 0.1 RCRpppω

Rs Capacitance series circuit: D ==ωCRss X s Inductor and Capacitor

I V R2 V RL222+ω L =⋅P L L = SS⋅ L SP222 PSω22 RLPP+ω LS LP 22 RS LS 222 ω LP I RLSS+ω RRSP=⋅222 R RRPS= 2 ⋅ RL+ P R PPω S

V V/RP ωL RP Q = S θ Q = IωLS ωLP RS V/ωLP θ I IRS

I V 1+ω 222CR V PP 1 CCSP=⋅222 CCPS= ⋅ ωCRPP C 222 P 1+ω CRSS I RS LS 1 222 RR=⋅ RP 1+ω CR SP222 RR= SS⋅ 1+ω CRPP IR PS222 S I ωCRSS I/ωC VωC DCR= ω S P 1 SS δ D = ωCRPP V V/RP δ AC Bridge: Balance Condition

B all four arms are considered as impedance Z Z1 2 (frequency dependent components) The detector is an ac responding device: I I 1 2 headphone, ac meter V A D C Source: an ac voltage at desired frequency

Z1, Z2, Z3 and Z4 are the impedance of bridge arms Z 3 Z4 At balance point: E=EBA BC or IZ=IZ 1 1 2 2 D VV I=12 and I= General Form of the ac Bridge Z+Z13 Z+Z 24

Complex Form: ZZ14 =ZZ 23

Polar Form: Magnitude balance: ZZ14 =ZZ 23

ZZ14()()∠+∠θ 1θθθ 4 =ZZ 23 ∠+∠ 2 3

Phase balance: ∠θ1423+∠θθ= ∠ +∠ θ Example The impedance of the basic ac bridge are given as follows:

o o Z1 =Ω∠100 80 (inductive impedance) Z3 =∠Ω400 30 (inductive impedance)

Z2 =Ω250 (pure resistance) Z4 = unknown Determine the constants of the unknown arm.

SOLUTION The first condition for bridge balance requires that

ZZ23 250× 400 Z4 = ==Ω1, 000 Z1 100

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

o ∠∠+∠−∠=+−=−θθ42= θ 3 θ 1 0 30 80 50

Hence the unknown impedance Z4 can be written in polar form as

o Z4 =Ω∠−1,000 50 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 SOLUTION Z Z1 2

I I Z1 = RjL+=+Ωω 200 j 100 V 1 2 Z = RjC+=−Ω1/ω 300 j 600 A D C 2 Z3 ==R 450 Ω

Z4 = unknown Z 3 Z4 D

The general equation for bridge balance states that ZZ14 =ZZ 23

ZZ23 450× (200+ j 100) Z=4 = =Ωj150 Z1 (300− j 600)

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

Measure an unknown inductance or R 1 R capacitance by comparing with it with a known 2 inductance or capacitance. At balance point: ZZ =ZZ Vs D 12x 3 C R 3 x where 1 ZZ=Z112=;RR 2 ; and 3=+ R 3 R Unknown jωC3 3 Cx capacitance 11 RR+=+ RR Diagram of Capacitance 123x jωωCx j C3 Comparison Bridge R R R Separation of the real and imaginary terms yields: R = 23 and CC= 1 x R x 3 1 R2

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

Measure an unknown inductance or R1 R capacitance by comparing with it with a known 2 inductance or capacitance.

D At balance point: ZZ =ZZ Vs 12x 3 L 3 Lx where ZZ=Z112=;R RRjL 2 ; and 3= 3+ ω 3 Unknown R3 R x inductance R RjLRRjL+=+ωω Diagram of Inductance 12()x xSS( ) Comparison Bridge R R R Separation of the real and imaginary terms yields: R = 23 and LL= 2 x R x 3 1 R1

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

Measure an unknown inductance in terms of R 1 a known capacitance R2

V D At balance point: Z=ZZYx 2 31

L 1 x where Z=R ; Z=+R ; and Y= jωC R 2233 1 1 3 Unknown R1 R x inductance 1 Z=xxRj+=ωωLRR x23 +j C 1 Diagram of Maxwell Bridge R1

R23R Separation of the real and imaginary terms yields: Rx = and Lx = RRC231 R1

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 At balance point: ZZ13x =ZZ2 C1 V D where j Z=11R − ; Z 2==RR 2; and Z 33 ωC1 Lx

R3 Unknown ω R 1 x inductance R123++=()RxxjωLRR jC1 Diagram of Hay Bridge Lx R123RRRx += (1) LjR C1 xxω which expands to R1123Rxx+− +jωLR = RR CC11 Rx ω = ωLx R1 (2) C1 Solve the above equations simultaneously Hay Bridge: continues

22 RRC ω CRRR1123 231 Lx = Rx = 222 and 222 1+ CR 1+ω CR11 ω 11

ωL Z R1 X ωL x tanθ = L ==x Q θC L RRx

X C 1 tanθC == R CR11 θ ω 1 L 1 tanθθ== tan or Q Rx ωC Z LC 1 CR11 Phasor diagram of arm 4 and 1

R231RCω Lx = 2 Thus, Lx can be rewritten as 1(1/)+ Q

For high Q coil (> 10), the term (1/Q)2 can be neglected Lx ≈ RRC231 Schering Bridge

C1 Used extensively for the measurement of capacitance R2 and the quality of capacitor in term of D

R1 At balance point: Z=ZZYx 2 31 V D 11 where Z=Rj; Z=+ ; and Y = ωC 223jCω 1 R C ωω31 C x 3 Unknown jj− 1 Rx capacitance Rx −=R21 +jωC CCRxx1 Diagram of Schering Bridge

j RC21jR 2 which expands to Rx −= − ωωCCx 331 CR

C1 R1 Separation of the real and imaginary terms yields: RRx = 2 and CCx = 3 C3 R2 Schering Bridge: continues

Rx Dissipation factor of a series RC circuit: D ==ωRCx x X x 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 = ωRCxx= ω RC11

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

Unknown R1 Measure frequency of the voltage source using series Freq. R2 RC in one arm and parallel RC in the adjoining arm C1 ω At balance point: ZZZY2143= V s D 1 1 Z = R + ;Z = R ;Y = + jωC ; and Z = R R 1 1 2 2 3 3 4 4 3 j C1 R3 R C 4 ω 3 j 1 R21=−RR 4 +jωC 3 CR13 Diagram of Wien Bridge RRC 21=+3 (1) RR jR R C 14 4 43 R431RC which expands to RjCRR2314=+ω − + R3131CR C 1 ωCR = (2) ω 31 CR13 1 In most, Wien Bridge, R ω= R and C = C Rearrange Eq. (2) gives f = 1 3 1 3 2π CC1313 RR 1 (1) R24= 2R (2) f = 2π RC Wagner Ground Connection

One way to control stray capacitances is by C Shielding the arms, reduce the effect of stray C6 C5 capacitances but cannot eliminate them R1 R2 completely. Rw 1 Stray across arm ABD 2 Cannot eliminate C1 C2

C C Rx w 3 Wagner ground connection eliminates some R3 effects of stray capacitances in a bridge circuit Cx D Simultaneous balance of both bridge makes the Wagner ground C4 point 1 and 2 at the ground potential. (short C1 and C2 to ground, C4 and C5 are eliminated from detector circuit) Diagram of Wagner ground The capacitance across the bridge arms e.g. C6 cannot be eliminated by Wagner ground. Capacitor Values

Ceramic Capacitor Capacitor Values

Film Capacitor Capacitor Values

Chip Capacitor Capacitor Values

Tantalum Capacitor Capacitor Values

Chip Capacitor