1

1

2

2

Component RLC• Impedance = Potential Difference Phasor

Current PhasorI d R i t + j R t j 1• Impedance = Resistance + j Reactance , j =–1

Z = R + j X= |Z|cos + j |Z|sin= |Z|ej

= |Z|

j Im

|Z|

X

3

where |Z| = R2 + X2

= tan–1(X/R)• Admittance, Y = 1/Z

Re0

R

AC Response

• Resistor Imsint Vm

I

Vmsint R

I V

ResistanceV = R I

V

Im

I

4

I V

Phasor Diagram

= R Im sint= Vm sint

(V and I in phase)

3

AC Response (Cont’d)• Capacitor is a passive component storing the

energy in an electric field charged by the voltage across the dielectric.

I = C dV/dt Vm

Capacitive ReactanceI = dq/dt = C dV/dtV = 1/C Imsint dt + V0

1/ C I tI

V = Vmsin(t)C

I C dV/dt= Imsin(t + /2)

m

V

Im I

/2

0

= –1/C Imcost= 1/C Imsin(t–/2)= XCIm sin(t–/2)= Vmsin(t–/2)(V lags I by 90)

XC = 1/C = 1/2fC = /2C , Higher f, Lower XC (Lowpass Filter) 5

V

Phasor Diagram

AC Response (Cont’d)• Note that capacitive reactance is an opposition to

the change of voltage across an element.

• There are two choices in the literature for defining greactance for a capacitor. One is to use a uniform notion of reactance as the imaginary part of impedance, in which case the reactance of a capacitor is a negative number: XC = – 1/C and ZC = j XC = – j/C .

• Another choice is to define capacitive reactance as a positive number, however, one needs to remember to add a negative sign for the impedance of a capacitor: XC = 1/C and ZC = – j XC = 1/jC . 6

4

AC Response (Cont’d)• Inductor is a passive component storing the energy

in a magnetic field induced by the electric current passing through it. Imsin(t) Vm VImV = L dI/dt

= Vmsin(t+/2) L

V

Inductive ReactanceV = d(N)/dt

= L d(Imsint)/dt= L I cost

m

I/2

7

IV

Phasor Diagram

= L Imcost= L Imsin(t+/2)= XLIm sin(t+/2)= Vm sin(t+/2)

(V leads I by 90)

XL = L = 2fL = 2L/ , Higher f, Higher XL (Highpass Filter)

Frequency Response• The resonance of a series RLC circuit occurs when

the inductive and capacitive reactances are equal in magnitude but cancel each other because they are

180 apart in phase, (|Z| = R2 + (XL – XC)2 = R) and1 / 2f0C = 2f0L

f0 = 1 / 2LC

8

5

• Q-factor is to express the quality of component in ability to store and release energy or quality of L L+R,

Q = Energy Stored / Power Loss

Q-Factor and D-Factor

Q = Energy Stored / Power Loss= Reactance / Resistance= L / R= tan

• D-factor is for a dissipation of C C + R,D = 1/Q

IR

IL

IV

9

D = 1/Q= Power Loss / Energy Stored= R / (1/C)= RC= tan

IR

I(1/C)

I

V

Capacitor Model• An ideal capacitor stores but does not dissipate

energy.

• Because the dielectric separating the capacitor• Because the dielectric separating the capacitor plates are not a perfect insulator, it causes a small leakage current flowing through the capacitors parallel model. D = V2/R / V2/XC = 1/RC

Pl t l d t th i t f th l t• Plate loss due to the resistances of the plates and leads can become quite significant in higher frequency case series model.D = I2R / I2XC = RC

10

6

Inductor Model• An ideal inductor stores but does not dissipate

energy.

• Time varying current in a ferromagnetic• Time-varying current in a ferromagnetic inductor, which causes a time-varying magnetic field in its core, causes energy losses in the core material that are dissipated as heat parallel model. Q = V2/X / V2/R = R/LQ = V2/XL / V2/R = R/L

• Resistance of the wire series model.Q = I2XL / I2R = L/R

11

AC Bridge

Z1 Z

B Balanced Bridge,I |Z | = I |Z | Z1 Z2

Z3 Z4

DetectorI1

I2

AC

D

I1 |Z1| 1 = I2 |Z3| 3

I1 |Z2| 2 = I2 |Z4| 4

|Z1|/|Z2| 1–2 = |Z3|/|Z4| 3–4

12

D | 1| | 2| 1 2 | 3| | 4| 3 4

R1+jX1 = R3+jX3

R2+jX2 R4+jX4Wheatstone Bridge

7

Inductance MeasurementThere is no pure components, e.g. an inductor can

be considered to be a pure inductance (L4) in series with a pure resistance (R4).

Maxwell-Wien Bridge (for medium Q = 1-10)

Impedances,1/Z1 = 1/R1 + 1/(1/jC1)

Z1 = R1 / (1+jR1C1)

R1 R2

DC1

13

Z2 = R2

Z3 = R3

Z4 = R4 + jL4R3

R4

D

L4

1

Maxwell-Wien Bridge (Cont’d)Balanced bridge,

Z1 / Z2 = Z3 / Z4

Z4 = Z2Z3 / Z1

R4+jL4 = R2 R3 (1+jR1C1)/R1

= R2R3/R1 + jR2R3C1

Real part: R4 = R2R3/R1

Imagination part: L4 = R2R3C1

James Clerk Maxwell (1831-1879), a Scottish

scientist in the field of mathematical physics.

14

g p 4 2 3 1

The balancing is independent of frequency.Adjust R1 and R2 to get the bridge balanced (Null)

Q = L4/R4 = (R2R3C1)/(R2R3/R1) = R1C1

8

Hay Bridge

C1

I d

For high Q 10

R1R2

R3

D

L4

Impedances, Z1 = R1 + 1/jC1

= R1 – j/C1

Z2 = R2

Z = R

15

R3R4

4 Z3 = R3

Z4 = R4 + jL4

Hay Bridge (Cont’d)Balanced bridge,

Z4 = Z2Z3 / Z1

R4+jL4 = R2 R3 / (R1 – j/C1)

(R R L /C ) j ( R L R / C ) R R(R1R4 + L4/C1) + j (R1L4 – R4/C1) = R2R3

Imagination part: R1L4 = R4/C1

L4 = R4 / 2R1C1

Real part: R1R4 + L4/C1 = R2R3

R R + R /2R C 2 = R R

16

R1R4 + R4/2R1C12 = R2R3

R4 ( R1 + 1/2R1C12 ) = R2R3

R4 (2R12C1

2 + 1)/(2R1C12) = R2R3

R4 = (2R1R2R3C12) / (2R1

2C12 + 1)

L4 = (R2R3C1) / (2R12C1

2 + 1)

9

Hay Bridge (Cont’d)

Q = L4/R4

1/ R C= 1/R1C1

Therefore, L4 = (R2R3C1) / ( (1/Q2) + 1 )

R2R3C1 if Q 10

17

and R4 2R1R2R3C12

Owen Bridge

C R2

R3

D

L4

C1

C

Impedances, Z1 = 1/jC1

Z2 = R2

Z3 = R3 – j/C3

Z4 = R4 + jL4

18

R4

4C34 4 j 4

10

Owen Bridge (Cont’d)

Balanced bridge,

Z4 = Z2Z3 / Z14 2 3 1

R4+jL4 = R2 (R3 – j/C3) jC1

= R2C1/C3 + jR2R3C1

Real part: R4 = R2C1/C3

I i ti t L R R C

19

Imagination part: L4 = R2R3C1

The balancing is independent of frequency.

Q = L4/R4 = R2R3C1C3 / R2C1 = R3C3

Series Capacitance BridgeCapacitor can be considered to be a pure

capacitance in series with, or sometimes in parallel with, a pure resistance.in parallel with, a pure resistance.

R2D

R1C2 Impedances,

Z1 = R1

Z2 = R2 – j/C2

Z3 = R3

20

R3

R4

C4

3 3

Z4 = R4 – j/C4

11

Series Capacitance Bridge (Cont’d)

Balanced bridge,

Z4 = Z2Z3 / Z14 2 3 1

R4 – j/C4 = (R2 – j/C2) R3 / R1

= R2R3/R1 – j(R3/C2R1)

Real part: R4 = R2R3/R1

I i ti t C C R /R

21

Imagination part: C4 = C2R1/R3

Used for low D = 0.001-0.1

D = 1/Q = R4C4 = R2C2

Parallel Capacitance Bridge

I d

Used for high D > 0.05

R2

R

R4

D

R1

C2

Impedances, Z1 = R1

Z2 = 1 / (1/R2 + jC2)= R2/(1+jC2R2)

Z3 = R3

Z = R /(1+jC R )

22

R3C4

Z4 = R4/(1+jC4R4)

12

Parallel Capacitance Bridge (Cont’d)

Balanced bridge,

Z4 = Z2Z3 / Z14 2 3 1

R4 / (1+jC4R4) = R2R3 / (1+jC2R2)R1

R1R4 + jC2R1R2R4 = R2R3 + jC4R2R3R4

Real part: R1R4 = R2R3

R R R /R

23

R4 = R2R3/R1

Imagination part: C2R1R2R4 = C4 R2R3R4

C4 = C2R1/R3

D = 1/R4C4 = 1/R2C2

Schering Bridge

R

Used for very low D

R

D

C4

C2R1

C1

Impedances, Z1 = R1/(1+jC1R1)Z2 = 1/jC2

Z3 = R3

Z4 = R4 – j/C4

24

R3R4

C4

Harald Schering (25 November 1880 – 10 April 1959), a German physicist

13

Schering Bridge (Cont’d)Balanced bridge,

Z4 = Z2Z3 / Z1

R j/C = R (1+jC R ) / jC RR4 – j/C4 = R3(1+jC1R1) / jC2R1

= (C1R1R3 – jR3) / C2R1

= R3C1/C2 – j(R3/R1C2)

Real part: R4 = R3C1/C2

Imagination part: 1/C4 = R3/R1C2

25

g p 4 3 1 2

C4 = C2R1/R3

D = R4C4 = R1C1

Wien Bridge

RR I d

Used as frequency-dependent circuit

R2

R

R4

D

R1

C3

Impedances, Z1 = R1

Z2 = R2

Z3 = R3 – j/C3

Z4 = R4/(1+jC4R4)

26

R3 C4Max Karl Werner Wien (25 December 1866 –22 February 1938), a German physicist.

14

Wien Bridge (Cont’d)Balanced bridge,

Z4 = Z2Z3 / Z1

R4 / (1+jC4R4) = R2(R3 – j/C3) / R1

R R /R = R + R C /C + j(C R R 1/C )R1R4/R2 = R3 + R4C4/C3 + j(C4R3R4 – 1/C3)

Imagination part: C4R3R4 = 1/C3

C4R4 = 1/2C3R3

Real part: R1R4/R2= R3 + R4C4/C3

R4 = (R3R2C3 + R2R4C4)/ R1C32

27

= (R3R2C3 + R2/2C3R3) / R1C3

= R2(2C32R3

2 + 1) / (2C32R1R3)

and C4 = 1/2C3R3R4

= C3R1 / R2(2C32R3

2 + 1) D = 1/R4C4 = R3C3

Wien Bridge OscillatorA type of electronic oscillator that generates sine waves

with a large range of frequencies. At balance, Imagination part: C4R4 = 1/2C3R3Imagination part: C4R4 = 1/ C3R3

2 = 1 / R3R4C3C4

f = 1/ 2R3R4C3C4

If R3 = R4 = R and C3 = C4 = C

28

Then f = 1/ 2RC

15

Stray ImpedanceThere are stray capacitances between the

various element and the ground and it mat affect bridge balanceaffect bridge balance.

Z1 Z2

A

B

CDCstray

Cstray

29

Z3 Z4D

Cstray

stray

Cstray

Stray Impedance (Cont’d)The stray capacitances can be reduced by

earthing one side of AC supply.

Z1 Z2

A

B

CD

Cstray

30

Z3 Z4DCstray

16

Stray Impedance (Cont’d)The minimize stray capacitances between

the detector terminals and earth, Wagner earth is used.

ZB

Z1 Z2

Z Z

ACD

To ensuring that the points B and

31

Z3 Z4

D

Z5 Z6

D of a balanced bridge are at ground potential

Ex1. Maxwell-Wein Bridge

Impedances, = 2f

= 2(1k)

D

L4

= 2000 rad/sZ1 = R1 / (1+jR1C1)

= 1k 1+j(2000)(1k)(0.5)

= 92 – j289

32

R4 92 j289

Z2 = R2 = 600 Z3 = R3 = 400 Z4 = R4 + jL4

= R4 + j2000L4

f = 1kHz

17

at balance,

Z4 = Z2Z3 / Z1

R4 + j2000L4 = (600)(400) / (92 – j289)

Ex1. (Cont’d)

R4 + j2000L4 = (600)(400) / (92 – j289)

= 240 + j754

R4 = 240 L4 = 754 / 2000

= 120 mH= 120 mH

Q = L4 / R4

= (2000)(120m)/(240)

= 3.14 33

check,

R4 = R2R3 / R1

= (600)(400) / (1k)

Ex1. (Cont’d)

= (600)(400) / (1k)

= 240 CORRECT

L4 = R2R3C1

= (600)(400)(0.5)

=120 mH CORRECT=120 mH CORRECT

Q = R1C1

= (2000)(1k)(0.5)

= 3.14 CORRECT 34

18

R4 = ?R4 = 240

L4 = 120 mHL4 = ?

Ex1. (Cont’d)

Parallel model,

Z4 = (R4)(jL4) = 2(R4)(L4)2 + j(R4)2(L4) R4 + jL4 (R4)2 + 2(L4)2

S i d l

4Q = 3.14 Q = ?

Series model,

Z4 = R4 + jL4

= (240) + j(120m)

= Z4 35

R4 = 12 k

C4 = 10 pF

Ex2.

R4 = ?

C4 = ?

Equivalent Series Resistance (ESR)

D = ?D = 132.63Parallel model at 10 kHz source,

Z4 = R4/(1+jC4R4)

= (12k) / (1 + j2(10k)(10p)(12k))

36

= (12000) – j(90.47)

Series model,

Z4 = R4 – j/C4 = Z4 therefore

R4 = 12 k, C4 = 0.18 F, D= 135.72

19

37

Transformer Ratio BridgesNot only varying the impedances, but bridge can be

also balanced by varying the turns ratio of a transformer. There is a small number of standard

i t d it d ff t fresistors and capacitors and no effect of temperature changes.

Z2

I2

N2V2

38

Vs D

Z1

I1

N1 V1

20

Single Ratio Transformer BridgeTap a transformer voltage divider of Vs

V1 = kN1 = I1Z1 I1 = kN1 / Z1

V2 = kN2 = I2Z2 I2 = kN2 / Z2

To balance the bridge or no current through th d t t D N ll

39

the detector, D = Null

I1 = I2 Z1 / Z2 = N1 / N2

ImpedanceRatio

TurnRatio

Single Ratio Transformer Bridge (Cont’d)

• Resistance Measurement

Z1 = Unknown resistor Rx

Z2 = Standard resistor Rs

Rx = Rs N1

N2

40

21

Single Ratio Transformer Bridge (Cont’d)

• Capacitance Measurement

Z1 = Unknown Cx || Rx (leakage resistance)

= 1/(1/Rx + jCx)

= Rx / (1+jRxCx)

Z2 = Standard Cs || Rs

41

= Rs / (1+jRsCs)

Single Ratio Transformer Bridge (Cont’d)

• Capacitance Measurement (Cont’d)

Balanced, Z1 / Z2 = N1 / N2

1/Z1 = (N2/N1) 1/Z2

(1+jRxCx)/Rx = (N2/N1) (1+jRsCs)/Rs

1/Rx + jCx = (N2 / N1Rs) + jCsN2/N1

42

Real part: Rx = RsN1/N2

Imagination part: Cx = CsN2/N1

22

Single Ratio Transformer Bridge (Cont’d)

• Inductance Measurement

Z1 = Unknown Lx || Rx

= 1/(1/Rx + 1/jLx)

= 1/(1/Rx – j/Lx)

Z2 = Standard Cs || Rs

43

= 1/(1/Rs + jCs)

Single Ratio Transformer Bridge (Cont’d)

• Inductance Measurement (Cont’d)

Balanced, 1/Z1 = (N2/N1) 1/Z2

1/Rx – j/Lx = (N2/N1) (1/Rs + jCs)

Real part:

Rx = RsN1/N2

44

Imagination part:

1/Lx = – (N2/N1)Cs , Reversed Current

Lx = N1 / N22Cs

23

Double Ratio Transformer BridgeTo measure the impedance of components in Situ.

Z1

I1

Vs DZ2

N2

N1 V2

V1n1

n2

45

I1 = V1 / Z1 = k(N1+N2)/Z1

I2 = V2 / Z2 = kN2/Z2

I2

Double Ratio Transformer Bridge (Cont’d)

Balanced, null current or zero magnetic flux,

n I = n In1I1 = n2I2n1(N1+N2)/Z1 = n2N2/Z2

Z1 = Z2 n1(N1+N2)

n2N2

46

24

Q-MeterRLC Series ResonanceZ = R + jL + 1/jC R L

= R + j(L – 1/C)

Resonant frequency(When the voltage across

C is a maximum )

CVsVC

XL = LIm

47

C is a maximum.)0L = 1/0C0

2 = 1/LC0 = 1/LC

RXC = 1/C

L

Re

Q-Meter (Cont’d)I0 = Vs/R

VC = I0XC

= (Vs/R)(1/0C) (Vs/R)(1/0C)

= (1/0RC) Vs

= Q Vs

Q

where Q = Reactance/Resistance

48

= 0L / R

= 0(1/02C) / R

Q = 1 / 0RC(Unloaded)

25

Q-Meter (Cont’d)

RxLx

Unknown Inductor

Tuning to resonance, 0 = 1/LC

CVsVC = QVs

across very lowresistance

Oscillator TuningCapacitor

V

49

Then the unknown is determined by C and Q.

Lx = 1 / 02C

Rx = 1 / Q0C

Q-Meter (Cont’d)

The circuit can be adjusted to the resonance byresonance by

• Preset or fix the source frequency, and then vary the tuning capacitor for maximum value under this condition

• Preset the tuning capacitor and then

50

g padjust the source frequency

• This measures value of Q in commonly regarded as the Q of the coil under test

26

Q-Meter: Low Impedance Measurement

RxL

Unknown Large Capacitor

Cx

CVs

VC = QVsOscillator

R1L1

C10 Q1

51

Short circuit and tune C C1, L1 = L, R1

Q1 = 1/0R1C1

or R1 = 1/0C1Q1

Q-Meter: Low Impedance Measurement (Cont’d)R2

L2

C20 Q2

Remove short circuit and then tune C C2

C2 & Cx = C1, L2 = L, R2 = R1 & Rx

Series capacitors, C1 = 1 / (1/Cx + 1/C2)

= C C / (C +C )

52

= C2Cx / (Cx+C2)

C1Cx + C1C2 = C2Cx

Cx = C1C2 / (C2 – C1)

Q2 = 1/0R2C2 R2 = 1/0C2Q2

27

Q-Meter: Low Impedance Measurement (Cont’d)

R2 = R1 + Rx

Rx = R2 – R1

= 1/0C2Q2 – 1/0C1Q1

Rx = (C1Q1 – C2Q2) / (0C1C2Q1Q2)

Qx = 1/0RxCx

53

= (0C1C2Q1Q2)(C2–C1)

0(C1Q1 – C2Q2)C1C2

Qx = Q1Q2(C2–C1) / (C1Q1–C2Q2)

If the unknown component is an inductor,

Lx = 1/02Cx

Q-Meter: Low Impedance Measurement (Cont’d)

Lx = (C2 – C1) / 02C1C2

If the unknown component is a pure resistor (no reactance),

54

Rx = (C1Q1 – C2Q2) / (0C1C2Q1Q2)

Rx = (Q1 – Q2) / 0C1Q1Q2 , C1 = C2

28

Q-Meter: High Impedance Measurement

Rx

L

CVsVCOscillator Cx

Unknown

Open circuit and tune C C1, L1 = L, R1

Then short circuit and tune C C2

C || C C L L R R || R

For high resistance, inductance > 100mH, or capacitance < 400 pF

Unknown Small Capacitor

55

C2 || Cx = C1, L2 = L, R2 = Rx || R1

Cx + C2 = C1 Cx = C1 – C2

R2 = Rx||R1

= RxR1/(Rx+R1) Rx = R1R2/(R1–R2)

Parallel-model C Q = 0RC (loaded)

R1 = Q1/0C1 and R2 = Q2/0C2

Q-Meter: High Impedance Measurement (Cont’d)

Therefore

Rx = Q1Q20C1C2 / 02C1C2(Q1C2 – Q2C1)

Rx = Q1Q2 / 0(Q1C2 – Q2C1)

56

Qx = 0RxCx

= 0Q1Q2(C1 – C2) / 0(Q1C2 – Q2C1)

Qx = Q1Q2(C1 – C2) / (Q1C2 – Q2C1)

29

For unknown inductance,

Lx = 1/02Cx2

Q-Meter: High Impedance Measurement (Cont’d)

Lx = 1/02(C1 – C2)

For pure resistance,

Rx = Q1Q2 / 0(Q1C2 – Q2C1)

57

Rx = Q1Q2 / 0C1(Q1 – Q2) , C1 = C2

References• http://www.faqs.org/docs/electric/DC/DC_8.html

• http://avstop.com/ac/Aviation_Maintenance_Technician_Handbook_General/10-74.html

http://www wisc online com/Objects/ViewObject• http://www.wisc-online.com/Objects/ViewObject. aspx?ID=DCE7104

• Hotek Technologies, Inc webpage : http://www.hotektech.com/

• Yokogawa webpage: http://tmi.yokogawa.com/us/

• MAGNET LAB – Wheatstone Bridge webpage: http://www.magnet.fsu.edu/education/tutorials/java/wheatstonebridge/index.html

• Electronics Demonstrations webpage:http://www.falstad.com/circuit/e-index.html 58