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CHAPTER CHAPTER 77

EMT 113: March 27, EMT 113: March 27, 20072007

School of Computer and School of Computer and Communication Engineering, UniMAPCommunication Engineering, UniMAP

Prepared By: Prepared By: Amir Razif b. Jamil AbdullahAmir Razif b. Jamil Abdullah

Alternating Alternating Current Current Bridge.Bridge.

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7.1 Introduction to AC Bridge.7.1 Introduction to AC Bridge.7.2 Similar-Angle Bridge.7.2 Similar-Angle Bridge.7.3 Maxwell-Wein Bridge.7.3 Maxwell-Wein Bridge.7.4 Opposite Angle Bridge.7.4 Opposite Angle Bridge.7.5 Wein Bridge.7.5 Wein Bridge.7.6 Scherning Bridge.7.6 Scherning Bridge.

7.0 AC Bridge.7.0 AC Bridge.

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7.1 Introduction to AC 7.1 Introduction to AC Bridge.Bridge. AC bridges are used to measure inductance and

capacitance. All the AC bridges are based on the Wheatstone

bridge. In the AC bridge the bridge circuit consists of

four impedances and an ac voltage source. The impedances can either be pure resistance or

complex impedance. Other than measurement of unknown impedance,

AC bridge are commonly used for shifting phase.

Figure 7.1: General AC Bridge Circuit.

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Operation of AC Bridge:Operation of AC Bridge: When the specific circuit conditions

apply, the detector current becomes zero, which is known as null or balance condition.

Since zero current, it means that there is no voltage difference across the detector, Figure 7.2.

Voltage at point b and c are equal.

The same thing at point d.

From two above equation yield general bridge equation;

Figure 7.2: Equivalent of Balance (nulled) AC Bridge.

2211 ZIZI

4231 ZIZI

Cont’d…Cont’d…

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Figure 7. 3(a) and 7.3 (b) is a simple AC Bridge circuit.

Figure 7.3: (a) and (b) are Simple AC Bridge Circuit.

Cont’d…Cont’d…

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Example 7.1:Example 7.1: AC Bridge.AC Bridge.The impedances of the AC bridge in The impedances of the AC bridge in Figure 7.4Figure 7.4 are given as follows, are given as follows,

Determine the constants of the unknown Determine the constants of the unknown arm.arm.Solution:Solution:

The first condition for bridge balance requires thatZ1Zx=Z2Z3

Zx =(Z2Z3/Z1) = [(150 * 250)/200]

= 187.5

Figure 7.4: Circuit For Example 7.1.

01 30200Z

02 0150Z

03 40250Z

unknownZZ x 4

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

1+ x = 2 + 3

x = 2 + 3 - 1

= 0 + (-40o) – 30o

= -70o

The unknown impedance Zx, can be written as,

Zx = 187.5 / -70 = (64.13 – j176.19)

This indicate that we are dealing with a capacitive element, possibly consisting of a series resistor and a capacitor .

                   

Cont’d…Cont’d…

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Figure 7.5 is a simple form of Similar–Angle Bridge, which is used to measure the impedance of a capacitive circuit.

Sometimes called the capacitance comparison bridge or series resistance capacitance bridge.

7.2 Similar-Angle Bridge7.2 Similar-Angle Bridge

Figure 7.5: Similar-Angle Bridge.

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The impedance of the arm can be written,

Substitute in the balance equation,

Further simplification,                    

cxx

c

jXRZ

jXRZ

RZ

RZ

4

333

22

11

2331 RjXRjXRR ccxx

32

1

31

2

231

321

321

321

323211

11

CR

RC

RR

RR

CRCR

CjR

CjR

XjRXjR

RRRR

XjRRRXjRRR

x

x

x

x

ccx

x

ccxx

Cont’d…Cont’d…

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It is used to measure unknown inductances with capacitance standard.

Because the phase shifts of inductors and capacitors are exactly opposite each other, a capacitive impedance can balance out an inductive impedance if they are located in opposite legs of a bridge

Figure 7.6 is the Maxwell-Wein Bridge or sometimes called a Maxwell bridge.

7.3 Maxwell-Wein Bridge7.3 Maxwell-Wein Bridge

Figure 7.6: Maxwell-Wein Bridge.

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The impedance of the arm can be written as,

Substitute in the balance equation,

Set real and imaginary part to zero,

                   

LXx

c

jXRZ

RZ

RZ

CjRZ

4

33

22

11 /1

1

132

1321

32

3211

)(/1

1

CRRL

CRRjR

RRXjR

RRjXRCjR

x

LXx

LXx

Cont’d…Cont’d…

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This bridge is from Similar-Angle Bridge but the capacitance is replace with the inductance, Figure 7.7.

It is used to measure inductance.

Sometimes called a Hay bridge.

7.4 Opposite-Angle Bridge7.4 Opposite-Angle Bridge

Figure 7.7: Opposite-Angle Bridge.

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Equivalent series of inductance,

Equivalent series of resistance,

For the opposite angle bridge, it can be seen that the balance conditions depend on the frequency at which the measurement is made.

                   

21

21

2

21321

2

1 CR

CRRRRx

21

21

2

132

1 CR

CRRLx

Cont’d…Cont’d…

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Example 7.2 (T2 2005):Example 7.2 (T2 2005): Opposite Angle Opposite Angle Bridge.Bridge.

Given the Opposite-Angle bridge of Given the Opposite-Angle bridge of Figure 5Figure 5. Find,. Find,

(i) The equivalent series resistance, R(i) The equivalent series resistance, Rxx..

(ii) The inductance,(ii) The inductance, L Lxx..

Solution:Solution:

HLKHz

XL

LX

X

R

jZ

FKHz

jK

C

jR

RR

Z

ZZZ

ZZZZ

x

Lxx

xLx

Lx

x

x

x

x

04.2471**2

552.1

552.1

75.9

)1(552.175.9

1*1**21

100*1001

1

32

1

32

321

75.9

)1()1(*)1*2(1

)1(*100*100*1*)1*2(

1

045.247

)1()1(*)1*2(1

1*100*100

1

222

22

21

21

2

21321

2

222

21

21

2

132

FKK

FKKR

CR

CRRRR

L

FKK

FL

CR

CRRL

or

x

x

x

x

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The Wein Bridge is versatile where it can measure either the equivalent –series components or the equivalent-parallel components of an impedance, Figure 7.8.

This bridge is used extensively as a feedback for the Wein bridge oscillator circuit.

7.5 Wein Bridge7.5 Wein Bridge

Figure 7.8: Wein Bridge.

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The Scherning Bridge is useful for measuring insulating properties, that is for phase angles of very nearly 90o.

Figure 7.9 is the Scherning Bridge. Arm 1 contains only a capacitor C3. This capacitor

has very low losses (no resistance) and therefore the phase angle of approximately 90o.

7.6 Schering Bridge.7.6 Schering Bridge.

Figure 7.9: Scherning Bridge.

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The impedance of the arm of the Schering bridge is,

Substitute the value,

Expand,

Equating the real and imaginary terms,

xx

c

c

jXRZ

jXZ

RZ

jXRZ

4

33

22

111 /1/1

1

13

2

3

12

3

113

2

1132

11

32

1

324

1

11)(

/1/1

1)(

RC

jR

C

CR

C

jR

CjRC

jR

jXRjXR

jXR

jXR

Z

ZZZ

x

cc

c

c

2

13

2

12

R

RCC

C

CRR

x

x

Cont’d…Cont’d…

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Example 7.3:Example 7.3: Schering Bridge.Schering Bridge.Find the equivalent series element for the unknown Find the equivalent series element for the unknown impedance of the Schering bridge network whose impedance of the Schering bridge network whose impedance measurements are to be made at null.impedance measurements are to be made at null.

RR1 1 = 470 k= 470 k CC1 1 = 0.01 mF= 0.01 mF

RR2 2 = 100 k= 100 k CC3 3 = 0.1 mF= 0.1 mF

Solution:Solution:

Find Rx and Cx ,

FFR

RCC

KC

CRR

x

x

47.010*47.010*100

)10*470(*)10*01.0(

1010*01.0

)10*01.0(*)10*100(

63

36

2

13

6

63

2

12

.