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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
.