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CHAPTER 7CHAPTER 7
EMT 113: March 27, 2007EMT 113: March 27, 2007
School of Computer and CommunicationSchool of Computer and Communication
Engineering, UniMAPEngineering, UniMAP
Prepared By:Prepared By:
Amir Razif b. Jamil AbdullahAmir Razif b. Jamil Abdullah
AlternatingAlternating
Current Bridge.Current Bridge.
2
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.
3
7.1 Introduction to AC Bridge.7.1 Introduction to AC 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.
4
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…
5
 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…
6
Example 7.1:Example 7.1: AC Bridge.AC Bridge.
The impedances of the AC bridge inThe impedances of the AC bridge in Figure 7.4Figure 7.4 are given as follows,are given as follows,
Determine the constants of the unknownDetermine the constants of the unknown
arm.arm.
Solution:Solution:
The first condition for bridge balance requires that
Z1Zx=Z2Z3
Zx =(Z2Z3/Z1)
= [(150 * 250)/200]
= 187.5 Ω
Figure 7.4: Circuit For Example 7.1.
Ω∠= 0
1 30200Z
Ω∠= 0
2 0150Z
Ω−∠= 0
3 40250Z
unknownZZx == 4
7
 The second condition for balance requires that the sums of the
phase angles of opposite arms be equal,
θ1+ θ x = θ2 + θ3
θ x = θ2 + θ3 - θ1
= 0 + (-40o
) – 30o
= -70o
 The unknown impedance Zx, can be written as,
Zx = 187.5 Ω / -70
= (64.13 – j176.19)Ω
 This indicate that we are dealing with a capacitive element, possibly
consisting of a series resistor and a capacitor.
Cont’d…Cont’d…
8
 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.
9
 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 −=−
3
2
1
3
1
2
231
3
21
321
321
323211
11
C
R
R
C
R
R
R
R
CRCR
C
jR
C
jR
XjRXjR
RRRR
XjRRRXjRRR
x
x
x
x
ccx
x
ccxx
=
=
=
−=−
−=−
=
−=−
ωω
Cont’d…Cont’d…
10
 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.
11
 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
CjR
Z
+=
=
=
+
=
4
33
22
1
1
/1
1
ω
132
132
1
32
32
11
)(
/1
1
CRRL
CRRj
R
RR
XjR
RRjXR
CjR
x
LXx
LXx
=
+=+
=+
+
ωω
ω
Cont’d…Cont’d…
12
 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.
13
 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.
2
1
2
1
2
2
1321
2
1 CR
CRRR
Rx
ω
ω
+
=
2
1
2
1
2
132
1 CR
CRR
Lx
ω+
=
Cont’d…Cont’d…
14
Example 7.2 (T2 2005):Example 7.2 (T2 2005): Opposite Angle Bridge.Opposite Angle Bridge.
Given the Opposite-Angle bridge ofGiven 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, LLxx..
Solution:Solution:
HL
KHz
X
L
LX
X
R
jZ
FKHz
j
K
C
j
R
RR
Z
ZZ
Z
ZZZZ
x
Lx
x
xLx
Lx
x
x
x
x
µ
π
ω
ω
µπ
ω
04.247
1**2
552.1
552.1
75.9
)1(552.175.9
1*1**2
1
100*100
1
1
32
1
32
321
=
Ω
=
=
=
Ω=
Ω=
−−−−−Ω+=






−Ω
ΩΩ
=






−
==
=
Ω=
Ω+
ΩΩΩ
=
+
=
=
Ω+
ΩΩ
=
+
=
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
2
1
2
1
2
2
1321
2
222
2
1
2
1
2
132
FKK
FKK
R
CR
CRRR
R
L
FKK
F
L
CR
CRR
L
or
x
x
x
x
µπ
µπ
ω
ω
µ
µπ
µ
ω
15
 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.
16
 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.
17
 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
jXR
Z
−=
−=
=
−+
=
4
33
22
11
1
/1/1
1
13
2
3
12
3
1
13
2
11
32
11
32
1
32
4
1
11
)(
/1/1
1
)(
RC
jR
C
CR
C
j
R
Cj
RC
j
R
jXR
jXR
jXR
jXR
Z
ZZ
Z
x
c
c
c
c
ωω
ω
ω
−=−






−




 −
=






−−=
−+
−
==
2
1
3
2
1
2
R
R
CC
C
C
RR
x
x
=
=
Cont’d…Cont’d…
18
Example 7.3:Example 7.3: Schering Bridge.Schering Bridge.
Find the equivalent series element for the unknown impedance of theFind the equivalent series element for the unknown impedance of the
Schering bridge network whose impedance measurements are to beSchering bridge network whose impedance measurements are to be
made at null.made at null.
RR11 = 470 k= 470 kΩΩ CC11 = 0.01 mF= 0.01 mF
RR22 = 100 k= 100 kΩΩ CC33 = 0.1 mF= 0.1 mF
Solution:Solution:
Find Rxand Cx ,
FF
R
R
CC
K
C
C
RR
x
x
µ47.010*47.0
10*100
)10*470(*)10*01.0(
10
10*01.0
)10*01.0(*)10*100(
6
3
36
2
1
3
6
63
2
1
2
====
Ω===
−
−
−
−
.

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Bridge 1

  • 1. 1 CHAPTER 7CHAPTER 7 EMT 113: March 27, 2007EMT 113: March 27, 2007 School of Computer and CommunicationSchool of Computer and Communication Engineering, UniMAPEngineering, UniMAP Prepared By:Prepared By: Amir Razif b. Jamil AbdullahAmir Razif b. Jamil Abdullah AlternatingAlternating Current Bridge.Current Bridge.
  • 2. 2 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.
  • 3. 3 7.1 Introduction to AC Bridge.7.1 Introduction to AC 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.
  • 4. 4 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…
  • 5. 5  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…
  • 6. 6 Example 7.1:Example 7.1: AC Bridge.AC Bridge. The impedances of the AC bridge inThe impedances of the AC bridge in Figure 7.4Figure 7.4 are given as follows,are given as follows, Determine the constants of the unknownDetermine the constants of the unknown arm.arm. Solution:Solution: The first condition for bridge balance requires that Z1Zx=Z2Z3 Zx =(Z2Z3/Z1) = [(150 * 250)/200] = 187.5 Ω Figure 7.4: Circuit For Example 7.1. Ω∠= 0 1 30200Z Ω∠= 0 2 0150Z Ω−∠= 0 3 40250Z unknownZZx == 4
  • 7. 7  The second condition for balance requires that the sums of the phase angles of opposite arms be equal, θ1+ θ x = θ2 + θ3 θ x = θ2 + θ3 - θ1 = 0 + (-40o ) – 30o = -70o  The unknown impedance Zx, can be written as, Zx = 187.5 Ω / -70 = (64.13 – j176.19)Ω  This indicate that we are dealing with a capacitive element, possibly consisting of a series resistor and a capacitor. Cont’d…Cont’d…
  • 8. 8  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.
  • 9. 9  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 −=− 3 2 1 3 1 2 231 3 21 321 321 323211 11 C R R C R R R R CRCR C jR C jR XjRXjR RRRR XjRRRXjRRR x x x x ccx x ccxx = = = −=− −=− = −=− ωω Cont’d…Cont’d…
  • 10. 10  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.
  • 11. 11  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 CjR Z += = = + = 4 33 22 1 1 /1 1 ω 132 132 1 32 32 11 )( /1 1 CRRL CRRj R RR XjR RRjXR CjR x LXx LXx = +=+ =+ + ωω ω Cont’d…Cont’d…
  • 12. 12  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.
  • 13. 13  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. 2 1 2 1 2 2 1321 2 1 CR CRRR Rx ω ω + = 2 1 2 1 2 132 1 CR CRR Lx ω+ = Cont’d…Cont’d…
  • 14. 14 Example 7.2 (T2 2005):Example 7.2 (T2 2005): Opposite Angle Bridge.Opposite Angle Bridge. Given the Opposite-Angle bridge ofGiven 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, LLxx.. Solution:Solution: HL KHz X L LX X R jZ FKHz j K C j R RR Z ZZ Z ZZZZ x Lx x xLx Lx x x x x µ π ω ω µπ ω 04.247 1**2 552.1 552.1 75.9 )1(552.175.9 1*1**2 1 100*100 1 1 32 1 32 321 = Ω = = = Ω= Ω= −−−−−Ω+=       −Ω ΩΩ =       − == = Ω= Ω+ ΩΩΩ = + = = Ω+ ΩΩ = + = 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 2 1 2 1 2 2 1321 2 222 2 1 2 1 2 132 FKK FKK R CR CRRR R L FKK F L CR CRR L or x x x x µπ µπ ω ω µ µπ µ ω
  • 15. 15  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.
  • 16. 16  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.
  • 17. 17  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 jXR Z −= −= = −+ = 4 33 22 11 1 /1/1 1 13 2 3 12 3 1 13 2 11 32 11 32 1 32 4 1 11 )( /1/1 1 )( RC jR C CR C j R Cj RC j R jXR jXR jXR jXR Z ZZ Z x c c c c ωω ω ω −=−       −      − =       −−= −+ − == 2 1 3 2 1 2 R R CC C C RR x x = = Cont’d…Cont’d…
  • 18. 18 Example 7.3:Example 7.3: Schering Bridge.Schering Bridge. Find the equivalent series element for the unknown impedance of theFind the equivalent series element for the unknown impedance of the Schering bridge network whose impedance measurements are to beSchering bridge network whose impedance measurements are to be made at null.made at null. RR11 = 470 k= 470 kΩΩ CC11 = 0.01 mF= 0.01 mF RR22 = 100 k= 100 kΩΩ CC33 = 0.1 mF= 0.1 mF Solution:Solution: Find Rxand Cx , FF R R CC K C C RR x x µ47.010*47.0 10*100 )10*470(*)10*01.0( 10 10*01.0 )10*01.0(*)10*100( 6 3 36 2 1 3 6 63 2 1 2 ==== Ω=== − − − − .