Download chapter 1 - UniMAP Portal

CHAPTER 7
Alternating
Current Bridge.
School of Computer and Communication
Engineering, UniMAP
Prepared By:
Amir Razif b. Jamil Abdullah
EMT 113: March 27, 2007
1
7.0 AC Bridge.
7.1 Introduction to AC Bridge.
7.2 Similar-Angle Bridge.
7.3 Maxwell-Wein Bridge.
7.4 Opposite Angle Bridge.
7.5 Wein Bridge.
7.6 Scherning Bridge.
2
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.
3
Cont’d…
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.
I 1 Z1  I 2 Z 2
 The same thing at point d.
I1 Z 3  I 2 Z 4
Figure 7.2: Equivalent of Balance
(nulled) AC Bridge.
 From two above equation yield
general bridge equation;
4
Cont’d…
 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.
5
Example 7.1: AC Bridge.
The impedances of the AC bridge in Figure 7.4 are given as follows,
Z1  20030 0 
Z 2  1500 0 
Z 3  250  40 0 
Z x  Z 4  unknown
Determine the constants of the unknown
arm.
Solution:
Figure 7.4: Circuit For Example 7.1.
The first condition for bridge balance requires that
Z1Zx=Z2Z3
Zx =(Z2Z3/Z1)
= [(150 * 250)/200]
= 187.5 
6
Cont’d…
 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 .
7
7.2 Similar-Angle Bridge
 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.
Figure 7.5: Similar-Angle Bridge.
8
Cont’d…
 The impedance of the arm can be
written,
Z 1  R1
Z 2  R2
Z 3  R3  jX c 3
Z 4  R x  jX cx
 Substitute in the balance equation,
R1 Rx  jX cx   R3  jX c3 R2
 Further simplification,
R1 R x  jR1 X cx  R2 R3  jR2 X c 3
R1 R x  R2 R3
 jR1 X cx   jR2 X c 3
 jR1
1
1
  jR2
C x
C 3
R1C 3  R2 C x
Rx 
R2
R3
R1
Cx 
R1
C3
R2
9
7.3 Maxwell-Wein Bridge
 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.
Figure 7.6: Maxwell-Wein Bridge.
10
Cont’d…
 The impedance of the arm can be written as,
Z1 
1
1 / Rc  jC1
Z 2  R2
Z 3  R3
 Substitute in the balance equation,
Z 4  R x  jX LX
 Set real and imaginary part
to zero,
1
( R x  jX LX )  R2 R3
1 / R1  jC1
R x  jX LX 
L x  R2 R3 C1
R2 R3
 jR2 R3 C1
R1
11
7.4 Opposite-Angle Bridge
 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.
Figure 7.7: Opposite-Angle Bridge.
12
Cont’d…
 Equivalent series of inductance,
Lx 
R2 R3C1
1   2 R1 C1
2
2
 Equivalent series of resistance,
 R1 R2 R3C1
Rx 
2
2
1   2 R1 C1
2
2
 For the opposite angle bridge, it can be seen that the balance
conditions depend on the frequency at which the measurement is
13
made.
Example 7.2 (T2 2005): Opposite Angle Bridge.
Given the Opposite-Angle bridge of Figure 5. Find,
(i) The equivalent series resistance, Rx.
(ii) The inductance, Lx.
Solution:
Z1 Z x  Z 2 Z 3
Zx 
Z2Z3
R2 R3

Z1

j 
 R1 


C
1 

100 * 100



j
1K 

2
*

*
1
KHz
*
1

F


Z x  9.75  j1.552
R x  9.75
     (1)
or
Lx 
Lx 
X Lx  1.552
X Lx  L x
Lx 
X Lx

1.552
2 *  * 1KHz
L x  247.04 H

R2 R3 C1
1   2 R1 C1
2
2
100 *100 *1F
1  (2 * 1K ) 2 * (1K) 2 (1F ) 2
 247.045L
 2 R1 R2 R3C1 2
Rx 
2
2
1   2 R1 C1
(2 * 1K ) 2 * 1K *100 *100 * (1F ) 2
Rx 
1  (2 *1K ) 2 * (1K) 2 (1F ) 2 14
 9.75
7.5 Wein Bridge
 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.
Figure 7.8: Wein Bridge.
15
7.6 Schering Bridge.
 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.
Figure 7.9: Scherning Bridge.
16
Cont’d…
 The impedance of the arm of the
Schering bridge is,
 Substitute the value,
 Expand,
Z1 
1
1 / R1  1 /  jX c1
Z 2  R2
Z 3   jX c 3
Z 4  R x  jX x
Z4 
Z2Z3

Z1
R2 ( jX c 3 )
1
1 / R1  1 /  jX c1
 1
1 

 R2 ( jX c 3 ) 
R
jX
c1 
 1
  j  1

  jC1 
 R2 

 C 3  R1
Rx 
RC
jR2
j
 2 1
C 3
C3
C 3 R1
 Equating the real and imaginary terms,
R x  R2
C1
C2
R1
C x  C3
R2
17
Example 7.3: Schering Bridge.
Find the equivalent series element for the unknown impedance of the
Schering bridge network whose impedance measurements are to be
made at null.
R1 = 470 k
C1 = 0.01 mF
R2 = 100 k
C3 = 0.1 mF
Solution:
Find Rx and Cx ,
C1 (100 *10 3 ) * (0.01 *10 6 )
R x  R2

 10 K
6
C2
0.01 *10
R1 (0.01 *10 6 ) * (470 *10 3 )
6
C x  C3


0
.
47
*
10
F  0.47 F
3
R2
100 *10
.
18