Download Lecture AC-DC Bridges

CHAPTER 5
DC AND AC BRIDGE
13 Mac 2007
NURJULIANA JUHARI
INTRODUCTION
 DC
& AC Bridge are used to measure
resistance, inductance, capacitance and
impedance.
 Operate on a null indication principle. This
means the indication is independent of the
calibration of the indicating device or any
characteristics of it.
 Very high degrees of accuracy can be achieved
using the bridges
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Types of bridges
Two types of bridge are used in measurement:
1) DC bridge:
a) Wheatstone Bridge
b) Kelvin Bridge
2) AC bridge:
a) Similar Angle Bridge
b) Opposite Angle Bridge/Hay Bridge
c) Maxwell Bridge
d) Wein Bridge
e) Radio Frequency Bridge
f) Schering Bridge
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DIRECT - CURRENT (DC) BRIDGE
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a) WHEATSTONE BRIDGE
 Defination: Consists of two parallel resistance branches
branch containing two series elements.
with each
 Used for accurate measurements of resistance
A
D
C
B
Figure 5.1: Wheatstone Bridge Circuit
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OPERATION
1. The dc source, E is connected across the resistance network to
provide a source of current through the resistance network
2. null detector usually a galvanometer is connected between the
parallel branches (C-D) to detect a condition of balance.
3. No current, galvanometer= O pointer scale, have currentpointer scale deflect
4. Current flows and divides into the two arms at point A, ie I1 and
I2
5. Bridge is balance when no current through the galvanometer or
when the potential difference at points C & D is equal, potential
across the galvanometer is zero.
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 Lets say to measured R4, vary the remaining resistors until the
current through the null detector decrease to zero.
Then bridge is in balance condition, means voltage drop
across R3 and R4 is equal
We can say that,
I3R3= I4R4
(1)
At balance the voltage across R1 and R2 also equal, therefore
I1R1=I2R2
(2)
No current flows through galvanometer G when the bridge is
balance
I1 = I3
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and
I2=I4
(3)
NURJULIANA JUHARI
Cont…
Substitute (3) in Eq (1),
I1R3 = I2R4
(4)
Eq (2) devide Eq (3)
R1/R3 = R2/R4
Then rewritten as
R 1R 4 = R 2R 3
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Example 1
Figure 5.2
Find Rx?
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Sensitivity of the Wheatstone
Bridge
When the bridge is in an unbalanced
condition, current flows through the
galvanometer, causing a deflection of its
pointer. The amount of deflection is a
function of the sensitivity of the
galvanometer.
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Cont..
Deflection may be expressed in linear or angular units
of measure, and sensitivity can be expressed:
mi lim eters deg rees radians
S


A
A
A
TOTAL
DEFLECTION,
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D  SxI
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Unbalanced Wheatstone Bridge
- Interest to measure current
through the galvanometer
-Used Thevenin’s theorem
-Appling the voltage divider equation
at point a n b
Fig. 5-3: Unbalanced Wheatstone Bridge
Vth= Va-Vb
Vth = Va-Vb
 R3
R4 

Vth  Va  Vb  E

 R1  R3 R2  R4 
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Rb is internal resistance
Assumed voltage source
is low, so Rb = 0 Ω
We draw the bridge as
shown in Figure 5-3
Fig. 5-4:
Thevenin’s resistance
Therefore, the equivalent
resistance thevenin is:
Rth = R1//R3 + R2//R4
=
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R1R3/(R1 + R3) + R2R4(R2+R4)
NURJULIANA JUHARI
Thevenin’s Equivalent Circuit
Figure 5.5 : Thevenin’s Equivalent circuit
If a galvanometer is connected to terminal a and b,
the deflection current in the galvanometer is
Vth
Ig 
R th  R g
where Rg = the internal resistance in the galvanometer
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Example 2
R2 = 1.5 kΩ
R1 = 1.5 kΩ
Rg = 150 Ω
E= 6 V
G
R3 = 3 kΩ
R4 = 7.8 kΩ
Figure 5.6 : Unbalance Wheatstone Bridge
Calculate the current through the galvanometer ?
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Slightly Unbalanced Wheatstone
Bridge
If three of the four resistors in a bridge are equal to R and
the fourth differs by 5% or less, we can developed an
approximate but accurate expression for Thevenin’s
equivalent voltage and resistance. Consider the circuit in
Fig- 5.7
Figure 5.7 : Wheatstone Bridge with three equal arms
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Cont…
The voltage at point a is given as
R
 R  E
Va  E
 E

RR
 2R  2
The voltage at point b
R  r
Vb  E
R  R  r
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Cont..
Thevenin’s equivalent voltage is the difference in this voltage
1
r
 R  r


Vth  Vb  Va  E 
   E

 R  R  r 2 
 4 R  2r 
If ∆r is 5% of R or less, Thevenin equivalent voltage can simplifies to
 r 
Vth  E 

 4R 
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Cont..
Thevenin equivalent resistance, when ∆r is small compared to R, therefore
R R
Rth  
2 2
OR
Rth  R
We can draw the Thevenin equivalent circuit as shown in Figure 5.8
Figure 5.8: Approximate Thevenin’s equivalent circuit for a Wheatstone bridge
containing three equal resistors and a fourth resistor differing by 5% or less
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b) KELVIN BRIDGE
 is a modified version of the Wheatstone
bridge. The purpose of the modification is
to eliminate the effects of contact and lead
resistance when measuring unknown low
resistances.
Used to measure values of resistance
below 1 Ω .
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Fig. 5-9: Basic Kelvin Bridge showing a second set of ratio arms
The resistor Rlc shown in figure
represents the lead and contact resistance
present in the Wheatstone bridge. The
second set of ratio arms (Ra and Rb in
figure) compensates
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Cont..
It can be shown that, when a null exists, the value
for Rx is the same as that for the Wheatstone bridge,
which is
for this relatively low lead contact resistance. At balance
the ratio of Ra to Rb must be equal to the ratio of R1 to R3.
R2 R3
Rx 
R1
Therefore when a Kelvin Bridge is balanced
Rx R3 Rb


R2 R1 Ra
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ALTERNATING – CURRENT (AC) BRIDGES
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INTRODUCTION AC BRIDGE
AC bridges are used to measure inductance and
capacitances and all ac bridge circuits are based on the
Wheatstone bridge. The general ac bridge circuit consists
of 4 impedances, an ac voltage source, and detector as
shown in Figure below. In ac bridge circuit, the
impedances can be either pure resistance or complex
impedances.
Fig 5.10 : General AC bridge circuit
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Cont..
The operation of the bridge depends on the fact that when certain
specific circuit conditions apply, the detector current comes zero.
This is known as the null or balanced condition. Since the zero
current means that are is no voltage difference across detector.
Fig. 5-11: Equivalent of balanced ac bridge circuit
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Cont....
The voltages from point a to point b and point a o point c must be equal
I1 Z1 = I 2 Z2
(1)
The voltages from point d to point b and point d to point c must also equal,
therefore
I1 Z3 = I 2 Z4
equation (1) divide by equation (2)
Z1
Z2

Z3
Z4
…(3)
(2)
Which also can written as
Z1 Z 4  Z 2 Z 3
…(4)
Both magnitude and phase angle of each four impedance must
satisfy equation (3) and (4) for there to be a null or balance condition.
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Definition of electrical impedance
The impedance of a circuit element is defined as the ratio of
the phasor voltage across the element to the phasor current
through the element:
Vr
ZR 
Ir
It should be noted that although Z is the ratio of two
phasors, Z is not itself a phasor. That is, Z is not associated
with some sinusoidal function of time.
For DC circuits, the resistance is defined by Ohm's law to be
the ratio of the DC voltage across the resistor to the DC
current through the resistor:
VR
R
IR
where the VR and IR above are DC (constant real) values
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Definition of Reactance, X
Reactance is the imaginary part of impedance,
and is caused by the presence of inductors or
capacitors in the circuit. Reactance is denoted
by the symbol X and is measured in ohms.
A resistor's impedance is R (its resistance) and
its reactance, XR is 0.
A capacitance impedance:
XC = -1/ωC = -1/(2 πfC)
An inductive impedance:
XL = ωL = 2πfL
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a) Similar Angle Bridge
The similar angle bridge (refer figure below) is
used to measure the impedance of a capacitive
circuit. This bridge is sometimes called the
capacitance comparison bridge of the series
resistance capacitance bridge.
Fig 5-12 : Similar angle bridge
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The impedance of
the arms
Used equation
Z1 = R1
Z2 = R2
Z3 = R3- jXc3
Z4 = Rx -jXcx
Z 1 Z 4 = Z2 Z 3
(1)
Real terms and imaginary on each side must be equal
From Eq 2 we get
 jR1
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(2)
-jR1Xcx = -jR2X3
(3)
Solve Eq 2 and Eq 4
1
1
  jR2
C x
C 3
R1C3 = R2Cx
R1Rx=R2R3
R2
Rx 
R3
R1
……(5)
R1
Cx 
C3
R2
….. (6)
(4)
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b) Maxwell Bridge
• to
determine an unknown inductance with capacitance
standard
Fig 5-13 : Maxwell Bridge
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The impedance of the arms
Used equation
Z1=1/(1/R1 + jwC1)
Z2 = R2
Z3 = R3
Z4= Rx + jXLx
Z 1 Z 4 = Z2 Z 3
(1)
1
Rx  jX Lx   R2 R3
1 / R1  jC1
Rx  jX LX
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R2 R3

 jR2 R3C1
R1
Setting both imaginary and real
parts equal to zero, we get
R2 R3
Rx 
R1
Lx  R2 R3C1
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c) Opposite Angle Bridge
The Opposite Angle Bridge or Hay Bridge (see Figure below) is
used to measure the resistance and inductance of coils in which the
resistance is small fraction of the reactance XL, that is a coil having
a high Q, meaning a Q greater than 10.
R1 R2 R3C1
Rx 
2
2
2
1   R1 C1
2
Lx 
R2 R3 C1
1   R1 C1
2
2
2
Fig 5-14: Opposite angle bridge
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d) Wien Bridge
The Wein Bridge shown in Figure below has a series RC combination in
one arm and a parallel combination in the adjoining arm. It is designed to
measure frequency (extensively as a feedback arrangement for a
circuit). It can also be used for the measurement of an unknown
capacitor with great accuracy
Fig 5-15: Wein Bridge
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Equivalent parallel components
R1
R3 
R2

1
 R4 
2
2


R
C
4 4





R2
C3 
R1

1

 1   2 R 2C 2
4
4


C 4






R1
C4 
R2

1
 C3 
2
2
2


R
C
3
3

Equivalent series components
R2
R4 
R1
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
R3

 1   2 R 2C 2
3
3





NURJULIANA JUHARI
e) Schering Bridge
used for the precession measurement of capacitors and their insulating
properties for phase angle is nearly 90o.
Fig 5-16: Schering bridge
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Impedance of the arms
Z1= 1/(1/R1 + 1/-jXc1)
Z2= R2
Z3= -jXC3
Z4= Rx -jXX
Substituting to the general equation gives
balance equation
Z4 
Z 2Z3

Z1
R2  jX C 3 
1
1 / R1  1 /  jX C1
  j  1

  jC1 
 R2 

 C3  R1
Equating the real and imaginary
terms, we find that
C1
R x  R2
C3
R1
C x  C3
R2
And expanding
R2 C1
jR2
j
Rx 


C x
C3
C3 R1
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