Download Chapter 6 - UniMAP Portal

CHAPTER 6
Direct-Current
Bridge.
School of Computer and Communication
Engineering, UniMAP
Prepared By:
Amir Razif b. Jamil Abdullah
EMT 113: V-2008
1
6.0 Direct Current Bridge.
6.1 Introduction to Bridge.
6.2 The Wheatstone Bridge.
6.2.1 Sensitivity of the Wheatstone Bridge.
6.2.2 Unbalance Wheatstone Bridge.
6.3 Kelvin Bridge.
2
6.1 Introduction to Bridge.
 Bridge circuits are the instruments for making comparison
measurements, are widely used to measure resistance,
inductance, capacitance and impedance.
 Bridge circuits operate on a null-indication principle, the
indication is independent of the calibration of the indicating
device or any characteristics of it. It is very accurate.
3
6.2 The Wheatstone Bridge.
 The Wheatstone bridge consists of two parallel resistance
branches with each branch containing two series resistor elements,
Figure 6.1.
 A DC voltage source is connected across the resistance network to
provide a source of current through the resistance network.
 A nul detector is the galvanometer which is connected between the
parallel branches to detect the balance condition.
 The Wheatstone bridge is an accurate and reliable instrument and
heavily used in the industries.
Figure 6.1: Wheatstone Bridge Circuit.
4
Cont’d…
Operation:
 We want to know the value of R4, vary one of the remaining
resistor until the current through the null detector decreases to
zero.
 The bridge is in balance condition, the voltage across resistor R3
is equal to the voltage drop across R4,(R3 = R4).
5
Cont’d…
 At balance the voltage drop at R1 and R2 must be equal to.
I 3 R3  I 4 R4
 No current go through the galvanometer G, the bridge is in balance
so,
I1 R1  I 2 R2
I2  I4
I1  I 3
I 1 R3  I 2 R4
 This equation, R1R4 = R2R3 , states
the condition for a balance Wheatstone
bridge and can be used to compute the
value of unknown resistor.
R1 R2

R3 R4
or
R1 R4  R2 R3
6
Example 6.1: Wheatstone Bridge.
Determine the value of unknown resistor, Rx in the circuit of Figure
6.2 assuming a null exist, current through the galvanometer is zero.
Figure 6.2: Circuit For Example 6.1.
Solution:
From the circuit, the product of the resistance in opposite arms of the
bridge is balance, so solving for Rx
Rx R1  R2 R3
R2 R3
Rx 
R1
.

15K * 32 K
 40 K
12 K
7
Example 6.1A(T2 2005): Wheatstone Bridge.
Calculate the value of Rx in the circuit of Figure 4 if VTh = 24 mV and
Ig =13.6 uA.
Figure 6.2A: Circuit For Example
6.1A.
Solution:
Calculate Rth
Ig 
Vth
RTh  R g
RTh 
Vth
 Rg
Ig
24mV
RTh 
 100
13.6A
RTh  1.665K
8
Calculate Rx
RTh  Rab  R1 // R3   R2 // R x 
RTh 
R1 R3
R R
 2 x
R1  R3 R2  R x
RTh 
R1 R3
R2 R x

R1  R3 R2  R x

RR 
( R2  R x ) *  RTh  1 3   R2 R x
R1  R3 

( R2  R x )
R2

Rx

RR 
 RTh  1 3 
R1  R3 

R2
R2

1
Rx 
R1 R3 
 RTh 

R

R
1
3 

.

RR 
R2  RTh  1 3 
R1  R3 

Rx 

RR 
R2   RTh  1 3 
R1  R3 

1K * 5 K 

1K1.665 K 

1K  5 K 

Rx 
1K * 5 K 

1K  1.665 K 

1K  5 K 

R x  4.941K
9
6.2.1 Sensitivity of the Wheatstone
Bridge.
 When the bridge is in unbalance 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.
 Sensitivity is the deflection per unit current.
 The more sensitive the galvanometer will deflect more with the
same amount of current.
S
 Total deflection D is,
mi lim eters deg rees radian


A
A
A
D  S *I
10
6.2.2 Unbalanced Wheatstone
Bridge.
 The current flows through the galvanometer can determine by
using Thevenin theorem.
Figure 6.3: Unbalance Wheatstone
Bridge.
RTh  Rab  R1 // R3   R2 // R4 

R1 R3
R R
 2 4
R1  R3 R2  R4
VTh  Va  Vb
E
R3
R4
E
R1  R3
R2  R4
11
Cont’d…
Figure 6.4: Thevenin’s Equivalent Circuit for an Unbalanced Wheatstone Bridge.
 The deflection current in the galvanometer is,
Vth
Ig 
R th  R g
 Rg = the internal resistance in the galvanometer
12
Example 6.2: Unbalance Wheatstone Bridge.
Calculate the current through the galvanometer in the circuit Figure
6.5. Given that E=6V, R1= 1kΩ, R2= 1.6kΩ, R3 = 3.5kΩ, R4= 7.5kΩ
and Rg=200Ω.
Figure 6.5: Circuit for Example 6.2.
Solution:
(1) Find Thevenin equivalent circuit as seen from by the
galvanometer,Vth is,
 R3
R4 


VTh  E 

 R3  R1 R4  R2 
3.5 K
7.5K


 6V 


 3.5 K  1K 7.5K  1.6 K 
 6V 0.778  0.824  0.276V
13
Cont’d…
(2) Find Thevenin’s equivalent resistance (Rth )is,
 R1 R3
R2 R4 


RTh  

 R1  R3 R2  R4 
 1K * 3.5 K 1.6 K * 7.5K 
 6


 1K  3.5K 1.6 K  7.5K 
 2.097 K
.
Figure 6.6: Thevenin’s Equivalent Circuit for the Example 6.2 Unbalance
Bridge.
14
Example 6.3: Slightly Unbalanced Wheatstone Bridge.
Use the approximate equation to calculate the current through the
galvanometer in Figure 6.7. The galvanometer resistance, Rg is 125Ω
and is center-zero 200-0-200-uA movement.
E=10V, R1=500Ω, R2=500Ω, R3 = 500 Ω and R4=525 Ω.
Figure 6.7: Circuit for Example 6.3.
Solution:
From formula,
Ig 
VTh
RTh  R g
(1) Find Thevenin equivalent voltage (Vth) is,
25
 r 
VTh  
 0.125V
  10V *
4
R
2000



(2) Find Thevenin equivalent resistance (Rth )is,
RTh  R  500
15
Cont’d…
(3) The current through the galvanometer (Ig)is,
VTh
0.125V
Ig 

RTh  R g 500  125
 200A
Observation: If the deflector is a 200-0-200-uA galvanometer, the
pointer deflected full scale for a 5% change in resistance.
.
16
6.3 Kelvin Bridge.
 The Kelvin Bridge is the modified version of the Wheatstone
Bridge.
 The modification is done to eliminate the effect of contact and
lead resistance when measuring unknown low resistance.
 By using Kelvin bridge, resistor within the range of 1 Ω to
approximately 1uΩ can be measured with high degree of accuracy.
 Figure 6.8 is the basic Kelvin bridge. The resistor Ric represent the
lead and contact resistance present in the Wheatstone bridge.
Figure 6.8: Basic Kelvin Bridge.
17
Cont’d…
 The second set of Ra and Rb compensates for this relatively low
lead contact resistance .
 At balance the ratio of Ra and Rb must be equal to the ratio of R1
to R3.
R 2 R3
Rx 
R1
R x R3

R2 R1
R x R3 Rb


R2 R1 Ra
18
Example 6.4: Kelvin Bridge.
Figure 6.9 is the Kelvin Bridge, the ratio of Ra to Rb is 1000. R1 is 5
Ohm and R1 =0.5 R2.
Find the value of Rx.
Figure 6.9: For Example 6.4.
Solution:
Calculate the resistance of Rx,
Rx Ra
1


R2 5 1000
R1 =0.5 R2, so calculate R2
R2 
Calculate the value of Rx
.
R1 5

 10
0.5 0.5
 1 
R x  R2 

1000


 1 
 10
  0.01
 1000 
19