Transcript Chapter 5 DCACBRIDGES_1_0

DC & AC BRIDGES
Part 1 (DC bridge)
Objectives
• Ability to explain operation of Wheatstone
Bridge and Kelvin Bridge.
• Ability to solve the Thevenin’s equivalent
circuit for an unbalance Wheatstone
Bridge.
• Define terms null or balance.
• Define sensitivity of Wheatstone bridge.
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
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
DC BRIDGES
The Wheatstone Bridge
The Kelvin Bridge
Wheatstone Bridge
A
Wheatstone bridge is a measuring
instrument invented by Samuel Hunter
Christie (British scientist & mathematician)
in 1833 and improved and popularized by
Sir Charles Wheatstone in 1843. It is
used to measure an unknown electrical
resistance by balancing two legs of a
bridge circuit, one leg of which includes the
unknown component. Its operation is
similar to the original potentiometer except
that in potentiometer circuits the meter
used is a sensitive galvanometer.
Sir Charles Wheatstone (1802 – 1875)
Wheatstone Bridge
Definition: Basic circuit configuration consists of two parallel
resistance branches with each branch containing two series elements
(resistors). To measure instruments or control instruments
Basic dc bridge used for accurate measurement of resistance:
R1R4  R2 R3
R 2R 3
R4 
R1
Fig. 5.1: Wheatstone bridge circuit
How a Wheatstone Bridge works?
• The dc source, E is connected across the resistance
network to provide a source of current through the
resistance network.
• The sensitive current indicating meter or null detector
usually a galvanometer is connected between the
parallel branches to detect a condition of balance.
• When there is no current through the meter, the
galvanometer pointer rests at 0 (midscale).
• Current in one direction causes the pointer to deflect
on one side and current in the opposite direction to
otherwise.
• The bridge is balanced when there is no current through
the galvanometer or the potential across the
galvanometer is zero.
Cont.
At balance condition;
voltage across R1 and R2 also equal, therefore
(1)
I1R1  I 2 R2
Voltage drop across R3 and R4 is equal
I3R3= I4R4
(2)
No current flows through galvanometer G when the
bridge is balance, therefore:
I1 = I3
and
I2=I4
(3)
Cont.
Substitute (3) in Eq (2),
I1R3 = I2R4
(4)
Eq (4) devide Eq (1)
R1/R3 = R2/R4
Then rewritten as
R1R4 = R2R3
(5)
Example 5-1
Figure 5.2 consists of the following, R1 = 12k, R2 = 15 k,
R3 = 32 k. Find the unknown resistance Rx.
Assume a null exists(current through the galvanometer
is zero).
Fig. 5-2: Circuit For example 5-1
Solution 5-1
RxR1 = R2R3
Rx
= R2R3/R1 = (15 x 32)/12 k,
Rx
= 40 k
Sensitivity of the
Wheatstone Bridge
When the bridge is in 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.
Cont.
Deflection may be expressed in linear or angular
units of measure, and sensitivity can be expressed:
milimeters
degrees radians
S


A
A
A
Total deflection,
D  SI
Unbalanced Wheatstone Bridge
Fig. 5-3: Unbalanced Wheatstone Bridge
Vth = Eab
 R3
R4 


E ab  E

 R1  R 3 R 2  R 4 
Fig. 5-4: Thevenin’s resistance
Rth = R1//R3 + R2//R4
=
R1R3/(R1 + R3)
+ R2R4(R2+R4)
Thévenin’s Theorem
An analytical tool used to extensively analyze an unbalance bridge.
Hermann von Helmholtz (1821 – 1894)
German Physicist
Léon Charles Thévenin (1857-1926)
French Engineer
Thévenin's theorem for electrical networks states that any combination of
voltage sources and resistors with two terminals is electrically equivalent to a
single voltage source V and a single series resistor R. For single frequency AC
systems the theorem can also be applied to general impedances, not just
resistors. The theorem was first discovered by German physicist Hermann
von Helmholtz in 1853, but was then rediscovered in 1883 by French
telegraph engineer Léon Charles Thévenin (1857-1926).
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
Example 5-2
R2 = 1.5 kΩ
R1 = 1.5 kΩ
Rg = 150 Ω
E= 6 V
G
R3 = 3 kΩ
R4 = 7.8 kΩ
Figure 5.5: Unbalance Wheatstone Bridge
Calculate the current through the galvanometer ?
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.
Eth 
E  r  r 
  E
4R
 4R 
1
 R  r
 r 
Vth  Vb  Va  
  E  
 E
 R  R  r 2 
 4 R  2r 
Cont..
To find Rth:
R R
Rth    R
2 2
An approximate Thevenin’s equivalent circuit
Example 5-3
500 Ω
10 V
500 Ω
G
500 Ω
525 Ω
Use the approximation equation to calculate the current through the
galvanometer in Figure above. The galvanometer resistance, Rg is 125
Ω and is a center zero 200-0-200-μA movement.
Kelvin Bridge
The 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 .
Fig. 5-6: Basic Kelvin Bridge
showing a second set of ratio arms
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
R2 R3
Rx 
R1
Therefore when a Kelvin Bridge is balanced
R x R3 Rb


R2 R1 Ra
Cont.
Fig. 5-6: 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
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
Example 5-4
If in Figure 5-6, the ratio of Ra and Rb is 1000, R1
is 5 and R1 =0.5R2. What is the value of Rx.
Solution
The resistance of Rx can be calculated by
using the equation,
Rx/R2=R3/5=1/1000
Since R1=0.5R2, the value of R2 is calculated as
R2=R1/0.5=5/0.5=10
So, Rx=R2(1/1000)=10 x (1/1000)=0.01
• CONTINUE…AC BRIDGE