Transcript Lect 6 transdcuer 1

Displacement Transducers/Sensors
Potentiometer for angular displacement
Wire Wound Potentiometers
Longitudinal Displacement
Potentiometer or longitudinal displacement
Theory of Operation:
The sensor consists of a length “L” of
resistance wire attached across a
voltage source “Vs”. The wiper is
pushed up or down by moving target, for
which displacement “x” is required to be
measured. Vo is the output voltage
representing displacement in terms of
volts and is given by:
x
vo =
vs
L
x
vo = vs
L
Wire-Wound Potentiometer
The resistance of the wire wound
potentiometer increases in step manner as
the wiper moves from one position to the
adjacent turn. This step change in resistance
limits the resolution of the potentiometer to L/n,
where n is the numbers of turns. The resolution
ranges from 0.05 to 1 percent are common.
Therefore such potentiometer are not suitable for
precise and finer movements.
Thin film potentiometer
The film resistance on an insulating substrate exhibits high
resolution, lower noise, and longer life. For example a
resistance of 50 to 100 Ohm/mm can be obtained with the
conductive plastic film
Thin Film Potentiometer
Thin Film potentiometer are introduced to
improve resolution. Movement can be nearly
continuous rather than in steps.
Thin Film
Potentiometer
For angular
Movements
Potentiometer …. slower
The dynamic response of both the linear and
the angular potentiometer is severely limited
by the inertia of the shaft and wiper
assembly. Since this inertia is large, the
potentiometer are used for static or quasistatic measurement where a high
frequency response is not required.
Application of Potentiometer
 Potentiometer
are normally used for:
Large displacement; i.e. 10mm or more
 15 deg or more in angular displacement

 Primary
advantage is its simplicity while
the disadvantage is its slower response to
fast changing.
Tutorial
1.
A slide wire potentiometer of 100mm length is
fabricated by wire wound with wire of diameter
0.1mm around a cylindrical insulating core.
Determine the resolution limit.
Solution:
Resolution: L/n
n= 100mm/0.1mm = 1000 turns
Therefore resolution= 100/1000 =0.1mm
Tutorial ….

If the potentiometer of previous example has
resistance of 2000 Ohms and can dissipate 2
watts of power, determine the voltage required
to maximize the sensitivity. What voltage change
corresponds to resolution limit?
Solution:
R per turn=2000/1000= 2
P=v2/R and v2 = 2/2 =1
Resolution = 1 v/ turn
Assignment….2
Q.No 1: A 20-turn potentiometer with calibrated dial (100
division per turn) is used to balance resistor in
Wheatstone Bridge. If the potentiometer has resistance
of 20 k and a resolution of 0.05 percent, what is the
minimum increment change in resistance R that can
be read from the calibrated dial?
Q.No 2: Design a displacement transducer that utilizes a
20-turn potentiometer to monitor the position of the
elevator over its 100-m range of travel; ie.( Wiper length
and resolution)