note2_measuring_displacement
3. Displacement Measurement
In this section we will introduce potentiometer and LVDT for linear displacement, and optical encoder for rotary displacement. As load resistance greatly influences the measurement, we will introduce two-port network for calculating the effects of load disturbance.
3.1 Potentiometers
Potentiometers are electrical devices that can convert displacement to change of resistance. A typical linear potentiometer and its electric diagram are shown in Fig.1.
Fig.1 A linear potentiometer
The potentiometer consists of a sliding contact that moves over the length of a resistance element. Its non-linear error can be designed less than 1% for a wire wound potentiometer. It can measure a big length, say, a few hundred millimeters.
The potentiometer input is the distance of the contacting point from the bottom end, l , whilst the output is the voltage, U . When it is not connected with a load, the input/output relationship is linear. This can be seen in Fig.2.
Fig.2 Load effects of a potentiometer
Firstly, we have
L
l R r (1)
Which can be used to calculate r from R .
When there is no load,
L
l E U *
= (2) It is obvious that U is directly proportional to l and the gain is E /L .
When the load L R is connected, the R L is parallel connected with r and the effective resistance is
L
L L effect R r rR R r R +==// (3) Then, the potentiometer output (voltage on the load) is
effect effect
effect R R r R E R i U *)(*+-== (4)
The input/output relationship becomes nonlinear. This nonlinearity is caused by the load and therefore is called the load effects. Usually the load of the potentiometer is a meter for displacement or a computer interface. In both cases the load resistance is very big. Otherwise, a voltage follower (buffer) can be used to increase the load resistance.
Example 3.1 A potentiometer connecting to a voltage source of E=10V has Ω=k R 1 and the input range is 0~300 mm. Determine the nonlinear error caused by the load when the potentiometer measures 150 mm and the load is Ω=k R L 1. Further determine the error of f.s.d.
Solution.
According to (1) it is obtained
r = R*(l / L) = R*(150/300) = R/2 = 0.5 k Ω
Then, we calculate the linear output, U L (no load) according (2).
)(5300
150*10*V L l E U L ===
The nonlinear output, U N (with load) is calculated according to (4), where the effective resistance is calculated according (3).
Ω=+=+=
=3333.01
5.01*5.0//L L L effect R r rR R r R . )(43333.0*3333
.05.0110*)(*V R R r R E R i U effect effect effect N =+-=+-==
The nonlinear error is the difference between U N and U L ,
)(154V U U e L N N -=-=-=
The nonlinear error of f.s.d is
%10%10010
1%1000.)..(-=?-=?-=E e e N d s f N
3.3 Linear Variable Differential Transformer (LVDT)
LVDT is the most commonly used transducer for accurately measuring displacement up to about 300 mm. An LVDT has one primary winding and two identical secondary windings. A soft iron core is connected to the body for which the displacement is to be measured. The two secondary winding are such connected as shown in Fig.4.1 that the difference of their output is generated as the output of the LVDT.
Fig.4 LVDT and characteristics
When an a high frequency (5 kHz) AC voltage sine signal is applied to the primary side and the core is moved from one side (x=0) to the other (x=l), the output is a sine wave with its amplitude proportional to the input, the displacement of the core as shown in Fig.4.
Input: t V s ωsin , Output: )sin(?ω+t V o and 21V V V o -=
At x = 0: V 1 < V 2, therefore, V 0 = V 1 – V 2 < 0
At x = l/2: V 1 = V 2, therefore, V 0 = V 1 – V 2 = 0
At x = l: V 1 > V 2, therefore, V 0 = V 1 – V 2 > 0
as shown in the lower graph in Fig.4. This means that the LVDT has no ability to identify the direction of the core displacement, or to which side the core deviates from the neutral point x=l/2.
Phase-sensitive demodulator
A phase-sensitive demodulator is used to restore the direction, and then it followed by a low-pass filter to transfer the sine wave to a smooth DC signal. These are shown in Fig.5 and Fig.6.
Fig.5 Phase-sensitive demodulator
Fig.6 working principles of the phase-sensitive demodulator
Phase-sensitive demodulation circuit:
Fig.6 shows the circuit arrangement for phase-sensitive demodulation using semiconductor diodes. Ideally, these diodes pass current only in one direction; thus when f is positive and e is negative, the current path is e-f-g-c-d-h-e, while when f is negative and e is positive, the path is e-h-c-d-g-f-e. The current through R is therefore always from c to d. A similar situation exists in the lower diode bridge. In this way, the position of the core (higher or lower than the
e as shown in Fig.6.
central position) can be reflected by the sign of
A typical LVDT has the following static characteristics:
Range = 0.1 ~ 300 mm, Resolution = 0.05 mm, Accuracy (F.S.D.) = 0.5%
Tutorial Questions:
1. A potentiometer has a resistance element of Ω=k R 2, with the length of 200 mm and the applied voltage power of 10 V.
(i) When there is no load connected to the potentiometer output, calculate the output voltage for the contacting point being positioned at x= 25mm, 138mm, and 169mm. (ii) When a load of Ω=k R L 1is connected, calculate the output voltage for the contacting point being positioned at x = 50mm, 100mm, and 150mm.
2. A potentiometer is with a voltage power of E = 10 V, it has Ω=k R 1 and the input range is 0~100 mm. When the potentiometer measures 50mm and is with a load Ω=k R L 2,5,10, Determine the nonlinear error caused by the load and the errors of f.s.d.