# TKim

```BASIC SENSORS AND PRINCIPLES
Introduction of Resistive Sensor
① Strain gages
• Measurement of extremely small displacement
① Potentiometers
• Translational and Rotational displacement
② FSR
and so on…
(Axial) Stress
On the surface,
The average force per area
is denoted as

A = xy
F

A

F = mg
(N )
(m 2 )
F is sometimes called as “LOAD”
; Stress
-solid
(cf. Pressure)
-liquid, gas
Strain
L
F

( m)
 
L ( m)

F
; Strain
(unitless)
 &amp;  CURVE
① Brittle material (ex. glass)

n
rupture
Broken : Brittle material’s property
; Non-linear

Linear region
; elastic region (like spring)
Not linear ; over a wide range
 &amp;  CURVE

 ys
② Ductile material (ex. Al, Steel)

n
 PL
rupture
 PL
rupture
rupture
; Al
elastic region
plastic
region
For elastic region ( Linear region )
(  &lt; PL )
; Steel
elastic
region
plastic
region
  E
Young’s modules
(cf. y = kx)
(modules of elasticity)
In Summary, so far...
F

A

  E

L
Cantilever
 E
L
F
F

E
A
L

E : constant
A &amp; L are almost constant
AE
F
  AE 
L
; Provided that you know  ,
it is possible to measure F
Strain gage is widely used to measure F
The Principle of Weighing Machine
F   
in terms of a variety of structures.
F  mg    
Able to derive ‘m’
Strain Gages
L
Substrate

A
: resistivity
Electrical wire
8XL
L
R
A
All variables changed
for volume constant
Eight
Electrical
X
times length
; The relation between
Strain and Resistor
F
F
Length
Length
( elastic region )
Strain Gages
Partial Derivative in order to know the relation to each component
R L
L
  R  
 A
A
L
R
A
R 

  RL  L
L A
A
R   L
L
 2  RA   2 A
A
A
A
Strain Gages
R  R  RL  RA
Measurable
size
L

L
 d   dL   2 dA
A
A
A
R

L
A



R

L
A
Resistor
Strain
Strain Gages

Poisson’s ratio
with
D
L
 
D
L
L
D
D D
R

L

 (1  2 )
R

L
Piezo-resistive
effect
Dimensional
effect
Strain Gages
R
Gage factor
R


R
R
G

 1  2 
L
L

L
L
• For metal strain gage
• For semiconductor strain gage
G : ~1.6
G : 100 ~ 170
(High temperature coefficient)
Problem (3)
 Four metal strain gages which gage factor is 10 are attached
on a plain. By forcing F to the plain, Gage1 and 2 are
expanded as long as ∆L, whereas Gage3 and 4 are shorten
in the same length.
It has a relation that ∆L/L = kf ,k is constant.
Design a bridge circuit getting output voltage in proportion to
F, describe output voltage as F.
Voltage source of the bridge circuit is dc 5[V].
Problem (3)
G
① ②
R
R  10  R  10  L
L
R
L
L
Gage1&amp; 2 : L  L  L
Gage3&amp; 4 : L  L  L
③ ④
(Structure + Strain gage)
①
②
Top view
L

k f
L
f   AE
1

f
AE
Problem (3)
③
②
Vo  Av(Va  VB )
①
④
And LPF
Register
Variable
Register
Resistive
Sensor
R1
R R
R R
Va 
5 
5 
5
R1  R3
R RR R
2R
R4
R R
R R
Vb 
5 
5 
5
R2  R4
R RR R
2R
10  R
L
Vo  Av 
 5 Av 10 
2R
L
Vo  50 Avkf
Part of it
5 X 10
Your
design
Given by
structure
and material
Problem (4)
Consider to design a system measuring force by using both two Ptype Si strain gages which gage factor is 100 and two N-type Si strain
gages which one is -100.
(a)Design a circuit including a bridge circuit having four strain
gages as well as instrumentation amplifier in order to magnify
output. Specify the type of each strain gage composing the bridge
circuit.
(b) Assuming that both top and bottom of cantilever is changed in the
same length in case that forced. By forced F, maximum change of the
length of strain gage is +0.05%, resistor is 200 without
Specify gain in order output to vary in the range between -5V to +5V.
(c) Derive to calibrate this kind of instrument.
Problem (4)
P-type Si Strain gage S1 &amp; S2 : G = 100
N-type Si Strain gage S3 &amp; S4 : G = -100
G
S3
S1
S1
S3
&lt;Top View&gt;
R
L
R
L
S4
S2
S2
S4
&lt;Bottom View&gt;
Problem (4)
S3
S4
Vo  Av(Va  VB )
S1
S2
And LPF
Va 
R1
R R
R R
E 
E 
E
R1  R3
R RR R
2R
Vb 
R4
R R
R R
E 
E 
E
R2  R4
R RR R
2R
Problem (4)
R
R
L
 Av  5  G 
L
Vo  AvE
Vomax  Av  5 100  0.0005
 Av  0.25  5V
5
 Av 
 20
0.25
 max 
L
 0.0005
L
R  200
Problem (4)
 Calibration
F : 0 – 100N
Change f by using different mass
to measure V0
Use Least Square Method to
find the Calibration Eqution.
Problem (6)
2 P-type Si strain gages and 2 N-type Si strain gages are
attached below diaphragm of Pressure sensor.
In case of pressure on diaphragm, same strain occur at the each strain
gage with its sensitivity of 105 %/mmHg, its resistor is 50  without
any pressure.
Assuming that it is linear between pressure and strain.
Catheter
Liquid
Sensor
S3
S4
Diaphragm
S2
S1
Problem (6)
(a) In variation of pressure from 0 to 500mmHg , how each resistor of
P-type and N-type Si strain gages changed.
R  Ro  G  Sensitivity  pressure
RP  50  100  0.0000001 500
 50.005
RN  50 100  0.0000001 500
 49.995
Problem (6)
(b) Design a bridge circuit included 4 strain gages and specify each
strain gage in the circuit.
S3
S1
S4
S2
Problem (6)
(c) Add instrumentation Amplifier to vary from 0V to 1V with dc 1V.
Specify its gain.
S3
S4
Vo  Av(Va  VB )
S1
S2
And LPF
Va 
R1
R R
R R
E 
E 
E
R1  R3
R RR R
2R
Vb 
R4
R R
R R
E 
E 
E
R2  R4
R RR R
2R
Problem (6)
R
Vo  AvE
R
L
 Av 1 G 
L
 max 
Vomax  Av 1100  0.0000001
 Av  0.00001  1V
1
 Av 
 100000
0.00001
L
 0.00005(@ 500mmHg )
L
R  50
```