R - DIEES

Conditioning electronics for resistive sensors
It’s not convenient to adopt a voltage divider
in case of resistive sensors!
•Low resolution: a small dR will produce a
negligible dVout;
•Low amplification (gain) can be used due to
voltage offset which saturates the system
output.
1
Conditioning electronics for resistive sensors
The Wheatstone bridge
C
R2
R1
V
- Vo +
A
R4
B
R3
D
2
Conditioning electronics for resistive sensors
The Wheatstone bridge
Zero balancement method
C
R2
R1
V
- Vo +
A
R4
B
R3
D
When the bridge is balanced (Vo=0) it is:
 R3
R4 
V
V0  

 R2  R3 R1  R4 
R1 R 2

R 4 R3
Which allows for the estimation of R3.
3
Conditioning electronics for resistive sensors
The Wheatstone bridge
C
R2
R1
V
- Vo +
A
R4
B
RR33=R0(1+x)
D
Eg. By tuning R2 it is possible to obtain
V0=0 and hence:
R4
R3 
R2
R1
4
Conditioning electronics for resistive sensors
The Wheatstone bridge, the effect of RW
C
R2
R1
V
- Vo +
A
B
Rw
R4
R3
D
Rw
R3  2 RW R4

R2
R1
R4
R3 
R2  2 RW
R1
RW will affect the R3 estimation
5
Conditioning electronics for resistive sensors
The Wheatstone bridge, three wires connection
C
R2
R1
V
Rw
- Vo +
A
B
R4
Rw
Rw
R3
D
R3  RW R4

R2  RW R1
R4
R2  RW 
R3  RW 
R1
assuming R1  R4
it will be : R3  R2
6
Conditioning electronics for resistive sensors
The automatic zero Wheatstone bridge
C
R2
R1
V
- Vo +
A
R4
B
RR3 3=R0(1+x)
D
•Practically, different solutions exist to
implement
automatic
balancement
operation, where R3 is indirectly estimated
by the action required to balance the
bridge.
•Mechanical balance system or digital
potentiometer
•Drawbacks: time consuming and
complex implementation
7
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge
C
R2
R1
V
- Vo +
A
R4
B
R3=R0(1+x)
D
 R3
R4 
V
V0  

 R2  R3 R1  R4 
 R0 1  x 
R4 
V
Vo  

 R2  R0 1  x  R1  R4 
1 
 1 x
Vo  

V
 k 1 x k 1
kx
Vo 
V
k  1k  1  x 
R1 R2
k 
R4 R0
8
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: linearity
C
R2
R1
V
- Vo +
A
R4
B
R3=R0(1+x)
D
(1)
kx
Vo 
V
k  1k  1  x 
(2)
R1 R2
k 
R4 R0
The linearity condition is: x<<k+1 !!!!
 kx 
V
V0  
2
  k  1 


9
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: responsivity
A variation of x (i.e. R3), as respect to x=0, will lead
to a variation of Vo:
S
DV0
V
kV
1
 0 
DR3 xR0 R0 k  1k  1  x 
Which is the k value assuring the maximum responsivity?
S
dS
 0  k2  1 x
dk
x= 0.01
k
10
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Two main conditions:
x  1  k
(1)
k  1 x
(2)
2
Strain Gauge conditioning

x  G  500010
6
2.0  0.01
In case of SG conditions (1) e (2) are fulfilled!!!
11
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Two main conditions:
x  1  k
(1)
k  1 x
(2)
2
PT100
x  DT
  0.00385
if DT  200C
x  0.8
hence by the (2) it follows that k  1.3
Eq . (1) 0.8  2.3 is not strictly fulfilled! ! 12
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design

x  G  5000 10
6
2.0  0.01
If the WB is powered by 5V,
in terms of voltage variation we can obtain:
R  R0 1  x   1201  0.01  121.2
DR  R  R0  1.2
5V * 0.01
V0 
 12.5mV
4
Signals must be amplified!
13
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
C
R2
R1
V
- Vo +
A
R4
B
R3=R0(1+x)
D
Maybe we can rise the driving voltage!!
Taking into account the maximum current
allowed for the SG is 30mA, the maximum supply
voltage is:
MAX
EXC
e
 2* R0 * 30 * 10
3
 7.2 V
which leads to an output voltage of:
V0 
7.2 * 0.01
 18mV
4
This quantity must be amplified!
14
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Resolution
Thermal noise define the minimum detectable
voltage variation!
Enoise,rms  4 kTRf
k  1.38 10 23 J / k
T  300K , f  100000Hz
Enoise,rms  0.45V
15
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Es. a temperature measurement in the
range [-10°C, 50°C] is required. The output
voltage span must be [-1V - 5V] and the
accuracy must be: 0.5% reading + 0.2%FS.
  0.4% / C
RTD specifications: R0 0C  100
5mW / C
Vo 
kx
V
k  1k  1  x 
k
R1 R2

R4 R0
Linearity condition:
The uncertainty due to the non linearity must be <0.005
kx
kx

k  12 k  1k  1  x   0.005
kx
k  1k  1  x 
Hence: x/(k+1)<<0.005 (*)
xmax  0.004 * 50  0.2
Considering xmax in (*): 0.2/(k+1)<<0.005
Which leads k>39. A good choice could be k=39
16
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Es. a temperature measurement in the range [-10°C, 50°C]
is required. The output voltage span must be [-1V - 5V] and
the accuracy must be: 0.5% reading + 0.2%FS.
RTD specifications:
  0.4% / C
R0 0C  100
5mW / C
k=39
k
R1 R2

R4 R0
Assuming R1=R2, R4=R0
R1=R2=3900, R4=R0=100 
The power voltage can be fixed taking into account the
constraint on the maximum power (self-heating)
0.2% 50C  0.1C
Hence a maximum power of 0.5 mW is admitted.
 V


P  R3I 2  R3 

 R2  R3 
Since R max  120
2
2
V


We can write 120
  0.5mW
 3900  120 
Thus obtaining V  8.2V
V=8 V
3
17
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Es. a temperature measurement in the range [-10°C, 50°C]
is required. The output voltage span must be [-1V - 5V] and
the accuracy must be: 0.5% reading + 0.2%FS.
RTD specifications:
  0.4% / C
R0 0C  100
5mW / C
Since we have fixed: k=39 , V=8 V
The responsivity is:
kV
S
 0.78 mV / C
2
k  1
We can fix the required Gain:
Vomax  0.78 * 50  39 mV
G AMPL  5 / 39 10 3  128.2
18
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
0.1
k=100
0
k=50
-0.1
k=10
Vo/V
k=5
k=1
-0.2
-0.3
-0.4
-0.5
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
x
19
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: design
Differential configuration
kR0
kR0
- Vo +
V
R0(1+x2)
R0(1+x1)
k ( x1  x2 )
Vo  V
(k  1  x1 )(k  1  x2 )
Per x1,x2<<(k+1)
k
Vo  V
( x  x2 )
2 1
(k  1)
20
Conditioning electronics for resistive sensors
Half bridge
R0(1-x)
R0
•Increased responsivity
•Linearity
- Vo +
V
R0(1+x)
R0
V0  V
R0 1  x 
2 R 1  x   2 R0
V
x
 V 0
V
R0 1  x   R0 1  x  2
4 R0
2
Full bridge
R0(1+x)
R0(1-x)
V0  Vx
- Vo +
V
R0(1-x)
R0(1+x)
21
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge:
SG applications
22
Conditioning electronics for resistive sensors
R2
R1
- Vo +
V
R4
R3
23
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: Dummy Gage
R0(1+y)
R0
- Vo +
V
R0(1+x)(1+y)
R0
R0 1  x 1  y 
V
V0  V
 
R0 1  x 1  y   R0 1  y  2

1  x V
21  x   2  x 
V
 V

1  x   1 2
22  x 
x
V
22  x 
Vo does not depend on y
24
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: Dummy Gage
25
Conditioning electronics for resistive sensors
The deflection Wheatstone bridge: linearization
V/R0
V
I3
V*
R0(1+x)
R0
Vo
I1
V
R0 R0
I2
R
V
Ix
V0   RI x   RI1  I 2 
I1 
V
V
, I3 
R0
R0
V *  R0 1  x I 3  V 1  x 
V*
V 1  x 
I2  

R0
R0
da cui :
 V V 1  x   R
  Vx
V0  -R(I1  I 2 )  -R  
R0  R0
 R0
27

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