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lecture 6
Dr. Ali Karimpour, Apr 2017
Instrumentation
In The Name of Allah
Dr. Ali Karimpour
Associate Professor
Ferdowsi University of Mashhad
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lecture 6
Dr. Ali Karimpour, Apr 2017
2
Signal Conditioners and Transmission
Topics to be covered
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
Lecture 6
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lecture 6
Dr. Ali Karimpour, Apr 2017
3
Introduction
In manufacturing process electric signals are interested, but:
Signals may be much too small. It goes down instead of up.
It has undesired dc offset. It may be nonlinear.
Series resistance in the wiring and connectors …
Interference from power sources will introduce 60-Hz and RFI
noises.
Failures, faults, and installation errors may introduce several
hundred volts into the signal.
Difference in earth ground potential can also cause erroneous
reading several times larger than the signal.
This problems combine to make virtually useless the signal
that arrives at the display and controller.
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lecture 6
Dr. Ali Karimpour, Apr 2017
4
Introduction
Fortunately, each of mentioned problems has a solution.
Proper signal conversion and transmission.
Current transmission rather than voltage.
Encoding the information in the frequency of the signal rather than
its amplitude.
Isolation amplifiers and couplers protect circuiting from fault
voltage.
Proper shielding and grounding break ground loop.
Signals may be much too small.
It goes down instead of up.
It has undesired dc offset.
It may be nonlinear.
Series resistance in the wiring
and connectors …
Interference from power sources
will introduce 60-Hz and RFI noises.
Failures, faults, and installation errors
may introduce several hundred volts
into the signal.Difference in earth ground potential
can also cause erroneous reading several
times larger than the signal.
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lecture 6
Dr. Ali Karimpour, Apr 2017
5
Introduction
Different parts of this lecture:.
Proper signal conversion and transmission.
Current transmission rather than voltage.
Encoding the information in the frequency of the signal rather than
its amplitude.
Isolation amplifiers and couplers protect circuiting from fault
voltage.
Proper shielding and grounding break ground loop.
• Instrumentation Amplifier• Zero and Span Circuits
• Current to Voltage Conversion• ……………………………….
• Isolation Circuits
• Cabling
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lecture 6
Dr. Ali Karimpour, Apr 2017
6
Introduction
PVSPe
Drawbacks:
3v ?
I ?
Drawbacks: Small signals buried in large common mode offsets or noise.
e.2u
We can use two voltage follower.
5 v ?
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lecture 6
Dr. Ali Karimpour, Apr 2017
7
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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lecture 6
Dr. Ali Karimpour, Apr 2017
It is the first
stage of IA.
8
Instrumentation Amplifier
3v ?
I ?
u ?
g
g
R
PVSPI
gI
)2( 10 gg RRIV
)2
1)(( 10
gR
RPVSPV
ground ? 02
2Output VR
R
• A very high
input impedance.
• Setting the gain
by a resistor.
Second stage of IA.
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
gI
0
2
2Output VR
R
0outputReal V
RR
R
wireL
L
wireL
L
RR
R
reductionGain
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lecture 6
Dr. Ali Karimpour, Apr 2017
gI
10
Instrumentation Amplifier
2
2gainNewR
RRwire
?LI
wireL
L
RR
R
reductionGain
By connecting the sense terminal at the load, any nonlinearity
and offsets between the output pin and the load are eliminated.
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
Is it possible to put an offset on the output?
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Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
Functional Block diagram of AD524 Instrumentation Amplifier
Three laser-trimmed resistors are closely matched to 20-kΩ resistors.
(Value and Temperature matched)
Gain=1
)40000
1(g
RGain
Gain=10
)4440
400001( Gain
Other gains?
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
AD524 Instrumentation Amplifier
dual in-line package (DIP)
Small Outline Integrated Circuit (SOIC)
Leadless chip carriers
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
AD524 Instrumentation Amplifier
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
AD524 Instrumentation Amplifier
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Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
Example 6-1: Derive Va, Vb and Vout if variable resistor is 349 ohms.
Answer: Vout=-715 mv
Note that 5 volts was removed from input.
Common-mode rejection ratiocA
GCMRR log20
modecommon
)modecommon (
e
VA
out
c
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Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
Example 6-2: For previous example derive the common mode error?
According to data sheet CMRR =100 db for gain of 100
100log20 cA
G 510cA
G001.0cA
001.0modecommon
)modecommon (
e
VA
out
c mvVout 5005.0)modecommon (
So common mode error is:
%7.0%100715
5
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Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
Any difference between IR1 and IR2 is called the offset current.
Input voltage at the inverting input is e1-IR1RS1
And the voltage at the nonnegative input is e2-IR2RS2
So, assure that the impedance from the inputs to ground(source
impedance) are equal.
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Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
So, assure that the impedance from the inputs to ground(source
impedance) are equal.
Any way there is difference between IR1 and IR2 so:
Try to keep the source resistance as low as possible.
Assure that the instrumentation amplifier and the sensor have a
common ground.
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Instrumentation Amplifier
The effects of offset voltage, bias currents, and offset current can be
eliminated by calibration of amplifier.
1. Connect both inputs to
same source.
2. Set the gain to maximum.
3. Adjust the left
potentiometer to give a
zero output.
4. Change the circuit to give
the gain of 1.
5. Adjust the right
potentiometer to give a
zero output.
6. Repeat from 2.
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lecture 6
Dr. Ali Karimpour, Apr 2017
21
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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Dr. Ali Karimpour, Apr 2017
22
Impedance to Voltage Conversion
• Voltage divider
ref
MF
F
outV
RR
RV
FMRR
ref
M
F
outV
R
RV
Drawbacks:
Source voltage
variation ?
ref
MF
M
outV
RR
RV
ref
MF
outV
RRV
1/
1
Load current ?
Output is not proportional with
the resistor deviation?
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Dr. Ali Karimpour, Apr 2017
23
Impedance to Voltage Conversion
• Voltage divider
• Wheatstone Bridge
refoutV
R
RRR
RV
1)1(1
1
1,2 RR
Drawbacks: Output is not proportional with the resistor deviation?
refoutVV
4
Output is proportional with the resistor deviation .
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Dr. Ali Karimpour, Apr 2017
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Impedance to Voltage Conversion
• Wheatstone Bridge
refoutV
R
RRR
RV
1)1(1
1
1,2 RR
refoutVV
4
1,2 RR
refoutVAV
4
Be sure about source voltage variation ?
Drawbacks: Resistance of wiring
and temperature variation.
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lecture 6
Dr. Ali Karimpour, Apr 2017
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Impedance to Voltage Conversion
• Wheatstone Bridge Drawbacks: Resistance of wiring and temperature variation.
refoutVV
4
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Impedance to Voltage Conversion
• Wheatstone Bridge Drawbacks: Resistance of wiring and temperature variation.
Dummy sensor
Placement of active and dummy gages for temperature compensation.
refoutVV
4
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Dr. Ali Karimpour, Apr 2017
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Impedance to Voltage Conversion
Example 6-3: Determine Vout if
R=240 ohms, Vref=10 V, and
a) Stress causes the upper resistor to
increase by 0.013 ohms.
b) Stress is zero but the upper resistor
increase by 9.4 ohms since of temperature.
c) Stress causes the upper resistor to increase by 0.013 ohms and the
upper resistor increase by 9.4 ohms since of temperature.
Answer of a) Vout= 0.135 mV
Answer of b) Vout= 96 mV ???!!!!
Answer of c) Vout= 96.2 mV ???!!!!
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lecture 6
Dr. Ali Karimpour, Apr 2017
28
Impedance to Voltage Conversion
Example 6-4:
Determine Vout if
R=240 ohms, E=10 V, and
b) Stress is zero but the right resistors increase by 9.4 ohms since of
temperature.
c) Stress causes the upper resistor to increase by 0.013 ohms and the
right resistors increase by 9.4 ohms since of temperature.
a) Stress causes the upper resistor to increase by 0.013 ohms.
Dummy sensor
Answer of a) Vout= 0.135 mV
Answer of b) Vout= 0 mV it is ok.
Answer of c) Vout= 0.13 mV it is ok.
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lecture 6
Dr. Ali Karimpour, Apr 2017
29
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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lecture 6
Dr. Ali Karimpour, Apr 2017
30
Zero and Span Circuits
The output of an amplifier rarely matches the levels you want to
provide to the controller, display, or computer.
For example one may need a 10 mV/ib inputs to a digital panel meter,
while a load cell provides a 0.02 mV/ib, and 18 mV output with no
load.
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Dr. Ali Karimpour, Apr 2017
31
Zero and Span Circuits
Temperature outputs 2.48 to 3.9 V
We need 0 to 5 V for ADC.
bmee inout
VR
Re
R
Re
os
f
in
f
u 1
1 VR
Re
R
Re
os
f
in
f
out 1
Zero and span with inverting amplifier
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Dr. Ali Karimpour, Apr 2017
32
Zero and Span Circuits
VR
Re
R
Re
os
f
in
i
f
out
Example 6-5: Suppose temperature outputs of a sensor is 2.48 to 3.9 V
derive a circuit that the output be 0 to 5 V.
52.348.29.3
05
i
f
R
Rm
current)(sensor high bemust iR
k330fR
terpotentiomek100k47,k7.93 iR
VR
R
os
f )48.2(52.30 73.8V
R
R
os
f
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Dr. Ali Karimpour, Apr 2017
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Zero and Span Circuits
VR
Re
R
Re
os
f
in
i
f
out
Example 6-5(Continue): Suppose temperature outputs of a sensor is
2.48 to 3.9 V derive a circuit that the output be 0 to 5 V.
73.8VR
R
os
f
12VLet 12/73.8os
f
R
R
terpotentiome500220,454 kandkkRos
kRLetkRRRR composifcomp 56,9.62||||
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Dr. Ali Karimpour, Apr 2017
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Zero and Span Circuits
refinout VeRg
e )40000
1(
Span
Zero
Example 6-6: A load cell changes 20 μV/ib
With an 18 mV output at no load.
Design a zero and span circuit which will
Output 0 V dc when there is no load and
will change 10 mV/ib
50002.0/10)40000
1( Rg
100332.80 Rg
refVmVV )18(5000 VVref 9
Zero and span with Instrumentation amplifier
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lecture 6
Dr. Ali Karimpour, Apr 2017
35
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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lecture 6
Dr. Ali Karimpour, Apr 2017
36
Voltage to Current Conversion
Voltage transmission have some problems:
* Series resistance between the output and load.
* The wire used.
* Temperature dependence.
Current transmission have some properties:
* None will be lost since of wiring resistance.
Types of V2C Conversion
* Floating Load * Grounded Load
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Dr. Ali Karimpour, Apr 2017
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Voltage to Current Conversion
R
eI in
Remark 1:
satin
loopVe
R
R )1(
Remark 2:
Deriving current?
* Floating Load
Negative current?
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Dr. Ali Karimpour, Apr 2017
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Voltage to Current Conversion
* Floating Load
Rload inf Vload +V
Rload 0 Vload ein
Remark 3:
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Dr. Ali Karimpour, Apr 2017
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Voltage to Current Conversion
* Floating Load
Bias
Remark 4: I=0 is natural or circuit failure?
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Dr. Ali Karimpour, Apr 2017
40
Voltage to Current Conversion
Suppose we need
Span
Zero
R
VI R
2
refin
R
eeV
R
eeI
refin
2
refeAeARI )()(2
refeBeBRI )()(2 ))()((2
)()(
AIBI
AeBeR
)()(2 AeARIeref
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Dr. Ali Karimpour, Apr 2017
41
Voltage to Current Conversion
Example 6-7: Suppose we need
VAeARIeref 8.8)5()004.0)(469(2)()(2
100430469001.0)420(2
)5(10
))()((2
)()(
AIBI
AeBeR
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Dr. Ali Karimpour, Apr 2017
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Voltage to Current Conversion
* Grounded Load
12 eeVV Lout
12 eeVVV LoutRs
sR
eeI 12
Remark 1: satload VeeIR 12
Remark 2: Deriving current?
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Dr. Ali Karimpour, Apr 2017
43
Voltage to Current Conversion
Example 6-8: Design a V2C that span 0 to 1 V to 4 to 20 mA.
For a current 20 mA and ±15 V supplies, what is the maximum load resistance.
s
in
R
eeI 1
sR
e10004.0
sR
e11020.0
Ve 25.01
100225.62 sR satload VeeIR 6.012
5586.012
I
eeVR satload
Span
Zero
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Dr. Ali Karimpour, Apr 2017
44
Voltage to Current Conversion
Buy a converter than build one? XTR110 precision V2C converter.
XTR110 precision V2C converter.
It converts 0-5 V or 0-10 V 4-20 mA or 5-25 mA
It can't give directly required current so it needs an external MOS transistor.
So, it keeps heat outside the XTR110 package to optimise its performance.
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Dr. Ali Karimpour, Apr 2017
45
Voltage to Current Conversion
V2C
C2C
Precision
resistor
divider
network
Precision
resistor
divider
network
Vin1 (10-V full scale)
Vin2 (5-V full scale)
Vref (for offsetting)
2416
21 ininref VVVV
span
ininref
RR
VVV
I 241621
8
Precision
10 V
reference
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Dr. Ali Karimpour, Apr 2017
46
Voltage to Current Conversion
V100
k67.6V5.20
mA6.10
mA20
mA200
mA200
0
V2C
C2C
mA4.00
0 to 20 mA
0
0
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Dr. Ali Karimpour, Apr 2017
47
Voltage to Current Conversion
Adding 4 mA
to output.
Vin 0 to 10 V Iout 4 to 20 mA
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Voltage to Current Conversion
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Dr. Ali Karimpour, Apr 2017
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Voltage to Current Conversion
0-10 V
0-10 A
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Dr. Ali Karimpour, Apr 2017
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Voltage to Current Conversion
Exercise 6-1 : Explain the performance of following system.
Span
Zero
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Dr. Ali Karimpour, Apr 2017
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Current to Voltage Conversion
Once the current signal gets to the place where it is to be used, it must
be converted back to voltage.
Types of C2V Conversion
* Grounded Load * Floating Load
But it needs suitable spanning.
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Current to Voltage Conversion* Ground Reference
Effect of load
on output? Zero and Span?Span
Zero
VR
RV
R
RV
os
f
R
i
f
a L V
R
RV
R
RV
os
f
R
i
f
out L
aV
Drawback?
There must be a ground return.
Voltage drop in ground return resistance.
Any change in ground return resistance …..
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Dr. Ali Karimpour, Apr 2017
53
Current to Voltage Conversion
* Floating
zspan
i
f
out VIRR
RV
Effect of load
on output?Span
Zero
z
f
f
b
i
f
a
i
f
out VR
RV
R
RV
R
RV
Common mode producing no difference across the load.
spani RR
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Dr. Ali Karimpour, Apr 2017
54
Current to Voltage Conversion
Example 6-9: Design a floating C2V converter that will convert a 4-20
mA current signal into a 0-10 V ground referenced voltage signal.
VbVmAbI
VaVmAaI
10)(20)(
0)(4)(
zspan
i
f VRaIR
RaV )()(
zspan
i
f VRbIR
RbV )()(
10/ Choose if
RR
5.62)004.002.0(10
010
spanR
kRspan
33ter potentiomemultiturn 50
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Dr. Ali Karimpour, Apr 2017
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Current to Voltage Conversion
What happen if we reduce Rf/Ri ?
625*20 mA=12.5 V ????
Rf/Ri=1, then Rspan I-V==625 Ω
Rf/Ri=10,
Rf=10Ri=22 kΩ
V
k
kVVz
5.2
)5.62(2.2
220
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Dr. Ali Karimpour, Apr 2017
56
Current to Voltage Conversion
625*20 mA=125 V ????
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Dr. Ali Karimpour, Apr 2017
57
Current to Voltage Conversion
Example 6-10: A V2C converter, has 12 V, Rspan V-I=312 Ω, Imax=20
mA. What is the maximum size Rspan I-V?
IVspanVIspan IRIRV 7.02
I
IRVR IVspan
VIspan
7.2(max)
153
02.0
)312)(02.0(7.212
(max)VIspanR
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58
Current to Voltage Conversion
Example 6-11: A modular transducer output 10 to 60 mA of current. The
manufacturer indicates that 100 Ω is the maximum allowable floating
load. Select a gain (Rf/Ri) and Rspan for the C2V converter of figure to
give a -10 to +10 V output.
)()(
)()(
aIbI
aVbVR
R
Rspan
i
f
400
01.006.0
)10(10span
i
fR
R
R
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lecture 6
Dr. Ali Karimpour, Apr 2017
59
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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lecture 6
Dr. Ali Karimpour, Apr 2017
60
Voltage to frequency conversion
Transmission of current rather than voltage:
It is eliminate the loop resistance error.
Differential nature of floating current transmission allows one to use
an instrumentation amplifier(IA) with high common-mode rejection.
Even best IA may not be able to reduce adequately errors due to
noise picked up in the transmission loop.
Using V2F:
Analog voltage Frequency Current Voltage Analog voltage
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61
Voltage to frequency conversion
Analog voltage Frequency Current Analog voltage Frequency Current Voltage Analog voltage Frequency Current Voltage Analog voltage
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62
Voltage to frequency conversion
A V2F circuit
1- Input comparator
2- One-shot timer
3- Transistor
4- Switch current source
What is in the chip?
Others are out of chip.
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63
Voltage to frequency
conversion
How a V2F works?
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64
Voltage to frequency
conversion
In this figure 1-10 V 0-10 kHz
Suggested by manufacturer
Vin1-10 V
fout0-10 kHz
tlow =1.1 RtCt must be less than the
period of the maximum output freq.
10 kΩ resistor at pin3 ?
i≈2/Rs < 200 μA
ttL
sin
out
CRR
RVf
2
47-Ω resistor in series with
CL provides hysteresis for
the comparator, improving
linearity. Why?
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Dr. Ali Karimpour, Apr 2017
65
Voltage to frequency conversion
Example 6-12: Design a V2F converter that the output be 20 kHz when
the input is 5 V.
sf
T 501
max
min
ttlowCRt 1.1 kRandFC tt 8.70047.0
kRandFCLettt
8.60047.0
stlow
35
kCRR
V
fR ttL
in
out
s6.25
1
2terpotentiome1022 kkR
s
min8.0 Tt
low
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66
Voltage to frequency conversion
Example 6-13: A pressure transducer measures a pressure from 0 to 100
psig. The output is 1-5 V for a 0-100 psig. It is desired to produce a count
every 20 ms. The count produced at 100 psig must be 100 larger than the
count in 0 psig.
a) What frequency span (in hertz) is needed?
b) What is the output frequency at 100 psig? At 0 psig?
c) What preset could be loaded into an 8-bit binary counter to yield a
count that goes from 0 to 100?
d) Why was a counting period of 20 ms chosen?
Solution: a)
b)
kHzms
countsf 5
20
100
VV 415 VkHzV
kHZ
V
f/25.1
4
5
.25.1,1,0 kHzfsoVVpsigat
.25.6,5,100 kHzfsoVVpsigat
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67
Voltage to frequency conversion
Example 6-13(Continue)
c)
d) 1/(20 ms)=50 Hz so:
25)ms20)(s
counts1250(counts
125)ms20)(s
counts6250(counts
23125256preset
1) Accurate timing pulses can be obtained from power line.
2) Any 50-Hz noise on the input voltage are eliminated.
Higher frequency in positive half-cycle and lower frequency in…..
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68
Frequency to voltage conversion
The LM131 can also be used as F2V converter.
The output frequency should
be tied to ground to reduce
the noise.
+ input of comparator is tied
to a reference level.
-input of comparator is driven
by the input frequency.
The average value of the
output voltage is proportional with frequency.
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69
Frequency to voltage conversion
The average value of the output voltage is proportional with frequency.
AR
is
2002
TR
CR
T
itI
s
ttlow
ave
1.12
in
s
Ltt
Laveavef
R
RCRRIV
2.2
Tf
in
1
This pulse waveform must be filtered to remove the ripple.
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70
Frequency to voltage conversion
Example 6-14: A reflective optical sensor is used to encode the velocity
of a shaft. There are six pieces of reflective tape. They are sized and
positioned to produce a 50% duty-cycle wave. The maximum speed is
3000 r/min. Design the frequency-to-voltage converter necessary to
output 10 V at maximum shaft speed. Provide filtering adequate to assure
no more than 10% ripple at 100 r/min(Let RL=100 kΩ)
Hzs
rev
rev
countsf 300
min
60
1
min30006
max
msTtmsHz
Tinpulse
67.15.033.3300
1minmin
msTTThighouthighout
664.28.0)(min)(
msTkRandFCLetCRThighouttttthighout
47.28.6331.1)()(
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71
Frequency to voltage conversion
Example 6-14(Continue):
nFCandkRmsCRDDDD
3.31067.15
kRfR
RCRV
sin
s
Ltt
ave48.110
2.2
ok. isit so2001352
AAR
is
Hzs
rev
rev
countsf 10
min
60
1
min1006
sHz
T 1.010
1min
At 100 r/min,
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72
Frequency to voltage conversion
Example 6-14(Continue):
fLCR
t
pkouteVv
For 10% ripple,
9.0ln%90
FL
CR
t
pk
out
CR
te
V
vfL
mstTthighoutrpm
53.9747.2100)(100
FR
tC
L
F2.7
)9.0ln( FCPick
F10
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73
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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vکاربردهای فیلترuحذف فرکانس های غیر ضروری برای بهبود نسبت سیگنال به نویزuبازیابی سیگنال حاملu به آنالوگو آنالوگکاهش خطای تداخل در مبدل های دیجیتال به
دیجیتالvانواع فیلترها
u نگذرپایین گذر، باال گذر، میان گذر و میانuفیلتر های غیر فعال و فعال
v و می توان از آن ها بهره خیلی کمتر استبارگذاریدر فیلترهای فعال اثرگرفت
74
Filters
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R
CVINVOUT
+
-
+
-
75
v 𝑓𝐻 =1
2𝜋𝑅𝐶
v𝑉𝑂𝑈𝑇(𝑠)
𝑉𝐼𝑁(𝑠)=
1
1+𝑅𝐶𝑠
vنکات طراحیu انتخاب خازن در محدودهµF-pFuبدست آوردن مقاومت مورد نیاز برای فرکانس قطع مورد نظرu 1در صورتی که مقاومت بدست آمده در محدودهk-1M نبود خازن را
این محدوده باعث.تغییر می دهیم تا مقاومت در محدوده فوق قرار گیردمی شود اثر بارگذاری در فیلتر کاهش یابد
Passive Low Pass Filter
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v𝑉𝑂𝑈𝑇(𝑠)
𝑉𝐼𝑁(𝑠)=
−𝑅2
𝑅1
1
1+𝑅2𝐶𝑠
v 𝑓𝐻 =1
2𝜋𝑅2𝐶
76
Active Low Pass Filter
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v 𝑓𝐿 =1
2𝜋𝑅𝐶
v𝑉𝑂𝑈𝑇(𝑠)
𝑉𝐼𝑁(𝑠)=
𝑅𝐶𝑠
1+𝑅𝐶𝑠
vن با سری کرد.برای افزایش مرتبه فیلتر می توان آن ها را با هم سری کرد.فیلترها معادالت مربوط به آن ها به توان طبقات می رسد
v کتر از طبقه اول بسیار کوچامپدانسبه علت اثر بارگذاری باید توجه نمود کهطبقه دوم باشدامپدانس
77
VIN VOUT
+
-
+
-
RC
Passive High Pass Filter
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v𝑉𝑂𝑈𝑇(𝑠)
𝑉𝐼𝑁(𝑠)=
−𝑅2𝐶𝑠
1+𝑅1𝐶𝑠
v 𝑓𝐿 =1
2𝜋𝑅1𝐶
78
Active High Pass Filter
C
R2
R1
+
-VIN
VOUT
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v 𝑓𝑐1 =1
2𝜋𝑅𝐻𝐶𝐻
v 𝑓𝑐2 =1
2𝜋𝑅𝐿𝐶𝐿
v𝑅𝐻
𝑅𝐿
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v𝑉𝑂𝑈𝑇(𝑠)
𝑉𝐼𝑁(𝑠)=
−1
1+𝑅𝐿𝐶𝐿𝑠∙−𝑅𝐻𝐶𝐻𝑠
1+𝑅𝐻𝐶𝐻𝑠
v 𝑓𝐿 =1
2𝜋𝑅𝐻𝐶𝐻
v 𝑓𝐻 =1
2𝜋𝑅𝐿𝐶𝐿
v 𝑓𝐿 ≫ 𝑓𝐻80
Active Band Pass Filter
CH
RH
RH
+
-
VIN
VOUT
RF
RF
+
-
RL
RL
+
-
CL
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Dr. Ali Karimpour, Apr 2017
v 𝑅1 =𝜋𝑅
10
v 𝐶1 =10𝐶
𝜋
v 𝑓𝐶 =1
2𝜋𝑅𝐶
v 𝑓𝑛 = 0.785𝑓𝑐81
R
C
R
C
R1C1
VINVOUT
v 𝑓𝐻 = 0.187𝑓𝑐v 𝑓𝐿 = 4.57𝑓𝑐
Passive Band Stop Filter
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82
v𝑉𝑂𝑈𝑇(𝑠)
𝑉𝐼𝑁(𝑠)=
−1
1+𝑅𝐿𝐶𝐿𝑠+
−𝑅𝐻𝐶𝐻𝑠
1+𝑅𝐻𝐶𝐻𝑠
v 𝑓1 =1
2𝜋𝑅L𝐶L
v 𝑓2 =1
2𝜋𝑅H𝐶H
v 𝑓1 ≪ 𝑓2
Active Band Stop Filter
VOUT
RF
+
-
RL
RL
+
-
CL
CH
RH
RH
+
-
VIN
RF
RF
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Active Filters
Sallen-Key Topology
83
vاختار پیاده سازی فیلترهای پایین گذر، میان گذر و باال گذر با سSallen-Key
v𝑉𝑂𝑈𝑇
𝑉𝐼𝑁=
𝑍3𝑍4
𝑍1𝑍2+𝑍3 𝑍1+𝑍2 +𝑍3𝑍4
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84
Low Pass High Pass
Band Pass
Active Filters
Sallen-Key Topology
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85
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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86
Isolation Circuits
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87
Isolation Circuits
Even with high-quality IA, with proper grounding and cabling and use
I or F transmission,
• Ground loops or common ground connections,
• Extremely high common-mode voltages,
• Very low failure current requirements
All of these can be solved by isolation circuits.
Common of IA and sensor common must be connected, but
ground connection are made primarily for personnel safety.
Common-mode voltages that exceed the power supply voltages
not only cause an IA to fail, but would pass it to ……………
In cardiac monitoring, the conditioner must be able to withstand a
defibrillator pulse of 5-kV and continue to process the patient’s
heartbeat properly.
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88
Isolation Circuits
More tolerate More accurate
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89
Isolation Circuits
Type of isolation Amplifiers
• Transformer coupled amplifiers
• Optically coupled amplifiers
• Optical coupling for On/Off applications
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90
Isolation CircuitsTransformer coupled amplifiers
Two or three port inputs. (Vin hi, Vin low and common)
Input
Output
Power
Three isolated ground.
Isolated output voltage
for sensor.
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91
Isolation CircuitsTransformer coupled amplifiers
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92
Isolation Circuits
Dc Ac
Ac DcAc Dc
Input modulation
Output
Transformer coupled amplifiers
EMI
Reduction
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93
Isolation Circuits
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94
Isolation Circuits
Transformer coupled amplifiers are expensive and bulky.
The carrier generates EMI if set at a high frequency.
The carrier limits the amplifier BW if set at a low frequency.
Optically coupled isolation amplifiers replace the transformer
with a LED and a pair of photodiodes.
The EMI is removed and frequency response is increased.
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95
Isolation Circuits
Comparing Transformer coupled versus optically coupled amplifiers.
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96
Isolation CircuitsOptically coupled amplifiers
You must provide separate power for input and output part.
i.e. use an IC with separate dc/dc converter.
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97
Isolation CircuitsOptically coupled amplifiers
Drawback?
Nonlinearity.
Two isolated ground.Vin Light in LED I2
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98
Isolation CircuitsOptically coupled amplifiers
D1 and D2 are
carefully matched.
1IRv
Gin
2IRv
kout
in
G
k
outv
R
Rv
This technique (an LED and a pair of photodiodes) reduce nonlinearity.
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99
Isolation CircuitsOptical coupling for On/Off applications
TIL112 Optical coupler is a much simpler and less expensive alternate
to isolation amplifier.
Phototransistor current is proportional to
LED current, but it is nonlinear.
So it is not suitable for analog transmission.
No current in LED phototransistor is off.
More than 20 mA phototransistor is on.
in LED
So digital data can transmit through this coupler.
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100
Isolation CircuitsOptical coupling for On/Off applications
How to derive R1 ?
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101
Isolation CircuitsOptical coupling for On/Off applications
1- ADC with optical isolation
There are two common ways of using optical couplers to transmit digital data.
± V1, and Vlogic 1must all be provided,
referenced to the
input common.
Vlogic 2 must be
provided,by a
different common.
It has a high
gathering rate. But it
is expensive since:
1- A/D converter.
2- A/D is in input
side.
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102
Isolation CircuitsOptical coupling for On/Off applications
2- V2F converter with optical isolation
There are two common ways of using optical couplers to transmit digital data.
If time is not an
important issue.
As the A/D
isolation scheme,
sensor, signal
conditioner and
appropriate power
Supplies are all
On the input side.
However, the expensive A/D converter has been replaced by much smaller,
much cheaper V2F converter.
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v The input is duty-cycle modulated and transmitted digitally
across the barrier
v The output section receives the modulated signal, converts
it back to an analog voltage and removes the ripple
component inherent in the demodulation
103
Capacitive Isolation Amplifier
Modulator DemodulatorInput
signal
Output
signal
Capacitiveisolation
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v ISO124
104
Ix
Iin Ic
D
B
C
Capacitive Isolation Amplifier
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v Ripple noises are removed
v It avoids device noise, radiation noise and conducted noise
v High immunity to magnetic noise
v Useful for analog systems
v high gain stability and linearity
v Supports faster data transition compared to optical isolation
105
Capacitive Isolation Advantages
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106
Signal Conditioners and Transmission
v Introduction
v Instrumentation Amplifier
v Z V Conversion
v Zero and Span Circuits
v V I and I V Conversion
v V F and F V Conversion
v Filter
v Isolation Circuits
v Cabling
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107
Cabling
Even with the best transducers, amplifiers, filters, transmission, and
isolation it is entirely possible that the signal received by controller
is junk.
Proper cabling can shield against interference from the large
magnetic and electric fields produced in all manufacturing
environment.
Proper grounding of the shields, circuits, and power supplies will
minimize the impact of differing ground potentials.
Without proper shielding and grounding even techniques, even the
most expensive …..
Proper cabling will allow even a cheap system to realize its full potential.
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108
Cabling
Current flow leads to magnetic field.
Magnetic field is 50 Hz or Dc but we have problem with 50 Hz and
switching in the Dc one.
Radio frequency of digital signal of the processor also produce
significant magnetic field.
So place small signal analog circuitry on a separate card from
computer and power electronics.
Separate this card from others or put it an a small magnetic shield box.
Absorbing magnetic field is in dB.
Never, never run 220-V ac power and low level analog cables in the
same …
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109
Cabling
Absorbing magnetic field is in dB. A 20 dB one change the ratio to 10.
In audio frequency steel is the choice but in radio frequency…..
In fact, there are several coating that can sprayed onto plastic cases
which effectively work (in high frequency).
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110
Cabling
For effective shielding against electrostatic field try to use a conductive
shield.
Shield must be connected to infinite source.
Do not use the shield as the signal common.
Use a shield for signal common as well.
Use a shielded pair or twin-axial, never simple coaxial cable.
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111
Grounding
Effective electrostatic shield must be grounded. But the signal common
must not be connected to the shield.
Connection to ground at some points, is it good or bad?
1- Shield is designed
for this current?
2- Magnetic field of this current?
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112
Grounding
Disconnect from some point.
Use a 100 pF parasitic capacitance.
MpFHz
Xc
32)100)(50(2
1
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113
Grounding
If disconnection is not possible?
Sometimes the computer input low signal, from the signal conditioner
is tied to earth by manufacturer for safety reasons.
Disconnect this
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114
Grounding
If Disconnection is not possible? Try it by following:
In fact the isolation amplifier (or digital optical coupler) breaks the
ground loop that had exited through electronics.
In fact, above configuration is probably the best general approach to
shielding and grounding.
The only remaining element in the data acquisition channel to be
connected is the power supply.
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115
Grounding
What about power supply?
In every sensitive systems,
this error may be a
significant part of the
signal from the transducer.
The solution is a power
supply transformer with
shielded secondary.
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116
Grounding
The solution is a
power supply
transformer with
shielded secondary.
In this configuration
any vnoise causes
current to flow
along the shield, not
along the signal
common.
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117
Grounding
An interesting problem.
Suppose it is
100 cm.
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118
Grounding
What do you do with ground connection of the other electronics?
Instead of -4 V
optional
?bv
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119
Grounding
Ideally, every stage throughout the data acquisition and control system
would have its own, separate ground return.
This is impractical, so group your circuits according to:
• The type and size of signal
• The magnitude of ground return currents
• The circuits sensitivity
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120
Grounding
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121
Appendix: Instrumentation Amplifier
AD620The AD620 is a monolithic
instrumentation amplifier based on
a modification of the classic three op
amp approach. Absolute
value trimming allows the user to
program gain accurately
(to 0.15% at G = 100) with only one
resistor. Monolithic
construction and laser wafer
trimming allow the tight matching
and tracking of circuit components,
thus ensuring the high level
of performance inherent in this
circuit.
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122
Appendix: Instrumentation Amplifier
The input transistors Q1 and Q2 provide a single differential pair bipolar input for high precision (Figure 36), yet offer 10× lower input bias current thanks to Super ϐeta processing.Feedback through the Q1-A1-R1 loop and the Q2-A2-R2 loop maintains constant collector current of the input devices Q1 and Q2, thereby impressing the input voltage across the external gain setting resistor RG. This creates a differential gain from the inputs to the A1/A2 outputs given by G = (R1 + R2)/RG + 1. Theunity-gain subtractor, A3, removes any common-mode signal, yielding a single-ended output referred to the REF pin potential.
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123
Appendix: Instrumentation Amplifier
The internal gain resistors,
R1 and R2, are trimmed to
anabsolute value of 24.7 kΩ,
allowing the gain to be
programmed accurately with
a single external resistor.
The gain equation is then
𝐺 =49.4𝑘Ω
𝑅𝑔+ 1
𝑅𝑔 =49.4𝑘Ω
𝐺 − 1
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124
Pressure MeasurementAlthough useful in many bridge applications, such as weigh scales, the AD620 is especially
suitable for higher resistance pressure sensors powered at lower voltages where small size
and low power become more significant. Figure 38 shows a 3 kΩ pressure transducer bridge
powered from 5 V. In such a circuit, the bridge consumes only 1.7 mA. Adding the AD620
and a buffered voltage divider allows the signal to be conditioned for only 3.8 mA of total
supply current.
Appendix: Instrumentation Amplifier
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125
Appendix: Instrumentation Amplifier
Furthermore, the low bias currents and low current noise, coupled with the low
voltage noise of the AD620, improve the dynamic range for better performance.
The value of capacitor C1 is chosen to maintain stability of the right leg drive loop.
Proper safeguards, such as isolation, must be added to this circuit to protect the
patient from possible harm.
Medical ECG
The low current noise
of the AD620 allows
its use in ECG
monitors (Figure 39)
where high source
resistances of 1 MΩ
or higher are not
uncommon.
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126
Precision V-I Converter
The AD620, along with another op
amp and two resistors,
makes a precision current source
(Figure 40). The op amp
buffers the reference terminal to
maintain good CMR. The
output voltage, VX, of the AD620
appears across R1, which
converts it to a current. This current,
less only the input bias
current of the op amp, then flows
out to the load.
Appendix: Instrumentation Amplifier
I L =Vx
R1=[(𝑽 𝑰𝑵+) – (𝑽 𝑰𝑵 – )] 𝑮
𝑹𝟏
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127
GROUNDING
Since the AD620 output voltage is
developed with respect to the
potential on the reference terminal, it
can solve many grounding problems
by simply tying the REF pin to the
appropriate “local ground.” To
isolate low level analog signals from
a noisy digital environment, many
data-acquisition components have
separate analog and digital ground
pins (Figure 45).
Appendix: Instrumentation Amplifier
It would be convenient to use a single ground line; however, current through ground
wires and PC runs of the circuit card can cause hundreds of millivolts of error.
Therefore, separate ground returns should be provided to minimize the current flow
from the sensitive points to the system ground. These ground returns must be tied
together at some point, usually best at the ADC package shown in Figure 45.
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References
v Industrial Control Electronics J.M. Jacob, Prentice-Hall,
1989
v AD620 and ISO124 datasheet
v Fundamentals of measurements in instrumentations by
prof hamid.taghirad and s.ali salamati
128
Some Useful websites for the course
http://saba.kntu.ac.ir/eecd/ecourses/instrumentation.htm
v http://profsite.um.ac.ir/~shoraka/Instrumentation.htm
v http://karimpor.profcms.um.ac.ir/index.php/courses/10328
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