Makalah Signal Conditioning
Short Description
Makalah tentang signal conditioning...
Description
2102-487 Industrial Electronics
Signal Conditioners and Transmission
Overview
4-20 mA Amp
V to I
I to V
Amp
Control Room
Field Noise Source
Interference
Motor/Power
Arcing
Lighting
Overview
4-20 mA Amp
V to I
I to V
Amp
Control Room
Field Noise Source
Interference
Motor/Power
Arcing
Lighting
Instrumentation Amplifier (IA) Instrumentation amplifiers a dedicate differential amplifier with extremely high input impedance. Its gain can be precisely se by a single internal or external resistor. The high common-mode rejection makes IA very useful in recovering small signals buried in large common-mode offsets or noise. •IA consists of Two stages: •The fist stage: high input impedance and gain control •The second stage: differential amplifier (change differential signal to common to ground R R-
R E
+
R
-
R+∆ R
+
V dm/2
V out
R
V cm
+
-
-
+
V dm/2
R+
Bridge circuit for sensor application
Equivalent circuit for a Load cell
IA: The First Stage For Ideal op amp, no voltage difference between the inverting and noninverting inputs
V 1 = e1 and V 2 = e2
+ -
e1
V 1
The voltage across Rg
R
V R g = e1 − e2 +
I Rg
R g
V o -
V 2 + e2
R g = gain setting resistor
R
I R g =
e1 − e2 R g
This current must flow through all three resistors because none of the current can flow into the op amp inputs
V o = I R g 2 R + R g
V o = (e1 − e2 )1 +
2 R
R g
• Changing R g will inversely alter the output voltage.
IA: First + Second stage The second stage of IA is a unit-gain differential amplifier V 1 + e1 R1
-
R
+ R g
R1
V o1 -
e2
+ R1
R
+
2 R V o = (e2 − e1 )1 + + R g
V o
R1
Gain = 1 +
2 R R g
V 2
V o1 = (e1 − e2 )1 +
2 R
V o = −V o1
R g
Without loading effect, we can applied the direct multiplication for the cascade system e1 − e2
2 R 1 + R g
V o1
-1
2 R V o = (e2 − e1 )1 + R g
IA with Offset (Reference) e1
+ R1
R2
V 1 R2
-
V o1
R g
-
e2
Output
V o = (e2 − e1 )1 +
+ R2
R1
2 R
R g
+ V ref
Reference
Load
V 2
+
Sense
R2 +V
V ref
2 R V o1 = (e1 − e2 )1 + R g
+
V ref
-V
V o = −V o1 + V ref e1 − e2
2 R 1 + R g
V o1
-1
+ V ref
V o = (e2 − e1 )1 +
2 R
R g
+ V ref
Commercial IA: AD524 Features: Low Noise: 0.3 Vp-p 0.1 Hz to 10 Hz Low Nonlinearity: 0.03% (G = 1) High CMRR: 120 dB (G = 1000) Low offset Voltage: 50 V Gain Bandwidth Product: 25 MHz Programmable Gain of 1, 10, 100, 1000
AD524 Functional Block Diagram
G = 1 +
40000 R g
± 20% Here R g = 40, 404, 4.44 k Ω or external resistor (connect: RG1 – RG2)
Commercial IA: AD524
V b
V a 349
V out
Ex For the circu circuit, it, calcula calculate te V a, V b, and V out as well as the effect on the output of the 5V common-mode signal. For a gain of 100, CMRR = 100 dB
V b = 12 V supply = 5 V
349Ω 10 V = 4.99285 V Ω + Ω 350 349
V a =
V out = G (V a − V b )
= 100(4.99285 V - 5 V) = -715 mV
CMRR = 20 log
Gdm Gcm
Gcm = 0.001
GcmV cm = 0.001× 5 V = 5 mV The common mode error is
5 mV
715 mV
×100% = 0.7%
Commercial IA: AD524
Error Budget Analysis To illustrate how instrumentation amplifier specification are applied, we will now examine a typical case where an AD524 is required to amplify the output an unbalance transducer. Figure above shows a differential transducer, unbalance by 100 supplying a 0 to 20 mV signal to an AD524C. The output of the IA feeds a 14-bit A/D converter with a 0 to 2 Voltage range. The operating temperature temperature is 25oC to 85oC. Therefore, the largest change in temperature T within the operating range is from ambient to +85oC (85oC - 25oC = 60oC)
Commercial IA: AD524 1000ppm = 0.1%
Zero and Span Circuits The zero and span circuit adjust the output of a transducer to match the levels you want to provide to the controller or display. Ex. You may need a 0.01 mV/lb input to a digital panel meter, while the load cell provides a 20 V/lb reading with 18 mV output with no load. Ex. A/D converter needs a 0- to 5-V signal, while the temperature transducer output 2.48 to 3.90 V.
V out
V out l e l b a u n i o i g d r O n a p S a l f
H a n p S
V in
y e l v t i s i o p d e i f t s h r o l e Z n a i i g O r l y e i v a t g n e d e i f t h s o r Z e
V in
Zero and Span: Inverting summer ±V
R f
ROS
eu1
R
ein
R
Ri
-
+
eu2
+ Rcomp
Inverting summer R f R f eu1 = − ein − V Ri ROS eu 2 =
Compare this to the eq. of straight line
R f Ri
ein +
R f ROS
R/2
Inverting amplifier eu 2 = −eu1
V
eout
y = mx + b
Here y = eu2, x = ein, m = R f / Ri and b = R f / Ros V
b=
R f ROS
V
m =
R f Ri
ein
Zero and Span: Instrument Amplifier Ex When the temperature in a process is at its minimum, the sensor outputs 2.48 V. At maximum temperature, it outputs 3.90 V. The A/D converter used to input these data into a computer has the range 0 to 5 V. To provide maximum resolution, you must zero and span the signal from the transducer so that it fills the entire range of converter. V out = 0-5 V V in = 2.48-3.90 V
V out
b=
R f ROS
m =
R f Ri
m=
V out (max) − V out (min) V in (max) − V in (min)
=
5V−0V 3.9 V − 2.48 V
= 3.52
b = V out − mV in = 0 − 3.52 × 2.48 V = -8.73 V
V in
V
Let R f = 330 k
The gain, m is set by R f / Ri , R f relatively large, so that Ri will not load down the sensors R f 330 k Ω = = 93.7 k Ω Ri = m 3.52
Select Rs as a 47 k
fixed resistor with a series 100 k potentiometer. (m ~ 2.25-7.02) R f V (330 k Ω)(- 12V) / ROS V select V = -12 V b is R f = = 454 k Ω ROS = − 8.73 V b Select Rs as a 220 k fixed resistor with a 500 k potentiometer. (b ~ -18 - -5.5 V)
Rcomp = R f // R1 // ROS = 62.9 k Ω The resistors in the 2nd stage should be in k pick R = 2.2 k and so R /2 = 1.1 k
Select Rcomp = 56 k range, this will not load the 1st stage,
Zero and Span: Instrument Amplifier +V
+ R2
8 2 16
+
ein
4
+
-
1
span
AD524 6
R g
R
-
Sense
R2
V out
9
11 3
O
10
12
R1
5
13
R g
-
S
-
span
Output
V out
+
7
+V
+V
R2
R1
Reference
R2
+
+V
-V
V ref
zero
V ref
zero
+
-
-V
-V
+
-V
V out = 1 +
40000 R g
(e2 − e1 ) + V ref
V ref
Zero and Span: Instrument Amplifier Ex The output from a load cell changes 20 V/lb with an output of 18 mV with no load on the cell. Design a zero and span converter using an instrumentation amplifier which will output 0 Vdc when there is no load, and will change 10 mV/lb. +V
V sensor = 20 V/lb x Load + 18 mV V out = 10mV/lb x Load = G V sensor + V ref
8 2 16
+
4
+
S 5
13
V sensor
R g
-
1
V out
9 6
11 3
O
10
AD524
12
V out = 20 V G x Load + 18mV G + V ref
R
-
G=
7
span
10 mV/lb 20 µ V/lb
= 500
+V +V -V
V ref
zero
G = 1+
40000 R g
= 500
Select R g as a 33 fixed resistor with a series 100 potentiometer. (G ~ 301-1213)
+ -
0 = 18 mVG + V ref = 18 mV × 500 + V ref -V
-V
V out = 1 +
40000
(e2 − e1 ) + V ref
R g
R g = 80.2 Ω
V ref = −9 V
So tie one end of the potentiometer to ground and the other end to –V supply. Select resistor and potentiometer values that will put approximately -10 V at one end and -8 V at the other end.
Voltage-To-Current Converters Signal voltage transmission presents many problems. The series resistance between the output of the signal conditioner and the load depends on the distance, the wire used, temperature, conditioner.
Rwire
Distance, wire type, temperature
Rwire
+ -
+
+ ein
-
R
R f
R f ein V o = 1 + R -
load
Load V o + R Load wire
V To I: Floating Load The most simple V to I actually is the non-inverting amplifier +V
Rwire
+ -
I +
-V
ein
-
load
+
ein
R
I
-
I =
ein R
Therefore, The resistance in the transmission loop ( Rloop = Rwire + Rload) does not affect the transmitted current. However, the output voltage of the op amp is affected by the Rloop Rloop ein < V sat R
V out = 1 +
Rloop must be kept small enough to keep the op amp out of the saturation.
V To I: Floating Load V to I with current booster
Add transistor for increase current output
+V
New op amp with high current output +
Q1
I
Rwire
Q2
load
+ ein
-V
-
I
R
I =
ein R
Many transmission standard call for either 20 or 60 mA current. These values are beyond the capabilities of most general-purpose op amps. However, the transistor can be added to increase the transmission current capability.
V To I: Floating Load The signal at the load is inherently differential. So we can use difference or instrumentation amplifier to reject any common-mode noise. Open and short circuit in the transmission loop can be detected by checking V out of the non-inverting amplifier (transmission side) open circuit : V out = V sat short circuit : V out = ein However, at the load side, with this circuit , if ein = 0, I L = 0, here it appears that the I L = 0 is the valid signal. If the open or short circuit fault occurs, I L will fall to zero too. The load would respond as if ein = 0. Therefore a method has been advised to allow the load to differentiate between no signal (circuit failure) I L = 0 and ein = 0 This can be achieved by adding an offset to I L when ein = 0 ein = 0: I L > 0 Ex. 4-20 mA standard in current transimission
V To I: Floating Load zero
An example of 4-20 mA V to I converter (span + zero adjust)
10 k Ω
-V
+V eref
+V
1 MΩ
+
I in
ein
Rwire
I L
Q1
-
1 MΩ
load
-V
R span
I b=
m=
1 2 R
eref 2 R
ein
I =
ein + eref 2 R
=
ein 2 R
+
eref 2 R
m = 1/2 R, b = eref /2 R
Zero and Span: Instrument Amplifier Ex Design an offset voltage-to-current converter that will produce 4 mA with an input of -5 V and 20 mA with an input of 10 V I (mA)
m=
15 10
1 2 R
=
I out (max) − I out (min) V in (max) − V in (min)
Select a 430
5
-5
I out = 4 - 20 mA
V in = -5 - 10 V
20
4 0
I =
10
5
V in 2 R
+
V ref 2 R
V in
=
20 mA − 4 mA 10 V − (−5 V)
fixed resistor with a series 100
R = 469 Ω
potentiometer.
V ref = 2 RI − V in = 2(469 Ω)(20 mA) − 10V = 8.8 V
V To I: Grounded Load This V to I actually is a derivative of difference amplifier R3 R1
I
+ e1
V out
R2
e2
Rs
+ V L
R4
load
I
R1=R2=R3=R4=R
Using superposition: V out due to e1 = -e1 V out due to e2 = e2 V out due to V L = V L
V out = V out |e1 +V out |e2 +V out |V L
= - e1 + e2 + V L = e2 - e1 + V L
The current drive into the load is approximately equal to the current in Rs if R Load IRload + e2 − e1
V To I: Grounded Load An example of 4-20 mA V to I converter (span + zero adjust) zero Ra
100 kΩ
+V
Rb eref
100 kΩ
-
I
R s
+ 100 kΩ
ein
span
-V
+ V L
load
100 kΩ
I L ≈
I m=
b=−
eref R s
ein
1 R s
ein − eref R s
=
ein R s
−
eref R s
m = 1/ R, b = -eref / R
V To I: XTR110 •4 to 20 mA Transmitter •Selectable input/output ranges: 0-5V or 0-10V Inputs 4-20mA, 0-20mA, 5-25mA Outputs • Required an external MOS transistor to transmit current to load Reference voltage I to I converter (current mirror)
Precision resistive network
V to I converter
V To I: XTR110 VCC 13.5-40V
XTR110
VIN1(10V)
R8 500
External MOS
R9 50
4
VREF IN 3
3
Io /10
V a
R5
Io
-
R3 20k
+
Io 4-20mA
Io /10
RL
V b VIN2(5V) 5
R6 1562.5 16mA span
R7 6250 4mA span 9
10
V a R6
Since there is no voltage difference between the op amp inputs
+ -
R4 10k
R6
=
I R 8 ≈ I R 6
14
16.25k R2 5k
V b
The current I R6 is approximately equal to I R8
1
R1 15k
I R 6 =
16
8
V R8 = V R 9 I o = I R 9 =
I R 8 R8
I o =
R9 10V a R6
= 10 I R8
V To I: XTR110 The voltage at V a is defined by precision divider network and the voltage at V in1, V in2 and V ref . Using superposition V a due to V in1 = V in1/4 V a due to V in2 = V in2/2 V a due to V ref = V ref /16 V a = V in1 / 4 + V in 2 / 2 + V ref / 16
Therefore, the relation of the output current can be shown as I o =
10 V in1 / 4 + V in 2 / 2 + V ref / 16 R span
This IC allows us to change the value of Rspan by using R6, R7 and external resistors so therefore the Io span can be set to 16mA, 4mA or arbitrary value. Ex if V in2 = 0, V ref = 10 V, V in1 varies from 0-10V and Use Rspan = R6 = 1562.5 At V in1 =0 V; Io = 4 mA at zero V in1
Zero = 4 mA
At V in1 =0 V; Io = 20 mA at V in1 = 10 V
Zero+span = 4mA + 16 mA
V To I: XTR110 VIN1
4
VREF IN
3
R1 15k
R3 20k
R5
R2 5k
1 + 2 + 3
V a
16.25k R4 10k
V a = 14 V IN1 + 12 V IN2 + 161 V REF
VIN2 5
Rin = 22 k Ω
Rin = 27 k Ω
Rin = 20 k Ω
VIN2
VIN1
VREF IN
1
V a
16.25k
V a =
R2 5k
R4 //( R5 + R1 // R2 ) R3 + R4 //( R5 + R1 // R2 )
R4 10k
V IN1 = 14 V IN1
V a
16.25k R1 15k
V a =
R2 5k
R3 //( R5 + R1 // R2 ) R4 + R3 //( R5 + R1 // R2 )
3
R5
R5
R5
R1 15k
R1 15k
R4 10k
2
R3 20k
R2 5k
R3 20k
V IN2 = 12 V IN2
V a
16.25k
V a =
R3 // R4
R4 10k
R3 20k
R2
( R5 + R1 // R2 ) + R3 // R4 R1 + R2
V REF = 161 V REF
V To I: XTR110 Pin connections for standard XTR110 input voltage/output current ranges
15 V 1 F 15 16 12
3
1
13
XTR110
14
4
5
9
A basic 4 to 20 mA transmitter 4 to 20 mA Out
2
0 to 10 V RL
VL
V To I: XTR110 I out =
V out = 400V in + 6 V V sensor = ±10 mV
10(V in1 / 4) 1250 Ω
I out = 4 − 20 mA
V sensor = 2 − 10 V
I out =
10( 400V sensor + 6) / 4 1250 Ω
Current-To-Voltage Converters Grounded Load I to V converter R f
Ros
Span and Zero circuit
±V
4 to 20 mA
R
+ -
R
R2
-
+
+
R L Rcomp
Current is converted into a voltage by R L
R/2
+ V out -
Voltage follower is inserted to avoid the loading effect
V out =
R f Ri
IR L +
R f ROS
V
m = R f R L/ R ,i b = R f / ROS V
The grounded load converter has many problems with the common ground (use by many device) especially the noise source. To reduce this problem, we use a floating load.
I To V Converters: Floating Load R f
4-20 mA
R span
Ri
R4 Ri
+
+V
+ V out -
R f R pot
+ V ref
-V
V out =
R f Ri
IR span + V ref
m = R f R L/ R ,i b = V ref
To prevent the loading effect be sure that Rspan Rspan I-V , pick Ri = 2.2 k Find V ref :
V ref = V out −
R f Ri
R span I = 0 −
22 k Ω 2.2 k Ω
(4 mA )(62.5Ω ) = −2.5 V
I To V Converters: Floating Load +V
+ 2V
+
+
-
0.7 V
-
+
I
IRspan I-V -V +
IRspan V-I
Rspan I-V
-
I
Rspan V-I
-
+ V = 2 + 0.7 + IR spanI −V + IR spanV − I R spanI −V (max) =
12 − 2 − 0.7 − (20 mA)(312) 20 mA
= 153 Ω
Therefore, 62.5 resistor would work. However, if we had chosen R f / Ri =1 then Rspan I-V = 625 , which is too large. The op Amp in the voltage-to-curren converter would saturate before 20 mA would be reached.
Voltage-To-Frequency Converters To provide the high noise immunity, digital transmission must be used. V to F will convert the analog voltage from the sensor or signal conditioner to a pulse train. The pulse width is constant but the frequency varies linearly with the applied voltage
f ∝ V in
V in
V to F
f out
f out
V in
Ideal V to F characteristics
V-To-F: LM131
LM131
Suitable for various applications: precision V to F converter simple low-cost circuits for A/D conversion precision F to V converter long-term integration linear frequency modulation and demodulation
Voltage-To-Frequency Converters Basic V to F block diagram Rt
C t
V CC 8 2
5
Switched current source
f o =
V logic
R S
R L
C L
Comparator 1
V x
6
7 Input V voltage 1
V in
3
-
One shot timer
+
Frequency output
4
a b
t
V x t
V out
t f out
High V in
high f out
Adjustment
Low V in
low f out
Adjust
High V in
high f out
t
V in
RS
1
2.09 V R L Rt C t
Voltage-To-Frequency Converters R
C
t
t
Determine the timer of one shot T OS = 1.1 Rt C t
V
CC
Determine I ref
8 2
5
Switched current source
V
logic
R
S
Comparator
R
L
C
L
1
6
7 Input V voltage in
-
3 One shot
Frequency output
timer
+
4
Phase I: Charge C L with I ref within the specific time T OS from the one shot timer. At the end of this phase, V C reaches the value assigned as V x Phase II: C L discharge through I ref until V C is equal to V in. The time in this phase is defined as T II We can found that 1/(T OS + T II ) = f
V in
Voltage-To-Frequency Converters Phase I: Charge C L with I ref within the specific time T OS determined from the one shot timer. At the end of this phase, V C reaches the value assigned as V x Initial condition: V C (0) = V in I ref
R L
C L
V C (t ) = (V in − I ref R L )e −t /τ + I ref RL
Assume R LC L >> T OS
e
V x = V C (T OS ) = V in + ( I ref R L − V in )
−T OS / τ
τ = R L C L
≈ 1 − T OS / τ
T OS
1
τ
Phase II: C L discharge through I ref until V C is equal to V in. The time in this phase is defined as T II Initial condition: V C (0) = V x R L
I
C L
τ = R L C L
V C (t ) = V x e −t /τ V in = V C (T II ) = V x e
−T II / τ
T II = τ ln
Since R LC L >> T OS , Therefore V x /V in ~ 1
V x
− 1 V in
T II = τ
V x V in ln
V x V in
2
≈
V x V in
−1
Voltage-To-Frequency Converters Combine eq.(1) and (2)
T = T OS + T II = f =
1
=
I ref R LT OS V in V in
T I ref R LT OS
Substitute I ref = 1.9 V/ Rs ,T OS = 1.1 Rt C t
f =
V in R s 1.9 V R L (1.1 Rt C t )
=
V in
R s
1
2.09 V R L Rt C t
V To F Converters Ex Design a voltage-to-frequency converter that will output 20 kHz when the input is 5 V. At maximum frequency, the minimum period is 1 1 = = 50 µ s T min = f max 20 kHz We have to set the pulse width no wider than about 80% of the minimum period, otherwise, at the higher frequencies the pulse width may approach or exceed the period, which will not work. t low = 1.1 Rt C t pick C t = 0.0047 F (somewhat arbitrary, but suggested by manufacturer)
Rt =
t low 1.1C t
=
(0.8)(50 µs ) = 7.7 k Ω (1.1)(0.0047 µF )
pick Rt = 6.8 k , since this is not a critical parameter, as long as it is not too big. This give t low = 35 s. Pick R L = 100 K .
R s =
(2 V ) f 0 R L Rt C t (2 V )(20 kHz )(100 k Ω )(6.8 k Ω )(0.0047 µF )
Select a 22 k
V in
=
5V
fixed resistor with a series 10 k
potentiometer.
= 25.6 k Ω
Voltage-To-Frequency Converters Low-pass filter
Simple V to F converter From
f o =
V in
RS
1
2.09 V R L Rt C t
I =
1.9 V RS
< 200 µA
Manufacturer’s recommended, R L = 100 k , Rt = 6.8 k , C t = 0.01 F and R S = 14.2 k f o =
1 kHz Volt
V in
V To F Converters Ex Design a voltage-to-frequency converter that will output 20 kHz when the input is 5 V. At maximum frequency, the minimum period is 1 1 T min = = = 50 µ s f max 20 kHz We have to set the pulse width no wider than about 80% of the minimum period, otherwise, at the higher frequencies the pulse width may approach or exceed the period, which will not work. t low = 1.1 Rt C t pick C t = 0.0047 F (somewhat arbitrary, but suggested by manufacturer)
Rt =
t low 1.1C t
=
(0.8)(50 µs ) = 7.7 k Ω (1.1)(0.0047 µF )
pick Rt = 6.8 k , since this is not a critical parameter, as long as it is not too big. This give t low = 35 s. Pick R L = 100 K .
R s =
(2 V ) f 0 R L Rt C t (2 V )(20 kHz )(100 k Ω )(6.8 k Ω )(0.0047 µF )
Select a 22 k
V in
=
5V
fixed resistor with a series 10 k
potentiometer.
= 25.6 k Ω
Frequency-To-Voltage Converters V CC
I out 1
R L
Connect to low pass filter
Comparator 7
R D
8
+
One shot timer
Rt V CC
C t
C D V in
V in i I out T
i=
The amplitude of pulse current output is The average voltage output is The average current output is
I ave =
1.9 V R s
it 1.9 V ×1.1 Rt C t T
=
R sT
R V ave = I av e R L = 1.9 V ×1.1 Rt C t L f in R s
Frequency-To-Voltage Converters
High pass filter for input frequency
Low pass filter for output current
Simple F to V converter From
R V ave = 1.9 V × 1.1 Rt C t L f in RS
R L = 100 k , Rt = 6.8 k , C t = 0.01 F and R S = 14.2 k V ave =
1V kHz
f in
F To V Converters Ex 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 shaft speed is 3000 r/min. Design the frequency-tovoltage converter necessary to output 10 V at maximum shaft speed. Provide filtering adequate to assure no more than 10% ripple at 100 r/min The maximum frequency is T min =
Select
1
f max =
= 3.33 ms
6 counts rev
×
3000 rev 1 min min
×
60 s
= 300 Hz
T pulse = 0.5T min = 1.67 ms
300 Hz T out ( high) ≤ 0.8T min = (0.8)(3.33 ms ) = 2.664 ms
Select C t = 0.33 F Rt = t low = (0.8)(50 µs ) = 7.7 k Ω 1.1C t (1.1)(0.0047 µF ) pick Rt = 6.8 k , This give t out(high) = 2.47 ms. We must set 5 R DC D
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