1.
2102487
Industrial Electronics
Signal Conditioners
and Transmission
2.
Overview
420 mA
Amp
V to I
I to V
Amp
Control Room
Field
Noise Source
Interference
Motor/Power
Arcing
Lighting
3.
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 commonmode rejection makes IA very useful
in recovering small signals buried in large commonmode 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
E
+
R

R+∆R
+
Vdm/2
Vout
R
Vcm
+


+
Vdm/2
R+
Bridge circuit for sensor application
Equivalent circuit for a Load cell
4.
IA: The First Stage
For Ideal op amp, no voltage difference
between the inverting and noninverting inputs
V1 = e1 and V2 = e2
+

e1
V1
V2
VRg = e1 − e2
e1 − e2
Vo
Rg
 This current must flow through all three
resistors because none of the current can flow
into the op amp inputs
+
IRg
Rg
The voltage across Rg
R

R
I Rg =
+
Vo = I R g (2 R + Rg )
Rg = gain setting resistor
2R
Vo = (e1 − e2 )1 +
R
g
e2
• Changing Rg will inversely alter the output voltage.
5.
IA: First + Second stage
The second stage of IA is a unitgain differential amplifier
V1
+
e1

R1
R
+
R1

Vo1
Rg

e2
+
+
R1
R
R1
Vo
2R
Vo = (e2 − e1 )1 +
R
+
g

Gain = 1 +
2R
Rg
V2
2R
Vo = −Vo1
Vo1 = (e1 − e2 )1 +
R
g
Without loading effect, we can applied the direct multiplication for the cascade system
e1 − e2
2R
1 +
R
g
Vo1
1
2R
Vo = (e2 − e1 )1 +
R
g
6.
IA with Offset (Reference)
e1
+
R1
R2
V1
R2

Vo1
Rg

e2
R1
+
Sense
2R
+V
Vo = (e2 − e1 )1 +
R ref
g
Output
+
R2
Reference
Load
V2
R2
+V
Vref
2R
Vo1 = (e1 − e2 )1 +
R
g
+
Vref
V
Vo = −Vo1 + Vref
e1 − e2
2R
1 +
R
g
Vo1
1
+
Vref
2R
+V
Vo = (e2 − e1 )1 +
R ref
g
7.
Commercial IA: AD524
Features:
Low Noise: 0.3 µVpp 0.1 Hz to 10 Hz
Low Nonlinearity: 0.03% (G = 1)
High CMRR: 120 dB (G = 1000)
Low offset Voltage: 50 µV
Gain Bandwidth Product: 25 MHz
Programmable Gain of 1, 10, 100, 1000
AD524 Functional Block Diagram
40000
± 20%
G = 1 +
Rg
Here Rg = 40, 404, 4.44 kΩ or external
resistor (connect: RG1 – RG2)
8.
Commercial IA: AD524
Vb
Va
349Ω
Vout
Ex For the circuit, calculate Va, Vb, and Vout as well as the effect on the output of the 5V commonmode signal.
For a gain of 100, CMRR = 100 dB
Vb = 1 Vsupply = 5 V
2
349Ω
Va =
10 V = 4.99285 V
350Ω + 349Ω
Vout = G (Va − Vb )
= 100(4.99285 V  5 V) = 715 mV
Gdm
Gcm = 0.001
Gcm
GcmVcm = 0.001× 5 V = 5 mV
CMRR = 20 log
The common mode error is
5 mV
× 100% = 0.7%
715 mV
9.
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
Ω supplying a 0 to 20 mV signal to an AD524C. The output of the IA feeds a 14bit A/D
converter with a 0 to 2 Voltage range. The operating temperature is 25oC to 85oC.
Therefore, the largest change in temperature ∆T within the operating range is from
ambient to +85oC (85oC  25oC = 60oC)
11.
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 µV/lb reading with 18 mV output with no load.
Ex. A/D converter needs a 0 to 5V signal, while the temperature transducer
output 2.48 to 3.90 V.
Vout
Sp
an
do
ub
le
Vout
l
na
i
r ig
O
a
Sp
a
nH
o
er
Z
lf
ed
ift
sh
ly
ve
ti
si
po
al
in
rig
O
Vin
o
er
Z
d
fte
i
sh
ly
ve
ti
ga
ne
Vin
12.
Zero and Span: Inverting summer
±V
ein
Rf
ROS
eu1
R
R
Ri
Rcomp

+
eu2
+
Inverting summer
Rf
Rf
eu1 = −
ein −
V
Ri
ROS
eu 2 =
Compare this to the eq. of straight line
Rf
Ri
ein +
R/2
Inverting amplifier
eu 2 = −eu1
Rf
ROS
V
eout
m=
Rf
y = mx + b
Here y = eu2, x = ein, m = Rf/Ri and b = Rf/Ros V
b=
Rf
ROS
V
ein
Ri
13.
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.
Vout = 05 V
Vin = 2.483.90 V
Vout
b=
Rf
ROS
V
m=
Rf
Ri
Vin
m=
Vout (max) − Vout (min)
5V−0V
=
= 3.52
Vin (max) − Vin (min) 3.9 V − 2.48 V
b = Vout − mVin = 0 − 3.52 × 2.48 V = 8.73 V
The gain, m is set by Rf/Ri , Rf relatively large, so that Ri
will not load down the sensors
R f 330 kΩ
Let Rf = 330 kΩ
Ri =
=
= 93.7 kΩ
m
3.52
Select Rs as a 47 kΩ fixed resistor with a series 100 kΩ potentiometer. (m ~ 2.257.02)
R f V (330 kΩ)( 12V)
b is Rf/ROSV select V = 12 V
ROS =
=
= 454 kΩ
b
− 8.73 V
Select Rs as a 220 kΩ fixed resistor with a 500 kΩ potentiometer. (b ~ 18  5.5 V)
Rcomp = R f // R1 // ROS = 62.9 kΩ
Select Rcomp = 56 kΩ
The resistors in the 2nd stage should be in kΩ range, this will not load the 1st stage,
pick R = 2.2 kΩ and so R/2 = 1.1 kΩ
14.
Zero and Span: Instrument Amplifier
+V
+
8
2
16
+
ein
4
+
5
13
Rg

AD524
1
R1
O
6

Rg
R

span
+V
Output
+
7
+V
Sense
R2
Vout
9
11
3
span
S
10
12
R2

R2
R1
Reference
R2
+
+V
V
Vref
zero
Vref
zero
+


V
V
+
V
40000
(e2 − e1 ) + Vref
Vout = 1 +
Rg
Vref
Vout
15.
Zero and Span: Instrument Amplifier
Ex The output from a load cell changes 20 µV/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
Vsensor = 20µV/lb x Load + 18 mV
Vout = 10mV/lb x Load = G Vsensor+ Vref
8
2
16
+
4
+
5
13
Vsensor
Rg
S
3
1
span
Vout
9
6
11

O
10
AD524
12

Vout = 20µV G x Load + 18mV G + Vref
R
G=
7
+V
+V
V
Vref
zero
G = 1+
ref
Vout
Rg = 80.2 Ω
40000
= 500
Rg
Select Rg as a 33 Ω fixed resistor with a series 100 Ω
potentiometer. (G ~ 3011213)
Vref = −9 V
0 = 18 mVG + V = 18 mV × 500 + V
+

V
10 mV/lb
= 500
20 µV/lb
V
40000
(e2 − e1 ) + Vref
= 1 +
Rg
ref
So tie one end of the potentiometer to ground and
the other end to –Vsupply. Select resistor and
potentiometer values that will put approximately 10 V
at one end and 8 V at the other end.
16.
VoltageToCurrent 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
Rf
Rf
ein
Vo = 1 +
R

load
Load
R + Load Vo
wire
17.
V To I: Floating Load
The most simple V to I actually is the noninverting amplifier
+V
Rwire
+
+
load
V
ein

I
+
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 < Vsat
Vout = 1 +
R
Rloop must be kept small enough to keep the op amp out of the saturation.
18.
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 generalpurpose op amps.
However, the transistor can be added to increase the transmission current
capability.
19.
V To I: Floating Load
The signal at the load is inherently differential. So we can use difference or
instrumentation amplifier to reject any commonmode noise.
Open and short circuit in the transmission loop can be detected by
checking Vout of the noninverting amplifier (transmission side)
open circuit : Vout = Vsat
short circuit : Vout = ein
However, at the load side, with this circuit , if ein = 0, IL = 0, here it appears
that the IL = 0 is the valid signal. If the open or short circuit fault occurs, IL
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)
IL = 0 and ein = 0
This can be achieved by adding an offset to IL when ein = 0
ein = 0: IL > 0
Ex. 420 mA standard in current transimission
20.
V To I: Floating Load
zero
+V
ein
10 kΩ
eref
An example of 420 mA V to I converter (span + zero adjust)
V
+V
1 MΩ
+
Iin

1 MΩ
IL
Q1
load
V
R span
I
b=
m=
1
2R
eref
2R
ein
Rwire
I=
ein + eref
2R
=
ein eref
+
2R 2R
m = 1/2R, b = eref/2R
21.
Zero and Span: Instrument Amplifier
Ex Design an offset voltagetocurrent converter that will produce 4 mA with an
input of 5 V and 20 mA with an input of 10 V
I (mA)
Vin = 5  10 V
20
m=
15
10
1
I (max) − I out (min) 20 mA − 4 mA
= out
=
2 R Vin (max) − Vin (min)
10 V − (−5 V)
R = 469 Ω
Select a 430 Ω fixed resistor with a series 100 Ω potentiometer.
5
5
Iout = 4  20 mA
4
0
I=
5
Vin Vref
+
2R 2R
10
Vin
Vref = 2 RI − Vin = 2(469 Ω)(20 mA) − 10V = 8.8 V
22.
V To I: Grounded Load
This V to I actually is a derivative of difference amplifier
R3
R1
I
+
e1
Vout
R2
e2
Rs
+
VL
R4
load
I

R1=R2=R3=R4=R
Using superposition:
Vout due to e1 = e1
Vout due to e2 = e2
Vout due to VL = VL
Vout = Vout e1 +Vout e2 +Vout VL
=  e1 + e2 + VL = e2  e1 + VL
The current drive into the load is approximately equal to the current in Rs if RLoad << R2+R4
I L ≈ I Rs =
As in the case of the floating load:
Vout − VL e2 − e1
=
Rs
Rs
VSat > IRload + e2 − e1
23.
V To I: Grounded Load
An example of 420 mA V to I converter (span + zero adjust)
zero
+V
Rb
eref
100 kΩ

Rs
+
100 kΩ
I
Ra
100 kΩ
span
ein
V
+
VL
load
100 kΩ
IL ≈
I
m=
b=−
eref
Rs
ein
1
Rs
ein − eref
Rs
ein eref
=
−
Rs Rs
m = 1/R, b = eref/R
24.
V To I: XTR110
•4 to 20 mA Transmitter
•Selectable input/output ranges: 05V or 010V Inputs
420mA, 020mA, 525mA 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
25.
V To I: XTR110
VCC 13.540V
XTR110
VIN1(10V)
16
R8
500Ω
4
Io/10
R1
15kΩ
R5
R2
5kΩ

Va
R4
10kΩ
Vb
Io
I R8 ≈ I R 6
14
+
+

Io
420mA
Io/10
R6
1562.5Ω
16mA
span
9
R7
6250Ω
4mA
span
10
8
Vb Va
=
R6 R6
The current IR6 is approximately
equal to IR8
3
R3
20kΩ
16.25kΩ
5
External
MOS
1
VREF IN 3
VIN2(5V)
R9
50Ω
I R6 =
Since there is no voltage
difference between the op
amp inputs
RL
VR 8 = VR 9
I o = I R9 =
Io =
I R 8 R8
= 10 I R 8
R9
10Va
R6
26.
V To I: XTR110
The voltage at Va is defined by precision divider network and the
voltage at Vin1, Vin2 and Vref . Using superposition
Va due to Vin1 = Vin1/4
Va due to Vin2 = Vin2/2
Va due to Vref = Vref/16
Va = Vin1 / 4 + Vin 2 / 2 + Vref / 16
Therefore, the relation of the output current can be shown as
Io =
10(Vin1 / 4 + Vin 2 / 2 + Vref / 16)
Rspan
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 Vin2 = 0, Vref = 10 V, Vin1 varies from 010V and Use Rspan = R6 = 1562.5Ω
At Vin1 =0 V; Io = 4 mA at zero Vin1
Zero = 4 mA
At Vin1 =0 V; Io = 20 mA at Vin1 = 10 V
Zero+span = 4mA + 16 mA
28.
V To I: XTR110
Pin connections for standard XTR110 input voltage/output current ranges
15 V
1 µF
15
16
1
12
3
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
29.
V To I: XTR110
I out =
Vout = 400Vin + 6 V
Vsensor = ±10 mV
10(Vin1 / 4)
1250 Ω
I out = 4 − 20 mA
Vsensor = 2 − 10 V
I out =
10(400Vsensor + 6) / 4
1250 Ω
30.
CurrentToVoltage Converters
Grounded Load I to V converter
Span and Zero circuit
Rf
Ros
±V
4 to
20 mA
R
+

R
R2

+
+
RL
Rcomp
Current is converted
into a voltage by RL
R/2
+
Vout

Voltage follower is inserted
to avoid the loading effect
Vout =
Rf
Ri
IRL +
Rf
ROS
V
m = RfRL/Ri, b = Rf/ROSV
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.
31.
I To V Converters: Floating Load
Rf
420 mA
Rspan
Ri
R4
Ri
+
Vout

+
+V
Rf
Rpot
+
Vref
V
Vout =
Rf
Ri
IRspan + Vref
m = RfRL/Ri, b = Vref
To prevent the loading effect be sure that Rspan << Ri
or Instrumentation amplifier can be used in stead
32.
I To V Converters: Floating Load
Ex Design a floating currenttovoltage converter that will convert a 4 to 20mA
current signal into 0 to 10V groundreferenced voltage signal. If voltagetocurrent
converter, has +V = 12 V, Rspan VI =312 Ω, Imax = 20 mA What is the maximum
allowable Rspan IV
Vout = 010 V
Iin = 4  20 mA
Vout =
Rf
Ri
IRspan + Vref
Rf
Ri
RspanI −V =
10 V − 0 V
Vout (max) − Vout (min)
=
= 625 Ω
20 mA − 4 mA
I in (max) − I in (min)
Choose Rf/Ri =10 This seem arbitrary now, but we’ll come to check it again
RspanI −V = 62.5 Ω
Since R i >> Rspan IV, pick Ri = 2.2 kΩ
Find Vref :
Vref = Vout −
Rf
Ri
Rspan I = 0 −
Rf = 22 kΩ
22 kΩ
(4 mA )(62.5Ω ) = −2.5 V
2.2 kΩ
33.
I To V Converters: Floating Load
+V
+
+
2V
+

0.7 V

I
+
IRspan IV
V
+
IRspan VI
Rspan IV

I
Rspan VI

+ V = 2 + 0.7 + IRspanI −V + IRspanV − I
RspanI −V (max) =
12 − 2 − 0.7 − (20 mA)(312)
= 153 Ω
20 mA
Therefore, 62.5 Ω resistor would work. However, if we had chosen R f/Ri =1 then
Rspan IV = 625 Ω, which is too large. The op Amp in the voltagetocurren converter
would saturate before 20 mA would be reached.
34.
VoltageToFrequency 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 ∝ Vin
Vin
V to F
fout
fout
Vin
Ideal V to F characteristics
35.
VToF: LM131
LM131
Suitable for various applications:
precision V to F converter
simple lowcost circuits for A/D conversion
precision F to V converter
longterm integration
linear frequency modulation and demodulation
36.
VoltageToFrequency Converters
Basic V to F block diagram
Rt
VCC
8
2
Ct
5
Switched
current
source
fo =
Vlogic
RS
RL
CL
Comparator
1
Vx
6
7
Input
V
voltage 1
Vin
+
3
One shot
timer
Frequency
output
4
a
b
t
Vx
t
Vout
t
fout
High Vin → high fout
Adjustment
Low Vin → low fout
Adjust
High Vin → high fout
t
Vin RS 1
2.09 V RL Rt Ct
37.
VoltageToFrequency Converters
Rt
VCC
Ct
8
Determine
Iref
2
Determine the timer of one shot
TOS = 1.1Rt Ct
5
Switched
current
source
Vlogic
RS
RL
CL
Comparator
1
6
7
Input
V
voltage in
+
3
One shot
timer
Frequency
output
4
Phase I: Charge CL with Iref within the specific time TOS from the one shot timer.
At the end of this phase, VC reaches the value assigned as Vx
Phase II: CL discharge through Iref until VC is equal to Vin. The time in this phase is
defined as TII
We can found that 1/(TOS + TII) = f ∝ Vin
38.
VoltageToFrequency Converters
Phase I: Charge CL with Iref within the specific time TOS determined from the one
shot timer. At the end of this phase, VC reaches the value assigned as Vx
Initial condition: VC(0) = Vin
Iref
RL
CL
VC (t ) = (Vin − I ref RL )e − t /τ + I ref RL
τ = RL C L
e −TOS /τ ≈ 1 − TOS / τ
Assume RLCL >> TOS
Vx = VC (TOS ) = Vin + (I ref RL − Vin )
TOS
1
τ
Phase II: CL discharge through Iref until VC is equal to Vin. The time in this phase is
defined as TII
Initial condition: VC(0) = Vx
RL
I
CL
τ = RL C L
VC (t ) = Vx e − t /τ
Vin = VC (TII ) = Vx e −TII /τ
TII = τ ln
Since RLCL >> TOS, Therefore Vx /Vin ~ 1
V
TII = τ x − 1
V
in
Vx
Vin
ln
Vx Vx
≈
−1
Vin Vin
2
39.
VoltageToFrequency Converters
Combine eq.(1) and (2)
T = TOS + TII =
f =
I ref RLTOS
Vin
1
Vin
=
T I ref RLTOS
Substitute Iref = 1.9 V/ Rs ,TOS = 1.1RtCt
f =
Vin Rs
Vin
Rs 1
=
1.9 V RL (1.1Rt Ct ) 2.09 V RL Rt Ct
40.
V To F Converters
Ex Design a voltagetofrequency converter that will output 20 kHz when the input
is 5 V.
At maximum frequency, the minimum period is
1
1
Tmin =
=
= 50 µs
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.
tlow = 1.1Rt Ct
pick Ct = 0.0047 µF (somewhat arbitrary, but suggested by manufacturer)
Rt =
tlow
(0.8)(50 µs ) = 7.7 kΩ
=
1.1Ct (1.1)(0.0047 µF)
pick Rt = 6.8 kΩ, since this is not a critical parameter, as long as it is not too
big. This give tlow = 35 µs. Pick RL = 100 KΩ.
Rs =
(2 V ) f 0 RL Rt Ct = (2 V )(20 kHz )(100 kΩ )(6.8 kΩ )(0.0047 µF) = 25.6 kΩ
Vin
5V
Select a 22 kΩ fixed resistor with a series 10 kΩ potentiometer.
41.
VoltageToFrequency Converters
Lowpass filter
Simple V to F converter
From
fo =
Vin RS 1
2.09 V RL Rt Ct
I=
1.9 V
< 200 µA
RS
Manufacturer’s recommended, RL = 100 kΩ, Rt = 6.8 kΩ, Ct = 0.01 µF and RS = 14.2 kΩ
fo =
1 kHz
Vin
Volt
42.
V To F Converters
Ex Design a voltagetofrequency converter that will output 20 kHz when the input
is 5 V.
At maximum frequency, the minimum period is
1
1
Tmin =
=
= 50 µs
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.
tlow = 1.1Rt Ct
pick Ct = 0.0047 µF (somewhat arbitrary, but suggested by manufacturer)
Rt =
tlow
(0.8)(50 µs ) = 7.7 kΩ
=
1.1Ct (1.1)(0.0047 µF)
pick Rt = 6.8 kΩ, since this is not a critical parameter, as long as it is not too
big. This give tlow = 35 µs. Pick RL = 100 KΩ.
Rs =
(2 V ) f 0 RL Rt Ct = (2 V )(20 kHz )(100 kΩ )(6.8 kΩ )(0.0047 µF) = 25.6 kΩ
Vin
5V
Select a 22 kΩ fixed resistor with a series 10 kΩ potentiometer.
43.
FrequencyToVoltage Converters
VCC
Iout
1
RL
Connect to low
pass filter
Comparator
7
RD
8
+
One shot
timer
Rt
VCC
Ct
CD
Vin
Vin
Iout
i
T
The amplitude of pulse current output is
The average voltage output is
The average current output is
i=
I ave =
1.9 V
Rs
it 1.9 V × 1.1Rt Ct
=
T
RsT
Vave = I av e RL = 1.9 V ×1.1Rt Ct
RL
f in
Rs
44.
FrequencyToVoltage Converters
High pass filter
for input
frequency
Low pass filter
for output
current
Simple F to V converter
From
Vave = 1.9 V ×1.1Rt Ct
RL
f in
RS
RL = 100 kΩ, Rt = 6.8 kΩ, Ct = 0.01 µF and RS = 14.2 kΩ
Vave =
1V
f in
kHz
45.
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 frequencytovoltage 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
Tmin =
Select
f max =
6 counts 3000 rev 1 min
×
×
= 300 Hz
rev
min
60 s
1
Tpulse = 0.5Tmin = 1.67 ms
= 3.33 ms
300 Hz
Tout ( high ) ≤ 0.8Tmin = (0.8)(3.33 ms) = 2.664 ms
Select Ct = 0.33 µF
Rt =
tlow
(0.8)(50 µs ) = 7.7 kΩ
=
1.1Ct (1.1)(0.0047 µF)
pick Rt = 6.8 kΩ, This give tout(high) = 2.47 ms. We must set 5RDCD << 1.67 ms, so
pick RD = 10 kΩ.
5 RD C D ≤ (0.1)tout ( high )
C D ≤ 3.3 nF
46.
F To V Converters
RL
f in
Rs
2 V × 1.1Rt Ct RL f in (2 V )(1.1)(6.8 kΩ )(0.33 µF)(100 kΩ )(300 Hz )
Rs =
=
= 14.8 kΩ
Vave
10 V
Vave = 2 V × 1.1Rt Ct
Select Rs as a 10 kΩ fixed resistor with a series 10 kΩ potentiometer.
i=
Check
At 100 r/min
1.9 V
1.9 V
=
= 135 µA < 200 µA
Rs
14.8 kΩ
f =
6 counts 100 rev 1 min
×
×
= 10 Hz
rev
min
60 s
Tpulse = 0.05 ms
= 0.1 s
1
10 Hz
Filtering is accomplished by RL and CF
Tmin =
For 10% ripple,
CF ≥ −
t
RL ln(0.9)
CF ≥ −
τ = RL C F
Vout = V pk e − t /τ
Vout
= 0 .9 ≥ e − t / R L C F
V pk
and, we have
97.53 ms
= 9.25 µF
(100 kΩ ) ln(0.9)
t = T10Hz − tout ( high ) = 100 ms  2.47 ms = 97.53 ms
Select CF = 10 µF
47.
Cabling
• Cables are important because they are the longest parts of the system
and therefore act as efficient antennas that pick up and/or radiate noise.
• There are three coupling mechanisms that can occur between fields
and cables, and between cables (crosstalk).
• Capacitive or electric coupling (interaction of electric field and
circuit)
• Inductive or magnetic coupling (interaction between of the magnetic
field of two circuits.
• Electromagnetic coupling or radiation (RF interference)
48.
Magnetic Coupling
• When a current flows in a closed circuit, it produce a magnetic flux, φ
which is proportional to the current.
φ = LI
• When current flow in one circuit produces a flux in a second circuit,
there is a mutual inductance M12 between circuits 1 and 2
M 12 =
φ12
I1
φ12 ∝ the current, geometry, and orientation
• The voltage VN induced in circuit to 2 due to the current I1
VN = M
di
= jωM 12 I1
dt
Assume i varies sinusoidal with time
Magnetic coupling between two circuits
49.
Magnetic Coupling
• Separate the sensitive, input signal condition from the other portions of
the electronics: Digital signal (high f), high power circuit, ac power (50
or 60 Hz line)
• place low level signal on the separate card
• board layout
• separate low level signal conduit or race way, from ac power line,
communication or digital cable conduits.
• Reduce loop area (twisted pair)
• Use magnetic shield (effective at high frequency)
51.
Capacitive Coupling
• The potential different between two conductors generates a proportional
electric field. This will result in the redistribution or movement (current)
of charge by external voltage source
*The shield must be
grounded.
VN =
VN =
jωRC12
V1
1 + jωR(C12 + C2G + C2 S )
jωRC12
V1
1 + jωR(C12 + C2G )
Capacitive coupling between two
conductors
Capacitive coupling with shield
52.
Grounding Problems
• Grounding is one of the primary ways of minimizing unwanted noise
and pick up. (also bad grounding can cause the serious problem of
interference)
• A welldesigned system can provide the protection against unwanted
interference and emission, without any addition perunit cost to the
product
• Grounds category:
safety grounds (earth ground)
signal grounds
A signal ground is normally defined as an equipotential point or plane that
serves as a reference potential for a circuit or system (can not be realized
in practical systems)
A signal ground (better definition) is a lowimpedance path for current to
return to the source.
53.
Ground System
Ground System can be divided into three categories.
singlepoint grounds
multipoint grounds
hybrid grounds (frequency dependence)
1
2
Series connection
3
1
2
3
Parallel connection
Two types of singlepoint grounding connection
1
2
3
Multipoint connection
54.
SinglePoint Ground System
1
R1
2
R2
I1
A
I1+ I2+ I3
I2+ I3
3
I2
B
R3
I3
VA = ( I1 + I 2 + I 3 ) R1
VB = V A + ( I 2 + I 3 ) R2
I3
VC = VA + VB + I 3 R3
C
A series ground system is undesirable from a noise standpoint but has the
advantage of simple wiring
1
3
VA = I1 R1
VB = I 2 R2
R1
I1
2
VC = I 3 R3
A
R2
I2
B
R3
C
I3
A parallel ground system provides good noise performance at low frequency
but is mechanical cumbersome.
56.
Single Point Grounding System
Ex Calculate Va and Vb resulting from the input signal and from the effects of the
ground returns current under the following circumstances
(a) Actuator off, Ic = 5 mA, Ip = 20 mA, Ia = 0;
(b) Actuator on, Ic = 8 mA, Ip = 35 mA, Ia = 1 A;
The voltage, Vb is determined by the 40mV input signal and any voltage developed
across the 50 mΩ resistance by the ground return current.
Vb = −
1 MΩ
1 MΩ
(40 mV) + 1 +
Va
10 kΩ
10 kΩ
Where Va = (Ic+Ip+Ia)50mΩ
(a) Actuator off
Va = (5 mA + 20 mA + 0)(50 mΩ) = 1.25 mV
Vb = −(100)(40 mV) + (101)(1.25 mV) = −3.87 V
The error is about 3% compare to the correct value (100)(40 mV) = 4 V
(b) Actuator on
Va = (5 mA + 20 mA + 1 A)(50 mΩ) = 52 mV
Vb = −(100)(40 mV) + (101)(52 mV) = 1.25 V
The large return current has raised the noninverting input to 52 mV, and this
causes the unacceptable error.
58.
MultiplePoint Ground System
Ground loops can occur when the multiple ground points are separated by a
large distance and connected to the ac power ground, or when lowlevel analog
circuits are used.
0100 m
signal wiring
Control
Building
Plant
02000V
Ground difference
59.
MultiplePoint Ground System
circuit
1
VN
circuit
2
Ground loop
VG
A ground loop between two circuits
VN
circuit
1
circuit
2
VG
A ground loop between two circuits can be broken
by inserting a commonmode choke
60.
MultiplePoint Ground System
VN
circuit
1
circuit
2
VG
A ground loop between two circuits can be broken
by inserting a transformer
VN
circuit
1
circuit
2
VG
A ground loop between two circuits can be broken
by inserting an optical coupler
61.
Isolation Amplifiers
General concepts: Isolation Amplifier provides three important advantages
over normal amplifiers.
Safety issues for some industrial and medical applications: signal
common isolation and common ground can be achieved
Extremely high commonmode voltage tolerance (in general amp
this is normally less than power supply)
Very low failure currents
62.
Transformercoupled Amplifiers
Symbol of transformercoupled amp.
Block diagram of AD289 isolation amp
63.
Optically Coupled Amplifiers
Transformercoupled isolation amplifiers are expensive, bulky, bandwidth
limited and slow response.
Opticallycoupled isolation amplifiers have less nonlinear and isolation.
64.
Optically Coupled Amplifiers
Symbol of optically coupled amp.
Simplified schematic diagram
I1 = I 2 =
Vin
RG
Vout = I 2 RK =
RK
Vin
RG
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