signal conditioning and transmission
TRANSCRIPT
Signal Conditioners and Transmission
2102-487 Industrial Electronics
Overview
Amp V to I I to V Amp
4-20 mA
Motor/Power Arcing Lighting
Noise Source
Interference
Field Control Room
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+∆R
R
E Vout+
Vdm/2
Vdm/2Vcm +- +-
+-
R-
R+
Equivalent circuit for a Load cell Bridge circuit for sensor application
e1
e2
R
R
Rg
+
-
+
-
Vo
+
-
IA: The First Stage
V1
V2
Rg = gain setting resistor
+−=
go R
ReeV 21)( 21
• Changing Rg will inversely alter the output voltage.
11 eV =
For Ideal op amp, no voltage difference between the inverting and noninverting inputs
22 eV =and
21 eeVgR −=
The voltage across Rg
gR R
eeIg
21 −=
This current must flow through all three resistors because none of the current can flow into the op amp inputs
( )ggRo RRIV += 2
IRg
IA: First + Second stage
+
-+
-Vo
R1
R
R
Rg
+
-
+
-
R1
R1
R1
+−=
go R
ReeV 21)( 211
e2
e1
Vo1
+
-
1oo VV −=
+−=
go R
ReeV 21)( 12-1
+
gRR2121 ee −
V2
V1
1oV
Without loading effect, we can applied the direct multiplication for the cascade system
gRR21Gain +=
+−=
go R
ReeV 21)( 12
The second stage of IA is a unit-gain differential amplifier
+
-
R2
Rg
+
-
+
-
+
-
+V
-V
VrefVref
LoadReference
Output
Sense
R1
R1
R2
R2
R2
IA with Offset (Reference)
+−=
go R
ReeV 21)( 211
e2
e1
refg
o VRReeV +
+−=
21)( 12
refoo VVV +−= 1
-1
+
gRR2121 ee − 1oV
+refV
refg
o VRReeV +
+−=
21)( 12
Vo1
V1
V2
Commercial IA: AD524
AD524 Functional Block Diagram
%20400001 ±
+=
gRG Here Rg = 40, 404, 4.44 kΩ or external
resistor (connect: RG1 – RG2)
Features:Low Noise: 0.3 µVp-p 0.1 Hz to 10 HzLow Nonlinearity: 0.03% (G = 1)High CMRR: 120 dB (G = 1000)Low offset Voltage: 50 µVGain Bandwidth Product: 25 MHzProgrammable Gain of 1, 10, 100, 1000
Commercial IA: AD524
349Ω
Ex For the circuit, calculate Va, Vb, and Vout as well as the effect on the output of the 5-V common-mode signal.
Vb Va
Vout
V 5supply21 == VVb
V 4.99285V 10349350
349=
Ω+ΩΩ
=aV
( )mV -715V) 5-V 4.99285(100 ==
−= baout VVGV
For a gain of 100, CMRR = 100 dB
cm
dm
GGCMRR log20=
mV 5V 5001.0 =×=cmcmVG
001.0=cmG
%7.0%100mV 715
mV 5=×
The common mode error is
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 14-bit 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)
Commercial IA: AD5241000ppm = 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 µV/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.
Vout
Vin
Vout
Vin
Zero shifted positiv
ely
Zero sh
ifted negati
vely
Original
Original
Span
dou
ble
Span Half
Zero and Span: Inverting summer
+
-Ri
Rf
Rcomp
ROS
+
-R
R
R/2
ein
±V
eu2
eu1
VRR
eRR
eOS
fin
i
fu −−=1 12 uu ee −=
VRR
eRR
eOS
fin
i
fu +=2
Inverting summer Inverting amplifier
Compare this to the eq. of straight line bmxy +=
Here y = eu2, x = ein, m = Rf/Ri and b = Rf/Ros Vein
eout
VRR
bOS
f=
i
f
RR
m =
Zero and Span: Instrument AmplifierEx 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.
Vin
Vout
VRR
bOS
f=
i
f
RR
m =
Vin = 2.48-3.90 V Vout = 0-5 V
52.3V 48.2V 9.3
V 0V 5(min)(max)(min)(max)
=−−
=−−
=inin
outout
VVVVm
V -8.73V 48.252.30 =×−=−= inout mVVb
The gain, m is set by Rf/Ri , Rf relatively large, so that Riwill not load down the sensors
Let Rf = 330 kΩ Ω=== k 7.9352.3kΩ 330
mR
R fi
Select Rs as a 47 kΩ fixed resistor with a series 100 kΩ potentiometer. (m ~ 2.25-7.02)
b is Rf/ROSV select V = -12 V Ω=−
Ω== k 454
V 73.812V) )(-k 330(
bVR
R fOS
Select Rs as a 220 kΩ fixed resistor with a 500 kΩ potentiometer. (b ~ -18 - -5.5 V)Ω== k 9.62//// 1 OSfcomp RRRR 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
Zero and Span: Instrument Amplifier
+
-
R2
Rg
+
-
+
-
+
-
+V
-V
VrefVref
Reference
Output
Sense
R1
R1
R2
R2
R2
Vout
8
4
5
10
96
7
2
16
13
12
11
3
1 -
+
AD524
span
ein
+
-
Rg
S
R
O
+
-
+V
-V
Vref
+V
-V
-V
Vout
zero
+V
span
zero
( ) refg
out VeeR
V +−
+= 12
400001
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.
Vsensor = 20µV/lb x Load + 18 mV
Vout = 10mV/lb x Load = G Vsensor+ Vref8
4
5
10
96
7
2
16
13
12
11
3
1 -
+
AD524
span
Vsensor
+
-
Rg
S
R
O
+
-
+V
-V
Vref
+V
-V
-V
Vout
zero
+V
Vout = 20µV G x Load + 18mV G + Vref
500V/lb 20
mV/lb 10==
µG
( ) refg
out VeeR
V +−
+= 12
400001
500400001 =+=gR
G
V 9−=refV
Select Rg as a 33 Ω fixed resistor with a series 100 Ωpotentiometer. (G ~ 301-1213)
refref VVG +×=+= 500mV 18mV 180
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.
Ω= 2.80gR
+
- +
-
load
Rwire
Rf
+
-ein
R
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.
inf
o eRR
V
+= 1
owire
VLoadR
Load
+
Rwire ∝ Distance, wire type, temperature
V To I: Floating Load
+
-load
Rwire
+
-ein
R I
I
ein
+
-
+V
-V
The most simple V to I actually is the non-inverting amplifier
ReI in=
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
satinloop
out VeR
RV <
+= 1
Rloop must be kept small enough to keep the op amp out of the saturation.
+
-
load
Rwire
R
I
+V
-V
+
-ein
I
Q1
Q2
V To I: Floating Load
V to I with current booster
ReI in=
Add transistor forincrease current output
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.
New op amp with high current output
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 Vout of the non-inverting amplifier (transmission side)
open circuit : Vout = Vsatshort 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, ILwill 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
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, ILwill 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 = 0ein = 0: IL > 0
Ex. 4-20 mA standard in current transimission
Rwire
R
Q1
-V+V10 kΩ
1 MΩ
1 MΩ
ein
eref
Iin IL
load
span
zero
-
+
-V
+V
V To I: Floating Load
An example of 4-20 mA V to I converter (span + zero adjust)
Re
Re
Ree
I refinrefin
222+=
+= m = 1/2R, b = eref/2R
ein
I
Re
b ref
2=
Rm
21
=
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
Vin = -5 - 10 V
V 8.810VmA) 20)( 469(22 =−Ω=−= inref VRIV
Select a 430 Ω fixed resistor with a series 100 Ω potentiometer.
Ω= 469R
I (mA)
Vin
5
10
15
20
4
- 5 5 100
RV
RVI refin
22+=
Iout = 4 - 20 mA
V) 5(V 01mA 4mA 20
(min)(max)(min)(max)
21
−−−
=−−
==inin
outout
VVII
Rm
V To I: Grounded Load
+
-
+
-
e1 e2
R1
R2
R4load
R3
Rs
I
I
VL
R1=R2=R3=R4=R
Vout
The current drive into the load is approximately equal to the current in Rs if RLoad << R2+R4
This V to I actually is a derivative of difference amplifier
Using superposition: Vout due to e1 = -e1Vout due to e2 = e2Vout due to VL = VL
LL
Vouteouteoutout
VeeV ee
VVVVL
+=++=
++=
1221 --
|||21
ss
LoutRL R
eeR
VVIIs
12 −=
−=≈
As in the case of the floating load: 12 eeIRV loadSat −+>
V To I: Grounded Load
s
ref
s
in
s
refinL R
eRe
Ree
I −=−
≈
An example of 4-20 mA V to I converter (span + zero adjust)
+
-
-V
+V
+
-ein load
100 kΩ
I
VL
RbRa
100 kΩ
100 kΩ
100 kΩ
Rs
eref
span
zero
m = 1/R, b = -eref/R
ein
I
s
ref
Re
b −=
sRm 1
=
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
Precision resistive network
V to I converter
I to I converter(current mirror)
Reference voltage
V To I: XTR110
R115kΩ
R320kΩ
R410kΩ
R25kΩ
16.25kΩ
R5
VIN1(10V)
VREF IN
R76250Ω4mAspan
R61562.5Ω
16mAspan
+
-
+
-
R8500Ω
R950Ω
VIN2(5V)
VCC 13.5-40V
RL
Io4-20mA
16
1
3
14
89 10
5
3
4
IoIo/10
Va
Io/10
Vb
XTR110External MOS
666 R
VRVI ab
R ==
The current IR6 is approximately equal to IR8
68 RR II ≈
Since there is no voltage difference between the op amp inputs
98 RR VV =
89
889 10 R
RRo I
RRIII ===
6
10RVI a
o =
V To I: XTR110
( )span
refinino R
VVVI
16/2/4/10 21 ++=
The voltage at Va is defined by precision divider network and the voltage at Vin1, Vin2 and Vref . Using superposition
Va due to Vin1 = Vin1/4Va due to Vin2 = Vin2/2Va due to Vref = Vref/16
Therefore, the relation of the output current can be shown as
16/2/4/ 21 refinina VVVV ++=
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 0-10V and Use Rspan = R6 = 1562.5Ω
At Vin1 =0 V; Io = 4 mA at zero Vin1
At Vin1 =0 V; Io = 20 mA at Vin1 = 10 V Zero+span = 4mA + 16 mA
Zero = 4 mA
V To I: XTR110
Va
R115kΩ
R25kΩ
16.25kΩ
R5
VIN1
VREF IN
VIN2
5
3
4
R320kΩ
R410kΩ
Va
R115kΩ R2
5kΩ
16.25kΩ
R5
VIN1
R320kΩ
R410kΩ
R115kΩ R2
5kΩ
16.25kΩ
R5
VIN2
R410kΩ
R320kΩ
Va
R115kΩ
R25kΩ
16.25kΩ
R5
VREF IN
R320kΩ
R410kΩ
Va
IN141
IN121543
2154
)////()////( VVRRRRR
RRRRVa =++
+= IN22
1IN2
21534
2153
)////()////( VVRRRRR
RRRRVa =++
+= REF16
1 REF
21
2
43215
43
//)//(// VV
RRR
RRRRRRRVa =
+++=
1 2 3
1 2 3+ +
REF161
IN221
IN141 VVVVa ++=
kΩ 27=inR kΩ 22=inR kΩ 20=inR
V To I: XTR110
XTR110
15
12
3
4
59 2
16
1
13
14
0 to 10 V
4 to 20 mA Out
VLRL
1 µF
15 V
A basic 4 to 20 mA transmitter
Pin connections for standard XTR110 input voltage/output current ranges
V To I: XTR110
V 6400 += inout VV
mV 10±=sensorV V 102 −=sensorV mA 204 −=outI
Ω=
1250)4/(10 1in
outVI
Ω+
=1250
4/)6400(10 sensorout
VI
Current-To-Voltage ConvertersGrounded Load I to V converter
+
-R2
Rf
Rcomp
Ros
+
-R
R
R/2
+
-
4 to20 mA
RL Vout
+
-
±V
Current is converted into a voltage by RL
Span and Zero circuit
Voltage follower is inserted to avoid the loading effect
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.
VRR
IRRR
VOS
fL
i
fout += m = RfRL/Ri, b = Rf/ROSV
I To V Converters: Floating Load
refspani
fout VIR
RR
V += m = RfRL/Ri, b = Vref
+
-
Rf
R4
+
-
Ri
RiRspan
Rf
Rpot
+V
-V
Vout
+
-
4-20 mA
Vref
To prevent the loading effect be sure that Rspan << Rior Instrumentation amplifier can be used in stead
Ex Design a floating current-to-voltage converter that will convert a 4- to 20-mA current signal into 0- to 10-V ground-referenced voltage signal. If voltage-to-current converter, has +V = 12 V, Rspan V-I =312 Ω, Imax = 20 mA What is the maximum allowable Rspan I-V
Iin = 4 - 20 mA Vout = 0-10 V
Ω=−−
=−−
=− 625mA 4mA 02V 0V 01
(min)(max)(min)(max)
inin
outoutVspanI
i
f
IIVVR
RR
Choose Rf/Ri =10 This seem arbitrary now, but we’ll come to check it again
Find Vref : ( )( ) V 5.25.62mA 4kΩ 2.2kΩ 220 −=Ω−=−= IR
RR
VV spani
foutref
refspani
fout VIR
RR
V +=
Ω=− 5.62VspanIR
Since R i >> Rspan I-V, pick Ri = 2.2 kΩ Rf = 22 kΩ
I To V Converters: Floating Load
IspanVVspanI IRIRV −− +++=+ 7.02
Therefore, 62.5 Ω resistor would work. However, if we had chosen R f/Ri =1 then Rspan I-V = 625 Ω, which is too large. The op Amp in the voltage-to-curren converter would saturate before 20 mA would be reached.
I To V Converters: Floating Load
Rspan I-V
I-
+
-V
+V
Rspan V-I
+-
+- +
-
IRspan I-V
IRspan V-I
+
-
I
0.7 V
2 V
Ω=−−−
=− 153mA 20
)312)(mA 20(7.0212(max)VspanIR
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
inVf ∝
V to FVin fout
fout
Vin
Ideal V to F characteristics
V-To-F: LM131
Suitable for various applications:precision V to F convertersimple low-cost circuits for A/D conversionprecision F to V converterlong-term integrationlinear frequency modulation and demodulation
LM131
Voltage-To-Frequency Converters
RS
3
Comparator
+
- One shottimer
Switchedcurrentsource
RL
CL
Rt Ct
Frequencyoutput
Vlogic
8VCC
V1Input
voltage7
1 6
25
4
ab
Vin
Vx
Vout
fout High Vin → high fout Adjustment Low Vin → low foutAdjust High Vin → high fout
t
t
t
t
Basic V to F block diagram
ttL
Sino CRR
RVf 1V 09.2
=ttL
Sino CRR
RVf 1V 09.2
=
Vx
Voltage-To-Frequency Converters
RS
3
Comparator
+
- One shottimer
Switchedcurrentsource
RL
CL
Rt Ct
Frequencyoutput
Vlogic
8VCC
VinInput
voltage7
1 6
25
4
Determine the timer of one shot
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
Determine Iref
ttOS CRT 1.1=
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
Voltage-To-Frequency 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
Phase II: CL discharge through Iref until VC is equal to Vin. The time in this phase is defined as TII
RL CLIref
Initial condition: VC(0) = Vin
( ) Lreft
LrefinC RIeRIVtV +−= − τ/)(
Assume RLCL >> TOS ττ /1/OS
T Te OS −≈−
LLCR=τ
( )τOS
inLrefinOSCxTVRIVTVV −+== )(
RLCL
I
Initial condition: VC(0) = Vxτ/)( t
xC eVtV −= LLCR=τ
Since RLCL >> TOS, Therefore Vx /Vin ~ 1
τ/)( IITxIICin eVTVV −==
in
xII V
VT lnτ=
1ln −≈in
x
in
x
VV
VV
−= 1
in
xII V
VT τ
1
2
Voltage-To-Frequency Converters
Combine eq.(1) and (2)
Substitute Iref = 1.9 V/ Rs ,TOS = 1.1RtCt
in
OSLrefIIOS V
TRITTT =+=
OSLref
in
TRIV
Tf ==
1
( ) ttL
sin
ttL
sin
CRRRV
CRRRVf
1
V .0921.1 V 9.1==
Ex Design a voltage-to-frequency converter that will output 20 kHz when the input is 5 V.
s 50kHz 2011
maxmin µ===
fT
At maximum frequency, the minimum period is
( ) ( )( )( )( )( )Ω=
ΩΩ== k 6.25
V 5µF .00470k .86k 100kHz 20V 2V 2 0
in
ttLs V
CRRfR
1.1 ttlow CRt =
pick Ct = 0.0047 µF (somewhat arbitrary, but suggested by manufacturer)
V To F Converters
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.
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.
Select a 22 kΩ fixed resistor with a series 10 kΩ potentiometer.
( )( )( )( ) Ω=== k 7.7
µF .004701.1µs 058.0
1.1 t
lowt C
tR
Voltage-To-Frequency Converters
ino VfVoltkHz 1
=
Simple V to F converter
ttL
Sino CRR
RVf 1V 09.2
=
Manufacturer’s recommended, RL = 100 kΩ, Rt = 6.8 kΩ, Ct = 0.01 µF and RS = 14.2 kΩ
From µA 200V 9.1<=
SRI
Low-pass filter
Ex Design a voltage-to-frequency converter that will output 20 kHz when the input is 5 V.
s 50kHz 2011
maxmin µ===
fT
At maximum frequency, the minimum period is
( ) ( )( )( )( )( )Ω=
ΩΩ== k 6.25
V 5µF .00470k .86k 100kHz 20V 2V 2 0
in
ttLs V
CRRfR
1.1 ttlow CRt =
pick Ct = 0.0047 µF (somewhat arbitrary, but suggested by manufacturer)
V To F Converters
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.
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.
Select a 22 kΩ fixed resistor with a series 10 kΩ potentiometer.
( )( )( )( ) Ω=== k 7.7
µF .004701.1µs 058.0
1.1 t
lowt C
tR
Frequency-To-Voltage Converters
Comparator
RL
8
+
-One shot
timer
Rt
Ct
7
Iout
VCC
VCC
1
Vin
RD
CD
Connect to low pass filter
The amplitude of pulse current output is
Vin
Iout
i
T
The average voltage output is
ins
LttLeavave f
RRCRRIV 1.1 V 9.1 ×==The average current output is
sRi V .91
=
TRCR
TitI
s
ttave
1.1V .91 ×==
Frequency-To-Voltage Converters
inave fVkHz
V 1=
Simple F to V converter
inS
Lttave f
RRCRV 1.1V 9.1 ×=
RL = 100 kΩ, Rt = 6.8 kΩ, Ct = 0.01 µF and RS = 14.2 kΩ
From
Low pass filter for output current
High pass filter for input frequency
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-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
The maximum frequency is Hz 300s 60
min 1min
rev 0003rev
counts 6max =××=f
Select
F To V Converters
ms 33.3Hz 300
1min ==T
( )( ) ms 664.2ms 33.38.08.0 min)( ==≤ TT highout
Select Ct = 0.33 µF ( )( )( )( ) Ω=== k 7.7
µF .004701.1µs 058.0
1.1 t
lowt C
tR
ms 67.15.0 min == TTpulse
pick Rt = 6.8 kΩ, This give tout(high) = 2.47 ms. We must set 5RDCD << 1.67 ms, so pick RD = 10 kΩ.
)()1.0(5 highoutDD tCR ≤ nF 3.3≤DC
At 100 r/min
F To V Converters
µA 200µA 135k 4.81V .91V .91
<=Ω
==sR
i
Filtering is accomplished by RL and CF
For 10% ripple,
FLCR=τ
( ) µF 25.9)9.0ln(k 100
ms 53.97=
Ω−≥FC
ins
Lttave f
RRCRV 1.1 V 2 ×=
( )( )( )( )( )( )Ω=
ΩΩ=
×= k 8.14
V 10Hz 003k 001µF .330k .861.1V 21.1 V 2
ave
inLtts V
fRCRR
Select Rs as a 10 kΩ fixed resistor with a series 10 kΩ potentiometer.
Check
Hz 10s 60
min 1min
rev 001rev
counts 6=××=f
s 1.0Hz 011
min ==T ms 05.0=pulseT
τ/tpkout eVV −=
FLCRt
pk
out eVV /9.0 −≥=
)9.0ln(LF R
tC −≥
Select CF = 10 µF
ms 97.53ms 2.47-ms 100)(o10Hz ==−= highuttTtand, we have
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)
Magnetic Coupling• When a current flows in a closed circuit, it produce a magnetic flux, φ
which is proportional to the current.
• When current flow in one circuit produces a flux in a second circuit, there is a mutual inductance M12 between circuits 1 and 2
• The voltage VN induced in circuit to 2 due to the current I1
LI=φ
1
1212 I
M φ=
112IMjdtdiMVN ω== Assume i varies sinusoidal with time
φ12 ∝ the current, geometry, and orientation
Magnetic coupling between two circuits
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)
Magnetic Coupling
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
Capacitive coupling between two conductors Capacitive coupling with shield
1212
12
)(1V
CCRjRCjV
GN ++
=ω
ω
12212
12
)(1V
CCCRjRCjV
SGN +++
=ω
ω
*The shield must be grounded.
• 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 well-designed system can provide the protection against unwanted interference and emission, without any addition per-unit cost to the product
• Grounds category: safety grounds (earth ground)signal grounds
A signal ground (better definition) is a low-impedance path for current to return to the source.
Grounding Problems
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)
Ground SystemGround System can be divided into three categories.
single-point groundsmultipoint groundshybrid grounds (frequency dependence)
1 2 3 1 2 3 1 2 3
Two types of single-point grounding connection
Series connection Parallel connection Multipoint connection
Single-Point Ground System
1 2 3R1 R2 R3
I3I1+ I2+ I3 I2+ I3
I3I2I1
A series ground system is undesirable from a noise standpoint but has the advantage of simple wiring
A B C
1 2 3R1
R2
I2
I1
I3
R3
A parallel ground system provides good noise performance at low frequency but is mechanical cumbersome.
A
B
C
33
232
1321
)()(
RIVVVRIIVV
RIIIV
BAC
AB
A
++=++=
++=
33
22
11
RIVRIVRIV
C
B
A
===
Single-Point Ground 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 40-mV input signal and any voltage developed across the 50 mΩ resistance by the ground return current.
Where Va = (Ic+Ip+Ia)50mΩ
Single Point Grounding System
ab VV
Ω++
Ω−=
k 01MΩ 11)mV 40(
k 01MΩ 1
(a) Actuator off
V 87.3)mV 25.1)(101()mV 40)(100( −=+−=bV
mV 1.25)m 0)(50mA 20mA (5 =Ω++=aV
The error is about 3% compare to the correct value (-100)(40 mV) = -4 V
(a) Actuator off
V 25.1)mV 52)(101()mV 40)(100( =+−=bV
mV 25)m A)(50 1mA 20mA (5 =Ω++=aV(b) Actuator on
The large return current has raised the non-inverting input to 52 mV, and this causes the unacceptable error.
Single-Point Ground System
Multiple-Point Ground SystemGround loops can occur when the multiple ground points are separated by a large distance and connected to the ac power ground, or when low-level analog circuits are used.
Plant ControlBuilding
0-2000VGround difference
signal wiring
0-100 m
Multiple-Point Ground System
circuit1
circuit2
Ground loop
VN
VG
circuit1
circuit2
VN
VG
A ground loop between two circuits
A ground loop between two circuits can be broken by inserting a common-mode choke
Multiple-Point Ground System
A ground loop between two circuits can be broken by inserting an optical coupler
A ground loop between two circuits can be broken by inserting a transformer
circuit1
circuit2
VN
VG
circuit1
circuit2
VN
VG
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 common-mode voltage tolerance (in general amp this is normally less than power supply)
Very low failure currents
Transformer-coupled Amplifiers
Block diagram of AD289 isolation amp
Symbol of transformer-coupled amp.
Optically Coupled Amplifiers
Transformer-coupled isolation amplifiers are expensive, bulky, bandwidth limited and slow response. Optically-coupled isolation amplifiers have less non-linear and isolation.
Optically Coupled Amplifiers
Symbol of optically coupled amp. Simplified schematic diagram
G
in
RVII == 21
inG
KKout V
RRRIV == 2