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Sig con

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SIG COn

SIG COn

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  • 1. 2102-487 Industrial Electronics Signal Conditioners and Transmission
  • 2. Overview 4-20 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 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 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 unit-gain 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 µVp-p 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 common-mode 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 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)
  • 10. Commercial IA: AD524 1000ppm = 0.1%
  • 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 5-V 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 = 0-5 V Vin = 2.48-3.90 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 ~ 2.25-7.02) 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 ~ 301-1213) 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. 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 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 non-inverting 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 general-purpose 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 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 = 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. 4-20 mA standard in current transimission
  • 20. V To I: Floating Load zero +V ein 10 kΩ eref An example of 4-20 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 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) 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 4-20 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: 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
  • 25. V To I: XTR110 VCC 13.5-40V 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 4-20mA 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 0-10V 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
  • 27. V To I: XTR110 VIN1 4 VREF IN 3 R1 15kΩ R3 20kΩ R5 R2 5kΩ 1 + 2 + 3 Va 16.25kΩ R4 10kΩ 1 Va = 1 VIN1 + 1 VIN2 + 16 VREF 4 2 VIN2 5 Rin = 22 kΩ Rin = 27 kΩ Rin = 20 kΩ VIN2 VIN1 VREF IN 1 Va 16.25kΩ R1 15kΩ R2 5kΩ R4 10kΩ R4 //( R5 + R1 // R2 ) VIN1 = 1 VIN1 4 R3 + R4 //( R5 + R1 // R2 ) R1 15kΩ R4 10kΩ R5 R5 Va = 2 R3 20kΩ Va = R2 5kΩ R5 Va 16.25kΩ R1 15kΩ Va 16.25kΩ R2 5kΩ R3 20kΩ R3 //( R5 + R1 // R2 ) VIN2 = 1 VIN2 2 R4 + R3 //( R5 + R1 // R2 ) 3 Va = R4 10kΩ R3 20kΩ R3 // R4 R2 1 VREF = 16 VREF ( R5 + R1 // R2 ) + R3 // R4 R1 + R2
  • 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. Current-To-Voltage 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 4-20 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 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 Vout = 0-10 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 I-V, 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 I-V -V + IRspan V-I Rspan I-V - I Rspan V-I - + 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 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.
  • 34. 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 ∝ Vin Vin V to F fout fout Vin Ideal V to F characteristics
  • 35. 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
  • 36. Voltage-To-Frequency 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. Voltage-To-Frequency 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. 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 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. Voltage-To-Frequency 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 voltage-to-frequency 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. Voltage-To-Frequency Converters Low-pass 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 voltage-to-frequency 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. Frequency-To-Voltage 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. Frequency-To-Voltage 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 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 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)
  • 50. Magnetic Coupling
  • 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 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 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 low-impedance path for current to return to the source.
  • 53. Ground System Ground System can be divided into three categories. single-point grounds multipoint grounds hybrid grounds (frequency dependence) 1 2 Series connection 3 1 2 3 Parallel connection Two types of single-point grounding connection 1 2 3 Multipoint connection
  • 54. Single-Point 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.
  • 55. Single-Point Ground System
  • 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 40-mV 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 non-inverting input to 52 mV, and this causes the unacceptable error.
  • 57. Single-Point Ground System
  • 58. Multiple-Point 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 low-level analog circuits are used. 0-100 m signal wiring Control Building Plant 0-2000V Ground difference
  • 59. Multiple-Point 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 common-mode choke
  • 60. Multiple-Point 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 common-mode voltage tolerance (in general amp this is normally less than power supply) Very low failure currents
  • 62. Transformer-coupled Amplifiers Symbol of transformer-coupled amp. Block diagram of AD289 isolation amp
  • 63. 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.
  • 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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