Unit2-MS

PROCESS CONTROL (IT62)

Chapter 2

 ANA  A NAL L OG CONTROLL ERS

by,

Dr. M. Meenakshi Professor and Head Department of Instrumentation Technology Dr. AIT, Bangalore - 560056

Chapter 2  An  A n alo al o g Con t r o ll ers er s 2.1 Introduction Selection of the controller mode to use and appropriate gains depend on many factors involved in the process operation. This decision is made by engineers who are familiar with the process itself and who are aided by process control technique experts who understand the characteristics of each mode. Chapter 1 presented the various characteristics of the process to be considered while selecting a suitable controller mode,  principles of various controller modes, advantages, disadvantages and applications of each controller modes.

This chapter provides a detailed description of how controller modes are implemented using analog techniques. Both electronic techniques and pneumatic techniques are discussed. More emphasis is given on the electronic controllers, realized using op amps as the active element because of its wide spread use. Pneumatic techniques are also discussed because there are many operations where a complete implementation of  process control loop uses pneumatic methods. Specific methods of controller mode realization, using either electronics or pneumatics are as varied as the manufacturers of this equipment. equipment. Hence, the materials presented in this chapter are in a general fashion, using op amps in electronics and general principles in pneumatics. 2.2 Objectives The objectives of this chapter are to provide an understanding of how controller modes are implemented using electronic and pneumatic techniques. After the detailed study of this chapter, you should able to: 1. Diagram the physical appearance of an analog controller. 2. Diagram and describe how two position and three position control can be implemented using comparators and a flip flop 3. Draw a schematics and describe how op amps can be used to implement the  proportional, integral and derivative modes of analog control 4. Design an op amp circuit that will implement the proportional integral,  proportional derivative, and proportional integral derivative control modes. 5. Diagram and describe how proportional, proportional integral, proportional derivative, and proportional integral derivative control modes can be implemented using pneumatic Nozzle/ Flapper system 2.3. General Features An analog controller is a device that implements different controller modes using analog signals to represent the loop parameters. The analog signal may be in the form of an electric current or a pneumatic air pressure. The controller accepts a measurement expressed in terms of one of these signals, calculates an output for the mode being used, and outputs an analog signal of the same type. Since controller does solve equations, it

Chapter 2  An  A n alo al o g Con t r o ll ers er s 2.1 Introduction Selection of the controller mode to use and appropriate gains depend on many factors involved in the process operation. This decision is made by engineers who are familiar with the process itself and who are aided by process control technique experts who understand the characteristics of each mode. Chapter 1 presented the various characteristics of the process to be considered while selecting a suitable controller mode,  principles of various controller modes, advantages, disadvantages and applications of each controller modes.

This chapter provides a detailed description of how controller modes are implemented using analog techniques. Both electronic techniques and pneumatic techniques are discussed. More emphasis is given on the electronic controllers, realized using op amps as the active element because of its wide spread use. Pneumatic techniques are also discussed because there are many operations where a complete implementation of  process control loop uses pneumatic methods. Specific methods of controller mode realization, using either electronics or pneumatics are as varied as the manufacturers of this equipment. equipment. Hence, the materials presented in this chapter are in a general fashion, using op amps in electronics and general principles in pneumatics. 2.2 Objectives The objectives of this chapter are to provide an understanding of how controller modes are implemented using electronic and pneumatic techniques. After the detailed study of this chapter, you should able to: 1. Diagram the physical appearance of an analog controller. 2. Diagram and describe how two position and three position control can be implemented using comparators and a flip flop 3. Draw a schematics and describe how op amps can be used to implement the  proportional, integral and derivative modes of analog control 4. Design an op amp circuit that will implement the proportional integral,  proportional derivative, and proportional integral derivative control modes. 5. Diagram and describe how proportional, proportional integral, proportional derivative, and proportional integral derivative control modes can be implemented using pneumatic Nozzle/ Flapper system 2.3. General Features An analog controller is a device that implements different controller modes using analog signals to represent the loop parameters. The analog signal may be in the form of an electric current or a pneumatic air pressure. The controller accepts a measurement expressed in terms of one of these signals, calculates an output for the mode being used, and outputs an analog signal of the same type. Since controller does solve equations, it

can be treated as an analog computer. The controller must be able to add, subtract, multiply, integrate and find derivatives of analog voltages or pressures. 2.3.1 Typical Physical Layout Analog controllers are usually designed to fit into a panel assembly as a slide in/out module, as shown in Fig. 2.1. The front panel displays all necessary information and  provides adjustment capability for the operator. When the unit is pulled out partway but still connected, other, less frequently required adjustments are available. When the controller is pulled still further out, an extension cable can be disconnected and the entire unit removed from the panel for replacement, if necessary.

Fig. 2.1: Typical physical appearance of a controller 2.3.2 Front panel The front panel of an analog controller displays information for operators and allows adjustment of the setpoint. The setpoint knob moves a sliding scale under the fixed setpoint indicator. Thus, a fixed span span of measurement above and below the setpoint is visible, as indicated by the measurement value indicator. The error is the difference  between the set point indicator and the measurement meter. The display is typically expressed in percent of span (4 to 20mA or 3 to 15 psi). The lower meter shows the controlled output, expressed in percentage of span. The output range is actually 4 to 20 mA or 3 to 15 psi, so that 0 % would mean 4 mA for example. There is often a switch on the front panel by which the controller can be placed in a manual control, which means that the output can be adjusted independently of the input using the output adjust knob. In automatic mode, this knob has no effect on the output. Connections to the controller are made through electrical or pneumatic cables connected to the rare of the unit. 2.3.3 Side panel On this side of the controller, when partially pulled out, knobs are available to adjust operation of the controller modes. On this panel as shown in the Fig 2.1, the proportional, integral and derivative gains can be adjusted. In addition, filtering action and reverse/ direct operation can often be selected.

2.4 Electronic controllers Electronic method of realizing controller modes necessitates the use of OPAMPs as the  primary circuit elements. 2.4.1 Error detector The detection of an error signal is done in electronic controllers by taking the difference  between voltages. One voltage is generated by the process signal current passed through a resistor. The second voltage represents the set point, which is usually generated by a voltage divider using a constant voltage as a source.

e.g1: Fig 2.2 shows error detection for systems using a floating power supply. Assume a two wire system is in use so that the current drawn from the floating power supply is the 4 to 20 mA signal current. The signal current is used to produce a voltage voltage IR across the resistor R. This is placed in series opposition to a voltage, Vsp tapped from a variable resistor, R sp sp connected to a constant positive source, V0. The result is an error voltage VE, where, VE = Vsp – IR. (2.1) VE is then used in the process controller to calculate controller output.

Fig. 2.2 Error detection for systems using a floating power supply e.g. 2: An error detector also can be made from a differential amplifier as shown in Fig. 2.3. Such a system can be used only if the current from the transducer is referenced to ground. The sensor signal signal current passes to ground through R L providing a signal voltage Vm = IR L. L. The differential amplifier then subtracts this from the setpoint voltage.

Fig. 2.3 Error detector using a ground based current and a differential amplifier.

Example2.1: A sensor converts from 0 to 2.0 m in to a 4 to 20 mA current. An error detector such as shown in Fig. 2.2 is used with R = 100 Ω, V0 =5.0 V and R sp = 1K Ω pot. a). If the setpoint is 0.85 m what is V sp?  b). If Vsp  = 1.5 V, what is the range of error voltage as position varies from 0 to 2.0m? Solution: Given: Sensor converts 0 to 2.0 m in to 4 to 20 mA, R = 100 Ω, V0 =5.0 V and R sp = 1K Ω pot  Now find the slope K and intersect I0 of the linear relation between the Position H and the current I, where, I = KH+ I0 Knowing H and I 4 mA = 0 K + I0 20 mA = 2.0K+ I0 Solving these simultaneous equations, I0 = 4 mA K = 8 mA/m So the relation between position and current becomes I = (8mA/m)H+4mA a). Given setpoint = 0.85 m, Vsp = ? Calculating corresponding current signal, gives: I = (8mA/m)0.85 m+4mA = 10.8 mA Since R = 100 Ω is used to convert the measurement signal current in to voltage signal, use same R sp to convert setpoint current signal in to equivalent voltage signal Vsp Thus Vsp = 10.8 mA× 100 Ω = 1.08 V  b). Given Vsp = 1.5V  Now to calculate the error if the position varies between 0 to 2.0m, first find the corresponding current signal and then its voltage equivalent. As position varies between 0 to 2.0 m current varies between 4 to 20 mA. Corresponding voltage equivalent is : 4 mA × 100 Ω to 20 mA × 100 Ω i.e, 0.4V to 2V. So the error signal is: Ve = 1.5 – 0.4 = 1.1 V and Ve = 1.5 – 2 = - 0.5 V 2.4.2. Single mode This section illustrates the implementation of pure modes of controller action. 2.4.2.1 Two Position A two position controller can be implemented by electronic and electromechanical designs. Many house hold air-conditioning and heating systems employ a two position controller constructed from bimetal strip and mercury switch as shown in F ig. 2.4.

Fig. 2.4. A mercury switch on a bimetal strip used as a two position controller As the temperature decreases, bimetal strip bends and it reaches a point where the mercury slides down to close an electrical contact. The inertia of the mercury tends to keep system in that position until the temperature increases to a value above the set point temperature. This provides the required neutral zone to prevent excessive cycling of the system. 2.4.2.2 Two Position controller implementation using OPAMP Fig. 2.5 represents the OPAMP implementation of ON/OFF controller with adjustable neutral zone.

Fig. 2.5 A two position controller with neutral zone made from op amps and a comparator. Assume that, if the controller input voltage, Vin reaches a value VH then the comparator output should go to the ON state, which is defined as some voltage V0. When the input voltage falls bellow a value VL the comparator output should switch to the OFF state, which is defined as 0 V. This defines a two position controller with a neutral zone of:  NZ = VH - VL as shown in the Fig. 2.6

Fig. 2.6. Two position controller response in terms of voltages Assume that, in the beginning, the comparator is in the OFF state. i.e. the voltage, V1 at the input of the comparator is less than the setpoint voltage, Vsp. Hence, Vout = 0 (2.2) The comparator output switches states when the voltage on its input, V1 is equal to the set  point value Vsp Analyzing this circuit, R  V1 = Vin + 1 Vout   (2.3) R 2 Substituting Eq. 2.2, in Eq. 2.3, yields V1 = Vin The comparator changes to ON state when V1 = Vin = VH. Thus, the high (ON) switch voltage is VH = Vsp (2.4) and the corresponding output voltage Vout is Vout = V0 (2.5) With this V1 changes to R  V1 = Vin + 1 V0   (2.6) R 2 If Vin = VL the comparator changes to OFF state, giving the relation, R  V1 = V sp = VL + 1 V0   R 2 This gives the low (OFF) switching voltage of R  VL = VSP − 1 Vo   R 2

(2.7)

(2.8)

As mentioned before, Fig 2.6 shows typical two position relationship between input and output voltage for the circuit. The width of the neutral zone between VL and VH can be adjusted by variation of R 2. The relative location of the neutral zone is calculated from the difference between the equations (2.4) and (2.8). The inverter resistance value in Fig. 2.5 can be chosen as any convenient value. Typically it is in the 1 to 100 K Ω range. Example 2.2 Level measurement in a sump tank is provided by a transducer scaled as 0.2 V/m. A  pump is to be turned on by application of + 5 V when the sump level exceeds 2.0 m. The  pump is to be turned back off when the sump level drops to 1.5 m. Develop a two  position controller.

Solution: Use the circuit shown in Fig. 2.5 Given: Transducer specification: 0.2 V/m Pump Specification: Pump turns on with + 5 V input Pump should be turned on when the level exceeds 2.0 m Pump should be turned off when the level falls below 1.5 m. Calculations: VH = (0.2 V/m) × (2.0 m) = 0.4 V VL = (0.2 V/m) × (1.5 m) = 0.3 V So, using the relations given in Eqs. 2.4 and 2.8, 0.4 V = VSP R  0.3V = VSP − 1 V0 R 2

Therefore, VSP = 0.4 V R  0.3V = 0.4 − 1 5 R 2 Or,

R 1 R 2

 = 0.02

There are two unknowns and only condition to be satisfied. So, one unknown can be selected. Let R 1 = 5 K Ω then, R 2 = 250 K Ω Example 2.3 Using a system of Fig. 2.5 design a two position controller with a 0 to 10 V input and a 0 or 10 V output. The setpoint is 4.3 V and the neutral zone is to be + 1.1 V about this setpoint. Solution: Given Setpoint = 4.3 V and NZ = + 1.1 V about this setpoint. V0 = 10 V So, VH = 4.3 + 1.1 V = 5.4 V VL = 4.3 -1.1V = 3.2 V Using Eq. 2.4 and 2.8, Set Vsp = VH = 5.4 V ( Can be adjusted with a voltage divider circuit with 15 V Supply) R  VL = 3.2V = VSP − 1 V0 R 2

So, 3.2V = 5.4 −

R 1 R 2

10   i.e

Selecting R 2 = 10 K Ω

R 1 R 2

 = 0.22

gives R 1 = 2.2 K Ω

2.4.2.2.1 Reverse action The two position controller in Fig. 2.5 can be made reverse acting by placing an inverter in the feedback and reversing the comparator. It is possible to devise other circuits that automatically incorporate the reverse action.

2.4.2.3 Three position Fig. 2.7 shows how a simple three position controller can be realized with op amps and comparators.

Fig. 2.7 A three position controller using two comparators and op amps Assume that, the output of the comparators is 0 V for the OFF state and V0 volts for the ON state. The summing amplifier also includes a bias voltage input, VB which allows the three position mode response to be biased up or down in voltage to suit particular needs. The inverter is needed to convert the sign of the inverting action of the summing amplifier. When Vin < VSP1 , Comparator C1 is OFF, C2 is OFF (Because VSP1< VSP2) Outputs of both comparators are 0 V. Thus, Vout = VB When VSP1 < Vin< VSP2 , Comparator C1 is OFF,

C2 is ON

Outputs of comparator C1 = 0 and Output of Comparator C2 = V out  = V  B +

 R3  R1

 R3  R1

V 0  Volts. Thus,

V 0

When Vin > VSP2, Comparator C1 is ON, C2 is ON  R  R Outputs of comparator C1 = 3 V 0  and Output of Comparator C2 = 3 V 0  Volts. Thus,  R2  R1 V out  = V  B + Thus, When Vin < VSP1 , VSP1 < Vin< VSP2 ,

 R3  R1

V 0 +

 R3  R2

V 0

Vout = VB V out  = V  B +

(2.9)  R3  R1

V 0

(2.10)

Vin > VSP2,

V out  = V  B +

 R3

V 0 +

 R3

V 0

(2.11)

 R1  R2 Here, the output need not be symmetric. (e.g. 0%, 50% and 100%). Fig. 2.8 shows the response of this circuit for a particular case VB = 0

Fig. 2.7. Response of the three position controller with VB = 0 Example 2.4 The level in the sump tank shown in Fig2.8 is to be controlled with a three position controller. The following specifications are given: 1. The level measurement sensor has a transfer function of 0.45 V/m 2. The valve is linearly driven from fully closed to fully opened by a voltage ranging from 0 to 8 V. 3. The following specifications are given: Level < 2 m Valve 20 % open Level > 2 m but < 3 m Valve 65 % open Level > 3 m Valve 100 % open Comparators are available that have outputs 0 V for OFF and 5 V for ON. Design a three  position OPAMP controller.

Fig.2.8 System for example 2.2 Solution: VSP1 = (2.m) (0.45 V/m) = 0.9 V VSP2 = (3.m )(0.45 V/m) = 1.35 V 20 % open is Vout = (0.2)(8V) = 1.6 V 65 % open is Vout = (0.65)(8V) = 5.2 V 100 % open is Vout = 8 V  Now, Vin < 0.9 Vout = VB = 1.6 V

Vout = VB +

0.9 < Vin < 1.35

R 3 R 1

Vout = VB +

Vin > 1.35

(5V ) = 5.2V

R 3 R 1

(5V ) +

R 3 R 2

(5V ) = 8V

The first equation gives VB = 1.6 V Second and third equation gives, R 3 5.2 − 1.6 R 3 8 − 5.2 = = 0.72  , = = 0.56 R 1 5 R 2 5 If R 3 = 10 K Ω, then R 1 = 10/0.72 = 13.89 K Ω, and R 2 = 10/0.56 = 17.86 K Ω Dividers would be used to obtain VB and the two comparator reference voltages of 0.9 V and 1.35 V. 2.4.2.4 Floating Controller The floating controller can be generated by connecting the output of a three position controller into an integrator as shown in Fig. 2.10

Fig. 2.10 A floating controller made with three position controller and an integrator In Fig. 2.10, Vs1 and Vs2 are the lower and upper set point voltages. Also it is assumed that three position controller is designed to provide outputs V1, Zero, and –V1 depending on input. With this input, the output of the integrator produces the outputs of: When Vin < Vs1 V out  =

t 

1

∫ (− V  )dt  + V  1

 RC  t a

= −

a

V1

(t − t a ) + Va RC When Vs1 Vs2

V out  =

1

t 

∫ (V  )dt  + V 

 RC  t c V1

1

c

(t − t c ) + Vc RC Note: Vout is the output that can be obtained by adding an inverter circuit to the output terminal of integrator. =

Thus,

Vout

⎧ V1 ⎪− RC (t − t a ) + Va ⎪ = ⎨V b ⎪V ⎪ 1 (t − t c ) + Vc ⎩ RC

Vin < Vs1 Vs1 < Vin < Vs2

(2.12)

Vin > Vs2

where, Vs1 = lower set point voltage Vs2 = upper set point voltage Va, V b, Vc = Values of output when the input condition occurs ta, t b, tc = times at which input reaches the set points The trip voltages can be set to provide the desired band of inputs producing no output, a  positive rate, or a negative rate. The actual rate of output change depends on the values of the resistor and capacitor in the integrator and the output level of the three position circuit  preceding the integrator. Note that, the output floats at whatever the latest value of output is when the input falls within the neutral zone. Example 2.5: A control signal varies from 0 to 5V. A floating controller, such as that shown in Fig.2.10 has trip voltages of 2 V and 4 V, and a three position controller has outputs of 0 and +2V. The integrator consists of a 1 MΩ resistor and a 1 µF capacitor. Plot the controller output in response to the input of Fig. 2.11a. Solution: 1. From 0 to 1 s, the output is zero (an assumed starting point) and remains so  because the input is within the neutral zone. 2. From 1 to 3 s, the lower setpoint has been reached, and the output is given by V Vout = − 1 (t − t a ) + Va = -2( t-1) volts RC At t = 3 s, the output is -4 V 3. From 3 to 4 s, the output remains at - 4 because the input is in the neutral zone 4. From 4 to 7 s, the input reaches the upper setpoint, and the output becomes V Vout = 1 (t − 4 ) − 4 V RC Vout = 2 ( t - 4) – 4V At t = 7 , the input again falls within the neutral zone, and the output becomes Vout = 2 V 5. The output will remain at + 2 until the input again hits a setpoint value.

Fig. 2.11 Input and output voltages for the floating controller 2.4.2.5 Proportional Mode: Implementation of this mode requires a circuit that has the response given by: P = K  pe p + P0  Where P = controller output 0 – 100 % K  p = Proportional gain e p = error in percent of variable range P0 = Controller output with no error

(2.13)

2.4.2.5.1 Implementation of P – Mode controller using OPAMP If both the controller output and error expressed in terms of voltage, then the above Eq. 2.13 is a summing amplifier. Fig.2.12 shows such an electronic proportional controller.

Fig.2.12. An op amp proportional mode controller

 Now, the analog electronic equation for the output voltage is: Vout = G pVe + V0  (2.14) Where, Vout = output voltage G p = R 2/R 1 = gain Ve = Error voltage V0 = output with zero error To use the circuit of Fig.2.12 for proportional mode, a relationship must be established with the characteristics of the mode, defined already, in chapter 1. In Eq. 2.13, the error is expressed as the percent of measurement range, and the output is simply 0 % to 100%. Yet Fig. 2.12 deals with voltage on both the input and output. Thus, first identify that the output voltage range of the circuit, whatever it is, represents a swing of 0% to 100%. Thus, if a final control element needs 0 to 5 V, then a Zener is added as shown in the Fig.2.13 so that the op amp output can swing only between 0 and 5V.

Fig. 2.13 A zener diode used to clamp the output swing of an op amp controller Design Procedure: First, for the input, determine the range of measurement from which the error voltage was  produced. e.g. Consider that, the temperature is to be measured and controlled from 100°C to 200°C and this is converted to 2.0 to 8.0 V. Hence, the measurement range is (8.0 – 2.0) = 6.0V. Then error voltage can be expressed as a percent of this range. This will be correlated with the error in equation 2.15. i.e, r  − b × 100%   e p = (2.15) bmax − bmin

Where, e p = error expressed as percentage of span r = setpoint value (Reference)  b = measured indication of variable  bmax = maximum of measured value  bmin = minimum of measured value Finally, calculate the gain (G p) of Fig.2.12, which is not simply K  p. The actual value of G p must be determined so that its effect in voltage is the same as that required by K  p in terms of percent. e.g. K  p = 4 %/% The output voltage range = 0 to 5 V Input voltage span = 2 to 8 V Find corresponding G p Solution: Input voltage range = (8-2) V = 6 V.

Method 1 K  p = 4 %/% means: if the error constitutes 1 % of the input range, then the output must  be changed by 4 % of its range. 1 % of the input range = (0.01) (6V) = 0.06 V 4 % of the output = ( 0.04) ( 5 V) = 0.2 V  Now, G p = (K  p % of the output)/ (1 % of the input range) G p = 0.2/0.06 = 3.33 Method2: PB = 100/K  p PB = Percent of error that will cause a 100 % change of output.  Now, with the given data, PB = 100/4 = 25 % Therefore, 25 % change of input error must produce a 100 % change of output.  Now, 25 % change of input is = (0.25) (6 V) = 1.5 V 100 % of the output = 5 V So, G p  = (5V)/(1.5 V) = 3.33. Example 2.6 A controller is shown in Fig. 2.12 with scaling so that 0 – 10 V corresponds to 0 – 100 % output. If R 2 = 10 K Ω and full scale error range is 10 V, find the value of V 0 and R 1 to support a 20 % proportional band about a 50 % zero error controller output. Solution: The value of V0 to provides the zero error controller output is calculated as, V0 = 50 % of 10 V = 5 V G p = (100 % controller output)/ (PB % change if input range) = 10V/ (20% of 10V) = 10/2 =5 The P- Controller output is given by: Vout = G pVe + V0 Where, G p = R 2/R 1 i.e., 5 = 10 K Ω / R 1 Thus,R 1= 2 K Ω Example 2.7 If the load in the above example changes such that a new controller output of 40 % is required, find the corresponding offset error. Solution Vout = G pVe + V0 40 % of the output = 40% of 10 V = 4V 4 = 5 Ve + 5 Ve = -1/ 5 = -0.2 V Because the full scale error signal is 10 V, the error expressed in percentage is: (- 0.2/ 10)(100) = - 2%

Example 2.8 A type - J thermocouple (TC) with a 0°C reference is used in a proportional mode temperature control system with a 140°C setpoint and a range of 100-180°C. The zero error output should be 45% and the PB = 35%. The output is 0- 10 V, and the full scale input range is 0 to 1V. Design a controller according to the circuit of Fig. 2.12. Solution a. In this problem we must perform the following steps. 1. Amplify the low TC voltage to a more convenient value than the TC mV output 2. Use this amplifier output as input to the proportional controller and pick a  proportional gain that gives the swings the output 0-10V as the input swings 35% of full scale.  Note that, type J TC produces a voltage of 5.27 mV at 100°C and 9.67mV at 180°C. An amplifier with a gain of 225 will convert these to 1.19V and 2.18V respectively. ~ So, the input swing is (2.18V – 1.19V) = 0.99 V 1 V At 140°C the TC produces a voltage of 7.45 mV. Amplifying this with a gain of 225 will give 1.63 V. Determination of Gp: 35 % of the input swing (i.e. 0.35× 1V = 0.35V ) must produce full scale, i.e 10 V output So, G p = 10/0.35 = 28.57. i.e R 2/R 1 = 28.57 If R 1 = 1K Ω then R 2 = 28.57 K Ω Calculation of Vo (Zero error controller output): Vo= 45% of full scale output = 0.45(10V) = 4.5V So, Vout=28.57(Vsp – VTC) + 4.5 2.4.2.6 Integral Mode The general representation of integral controller is: t

∫

P(t ) = K I e p (t )dt + PI (0 )  

(2.16)

0

Where, P( t) = controller output in percent of full scale -1 K I = integration gain (s ) e p(t) = deviations in percent of full scale variable value PI(0) = Controller output at t = 0 2.4.2.6.1. Implementation Using OPAMP Integral controller implemented using OPAMPs is shown in Fig. 2.1 4 Analysis of the circuit gives, t

∫

Vout = G I Ve dt + Vout (0 )   0

Where, Vout = Output voltage GI = 1/RC = Integration gain Ve = error voltage Vout(0) = Initial output voltage.

(2.17)

Fig. 2.14. An op amp integral mode controller The values of R and C can be adjusted to obtain the desired integration time. The initial controller output is the integrator output at t = 0. If K I  is made too large, the output rises so fast that overshoots of the optimum setting occur and cycling is produced. Determination of GI The actual value of GI and therefore R and C, is determined from K I  and the input and output voltage ranges. Integral gain says that, an input error of 1 % must produce an output that changes as K I %  per second. Or if an error of 1 % lasts for 1 s, the output must change by K I percent. e.g. Consider an input range of 6 V Output range of 5V K I = 3.0 %/(%-min)  Note: Integral gain is often given in minutes because industrial processes are slow, compared to a time of seconds. This gain is often expressed as integration time, T I, which is just the inverse of the gain. Solution First convert the time units to seconds. Therefore, [(3 %)/(%-min)][( 1min/60s)] = 0.05%/(%-s) Error of 1 % for 1 sec = (0.01)(6V)(1s) = 0.06 V-s K I % of the output = (0.0005)( 5V) = 0.0025 V The integral gain GI = (K I % of the output ) / (Error of 1 % for 1 sec) = (0.0025V)/(0.06 V-s) -1 = 0.0417 s Values of R and C can be selected from this. Example 2.9 An integral control system will have a measurement range of 0.4 to 2.0 V and an output range of 0 to 6.8 V. Design an op amp integral controller to implement a gain of K I = 4.0 %/(%-min). Specify the values of GI, R and C Solution The input range = (2.0 - 0.4) = 1.6 V The output range = 6.8 V Converting K I to units of seconds: [4.0 %/(%-min)] [( 1min/60s)] = 0.0667 % /(%-s)

1 % of the input for 1 sec = (0.01) ( 1.6V) ( 1s) = 0.016 V-s 0.0667 % of the output = ( 0.000667) ( 6.8 V) = 0.00454 V -1 Thus the gain is, GI = (0.00454V)/(0.016 V-s) = 0.283 s GI = 1 /( RC) so RC = 3.53 s If C = 100 µF, then R 35.3 K Ω Example 2.10 An integral controller has an input range of 1 to 8 V and an output range of 0 to 12 V. If K I = 12%/(%-min), find GI, R and C. Solution -1 K I = 12%/(%/min) = K I = [12%/(%-min)] [ 1min/60s)] = 0.2 s 1 % of the input for 1 sec = ( 0.01)(8-1) V( 1s) = 0.07 V-s 0.2 % of the output = (0.002)12V = 0.024 V -1 Thus the gain is, GI = 0.024V/0.07V-s = 0.3428 s Or, RC = 2.92 s If C = 10 µF, R = 292 K Ω 2.4.2.7 Derivative mode The derivative mode is never used alone because it can not provide a controller output when the error is zero or constant. The control mode equation is given by: de p P(t ) = K D   (2.18) dt where, P = Controller output in percent of full output K D = Derivative time constant ( s) e p = error in percent of full scale range Implementation of derivative controller using OPAMP: Consider an OPAMP differentiator circuit shown in Fig. 2.15 The theoretical transfer function for this circuit will be given by: dV Vout = −RC e   (2.19) dt where, the input voltage has been set equal to the controller error voltage.

Fig 2.15 OPAMP differentiator circuit

From a practical perspective, this circuit can not be used because it tends to be unstable, that is, it may begin to exhibit spontaneous oscillations in the output voltage. The reason for this instability is the occurrence of very large gain at high frequencies where the derivative is very large. To study this effect, consider the input voltage given by a sinusoidal voltage oscillating with some frequency f. then Ve = V0 sin (2π ft ). Using ac analysis to study the transfer function of the circuit, The current at the summing point is I1 + I2 = 0 (2.20) From ac analysis, we can write these currents in terms of voltage and impedance as, Ve V + out = 0   (2.21) −  jX c R  Where, Xc  = 1/(2π f C ) Therefore, the output voltage is given by, R  Vout = Ve = − j2π  fRCVe   −  jX c The magnitude of the output is expressed as, Vout = 2π  fRC Ve  

(2.22)

(2.23)

Equation 2.23 shows that the magnitude of the output voltage increases linearly with frequency, so in principle, as the frequency goes to infinity, so does the output. Clearly this is unacceptable in control theory. A little high frequency noise will cause large excursions in output voltage. In order to make a practical circuit, a modification is provided that essentially clamps the gain above some frequency to a constant value. We make sure that the clamped frequencies are well above anything that could occur in the actual control system. This way, the circuit provides a derivative output in the frequencies of practical interest, but simply acts like a fixed gain amplifier at higher frequencies. Figure 2.16 shows that the simple modification is to place a resistor in series with the capac itor.

Fig. 2.16 A Practical derivative mode op amp controller Working principle: Assume a sinusoidal input voltage and use ac analysis. The input impedance is now given by:

X in = R 1 +

1

 

 j2π  fC  j Or, X in = R 1 −   2π  fC

(2.24) (2.25)

 Now Vout is given by

Vout = −

Vout = − V out  =

R 2 R 1 −

 j

Ve  

2π  fC 2π  fCR 2

2π  fCR 1 −  j

Ve

2π  fR2 C 

(2  fR1C ) + 1 π 

2

(2.26)

V e  

(2.27)

For frequencies for which, 2π   fR1C  << 1   (2.28) Above equation reduces to equation 2.23, and the circuit takes the derivative as required. But for frequencies for which, (2.29) 2π   fR1C  >> 1   The response reduces to  R (2.30) V out  = 2 V e    R1 showing no derivative action. Therefore when using a derivative action circuit, we must estimate the maximum physical frequency, f max  at which the system can respond. Then  pick R 1  so that for frequencies much higher than f max, the inequality of Eq. 2.29 is satisfied. Guidelines to design derivative mode controller 1. Estimate the maximum frequency at which the physical system can respond, f max. 2. Set 2π f max R 1C = 0.1 and solve for R 1( C is found from the mode derivative gain requirement) 3. Equation 2.27 shows that the following responses will result from this selection: f = 0.1 f max Vout = 0.995 (2π fR 2C)|Ve| Derivative action f = f max Vout = 0.707 (2π fR 2C)|Ve| Transition action f = 10f max Vout = 0.0995 (2π fR 2C)|Ve| No Derivative action Assuming this criterion has been met, we can ignore R 1  for the controller design and define the circuit derivative gain or derivative time in seconds as GD = R 2C. GD will be determined from the design controller derivative gain K D. Example 2.11: Derivative control action with a gain K D  =0.04%/(%/min) is needed to control flow through a pipe. The flow surges with a minimum period of 2.2 s. The input signal has a range of 0.4 to 2.0 V, and the output varies from 0.0 to 5.0 V. Develop the opamp derivative action circuit. Solution First find the circuit gain GD First convert the derivative gain K D to the units of seconds. K D = [0.04 %/( %/min)] [60 s/min] = 2.4 %/( %/s)

This result says that, for every 1 % /s rate of change of input, the output should change by 2.4 %. 1 % of the input per 1 sec= (0.01) (2.0 – 0.4) V/s = 0.016V/s 2.4 % of the output = (0.024)(5) = 0.12 V Thus, GD = (0.12 V)/ (0.016V/s) = 7.5 s So, R 2C = 7.5 s If C = 20 µ F, then R 2 = 375 K Ω To find R 1: Find the maximum frequency as: f max = 1/period = 1/2.2s = 0.45 Hz. Thus, using the relation, 2π f max R 1C = 0.1 2π f max R 1C = 2π (0.45) R 1C = 0.1 i.e R 1 = 1800 Ω Note:  A pure mode of controllers is seldom used in process control because of the advantages of composite modes in providing good control. Example 2.12: Rate (derivative) action is needed for steering a boat. The rate gain should be K D = 0.02 %/( %/min). The error voltage range is - 4.0 V to 4.0 V, and the output signal varies from 0 to 2.5 V. The fastest physical turning period is 0.4 min. find the component values of a derivative mode op amp circuit such as shown in Fig. 2.16. Solution First find the circuit gain GD For this, First convert the derivative gain K D to the units of seconds. K D = [0.02 %/( %/min)] [60 s/min] = 1.2 %/( %/s) K D says that, for every 1 % /s rate of change of input, the output should change by 1.2 %. 1 % of the input per second = (0.01)( 4.0 – (-4.0)) V/s = 0.08V/s 1.2 % of the output = (0.012)(2.5) = 0.03 V Thus, GD = (0.03 V)/ (0.08V/s) = 0.375 s So, R 2C = 0.375 s. If C = 10 µ F, then R 2 = 37.5 K Ω Find the maximum frequency as: f max = 1/period = 1/(0.4×60)s = 0.0417Hz. Thus, using the relation, 2π f max R 1C = 0.1 2π f max R 1C = 2π (0.0417) R 1C = 0.1. i.e. R 1 = 38.17K Ω 2.4.3. Composite controller modes. Composite modes combine the advantages of each mode and in some cases eliminate the disadvantages. Composite modes are implemented easily using o pamp techniques. 2.4.3.1 Proportional – Integral mode: PI controller is the combination of proportional and integral controller defined by: t 

∫

 P  =  K  p e p +  K  p K  I  e p dt  + P  I  (0 )   0

Where, P = controller output in percent of full scale e p = process error in percent of the maximum K  p = Proportional gain

(2.31)

K I= Integral gain PI(0) = initial controller integral output Implementation of PI controller using opamps Figure 2.17a shows one method of implementation of the PI controller using opa mps.

Fig. 2.17a. An op amp proportional integral (PI) mode controller To derive an expression for the output voltage of this circuit, first define nodes and currents as shown in the Fig. 2.17b.

Fig.2.17b. An op amp proportional integral (PI) mode controller  Note that, there is no current through op amp input terminals and no voltage across the input terminals. Therefore, Va = 0 and I1 + I2 = 0 (2.32) I3 – I2 = 0 (2.33) The relationship between the voltage across the capacitor and current through a capacitor is given by dV   I c = C  C    (2.34) dt  Where Vc is the voltage across the capacitor. Combining this with Ohm’s law allows the  preceding current equations (2.32 and 2.33) to be written in terms of voltage as: V e V b (2.35) + =0   R 1  R2 C 

d  dt 

[V 

out 1

−V b ] −

V b  R2

= 0 

The Eq. 2.35 can be solved for V b as:  R V b = − 2 V e    R1

(2.36)

(2.37)

Substituting this in to Eq. 2.36: dV out 1  ⎞ 1 ⎛   R  ⎞ d  ⎛   R − C  ⎜⎜ − 2 V e ⎟⎟ − ⎜⎜ − 2 V e ⎟⎟ = 0   C  dt  dt  ⎝   R1  ⎠  R2 ⎝   R1  ⎠

C 

dV out 1

Or,

dt 

(2.38)

 R2 d 

1 V e + V e = 0  R1 dt   R1

+ C 

dV out 1 dt 

+

 R2 d 

1 V e + V e = 0  R1 dt   R1C 

In order to solve for Vout, integrate this equation to eliminate the derivative on Vout. i.e.: V out 1 = −

 R2

V out 1 = −

 R2

 R1  R1

V e − V e −

t 

1

∫ V  dt  + V (0) e

 R1C  0  R2

t 

1

∫ V  dt  + V (0)  

 R1  R2 C  0

e

(2.39)

After inverting, V out  =

 R2  R1

V e +

 R2

1

t 

∫ V  dt  + V (0)   e

 R1  R2 C  0

(2.40)

t 

∫

Or, V out  = G pV e + G p G I  V e dt  + V (0 )  

(2.41)

0

Where, Proportional gain, G p = R 2/R 1 and integral gain GI = 1/ (R 2C) Example 2.13: Design a proportional integral controller with a proportional band of 30 % and integration gain of 0.1 %/(%-s). The 4 to 20 mA input converts to a 0.4 to 2.0 V signal, and the output is to be 0 – 10 V. Calculate values of G p, GI , R 2 , R 1 and C respectively. Solution Proportional band = 30 %. 30 % of input range = 0.48 V 100 % 0f output = 10 V  R 10V  = 20.83 So G p = 2 =  R1 0.48V 

K I = 0.1 %/( %-s) indicates that a 1 % error for 1 sec, should produce an output change of 0.1 % 1 % of 1.6 V for 1 sec = 0.016 V-s 0.1 % of 10 V = 0.01 V 1 0.01V  So, G I  = = = 0.625s −1  R2 C  0.016V  −  s Or, R 2C = 1.6 s -5 If C = 10 µF then, R 2 = 1.6 s/ 10 = 160 K Ω Then from GP, R 1 = 160 K Ω/20.83 = 7.68 K Ω

2.4.3.2 Proportional Derivative Mode of controller PD controller is the combination of proportional and derivative mode of controllers. The general definition of PD controller is: de p P(t ) =  K  p e p + K  p K D (2.42) + P (0 )   dt Where, P = Controller output in percent of full output K  p = Proportional gain K D = Derivative time constant (s) e p = error in percent of full scale range P(0) = Zero error controller output Implementation of PD controller using opamps. Fig. 2.18a shows how a PD controller can be implemented using op amps. Where the quantities are defined in the figure and the output inverter has been included. This circuit includes the clamp to protect against high gain at high frequency in the derivative term. In this case the condition represented by Eq. 2.29 is modified slightly to use an effective resistance given by:  R1 R3  R =  R1 +  R3

Then the condition becomes as usual, 2π f max RC = 0.1. Assuming this criterion has been met, while deriving the equation for the PD response given below:

⎛   R2  ⎞ ⎛   R2  ⎞ dV  ⎟⎟V e + ⎜⎜ ⎟⎟ R3C  e + V 0 V out  = ⎜⎜ dt  ⎝  R1 +  R3  ⎠ ⎝  R1 +  R3  ⎠ V out 1 = G pV e + G p G D

dV e dt 

+ V 0

⎛   R2  ⎞ ⎟⎟ Where the proportional gain is G p = ⎜⎜ +  R  R 3  ⎠ ⎝  1 And the derivative gain is G D =  R3C 

Fig.2.18a An op amp Proportional Derivative (PD) mode controller Derivation of PD controller response: Analysis of PD circuit can be performed using the circuit shown in Fig. 2.18b showing currents and nodes. The voltage across the op amp input terminals, V b = 0. Also there is no current in to the op amp inputs.

Fig.2.18b An op amp Proportional Derivative (PD) mode controller Application of KCL, to the two active nodes provides the equations: I1 + I2 -I3 = 0 (2.43) I4+ I3 = 0 (2.44) Ohm’s law and the differential relation between current and voltage for a capacitor can be used to express these equations in terms of voltage. V e − V a V  d  (2.45) + C  [V e−V a ] − a = 0    R 3 dt   R1 V out  1  R2

+

V a  R1

= 0 

(2.46)

Eq. 2.46 reduces to:  R V a = − 1 V out 1    R2

(2.47)

Substituting Eq. 2.47 in Eq. 2.45 dV out 1 V e  R dV   R 1 + 1 V out 1 + C  e + 1 C  + V out 1 = 0  R 3  R2 R3 dt   R2 dt   R2 Or, V e +

 R1  R2

V out 1 +  R3 C 

dV e dt 

+

 R3 R1  R2

C 

dV out 1 dt 

+

 R3  R2

V out 1 = 0

After rearranging and some more algebra, this reduce s to: dV  ⎛   R1  ⎞ ⎛   R2  ⎞ ⎛   R2  ⎞ dV  ⎟⎟ R3 C  out 1 = −⎜⎜ ⎟⎟V e − ⎜⎜ ⎟⎟ R3 C  e   V out 1 + ⎜⎜ dt  dt  ⎝  R1 +  R3  ⎠ ⎝  R1 +  R3  ⎠ ⎝  R1 +  R3  ⎠

(2.48)

After inverting dV  ⎛   R2  ⎞ ⎛   R1  ⎞ ⎛   R2  ⎞ dV  ⎟⎟ R3 C  out  = ⎜⎜ ⎟⎟V e + ⎜⎜ ⎟⎟ R3 C  e dt  dt  ⎝  R1 +  R3  ⎠ ⎝  R1 +  R3  ⎠ ⎝  R1 +  R3  ⎠ ⎛   R2  ⎞ ⎛   R2  ⎞ dV  ⎟⎟V e + ⎜⎜ ⎟⎟ R3 C  e + V 0   Or, V out  = ⎜⎜ dt  ⎝  R1 +  R3  ⎠ ⎝  R1 +  R3  ⎠ V out  + ⎜⎜

V out  = G pV e + G p G D

1

dV e dt 

+ V 0  

⎛   R2  ⎞ ⎟⎟ Where the proportional gain is G p = ⎜⎜ +  R  R 3  ⎠ ⎝  1 And the derivative gain is G D =  R3C 

(2.49) (2.50)

PD controller still has the offset error of a proportional controller because the derivative term cannot provide reset action. Example 2.14. A PD controller has a 0.4 to 2.0V input measurement range, a 0 to 5 V output, K  p = 5 %/%, and K D = 0.08 % /(%/min). The period of the fastest expected signal change is 1.5s. Implement this controller with an OPAMP circuit. Solution To find G p Given K  p = 5 %/% means PB = 20 %. ⎛ 100%ofoutput  ⎞ 5 ⎟⎟ = So, G p = ⎜⎜ = 15.625 20 % of   1 . 6 V  0 . 32 ( ) ⎝   ⎠

To find GD Given, K D = 0.08 % /( %/min) = [0.08 % /( %/min)][60 s/min] = 4.8%/(%/s). ⎛ 4.8%ofoutput  ⎞ 4.8%(5V ) ⎟⎟ = = 15s  Now, G D = ⎜⎜ 1 % of   1 . 6 V  1 %( 1 . 6 V  /  s ) ( ) ⎝   ⎠ The period limitation is:  R1 0.1  R3C  = (1.5 s ) = 0.024s  R1 +  R3 2π    Now,

⎛   R2  ⎞ ⎟⎟ = 15.625 G p = ⎜⎜ +  R  R 3  ⎠ ⎝  1 G D =  R3C  = 15 s  R1  R1 + R 3

= 0.0016

Three relations and four unknowns Let C = 100 µF Then R 1 = 240Ω, R 2 = 2.35 M Ω, R 3 = 150K Ω 2.4.3.3. Three Mode Controller: Three mode controllers is the combination of proportional, integral and derivative mode of controllers. Characterized by: t  de p  P  =  K  p e p +  K  p K  I  e p dt  +  K  p K  D (2.51) + P  I  (0 )   dt  0

∫

Where, P = controller output in percent of full scale e p = process error in percent of the maximum K  p = Proportional gain K I= Integral gain K D= Derivative gain PI(0) = initial controller integral output

The zero error term of the proportional mode is not necessary because the integral automatically accommodates for offset and nominal setting. Implementation of three mode controller using op amps Three mode controller can be implemented by a straight application of op amps as shown in Fig. 2.19a.

Fig. 2.19a. Implementation of a three mode (PID) controller with op amps For the analysis, assume the voltages as indicated in Fig. 2.19b.

Fig. 2.19b. Implementation of a three mode (PID) controller with op amps  R V  p1 = − 2 V e   (2.52)  R1 V  p =

 R2  R1

V e  

(2.53)

V  I  = −

1  R I C  I 

V  D = − R D C  D

t 

∫ V 

 p1

dt    or, V  I  =

0

 R2

1

 R1  R I C  I 

t 

∫ V  dt   e

(2.54)

0

 R d  V  p1   or, V  D = 2  R D C  D V e   dt   R1 dt  d 

(2.55)

- Vout = V p + VI + VD 

(2.56)

⎛  R  ⎞ ⎛  R  ⎞ 1 ⎛  R2  ⎞ dV e ⎜ ⎟ V  dt   R C  −V out = ⎜⎜ 2 ⎟⎟V e + ⎜⎜ 2 ⎟⎟ + ∫ e ⎜⎝  R1  ⎠⎟  D  D dt  + V out  (0)   ⎝  R1  ⎠ ⎝  R1  ⎠  R I C  I 

(2.57)

R 3 has been chosen from 2π f max R 3CD = 0.1 for stability. Comparing equations 2.57 and 2.51, gives

⎛  R  ⎞ ⎛  1  ⎞ ⎟⎟   and G D =  R D C  D G p = ⎜⎜ 2 ⎟⎟ , G I  = ⎜⎜ ⎝  R1  ⎠ ⎝  R I C  I   ⎠ dV  −V out = G pV e + G p G I  V e dt  + G p G D e + V out  (0 )   dt  Adding an inverter at the output stage, dV e + V out  (0)   V  out = G pV e + G p G I  V e dt  + G p G D dt 

∫

∫

(2.58)

(2.59)

Example 2.15 A temperature control system inputs the controlled variable as a range from 0 to 4 V. The output is a heater requiring 0 to 8 V. A PID is to be used with K   p = 2.4 %/%, K I = 9%/(% -min), K D = 0.7 %/(%/min). The period of the fastest expected change is estimated to be 8 s. Develop PID circuit. Solution The input range = (4 -0) = 4 V The output range = 8 V For Proportional mode: 1 % error = (0.01)( 4 V) = 0.04 V 2.4 % of output = (0.024)( 8 V) = 0.192 V Thus, G p = ( 0.192V)/(0.04 V) = 4.8 For the integral term: An error of 1 % should cause the output to change by 9 %/min which is ( 9/60) = 0.15%/s -1 Thus, GI = (0.0015 s )(8V)/( 0.04 V) -1 GI = 0.3 s For the derivative term: -4 An error change of 1 % per min or ( 0.04 V/60) = 6.67 ×10 V/s should cause an output change of 0.7 % or ( 0.007 ) (8V) = 0.056 V. Thus, -4 GD = (0.056 V/ 6.67 ×10 V/s) = 84 s

⎛  R  ⎞ G p = ⎜⎜ 2 ⎟⎟ = 4.8 , ⎝  R1  ⎠ ⎛  1  ⎞ ⎟⎟ = 0.3s −1 G I  = ⎜⎜ ⎝  R I C  I   ⎠

and G D =  R D C  D = 84 s from the fastest period specification, 2π f max R 3CD = 0.1 2π R 3CD = 0.1(8s) = 0.8 Seven unknowns and four equations. Let, R 1 = 10 K Ω , CI = CD = 10µF This gives R 2 = 4.8/R 1 = 48 K Ω R I = 1/(0.3 CI) = 333 K Ω R D = 84/CD = 8.4 M Ω R 3 = 0.8/( 2π CD) = 12. 7 K Ω 8.4 M Ω is too large for practical consideration. So change CD to 100µF, Then, R D = 84/CD = 840 K Ω R 3 = 0.8/( 2π CD) = 1270 Ω which seems to be more reasonable. 2.5 Pneumatic controllers Reason for using pneumatic controllers: 1. Competitive in cost and reliability 2. Safety (Danger of explosion from electrical malfunctions exists.) 3. Final control element is often pneumatically or hydraulically operated, which suggests that an all pneumatic process control loop might be advantageous.

It appears that analog or digital electronic methods will eventually replace most  pneumatic installations. But we will still have pneumatic equipment for many years until these are depreciated in industry. 2.5.1. General features The outward appearance of a pneumatic controller is typically the same as that for the electronic controller. The same readout of set point, error and controller output appears, and adjustments of gain, rate and reset are available. The working signal is most typically 3- 15 psi standard pneumatic process control signal usually derived from a regulated air supply of 20 to 30 psi. The pneumatic controller is based on the principles of nozzle/flapper system shown in Fig. 2.20.

a). Flapper/nozzle System

 b). Signal pressure versus gap distance Fig. 2.20 Principles of Nozzle/Flapper System  Nozzle Flapper system (also known as nozzle/baffle system) provides an important signal conversion i.e from pressure to mechanical motion and vice versa. A diagram of this is shown in Fig. 2.20 a. A regulated supply of pressure, usually over 20 psig provides a source of air through the restriction. The nozzle is open at the end where the gap exists  between the nozzle and flapper, and air escapes in this region. If the flapper moves down and closes off the nozzle opening so that no air leaks, the signal pressure will rise to the supply pressure. As the flapper moves away, the pressure will stabilize at some value determined by the maximum leak through the nozzle. Figure 2.20b shows the relationship  between signal pressure and gap distance. Note the great sensitivity in the central region. A nozzle/Flapper system is designed to operate in the central region, where the slope of this line is greatest. In this region, the response will be such that a very small motion of the flapper can change the pressure by an order of magnitude. 2.5.2 Mode implementation Here, the essential features of controller mode implementation using pneumatic techniques are discussed. 2.5.2.1 Proportional Mode A proportional mode of operation can be achieved with the system shown in Fig.2.21.

Fig.2.21. The pneumatic Proportional mode controller

If the input pressure increases, then the input bellows forces the flapper to rotate to close off the nozzle. When this happens, the output pressure increases so that the feedback  bellows exerts a force to balance that of the input bellows. A balance condition then occurs when torques exerted by each about the pivot are equal, or ( P out  −  P 0 ) A2 x2 = ( P in −  P  sp ) A1 x1   (2.60) Solving for output pressure,  x  A  P out  = 1 1 ( P in −  P  sp ) +  P 0    x 2  A2

(2.61)

Where, P0 = pressure with no error Pin = Input pressure (pa) 2 A1 = input and setpoint bellows effective area (m ) x1 = lever arm of input (m) Pout = Output pressure (pa) 2 A2 = feedback bellows effective area (m ) x2 = feedback lever arm (m) Psp = Setpoint pressure. This relation is based on the notion of torque equaling force times lever arm, and the fact that a pressure in a bellows produces a force that is effectively the pressure times bellows area, much like a diaphragm. From Eq.2.61, the proportional gain is given by:  x  A  K  p = 1 1   (2.62)  x 2  A2 Because the bellows are usually of fixed geometry, the gain is varied by changing the lever arm length. In this simple representation, the gain is established by the distance  between the bellows. If this separation is changed, the forces are no longer balanced, and for the same pressure a new controller output will be formed, corresponding to the new gain. Example2.16 2 Suppose a proportional pneumatic controller has A1 = A2 = 5 cm , x 1 = 8 cm, and x2 = 5 cm. The input and output pressure ranges are 3 to 15 psi. Find the input pressures that will drive the output from 3 to 15 psi. The setpoint pressure is 8 psi and P 0 = 10 psi. Find the proportional band. Solution

5cm 2  ⎞ ⎛ 8cm ⎞⎛  ⎟ =⎜  K  p = ⎟⎜  x 2  A2 ⎝ 5cm ⎠⎜⎝ 5cm 2  ⎠⎟  x1  A1

K  p = 1.6  Now we have,  P out  =  K  p  P in −  P  sp +  P 0  P out  = 1.6(P in − 8) + 10 The low input occurs when, Pout = 3psi. So, 3 = 1.6( P L − 8) + 10

Which gives, PL = 3.625 psi The high is found from, 15 = 1.6( P H  − 8) + 10 which gives, PH = 11.125psi The proportional band (PB) is,

⎛ 11.125 − 3.625 ⎞ ⎟100 15 − 3 ⎝   ⎠

 PB = ⎜

PB = 62.5 % Or, PB = 100/K  p = 100/1.6 = 62.5% Which could be used because the input and output ranges are the same. 2.5.2.2 Proportional integral This control mode also implemented using pneumatics by the system shown in Fig.2.22.

Fig.2.22. Pneumatic Proportional Integral Controller In this case, an extra bellows with a variable restriction is added to the proportional system. Suppose the input pressure shows a sudden increase. This drives the flapper toward the nozzle, increasing output pressure until the proportional bellows balances the input as in the previous case. The integral bellows is still at the original output pressure,  because the restriction prevents pressure changes from being transmitted immediately. As the increased pressure on the output bleeds through the restriction, the integral bellows slowly moves the flapper closer to the nozzle, thereby causing a steady increase in output  pressure. The variable restriction allows for variation of the leakage, rate, and hence the integration time. 2.5.2.3 Proportional Derivative This controller action can be accomplished pneumatically by the method shown in Fig.2.23.

Fig. 2.23. Pneumatic Proportional Derivative Controller A variable restriction is placed on the line leading to the balance bellows. Thus, as the input pressure increases, the flapper is moved toward the nozzle with no impedance,  because the restrictions prevent an immediate response of the balance bellows. Thus the output pressure rises very fast and then, as the increased pressure leaks in to the balance  bellows, decreases as the balance bellows moves the flapper back away from the nozzle. Adjustment of the variable restriction allows for changing the derivative time constant. 2.5.2.4 Three mode Controller Three mode controller is the most common type, because it can be used to accomplish any of the previous modes by setting of restrictions. This mode is implemented as shown in Fig.2.24. By opening or closing restrictions, the three mode controller can be used to implement the other composite modes. Proportional gain, reset time, and rate are set by adjustment of bellows separation and restriction size.

Fig. 2.24. Pneumatic Proportional Integral Derivative (PID) Controller 2.6 Design Considerations To illustrate some of the facets involved in setting up a process control loop, it would be valuable to follow through some hypothetical examples as illustrated below: Example 2.17 Design a process control system that regulates light level outputting a 0-10 V signal to a lighting system that provides 30 -180 lux. The sensor has a transfer function of -120Ω/lux with a 10 K Ω resistance at 100 lux. The setpoint is to be 75 lux, and proportional control with a 75 % proportional band has been selected.

Solution: First establish the characteristics of each part of the system. 1. The illumination varies from 30 to 180 lux. Find the resistance changes according to: R = 10 K Ω – 0.12 K Ω ( I-100) where I is the illumination in lux. 2.  Now find the resistance at 30 lux as: R = 10 K Ω – 0.12 K Ω (30 -100) R = 18.4 K Ω and at 180 lux: R = 10 K Ω – 0.12 K Ω (180 -100) R = 0.4 K Ω. The setpoint of 75 lux has a resistance of 13 K Ω 3.  Next convert this resistance variation to voltage using the photocell in an opamp circuit. In Fig.2.25 an inverting amplifier is used with a gain of 1 at the setpoint and a constant -1 V input. The resistance to voltage conversion gives,  R  R V  = − (− 1V ) = 13 K Ω 13 K Ω Using this equation, find the output voltage at 18.4 K Ω as: 18.4 K Ω (− 1V ) = +1.42V  V  = − 13 K Ω And at 0.4 K Ω, we get 0.4 K Ω V  = − (− 1V ) = 0.031V  13 K Ω So the input voltage range is 1.42 -0.031 = 1.389 V. 4.  Now, use a summing amplifier to find the error in Fig 2.25 as:  R V e = −1 13 K Ω A 75 % proportional band controller with a 75 lux setpoint requires a zero error output of 75 − 30 V 0 = 10V  = 3V  180 − 30 5. The 75 % band means that when the illumination changes by 75 % of (180-30) = 112.5 lux, the output should swing by 10 V. Thus, in terms of resistance, this corresponds to 13.5 K Ω, and in terms of error voltage, it is 13.5 K Ω /13 K Ω. Or 1.038V. 6. Finally the gain must be, 10V  = 9.63 G p = 1.038 The overall response is, V out  = 9.63V e + 3

⎛   R  ⎞ − 1⎟ + 3 ⎝ 13 K Ω  ⎠

Or, V out  = 9.63⎜

The rest of the circuit in Fig.2.25 accomplishes this function. When Vout = 0, R = 8.9 K Ω, or 90.83 lux, and for Vout = 10 V, R = 22.4 K Ω or 203.33 lux, so that the output swings 100% as the input swings. 203.33 − 90.83 = 0.75, or 75%  as required. 180 − 30

Fig.2.25 Circuit of Example 2.17 Example 2.18 A type J thermocouple (TC) with a 0°C reference is used to control temperature between 100°C and 200°C. Design a proportional integral controller with a 40 % band and a 0.08 min reset (integral) time. The final control element requires a 0-10 V range. Solution: a. In this problem, we must perform the following steps: 1. Amplify the low TC voltage to a more convenient value than the TC mV output. 2. Use this amplifier output as input to the proportional integral controller and pick a  proportional gain that swings the output 0 – 10 V as the input swings 40 % of full scale. 3. Select values to provide 0.08 min (4.8 –s) integral time.  b. The solution is shown in Fig.2.26 1.  Note that, a type J TC produces a voltage of 5.27 mV at 100 °C and 10.78 mV at 200 °C. An amplifier with a gain of 100 will convert these to 0.527 and 1.078 V, respectively. 2.  Now, sum this output to a properly scaled setpoint voltage to get an error signal. The setpoint value is obtained from a voltage divider. To get the proper controller values, note that 40 % of the input swing is: 0.4(1.078 – 0.527) = 0.2204V. Thus, the proportional gain is, 10V  G p = = 45.37 0.2204V  So values are to be chosen to provide this gain 3. For the integral term, a 0.08 min reset means K I=2.5 %/( %-min) or (12.5/60)=0.21 %/( %-s). Thus an error of 1 % for 1 s must produce a change to output of 0.21%.

G I  =

(0.0021)(10V ) (0.01)(0.551)V  − s

= 3.81s −1

So, G p =

 R2  R1

= 45.37,

 R2 C  = 0.262 s

Let, C = 1µF, then, R 2 = 262K Ω, and R 1 = 5.77 K Ω The overall transfer function for the final circuit shown in Fig.2.26 is found to be,

∫

V out  = 45.3V e + 173.2 V e dt  where, V e = 100V TC  − V SP  The output diode and zener limit the swing from 0 to 10 V.

Fig.2.26 Circuit of Example 2.18 Example 2.19 A differential pressure gauge is used to measure flow that varies as the square root of the  pressure difference. The pressure signal is a 0-2V range of minimum to maximum flow. A square root extractor circuit is available that accepts from 0 to 10V and outputs the square root of the input. Design a proportional controller with a 15% proportional band having a 0 -10V output and a nominal (zero error) output of 5V. Solution The circuit of Fig.2.27 implements this function. Pressure signal i.e 0 – 2V is amplified to a range of 0 -10V by using non inverting amplifier with gain of 5.  Now, the controller input is a 0 – 3.162 V signal. A 15 % proportional band means that if the input changes by (0.15) (3.162 V) = 0.474 V, the output must change by 10 V. thus the gain is, 10 = 21.1 G p = 0.474 This is provided by the 1 K Ω and 21.1 K Ω resistors.

Fig.2.27 Circuit of Example 2.19

Exercise 2.20 Design a 45 % PB controller for motor speed control. The motor speed varies from 100 to 150 rpm for an input control voltage of 0 to 5 V. A speed sensor linearly changes from 2.0 to 5.0 K Ω over the speed range. A set point of 125 rpm is desired for which the motor control circuit input is 2.5 V. Suppose the set point is changed to 120 rpm with no other adjustments. What offset error will occur? Solution Given PB = 45% Speed range: 100rpm to 150 rpm Controller output voltage range: 0 to 5V Specification of speed sensor: Over the speed range, sensor linearly changes its resistance from 2.0 to 5.0 K Ω. Set point speed = 125 rpm V0 = 2.5V First find the relation between the speed and resistance as follows: R = K S+ R 0 At S = 100 rpm, 2.0K Ω = K×100rpm+ R 0 At S = 150 rpm, 5.0K Ω = K×150rpm+ R 0 Solving two equations, K = 0.06 K Ω/rpm R 0 = - 4K Ω. So, R = 0.06S-4 If S = 125 rpm (set point) R = 3.5K Ω.

 Next convert this resistance variation to voltage using the speed sensor in an opamp circuit. In Fig.2.29 an inverting amplifier is used with a gain of 1 at the setpoint and a constant -1 V input. The resistance to voltage conversion gives,  R  R V  = − (− 1V ) = 3.5 K Ω 3.5 K Ω Using this equation, find the output voltage at 2.0 K Ω and 5.0 K Ω as: 2.0 K Ω V  = − (− 1V ) = +0.57V  3.5 K Ω 5.0 K Ω V  = − (− 1V ) = 1.43V  3.5 K Ω So the input voltage range is 1.43 -0.57 = 0.86 V.  Now, use a summing amplifier to find the error in Fig 2.29 as:  R V e = −1 3.5 K Ω A 45 % proportional band controller with a 125rpm setpoint requires a zero error output of 125 − 100 V 0 = 5V  = 2.5V  150 − 100 As given in the problem statement. The 45 % band means that when the speed changes by 45 % of (150-100) = 22.5rpm, the output should swing by 5 V. Thus, in terms of resistance, this corresponds to -2.65K Ω, and in terms of error voltage, it is (-2.65 K Ω /3.5 K Ω) or -0.757V. Finally the gain must be, 5V  = −6.61 G p = − 0.757  Negative G p indicates direct action proportional controller.  R Let, G p = 2 = 6.61  R1 If R 1 = 10 K Ω then R 2 = 66.1 K Ω The overall response is, V out  = −6.61V e + 2.5

⎛   R  ⎞ − 1⎟ + 2.5 ⎝ 3.5 K Ω  ⎠

V out  = −6.61⎜

Fig. 2.29 Circuit of exercise 2.20 If the setpoint is changed to 120 rpm, this corresponds to 3.2K Ω or 0.9 V. So error resistance is: ⎛   R  ⎞ − 1⎟ + 2.5 = 4.35K Ω 0.9 = −6.61⎜ ⎝ 3.5 K Ω  ⎠ In terms of speed, S = 139.17rpm Thus offset error is: 139.17 − 120 × 100 = 38.34% 150 − 100 Exercise 2.21 Design a three position controller for the temperature of the drying oven. shown in Fig.2.30. The specifications are 1. For T < 35°C, the heat should be 100% 2. For 35°C 60°C, the heat should be 40 % The heater needs 0- 8V for 0 to 100 % heat production. The sensor is a thermistor with a 5 – mW/°C dissipation constant and resistance of 4.7 K Ω at 35°C and 1.4 K Ω at 60°C. Use comparators with a 5.0 V ON output, provide hysteresis of + 2% about the reference, and keep thermistor dissipation below 1°C.

Fig. 2.30. System for Problem 2.21

Solution Given that, the heater input needs 0- 8V for 0 to 100 % heat production. • For T < 35°C, the heat should be 100% • For 35°C 60°C, the heat should be 40 % So, For T < 35°C, the heater input = 100% = 8 V For 35°C 60°C, the heater input = 40 %of 8 V = 3.2 V Thermistor Specification: Dissipation constant = 5 – mW/°C, and keep thermistor dissipation below 1°C. Resistance of 4.7 K Ω at 35°C, and 1.4 K Ω at 60°C. Design a resistance to voltage signal converter as shown in Fig. 2.31: V  Using the relation, V  D = RTH   R1 +  RTH 

Let V = 10 V and R 1 = 10 K Ω then, At 35°C, R TH = 4.7 K Ω 10 V  D = 4.7 K Ω = 3.19V  10 K Ω + 4.7 K Ω At 60°C, R TH = 1.4 K Ω 10 V  D = 1.4 K Ω = 1.23V  10 K Ω + 1.4 K Ω

Fig.2.31 Divider Circuit  Now consider the effect of self heating: Given, Dissipation constant PD = 5 mW/°C The power dissipation in the thermistor will be given by:

 P  =

V  2  RTH 

So at 35°C V  2 3.19 2  P  = = = 0.0022W   RTH  4.7 K Ω At 60°C V  2 1.23 2  P  = = = 0.0011W   RTH  1.4 K Ω The temperature rise of the thermistor can be obtained as:  P  ∆T  =  P  D At 35°C  P  0.0022W  ∆T  = = = 0.44 C   P  D 5mW  o

At 60°C  P  0.0011W  ∆T  = = = 0.22 C   P  D 5mW  o

As specified, the thermistor dissipation lies below 1°C Design of three position controller: Comparator specification: Comparators ON voltage= 5.0 V Provide hysteresis of + 2% about the reference. Let VSP1 = 1.23 V VSP2 = 3.19 V When Vin < VSP1 , Vout = VB = 3.2V When VSP1 < Vin< VSP2 ,  R V out  = V  B + 3 V 0  R1 5.6 = 3.2 +  R3  R1

 R3  R1

5

= 0.48

When Vin > VSP2, V out  = V  B + 8 = 3.2 +

 R3  R1

 R3  R1

V 0 +

5+

 R3  R2

 R3  R2

8 = 3.2 + 0.48 × 5 +

V 0

5  R3  R2

5

 R3

= 0.48  R2 So, If R 3 = 10 K Ω then R 2 = 20.83 K Ω., and R 1 = 20.83 K Ω  Now to incorporate hysterisis effect of + 2%, consider hystersis comparator as shown in the Fig. 2.32.

Fig. 2.32 Hystersis comparator V  L = V ref   −

 R  R f  

V 0   and VH = Vref

 Now for the comparator with setpoint of VSP1 = 1.23V VH = 1.23+ 2% of 1.23V = 1.2546V VL = 1.23- 2% of 1.23V = 1.2054V So hystersis or deadband is 1.2546V-1.2054V = 0.0492V = i.e

 R  R f  

5 = 0.0492 .

If R f  = 120K Ω then R = 1.18K Ω Similarly for the comparator with setpoint VSP2 = 3.19V VH = 3.19+ 2% of 3.19V = 3.2538V VL = 3.19- 2% of 3.19V = 3.1262V

 R  R f  

V 0