Controller principle have 4 type controller :
- On Off Controller ( Two position / Multiposition Mode )
- Analog Controller ( P, I, D, PI, PD, PID )
- Digital Controller
- Fuzzy Controller
Controller modes refer to the methods to generate different types of control signals to final control element to control the process variable. Broad classifications of different controller modes used in process control are as follows:
(1) Discontinuous Controller Modes
- Two-position (ON/OFF) Mode
- Multiposition Mode
- Floating Control Mode: Single Speed and Multiple Speed
(2) Continuous Controller Modes
- Proportional Control Mode
- Integral Control Mode
- Derivative Control Mode
(3) Composite Controller Modes
- Proportional-Integral Control (PI Mode)
- Proportional-Derivative Control (PD Mode)
- Proportional-Integral-Derivative Control (PID or Three Mode Control)
2.2 Controller Modes
2.2.1 DISCONTINUOS CONTROL MODES
Two position (ON/OFF) Mode
The most elementary controller mode is the two-position or ON/OFF controller mode. It is the
simplest, cheapest, and suffices when its disadvantages are tolerable. The most general form can be
given by:
P = 0 % ep < 0
100 % ep > 0
The relation shows that when the measured value is less than the setpoint (i.e. ep > 0), the
controller output will be full (i.e. 100%), and when the measured value is more than the
setpoint (i.e. ep < 0), the controller output will be zero (i.e. 0%).
Neutral Zone : In practical implementation of the two-position controller, there is an
overlap as ep increases through zero or decreases through zero. In this span, no change in
the controller output occurs which is illustrated in Fig. 2.2.1.1
Fig. 2.2.1.1 Two-position controller action with neutral zone.
It can be observed that, until an increasing error changes by Δep above zero, the controller output
will not change state. In decreasing, it must fall Δep below zero before the controller changes to 0%. The range 2Δep is referred to as neutral zone or differential gap. Two-position controllers are purposely designed with neutral zone to prevent excessive cycling. The existence of such a neutral zone is an example of desirable hysteresis in a system.
(Example 1)
A liquid-level control system linearly converts a displacement of 2 to 3 m into a 4 to 20 mA control signal. A relay serves as the two-position controller to open and close the inlet valve. The relay closes at 12 mA and opens at 10 mA. Find (a) the relation between displacement level and current, and (b) the neutral zone or displacement gap in meters.
Solution
Given data : Liquid-level range = 2 to 3 m i.e. Hmin = 2m & Hmax = 3m
Control signal range = 4 to 20 mA i.e. Imin = 4mA & Imax = 20mA
(a) Relation between displacement level (H) and current (I)
(b) Neutral zone (NZ) in meters.
(a) The linear relationship between level and current is given by :
H = K I + Ho
The simultaneous equations for the above range are:
For low range signal 2 = K x 4 + Ho
For higher range signal 3 = K x 20 + Ho
Solving the above simultaneous equations we get:
K = 0.0625 m/mA, & Ho = 1.75 m
Therefore, the relation between displacement level (H) and current (I) is given by :
H = 0.0625 I + 1.75
(b) The relay closes at 12 mA, which is high level, HH
HH = 0.0625 x 12 + 1.75 = 2.5 m
The relay opens at 10 mA, which is low level, HL
HL = 0.0625 x 12 + 1.75 = 2.375 m
Therefore, the neutral zone, NZ = (HH - HL) = (2.5 - 2.375) = 0.125 m
(Example 2)
As a water tank loses heat, the temperature drops by 2 K/min when a heater is on, the system gains temperature at 4 K/min. A two-position controller has a 0.5 min control lag and a neutral zone of ± 4% of the setpoint about a setpoint of 323 K. Plot the heater temperature versus time. Find the oscillation period.
Solution
Given data : Temperature drops = 2 K/min
Temperature rises = 4 K/min
Control Lag = 0.5 min
Neutral zone = ± 4%
Setpoint = 323 K
± 4% of 323 = 13 K. Therefore, the temperature will vary from 310 to 336 K (without considering the lag)
Initially we start at setpoint value. The temperature will drop linearly, which can be expressed by
T1(t) = T(ts) – 2 (t – ts)
where ts = time at which we start the observation
T(ts) = temperature when we start observation i.e. 323.
The temperature will drop till - 4% of setpoint (323K), which is 310 K.
Time taken by the system to drop temperature value 310 K is 310 = 323 – 2 (t – 0) t = 6.5 min
Undershoot due to control lag = (control lag) x (drop rate) = 0.5 min x 2 K/min = 1 K
Due control lag temperature will reach 309 instead of 310 K. From this point the temperature will rise at 4 K/min linearly till +4% of setpoint i.e. 336
K, which can be expressed by :
T2(t) = T(th) + 2 (t – th)
where th = time at which heater goes on
T(th) = temperature at which heater goes on
336 = (310-1) + 4 [t – (6.5 +0.5)]
t = 13.75 min
Overshoot due to control lag = (control lag) x (rise rate) = 0.5 min x 4 K/min = 2 K
Due control lag temperature will reach 338 instead of 336 K.
The oscillation period is = 13.75 + 0.5 +0.5 + 6.5 = 21.25 ≈ 21.5 min
The system response is plotted as shown in Fig. 2.2.1.2 with undershoot and overshoot values.
p = pi e p > ei i =1, 2,...., n
As the error exceeds certain set limits ± ei, the controller output is adjusted to present values pi.
Three-position Control Mode : One of the best example for multiposition control mode is three position control mode, which can be expressed in the following analytical form:
P = 100% ep > +e1
= 50% -e1 < ep < +e1
= 0% ep < - e1
As long as the error is between +e1 and -e1 of the setpoint, the controller stays at some nominal setting indicated by a controller output of 50%. If the error exceeds the set point by +e1 or more, then the output is increased to 100%. If it is less than the setpoint by -e1 or more, the controller output is reduced to zero. Figure 2.2.2.1 illustrates three-position mode graphically.
The Control Units are in general build on the control principles
A liquid-level control system linearly converts a displacement of 2 to 3 m into a 4 to 20 mA control signal. A relay serves as the two-position controller to open and close the inlet valve. The relay closes at 12 mA and opens at 10 mA. Find (a) the relation between displacement level and current, and (b) the neutral zone or displacement gap in meters.
Solution
Given data : Liquid-level range = 2 to 3 m i.e. Hmin = 2m & Hmax = 3m
Control signal range = 4 to 20 mA i.e. Imin = 4mA & Imax = 20mA
(a) Relation between displacement level (H) and current (I)
(b) Neutral zone (NZ) in meters.
(a) The linear relationship between level and current is given by :
H = K I + Ho
The simultaneous equations for the above range are:
For low range signal 2 = K x 4 + Ho
For higher range signal 3 = K x 20 + Ho
Solving the above simultaneous equations we get:
K = 0.0625 m/mA, & Ho = 1.75 m
Therefore, the relation between displacement level (H) and current (I) is given by :
H = 0.0625 I + 1.75
(b) The relay closes at 12 mA, which is high level, HH
HH = 0.0625 x 12 + 1.75 = 2.5 m
The relay opens at 10 mA, which is low level, HL
HL = 0.0625 x 12 + 1.75 = 2.375 m
Therefore, the neutral zone, NZ = (HH - HL) = (2.5 - 2.375) = 0.125 m
(Example 2)
As a water tank loses heat, the temperature drops by 2 K/min when a heater is on, the system gains temperature at 4 K/min. A two-position controller has a 0.5 min control lag and a neutral zone of ± 4% of the setpoint about a setpoint of 323 K. Plot the heater temperature versus time. Find the oscillation period.
Solution
Given data : Temperature drops = 2 K/min
Temperature rises = 4 K/min
Control Lag = 0.5 min
Neutral zone = ± 4%
Setpoint = 323 K
± 4% of 323 = 13 K. Therefore, the temperature will vary from 310 to 336 K (without considering the lag)
Initially we start at setpoint value. The temperature will drop linearly, which can be expressed by
T1(t) = T(ts) – 2 (t – ts)
where ts = time at which we start the observation
T(ts) = temperature when we start observation i.e. 323.
The temperature will drop till - 4% of setpoint (323K), which is 310 K.
Time taken by the system to drop temperature value 310 K is 310 = 323 – 2 (t – 0) t = 6.5 min
Undershoot due to control lag = (control lag) x (drop rate) = 0.5 min x 2 K/min = 1 K
Due control lag temperature will reach 309 instead of 310 K. From this point the temperature will rise at 4 K/min linearly till +4% of setpoint i.e. 336
K, which can be expressed by :
T2(t) = T(th) + 2 (t – th)
where th = time at which heater goes on
T(th) = temperature at which heater goes on
336 = (310-1) + 4 [t – (6.5 +0.5)]
t = 13.75 min
Overshoot due to control lag = (control lag) x (rise rate) = 0.5 min x 4 K/min = 2 K
Due control lag temperature will reach 338 instead of 336 K.
The oscillation period is = 13.75 + 0.5 +0.5 + 6.5 = 21.25 ≈ 21.5 min
The system response is plotted as shown in Fig. 2.2.1.2 with undershoot and overshoot values.
Fig. 2.2.1.2 Plot of heater temperature versus time for Example 2
2.2.2 Multiposition Mode
It is the logical extension of two-position control mode to provide several intermediate settings of the controller output. This discontinuous control mode is used in an attempt to reduce the cycling behaviour and overshoot and undershoot inherent in the two-position mode. This control mode can be preferred whenever the performance of two-position control mode is not satisfactory. The general form of multiposition mode is represented byp = pi e p > ei i =1, 2,...., n
As the error exceeds certain set limits ± ei, the controller output is adjusted to present values pi.
Three-position Control Mode : One of the best example for multiposition control mode is three position control mode, which can be expressed in the following analytical form:
P = 100% ep > +e1
= 50% -e1 < ep < +e1
= 0% ep < - e1
As long as the error is between +e1 and -e1 of the setpoint, the controller stays at some nominal setting indicated by a controller output of 50%. If the error exceeds the set point by +e1 or more, then the output is increased to 100%. If it is less than the setpoint by -e1 or more, the controller output is reduced to zero. Figure 2.2.2.1 illustrates three-position mode graphically.
Fig. 2.2.2.1 Three-position controller action
The three-position control mode usually requires a more complicated final control element, because it must have more than two settings. Fig. 2.2.2.2 shows the relationship between the error and controller output for a three-position control. The finite time required for final control element to change from one position to another is also shown. The graph shows the overshoot and undershoots of error around the upper and lower setpoints. This is due to both the process lag time and controller lag time, indicated by the finite time required for control element to reach new setting.
Fig.
Fig. 2.2.2.2 Relationship between error and three-position controller action, including the
effects of lag.
2.2.3 CONTINUOS CONTROL MODES
In continuous controller modes the controller output changes smoothly in response to the
error or rate of change of error. These modes are an extension of discontinuous controller
modes. In most of the industrial processes one or combination of continuous controllers
are preferred.
Analog Controller ( P, I, D, PI, PD, PID )
- proportional controller
- integral controller
- derivative controller
- a linear relationship exits between the controller output and the error
- range of error to cover the 0% to 100% controller output is called proportional band
- p=Kp Ep + Po (2.0)
where Kp = proportional gain (% per %)
p0 = controller output with no error or zero error (%)
The equation (2.0) represents reverse action, because the term KpEp will be subtracted
from Po whenever the measured value increases the above setpoint which leads negative
error. The equation for the direct action can be given by putting the negative sign in front
of correction term i.e. - KpEp. A plot of the proportional mode output vs. error for
equation (2.0) is shown in Fig.2.0.
In Fig.2.0, Po has been set to 50% and two different gains have been used. It can be
observed that proportional band is dependent on the gain. A high gain (G1) leads to large
or fast response, but narrow band of errors within which output is not saturated. On the
other side a low gain (G2) leads to small or slow response, but wide band of errors within
which output is not saturated. In general, the proportional band is defined by the
equation:
PB = 100/Kp
The summary of characteristics of proportional control mode are as follows:
1. If error is zero, output is constant and equal to p0.
2. If there is error, for every 1% error, a correction of Kp percent is added or
subtracted from p0, depending on sign of error.
3. There is a band of errors about zero magnitude PB within which the output is not
saturated at 0% or 100%.
Offset: An important characteristic of the proportional control mode is that it produces a
permanent residual error in the operating point of the controlled variable when a load
change occurs and is referred to as offset. It can be minimized by larger constant Kp
which also reduces the proportional band. Figure 2.1 shows the occurrence of offset in
proportional control mode.
- image
Consider a system under nominal load with the controller output at 50% and error zero as
shown in Fig.2.1. If a transient error occurs, the system responds by changing controller
output in correspondence with the transient to effect a return-to-zero error.
Problem 6
For a proportional controller, the controlled variable is a process temperature with a range
of 50 to 130 o
C and a setpoint of 73.5 o
C. Under nominal conditions, the setpoint is
maintained with an output of 50%. Find the proportional offset resulting from a load
change that requires a 55% output if the proportional gain is (a) 0.1 (b) 0.7 (c) 2.0 and (d)
5.0.
Solution:
Given data: Temperature Range = 50 to 130 oC
Setpoint (Sp) = 73.5 oC
Po = 50%
P = 55%
ep = ?
Offset error = ? for Kp=0.1, 0.7, 2.0 & 5.0
For proportional controller: P = Kp ep + Po
ep = [p-Po] / Kp = [55 – 50] / Kp = 5 / Kp %
(a) when Kp = 0.1 Offset error, ep = 5/0.1 = 50%
(b) when Kp = 0.7 Offset error, ep = 5/0.7 = 7.1%
(c) when Kp = 2.0 Offset error, ep = 5/2.0 = 2.5%
(d) when Kp = 5.0 Offset error, ep = 5/5.0 = 1%
[It can be observed from the results that as proportional gain Kp increases the offset error
decreases.]
Integral Control Mode
- eliminates the offset error problem by allowing the controller to
- adapt to changing external conditions by changing the zero-error output.
- Integral action is provided by summing the error over time, multiplying that sum by a gain, and adding the result to the present controller output
- error becomes positive or negative for an extended period of time, the integral action will begin to accumulate and make changes to the controller output
