The document discusses integral and derivative control modes. It explains that integral control eliminates offset errors by allowing the controller to adapt to changing conditions over time. The integral term accumulates the error over time by summing the error and multiplying by a gain. Derivative control responds to the rate of change of error and is useful for anticipating changes, but can cause instability if not carefully tuned.

1. ICE401: PROCESS INSTRUMENTATION
AND CONTROL
Class 17
Integral and Derivative Control Mode
Dr. S. Meenatchisundaram
Email: meenasundar@gmail.com
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015

2. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• The offset error of the proportional mode occurs because
the controller cannot adapt to changing external
conditions—that is, changing loads. In other words, the
zero-error output is a fixed value.
• The integral mode eliminates this problem by allowing the
controller to adapt to changing external conditions by
changing the zero-error output.
• The need for integral action shows up when it is noted that
even with proportional action correction, the error does
not go to zero in time.

3. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• 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.
• You can see that if the error makes random excursions
above and below zero, the net sum will be zero, so the
integral action will not contribute.
• But if the 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.
• This mode is represented by an integral equation
0
( ) (0)
t
I pp t K e dt p= +∫

4. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• where p(0) is the controller output when the integral
action starts.
• The gain expresses how much controller output in percent
is needed for every percent-time accumulation of error.
• Another way of thinking of integral action is found by
taking the derivative of the above equation.
• In that case, we find a relation for the rate at which the
controller output changes,
I p
dp
K e
dt
=

5. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• This equation shows that when an error occurs, the
controller begins to increase (or decrease) its output at a
rate that depends upon the size of the error and the gain.
• If the error is zero, the controller output is not changed.
• If there is positive error, the controller output begins to
ramp up at a rate determined by the above equation.
• Figure 17.1 illustrates this for two different values of gain.

6. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Fig 17.1 Integral mode controller action - The rate of output change depends on error

7. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Fig 17.2 Integral mode controller action – an illustration of integral mode output and error

8. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Summary:
1. If the error is zero, the output stays fixed at a value equal
to what it was when the error went to zero.
2. If the error is not zero, the output will begin to increase
or decrease at a rate of KI percent/second for every 1% of
error.
Area Accumulation: The integral equation can be
interpreted as providing a controller output equal to the net
area under the error-time curve multiplied by . We often say
that the integral term accumulates error as a function of time.
Thus, for every 1%−s of accumulated error-time area, the
output will be KI percent.

9. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Try this one:
An integral controller is used for speed control with a
setpoint of 12 rpm within a range of 10 to 15 rpm. The
controller output is 22% initially. The constant KI = − 0.15%
controller output per second per percentage error. If the speed
jumps to 13.5 rpm, calculate the controller output after 2s for
a constant ep.
Solution:
max min
12 13.5
30%
15 10
p
r b
e
b b
− −
= = = −
− −

10. The rate of controller output change is then given by
The controller output for constant error will be found as
Since ep is constant,
After 2s,
Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
( )( )1 %
0.15 30% 4.5I p
dp
K e s
dt s
−
= = − − =
0
(0)
t
I pp K e dt p= +∫
(0)I pp K e t p= +
4.5 2 22 31%p = × + =

11. Integral-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• The integral gain, KI, is often represented by the inverse, which
is called the integral time, or the reset action, TI = 1/ KI.
• This is often expressed in minutes instead of seconds because
this unit is more typical of many industrial process speeds.
• The integral controller constant KI may be expressed in
percentage change per minute per percentage error, whenever
a typical process-control loop has characteristic response time
in minutes rather than seconds. Thus, an integral mode
controller with reset action at 5.7 minutes means that
( )( )
3 11
2.92 10
5.7min 60 / min
IK s
s
− −
= = ×

12. Derivative-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• Derivation controller action responds to the rate at which the
error is changing—that is, the derivative of the error.
• Appropriately, the equation for this mode is given by the
expression
• where the gain, KD, tells us by how much percent to change the
controller output for every percent-per-second rate of change of
error.
• Derivative action is not used alone because it provides no
output when the error is constant.
• Derivative controller action is also called rate action and
anticipatory control.
( ) p
D
de
p t K
dt
=

13. Derivative-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• For example, it is assumed that the controller output with no
error or rate of change of error is 50%.
• When the error changes very rapidly with a positive slope, the
output jumps to a large value, and when the error is not
changing, the output returns to 50%.
• Finally, when the error is decreasing—that is, has a negative
slope—the output discontinuously changes to a lower value.
• The derivative mode must be used with great care and usually
with a small gain, because a rapid rate of change of error can
cause very large, sudden changes of controller output.
• Such an event can lead to instability.

14. Derivative-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015

15. Derivative-Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
To summarize the characteristics of the derivative mode:
• If the error is zero, the mode provides no output.
• If the error is constant in time, the mode provides no output.
• If the error is changing in time, the mode contributes an output
of percent for every 1%-per-second rate of change of error.
• For direct action, a positive rate of change of error produces a
positive derivative mode output.

16. Assignment Questions:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
2.1 A controller outputs a 4 to 20 mA signal to control motor
speed from 140 to 600 rpm with a linear dependence.
Calculate (a) current corresponding to 310 rpm, and (b) the
value of (a) expressed as the percent of control output.
2.2 A 5m diameter cylindrical tank is emptied by a constant
outflow of 1.0 m3 /min. A two position controller is used to
open and close a fill valve with an open flow of 2.0 m3 /min.
For level control, the neutral zone is 1 m and the
setpoint is 12 m. (a) Calculate the cycling period (b) Plot
the level vs time.

17. Assignment Questions:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
2.3 A proportional controller has a gain of Kp = 2.0 and P0 =
50%. Plot the controller output for the error given in the Fig.

18. References:
• Process Control Instrumentation Technology, by Curtis D.
Johnson, Eighth Edition, Pearson Education Limited.
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015