This document discusses PID control modes and their parameters. It explains the function of the proportional, integral and derivative terms in PID control. The effects of changing the proportional gain Kp, integral gain Ki and derivative gain Kd are described. It also provides an example of calculating the controller output for a three-mode PID controller given an error signal over time.

1. ICE401: PROCESS INSTRUMENTATION
AND CONTROL
Class 20
Effect of Kp, Ki & Kd and PID Control Mode
Dr. S. Meenatchisundaram
Email: meenasundar@gmail.com
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015

2. Function of the Proportional Term:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• The plot below illustrates the variation in process variable
for a set point response.
• The error used in the proportional calculation is shown on
the plot:
• At time t = 25 min, e(25) = 60–56 = 4, at time t = 40 min,
e(40) = 60–62 = –2

3. Function of the Proportional Term:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Computing and plot e(t) at each point in time t

4. Function of the Integral Term:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• The integral sum of error is computed as the shaded areas
between the SP and PV traces as

5. Function of the Integral Term:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Since it is controller error that drives the calculation, we
get a direct view the situation from a controller error plot
as shown below

6. Function of the Integral Term:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• At time t = 60 min on the plots, the integral sum is 135 –
34 = 101. The response is largely settled out at t = 90 min,
and the integral sum is then 135 – 34 + 7 = 108.
• The response is largely complete at time t = 90 min, yet
the integral sum of all error is not zero.
• In this example, the integral sum has a final or residual
value of 108.
• It is this residual value that enables integral action of the
PI controller to eliminate offset.

7. Effects of Kp, Ki and Kd:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Effects of Kp:

8. Effects of Kp, Ki and Kd:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Effects of Ki:

9. Effects of Kp, Ki and Kd:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• Effects of Kd:

10. Three-Mode Controller (PID):
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• One of the most powerful but complex controller mode
operations combines the proportional, integral, and
derivative modes.
• This system can be used for virtually any process
condition. The analytic expression is
• This mode eliminates the offset of the proportional mode
and still provides fast response.
0
( ) (0)
t
p
P P P I p P D I
de
p t K e K K e dt K K p
dt
= + + +∫

11. Three-Mode Controller (PID):
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
• For a three-mode controller with KP = 5, KI = 0.7 s-1, KD =
0.5 s and PI(0) = 20%, draw a plot of the controller output
for the error given in figure.
Solution:
• The error can be expressed in three time regions.
0 ≤ t ≤ 1 ep = t %
1 ≤ t ≤ 3 ep = 1 %
3 ≤ t ep = -0.5t +2.5 %

12. Three-Mode Controller (PID):
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Case 1: 0 ≤ t ≤ 1 ep = t %
Case 2: 1 ≤ t ≤ 3 ep = 1 %
1
1
0
5 3.5 2.5 20pde
p t t dt
dt
= + + +∫
1 5 1.75 2.5 20p t= + + +
( )2 3.5 1 26.75p t= − +
2 p
1
5 3.5 e 2.5 21.75
t
p
p
de
p e dt
dt
= + + +∫

13. Three-Mode Controller (PID):
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015
Case 3: 3 ≤ t ≤ 5 ep = -0.5t +2.5 %
( )
2
3
0.5 4.5
5 0.5 2.5 3.5 2.5 7.5 28.75
2 2
t
p t t
 −
= − + + + + − + 
 
3 p
3
5 3.5 e 2.5 28.75
t
p
p
de
p e dt
dt
= + + +∫
2
3 0.875 6.25 21.625p t t= − + +

14. Three-Mode Controller (PID):
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Jan – May 2015

15. 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, Jan – May 2015