Integral and Derivative Electronic Controllers

1. ICE401: PROCESS INSTRUMENTATION
AND CONTROL
Class 25
Integral, Derivative Electronic Controllers
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
• Suppose we have an input range of 6 V, an output range of 5 V, and
KI = 3.0%/(% ‒ min).
• 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, TI, which is just the
inverse of the gain, so TI = 0.33 min or 19.8 sec.
• We must first convert the time units to seconds. Therefore
• An error of 1% for 1 s is found from
(0.01)(6 V)(1 s) = 0.06 V ‒ s
3% 0.05%1min
% min 60sec % s
    =
− −   

3. Integral Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• Furthermore, KI percent of the output (using the seconds
expression for gain) is
(0.0005)(5 V) = 0.0025V
• Therefore, the integral gain in terms of voltage must be
GI=(0.0025V)/(0.06V‒s)=0.0417s-1
• The values of R and C can be selected from this.

4. Integral Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
Try:
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 KI = 4.0%/(% ‒ min). Specify the
values of GI, R1 and C.
Solution:
We must first convert the time units to seconds. Therefore
KI percent of the output (using the seconds expression for gain) is
(0.000667)(6.8V) = 0.00454 V
4% 0.0667%1min
% min 60sec % s
    =
− −   

5. Integral Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• An error of 1% for 1 s is found from
(0.01)(1.6 V)(1 s) = 0.016 V ‒ s
• Therefore, the integral gain in terms of voltage must be
GI=(0.00454V)/(0.016V‒s)=0.283s-1
• Since, GI = 1/RC,
• RC = 3.524. By selecting C = 100µF, then R = 35.3K .

6. Derivative Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
( ) p
D
de
p t K
dt
=
e
out
dV
V RC
dt
= −
• For a theoretical derivative op-amp circuit,

7. Derivative Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• For a practical derivative op-amp circuit,
• From the above equation, the output depends upon the derivative of
the input voltage, but there is now an extra term involving the
derivative of the output voltage.
• Essentially, we have a first-order differential equation relating input
and output voltage.
• For very high frequencies the impedance of the capacitor becomes
very small and can be neglected.
• Then the circuit becomes just an inverting amplifier with a gain is
‒ (R2/R1).
1 2
out e
out
dV dV
V R C R C
dt dt
+ = −

8. Derivative Control Mode:
Process Instrumentation and Control (ICE 401)
Dr. S.Meenatchisundaram, MIT, Manipal, Aug – Nov 2015
• At low frequency the impedance of the capacitor will be large so R1
can be neglected. Then, the response of can be given as,
• The circuit exhibits a derivative response provided the following
inequality is satisfied, i.e, 2 π f R1 C << 1.
• The following derivative mode design guidelines can be followed:
1. Estimate the maximum frequency at which the physical system
can respond, fmax.
2. Set 2πfmaxR1 C =0.1 and solve for R1. (C is found from the
mode derivative gain requirement.)
e
out
dV
V RC
dt
= −

9. 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