CACULATING P,I CONSTANTS Kp and Ki
George Younkin, P.E., Life FELLOW IEEE
PI compensation is used with most industrial servos. D compensation is used
when needed. An electric motor can be represented by a second order equation (1)
Vm
1/ K e
= 2
2
ei
s + 2δω n s + ω n
(1)
Where: Ke= [volts/rad/sec]
ω n = bandwidth of motor drive = [rad / sec ]
An index of performance can be written to equate with the characteristic equation of eq.1
such that:
(( K p / J ) s + ( K i / J ))
(2)
[s 2 + ( K p / J )s + K i / J ]
Equating coefficients of eq. 1 and eq. 2 :
2δω n =
Kp
(3)
J
and
Ki
J
Where:
J= Jmotor+Jload reflected to motor= [lb − in − sec 2 ]
ω n = motor drive bandwidth = rad / sec
δ = damping factor
It can be assumed that the drive will be well damped, thus δ = 1
ωn 2 =
(4)
The bandwidth should be chosen as a reasonable value of approximately 20 to 30 Hz
(125 to 188 rad/sec) for industrial machines.
Therefore, substituting these approximations into eq. 3 and eq. 4 yields
2ω n =
Kp
J
K p = 2ω n J [a/rpm]
and
ωn 2 =
Ki
J
K I = ω N j [a/sec/rpm]
2
and
1
EXAMPLE:
An industrial machine servo has the following characteristicsJtotal at motor=0.311 lb − in − sec 2
The drive response is desired to respond with a damping factor as close to critical
damping as possible making δ = 1 .
The normal servo bandwidth of this industrial servo will be about 30 Hz (188 rad.sec).
Therefore:
K p = 2 x188 x0.311 = 118 [a/rpm]
K i = 188 2 x0.311 = 11000 [a/sec/rpm]
The PI compensation can be represented as-
Kp

s + 1
Ki 
K
K  K p s + Ki
I 

= K p + i  =
=  i
s
s 
s
V2 
t2 =
ω2 =
Kp
Ki
w2 =
Ki
Kp
K 2 [t 2 s + 1]
(Corner frequency)
K i 11000
=
= 94 rad / sec
Kp
117
The frequency response for PI compensation is as follows:
2
=
s
100
75
50
Magnitude dB, Phase-deg
25
20⋅ log( c( j ⋅ w)
180
π
)
⋅ arg( c( j ⋅ w) )
0
25
50
75
100
1
10
3
1 .10
100
w
rad/sec
Magnitude
Phase
PI Compensation
3
4
1 .10
1 .10
5