Control Techniques Commander SE Advanced User's Manual page 157

Step 5
The maximum value of inductance is determined by the acceptable voltage drop at the working frequency. Calculate it from
the following expression:
--------------------------
L =
max
2πf
Where:
x is acceptable voltage drop fraction e.g. for 5% use 0.05
Vac is motor voltage rating (line to line)
f is maximum drive output frequency
o
Since the voltage drop is inductive, it does not subtract directly from the motor terminal voltage. A value of k of 0.05 is
generally acceptable. If the application is very critical to obtaining full rated torque at full speed then it may be advisable to
apply a lower value of k e.g. 0.02
Step 6
If 2L < L
min max
values to be held in stock to cover most applications.
If there is a need to minimise the high frequency current, for example to prevent premature operation of thermal relays, then
the highest value should be used.
If 2L > L
min max
value of L.
Step 7
Consideration must now be given to the high frequency losses in the chokes. The loss in each choke can be estimated from
the following expression:
P = 0.8f CV
s
F = switching frequency
s
The factor 0.8 is a rough estimate of the fraction of the total losses which are dissipated in the choke. Note that the loss is
proportional to the switching frequency so the lowest acceptable frequency should be selected.
Step 8
It is now necessary to decide whether the choke is able to tolerate this loss. This is a difficult judgement. As a crude rule, the
loss should not exceed 0.1 of the VA in the choke at maximum speed, i.e.
P = <0.2f
o(max)
Where:
f = maximum output frequency
o(max)
If the loss exceeds this limit, and it is not possible to reduce the switching frequency, then a resistor should be connected in
parallel with the choke to extract some of the power.
The resistor value is given by:
2L
R = ------ -
C
The value is not critical and variations of ±50% are acceptable.
The power rating of the resistor should be at least 0.8P.
Provision must be made for the resistor to dissipate this power without overheating itself or nearby equipment. Values of
100W per phase are not uncommon.
Example
A Commander SE 7.5kW with 8 motors of 0.75kW each connected by 140m of multi-core cable. 380VAC supply, maximum
output frequency 50Hz, switching frequency 6kHz. Thermal relays in all motor circuits.
Step 1
Cable capacitance 130 x 140 x 8 = 0.146uF = 0.154uF
Motor capacitance 1nF x 8 = 0.008uF
= 0.154uF
Step 2
Use k = 1.25 (note motor power total is only 6kW)
I = 16A
n
= 19.2A
Step 3
380VAC +10% x 1.41
= 591V
Commander SE Advanced User Guide
Issue Number: 4
xVac
3In
o
then any value between these limits can be used. This may be helpful in allowing a small number of standard
then the drive cannot operate with this length of cable and a higher rated drive must be used. Choose the
2
DC where
x L x In x In
157