Mitsubishi Electric FR-A800, Инструкция по эксплуатации

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Position control under vector control and PM sensorless vector control Parameters
5 - 178
Relationship between the position resolution
Δ
l and system accuracy
The system accuracy (the positioning accuracy of the machine) is the sum of electric deviation and
mechanical deviation. Normally try to prevent the total deviation from being affected by the elec-
tronic deviation. Refer to the following relationship as a reference.
<Motor stop characteristics>
When running the motor by parameter settings, the relationship between the internal command
pulse frequency and the number of motor rotations will be as shown in Figure page 5-150. Pluses as
much as the motor speed delay are accumulated in the deviation counter. These pulses are called
droop pulses (ε). The relationship between the command frequency (fo) and position control gain
(Kp: Pr. 422) is shown in the following formula.
The number of droop pulses (ε) will be 8192 with the initial value Kp = 25 s
-1
.
Since the inverter has droop pulses during operation, a stop settling time (ts), which is the time be-
tween the zero command output and the motor stop, is required. Set the operation pattern taking
into the account the stop setting time.
The stop settling time (ts) will be 0.12 s for the initial value Kp = 25 s
-1
.
The accuracy of positioning
Δ
ε will be (5 to 10) ×
Δ
l =
Δ
ε [mm]
Example
쑴
Setting example 1:
In a driving system whose ball screw pitch is PB =10 (mm) and the reduction ratio is 1/n =1, the elec-
tronic gear ratio is
Δ
s =10 (mm) when
Δ
l = 0.01 (mm) and Pf = 4000 (pulses/rev) is set as the number
of feedback pulses. Based on this, use the following formula:
Thus, set the parameters as follows: Pr. 420 = "4", Pr. 421 = "1".
쑶
Example
쑴
Setting example 2:
Find the internal command pulse frequency for the rated motor speed of the dedicated motor.
However, the command pulse ratio is Pr. 420/Pr. 421 = "1".
If the number of encoder pulses is 2048 (pulses/rev), (feedback pulse pf = 2048 × 4)
The internal command pulse will be 204800 (pps) in accordance with the above formula.
쑶
Δ
l < (
1
to
1
) ×
Δ
ε
Δ
ε: positioning accuracy
5 10
ε
=
fo
[pulse] ε =
204800
[pulse] (with the rated motor speed)
Kp 25
ts = 3 ×
1
[s]
Kp
Δ
l =
Δ
s
×
Pr. 420
Pf Pr. 421
Pr. 420
=
Δ
l ×
Pf
Pr. 421
Δ
s
= 0.01 ×
4000
=
4
10 1
fo = 2048 × 4 (multiplication) ×
No.
×
Pr. 421
60 Pr. 420
= 204800