
B-65150E/04 4. SERVO FUNCTION DETAILS
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The following table lists the number of positional deviation pulses for a
VCMD voltage of 5 V.
Table 4.19 (c) Number of positional deviation pulses for a VCMD voltage
of 5 V for semi-closed loop
VCM2 VCM1
Number of positional deviation pulses for a VCMD
voltage of 5 V
0 0 15,258 × FFG/Kp
0 1 244,133 × FFG/Kp
1 0 3,906,133 × FFG/Kp
1 1 62,498,133 × FFG/Kp
Kp: Position gain (s
−1
)
FFG: Flexible feed gear (numerator/denominator)
Table 4.19 (d) Number of positional deviation pulses for a VCMD voltage
of 5 V for full-closed loop
VCM2 VCM1 Number of positional deviation pulses for a VCMD
voltage of 5 V
0 0 0.0153 × (number of positional feedback occurrences per
motor revolution)/Kp
0 1 0.2441 × (number of positional feedback occurrences per
motor revolution)/Kp
1 0 3.96061 × (number of positional feedback occurrences per
motor revolution)/Kp
1 1 62.5 × (number of positional feedback occurrences per motor
revolution)/Kp
Kp: Position gain (s
−1
)
Table 4.19 (e) Number of positional deviation pulses for a VCMD voltage
of 5 V when a linear motor is in use
VCM2 VCM1 Number of positional deviation pulses for a VCMD
voltage of 5 V
0 0 1,250/(A × Kp)
0 1 20,000/(A × Kp)
1 0 320,000/(A × Kp)
1 1 5,120,000/(A × Kp)
Kp: Position gain (s
−1
)
A: Detection unit (µm)
(Example)
Assume the following conditions:
Position gain = 30 (s
−1
), semi-closed loop, detection unit of 1
µm/pulse, flexible feed gear = 1/100,
VCM2 = 0, VCM1 = 1 (VCMD waveform signal calculation
parameters)
If a waveform with E = 0.3 V and I/f = 20 ms is observed:
Number of positional deviation pulses for a VCMD voltage of 5 V
= 244133/100/30 = 81 pulses
Table vibration = 81 × 0.3/5 = 4.88 µm
Vibration frequency = 50 Hz