Datasheet

Vacuum Equipment Selection Example
7
Model Selection
 Transfer of Semiconductor Chips
Selection conditions:
(1) Workpiece: Semiconductor chips
Dimensions: 8 mm x 8 mm x 1 mm, Weight: 1 g
(2) Vacuum piping length: 1 m
(3) Adsorption response time: 300 msec or less
1. Vacuum Pad Selection
(1) Based on the workpiece size, the pad diameter is 4 mm (1 pc.).
(2) Using the formula on the front matter 13, confirm the lifting force.
W = P x S x 0.1 x 1/t W = 1 g = 0.0098 N
0.0098 = P x 0.13 x 0.1 x 1/4 S = π/4 x (0.4)
2
= 0.13 cm
2
P = 3.0 kPa t = 4 (Horizontal lifting)
According to the calculation, –3.0 kPa or more of vacuum pressure can adsorb the workpiece.
(3) Based on the work shape and type, select:
Pad type: Flat
Pad material: Silicone
(4) According to the results above, select a vacuum pad part number ZPT04US-
.
(Specify the vacuum entry port (
) from the pad mounting status.)
2. Vacuum Ejector Selection
(1) Find the vacuum piping capacity.
Assuming that the tube I.D. is 2 mm, the piping capacity is as follows:
V = π/4 x D
2
x L x 1/1000 = π/4 x 2
2
x 1 x 1/1000
= 0.0031 L
(2) Assuming that leakage (Q
L) during adsorption is 0, find the average suction flow rate to meet the adsorption response
time using the formula on the front matter 17.
Q = (V x 60) /T
1 + QL = (0.0031 x 60) /0.3 + 0 = 0.62 L
From the formula on the front matter 17, the maximum suction flow rate Q
max is
Q
max = (2 to 3) x Q = (2 to 3) x 0.62
= 1.24 to 1.86
L/min (ANR)
According to the maximum suction flow rate of the vacuum ejector, a nozzle with a 0.5 diameter can be used.
If the vacuum ejector ZX series is used, representative model ZX105
 can be selected.
(Based on the operating conditions, specify the complete part number for the vacuum ejector used.)
3. Adsorption Response Time Confirmation
Confirm the adsorption response time based on the characteristics of the vacuum ejector selected.
(1) The maximum suction flow rate of the vacuum ejector ZX105
 is 5 L/min. From the formula on the front matter 18,
the average suction flow rate Q
1 is as follows:
Q
1 = (1/2 to 1/3) x Ejector’s max. suction flow rate
= (1/2 to 1/3) x 5 = 2.5 to 1.7
L/min
(2) Next, find the maximum flow rate Q
2 of the piping. The conductance C is 0.22 from the Selection Graph (3).
From the formula on the front matter 18, the maximum flow rate is as follows:
Q
2 = 5 x C x 11.1 = 5 x 0.22 x 11.1 = 12.2 L/min
(3) Since Q
2 is smaller than Q1, Q = Q1.
Thus, from the formula on the front matter 18, the adsorption response time is as follows:
T = (V x 60)/Q = (0.0031 x 60)/1.7 = 0.109 second
= 109 msec
It is possible to confirm that the calculation result satisfies the required specification of 300 msec.
Front matter 24