Program Sequence
P
OG
M]
PROG NO . = 1
PROG CHA IN NO .= 0
RUN COUN T = 0
CLE AR SEQ . = NO
[ SEQUE NC E ]
SE Q NO . =1 SEQ . TY P E=AUTO
VOL T AGE=1 0 ( V ) V S . R . =1 ( V / mS )
CURRENT=20 (A) I S.R. =1 (A/mS)
TTL OUT=1 <B INARY = 00000001>
TIME =5 (S)
[ SEQUE NC E ]
SE Q NO . =2 SEQ . TYPE=MANUA L
VOL T AGE=3 0 ( V ) V S . R . =10 ( V / mS)
CU RR EN T = 1 5 ( A ) I S . R . =0 . 1 ( A / mS )
TTL OUT=2 <B INARY=00000100>
TIME =10 (S)
↓
↓
↓
[SEQUENCE ]
SEQ NO . = 9 SEQ . TYPE=AUTO
VOL TAGE= 0 (V ) V S . R . =1 . 0 ( V /mS )
CURRENT= 0 ( A ) I S . R . =1 . 0 ( A / mS )
TTL OUT= 0 <B INARY=00000000>
TIME =0 .1(S)
[SEQUENCE ]
SEQ NO. =10 SEQ. TYPE=AUTO
VOL TAGE= 0 (V ) V S . R . =1 . 0 ( V /mS )
CURRENT= 0 ( A ) I S . R . =1 . 0 ( A / mS )
TTL OUT= 0 <B INARY=00000000>
TIME =0 .1(S)
Figure 4-11
A5: Execution step:
(1) SEQ#1:
(1) Since SEQ TYPE = AUTO is set for SEQ#1, it begins to execute the settings
in SEQ#1.
(2) The actual current Slew_rate is 0.05(A/mS) that is smaller than the setting
(1A/mS); therefore SEQ#1 is in CV Mode during voltage raise.
(3) When the current reaches the set 10V, it remains 5 sec.
(4) Skip to SEQ#2.
(2) SEQ#2:
(1) Since SEQ TYPE = AUTO is set for SEQ#2, it begins to execute the settings
in SEQ#2.
(2) The actual current Slew_rate is 0.05(A/mS) that is smaller than the setting
(0.1A/mS); therefore, SEQ#2 is in CV Mode during voltage raise.
4-11