LT1215/LT1216
典型PERFOR一个CE特征
5V小信号响应
3V
20mV/DIV
50ns/DIV
V
S
= 5V
A
V
= 1
1215/16 G34
±15V
小信号响应
20mV/DIV
V
S
=
±15V
A
V
= 1
50ns/DIV
1215/16 G34
5V沉降
500mV/DIV
输出步（ V）
2V/DIV
50ns/DIV
V
S
= 5V
A
V
= 1
1215/16 G30
ü W
5V大信号响应
3V
5V大信号响应
0V
200ns/DIV
V
S
= 5V
A
V
= 1
1215/16 G28
0V
100ns/DIV
V
S
= 5V
A
V
= –1
R
F
= R
G
= 1k
C
F
= 20pF的
1215/16 G31
±15V
大信号响应
±15V
大信号响应
10V
10V
0V
0V
–10V
–10V
V
S
=
±15V
A
V
= 1
200ns/DIV
1215/16 G29
V
S
=
±15V
A
V
= –1
R
F
= R
G
= 1k
200ns/DIV
1215/16 G32
±15V
解决
10
8
6
4
2
0
–2
–4
–6
–8
1215/16 G33
建立时间0.01％
与输出步骤
V
S
= ±15V
250V/DIV
1mV/DIV
同相
反相
V
S
=
±15V
A
V
= –1
100ns/DIV
–10
200
400
300
稳定时间（纳秒）
500
1215/16 G36
11
