參數(shù)資料
型號(hào): CLC5612
廠商: National Semiconductor Corporation
英文描述: Dual, High Output, Programmable Gain Buffer
中文描述: 雙路,高輸出,可編程增益緩沖器
文件頁(yè)數(shù): 8/12頁(yè)
文件大?。?/td> 208K
代理商: CLC5612
http://www.national.com
8
Figure 6: AC Coupled, A
v
= +2V/V Configuration
Dual Supply Operation
The CLC5612 operates on dual supplies as well as sin-
gle supplies. The non-inverting and inverting configura-
tions are shown in Figures 7, 8 and 9.
Figure 7: Dual Supply, A
v
= -1V/V Configuration
Figure 8: Dual Supply, A
v
= +1V/V Configuration
Figure 9: Dual Supply, A
v
= +2V/V Configuration
Load Termination
The CLC5612 can source and sink near equal amounts
of current. For optimum performance, the load should be
tied to V
cm
.
Driving Cables and Capacitive Loads
When driving cables, double termination is used to
prevent reflections. For capacitive load applications, a
small series resistor at the output of the CLC5612 will
improve stability and settling performance.
Frequency Response vs. C
L
plot, shown below in
Figure 10, gives the recommended series resistance
value for optimum flatness at various capacitive loads.
The
Figure 10: Frequency Response vs. C
L
Transmission Line Matching
One method for matching the characteristic impedance
(Z
o
) of a transmission line or cable is to place the
appropriate resistor at the input or output of the amplifier.
Figure 11 shows typical inverting and non-inverting
circuit configurations for matching transmission lines.
1
7
6
8
5
3
4
2
1k
+
-
+
-
1k
1k
1k
CLC5612
V
o
0.1
μ
F
6.8
μ
F
+
V
in
R
t
V
CC
Note:
R
provides DC bias for the
t
to
non-inverting input. Select R
in
= R
t
||1k
.
Channel 2 not shown.
R
b
6.8
μ
F
0.1
μ
F
+
V
EE
V
Low frequency cutoff
2V
2.5
1
2 R C
R
source
whereR
R
2
R
o
C
in
=
+
=
=
>>
,
1
7
6
8
5
3
4
2
1k
+
-
+
-
1k
1k
1k
CLC5612
0.1
μ
F
6.8
μ
F
V
in
R
R
V
CC
V
CC
C
C
V
o
Note:
Channel 2 not shown.
C
1
7
6
8
5
3
4
2
1k
+
-
+
-
1k
1k
1k
CLC5612
V
o
0.1
μ
F
6.8
μ
F
+
V
in
R
t
V
CC
Note:
Channel 2 not shown.
6.8
μ
F
0.1
μ
F
+
V
EE
1
7
6
8
5
3
4
2
1k
+
-
+
-
1k
1k
1k
CLC5612
V
o
0.1
μ
F
6.8
μ
F
+
V
in
R
t
V
CC
Note:
Channel 2 not shown.
6.8
μ
F
0.1
μ
F
+
V
EE
y
M
Frequency (Hz)
1M
10M
100M
V
o
= 1V
pp
C
L
= 10pF
R
s
= 49.9
C
L
= 100pF
R
s
= 17.4
C
L
= 1000pF
R
s
= 6.7
C
L
1k
R
s
+
-
1k
1k
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