CA3193AE INTERSIL [Intersil Corporation], CA3193AE Datasheet - Page 7

• Manufacturer Part Number: CA3193AE
• Description: 1.2MHz, BiCMOS Precision Operational Amplifiers
• Manufacturer: INTERSIL [Intersil Corporation]

Typical Applications
FIGURE 7. TYPICAL TWO OP AMP BRIDGE-TYPE
FIGURE 9. USING CA3193 AS A BILATERAL CURRENT
2M
R
V
For Ideal Resistors with
V
A
FOR VALUES ABOVE V
1
OUT
OUT
=
ALL RESISTORS ARE 1%
IF R
I
FOR R
L
R
R
------------------- -
V
V
IS INDEPENDENT OF VARIATIONS IN R
b
=
=
V
OUT
1
–
= R
V+
–
V
I
V
L
L
DIFFERENTIAL AMPLIFIER
SOURCE
V
V ALUES OF0 :TO 3k:W ITH V = 1V
b
a
3
a
R
2M:
R
=
–
§
¨
©
R
R + 'R
=
3
S
AND R
V
R
------ -
R
-------------- -
R
VR
a
§
¨
©
2
1
3
§
¨
©
R
------ -
R
R
+
R
------ -
R
4
4
3
a
b
5
1
2
4
3
+
|R
· R
¸
¹
3
2
+
=
OUT
------ -
R
1
V a
V
1
·
¸
¹
CA3193
4
3
+
b
R
------ -
R
------------------------- -
-
4
·
¸
¹
2M
+ R
+
V 1M
1
2
2
3
= (V b - V a ) (I
3
2
V
=
1K
4
5
+
CA3193
-
+
b
7
CA3193
-
, THEN
§
¨
©
R
------ -
R
-15V
R
------ -
R
+15V
1M
1M
R
R
3
4
-15V
+15V
4
3
+15V
2
4
-15V
4
4
6
7
7
+
=
1
O
·
¸
¹
-------
2K
V
)
6
6
=
L
I
1K
R
L
9K
R
500PA
1
2
(0:T O 3.0k:)
1K
R
R
WITHV = 1V
1K
9K
R
R
5
CA3193, CA3193A
L
V
3
4
OUT
66
FIGURE 8. DIFFERENTIAL AMPLIFIER (SIMPLE
FIGURE 10. TYPICAL SUMMING AMPLIFIER APPLICATION
Then V
V
V
ALL RESISTANCE VALUES ARE IN OHMS.
V
V
V
ALL RESISTANCE VALUES ARE IN OHMS.
OUT
V
If R
THEN V
For values above V
If A
with R
Thus, the CMRR of this circuit is limited by
the matching or mismatching of this network
rather than the amplifier.
OUT
3
2
1
OCM
OUT
V
V
1
2
V
= - (2V
4
is To be made 1 and if R
SUBTRACTER) USING CA3193
10K
10K
10K
=
=
2
R
R
R
= 0.0005 V
=
3
2
1
= 0.999R (0.1% mismatch in R
–
R
10K
10K
OUT
R
R
§
¨
©
V
1
2
1
3
R
------- V
R
+ 2V
2
, R
F
1
§
¨
©
-------------------- -
R
=
2.8K
3
R
3
1
2
=
R
3
4
+ 2 V
20K
2
R
+
+
V
4
IN
4
R
R
------- V
3
R
R
2
2
CA3193
+
OUT
-
1
4
or CMRR = 66dB
F
2
–
3
· R
¸
¹
CA3193
+
and
)
-
V
§
¨
©
20K
2
R
-------------------- -
1
= 2(V
4
7
+
F
1
20K
§
¨
©
R
R
4
7
R
------ -
R
R
------ -V
R
+
R
------ -
R
2
+15V
1
-15V
2
1
F
3
R
2
1
6
1
2
·
¸
¹
+15V
= R
-15V
2
- V
=
3
6
·
¸
¹
·
¸
¹
–
R
------ -
R
1
3
V
4
3
):
= R
1
§
¨
©
2
R
------ -
R
)
4
2
1
= R
V
·
¸
¹
OUT
R
R
L
L