CA3094E Intersil, CA3094E Datasheet - Page 7

• Manufacturer Part Number: CA3094E
• Description: PROGRAMMABLE POWER SWITCH/AMP
• Manufacturer: Intersil

Specifications of CA3094E

• Rohs Status: RoHS non-compliant
• Amplifier Type: Transconductance
• Number Of Circuits: 1
• Slew Rate: 500 V/µs
• Gain Bandwidth Product: 30MHz
• Current - Input Bias: 200nA
• Voltage - Input Offset: 400µV
• Operating Temperature: -55°C ~ 125°C
• Mounting Type: Through Hole
• Package / Case: 8-DIP (0.300", 7.62mm)
• Current - Supply: -
• Output Type: -
• Current - Output / Channel: -
• Voltage - Supply, Single/dual (±): -
• -3db Bandwidth: -
• Other names: CA3094

Typical Applications
INPUT
Where
VOLTAGE A
2/3V+
VOLTAGE AT
TERMINAL 8
+18
A
V+
E
0
0
12V
IN
E OUT
---------------- -
DC
E
0.01µF
V+ = 18V
FIGURE 11A. INVERTING OP AMP
IN
Z
1
=
S
1N914
1
100
kΩ
100
kΩ
100
kΩ
B
f


NOTE: In single-ended output operation, the CA3094 may require a pull up or pull down resistor.
C
Z 2
------ -
Z
A
1



1N914
R
270
kΩ
100
kΩ
C
depends on the characteristics of Z
3-7
12V
-
+
CA3094
I
TIME DELAY (s) = RC (APPROX.)
I
FIGURE 13. RC TIMER TRIGGERED BY EXTERNAL NEGATIVE PULSE
D
R
1MΩ
R
2.2MΩ
R
1MΩ
R
100kΩ
R
220kΩ
1
2
3
1
2
Z
3
2
2
2
3
220kΩ
R
ABC
+
-
7
CA3094
CA3094, CA3094A, CA3094B
FIGURE 11. APPLICATION OF THE CA3094
+
-
5
CA3094
4
4
8
7
6
5
FIGURE 12. RC TIMER
E
(NOTE)
8
OUT
220kΩ
1
6
and Z
E
2kΩ
PULL UP
OUT
2
C
0.5µF
E
1
R
2kΩ
Where E
LOAD
Problem: To calculate the maximum value of R required to
Given:
I
I
I
curve.
Then: I
†
R
R
I
I
ABC
E
= 500nA at I
= 5µA can be determined by drawing a line on Figure 3 through
MAX
MAX
of I
OUT
Ratio of I
E
FIGURE 11B. NON-INVERTING MODE, AS A FOLLOWER
IN
ABC
= 100µA and I
OUT
I
=
=
I ABC
On a negative going transient at input (A), a negative
pulse at C will turn “on” the CA3094, and the output (E)
will go from a low to a high level.
At the end of the time constant determined by C
R
output will be pulled low by R
be independent of the interval when input (A) returns
to a high level.
= 33nA at I
switch a 100mA output current comparator
2
18V 12V
--------------------------- -
180MΩ
I
, R
= E
at T
33nA
A
B
C
D
E
ABC
3
=
IN
, the CA3094 will return to the “off” state and the
–
V+
V+
A
5µA,
0
0
0
0
0
= 25
= 100µA (from Figure 3)
×
B
ABC
2 3
= 500nA parallel to the typical T
=
o
⁄
R ABC
C to I
= 5µA
180 MΩ at T
†
+
-
CA3094
=
I
at T
120M Ω at T
=
3
/
4
3.6MΩ
A
V+
= -55
A
LOAD
=
o
≈
C for any given value
A
---------- -
5µA
25
18V
. This condition will
=
o
C
–
55
o
C
A
E
(NOTE)
OUT
= 25
1
, R
o
C
1
,