LT1001CJ8 Linear Technology, LT1001CJ8 Datasheet - Page 5

• Manufacturer Part Number: LT1001CJ8
• Description: IC PRECISION OP-AMP SINGLE 8-DIP
• Manufacturer: Linear Technology

Specifications of LT1001CJ8

• Amplifier Type: General Purpose
• Number Of Circuits: 1
• Slew Rate: 0.25 V/µs
• Gain Bandwidth Product: 800kHz
• Current - Input Bias: 700pA
• Voltage - Input Offset: 18µV
• Voltage - Supply, Single/dual (±): ±3 V ~ 18 V
• Operating Temperature: 0°C ~ 125°C
• Mounting Type: Through Hole
• Package / Case: 8-CDIP (0.300", 7.62mm)
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Current - Supply: -
• Output Type: -
• Current - Output / Channel: -
• -3db Bandwidth: -

TYPICAL PERFORMANCE CHARACTERISTICS
1200k
1000k
800k
600k
400k
200k
–0.2
–0.4
–0.6
–0.8
–1.0
+1.0
+0.8
+0.6
+0.4
+0.2
2.0
1.5
1.0
0.5
V
V
0
–50
+
–
–50
Common Mode Limit
vs Temperature
Supply Current vs Supply Voltage
Open Loop Voltage Gain
vs Temperature
–25
–25
3
0
6
SUPPLY VOLTAGE (V)
0
TEMPERATURE ( C)
TEMPERATURE C
V
V
V
V
–
–
9
25
25
+
+
= –12 to –18V
= –1.2 to –4V
= 1.2 to 4V
= 12 to 18V
–55 C
V
V
12
S
50
S
50
= 15V, V
= 3V, V
125 C
15
W
75
75
O
18
O
= 1V
100
= 12V
100
25 C
U
1001 G16
1001 G10
21
1001 G13
125
125
140
120
100
140
120
100
–20
16
12
80
60
40
20
80
60
40
20
8
4
0
0
0.1
100
1
Common Mode Rejection Ratio
vs Frequency
Output Swing vs Load Resistance
Open Loop Voltage Gain
Frequency Response
NEGATIVE SWING
1
10
V
T
S
A
300
= 15V
= 25 C
10
LOAD RESISTANCE ( )
100
FREQUENCY (Hz)
FREQUENCY (Hz)
POSITIVE SWING
100
V
S
1000
= 3V
1k
V
1k
S
= 15V
T
10k
A
10k
= 25 C
3k
100k
V
T
A
S
= 25 C
= 15V
100k
1M
1001 G17
1001 G14
1001 G11
10M
10k
1M
–4
–8
20
16
12
8
4
0
0.1
Gain, Phase Shift vs Frequency
PHASE MARGIN –55 C = 63
–10
–20
–30
–40
–50
V
140
120
100
50
40
30
20
10
S
80
60
40
20
GAIN 25 C & –55 C
0
= 15V
0.1
0
Power Supply Rejection Ratio
vs Frequency
Output Short-Circuit Current
vs Time
GAIN 125 C
TIME FROM OPUTPUT SHORT (MINUTES)
0.2
POSITIVE SUPPLY
FREQUENCY (MHz)
1
125 C = 57
1
10
FREQUENCY (Hz)
0.5
PHASE 25 C
MARGIN
V
T
100
PHASE
= 60
A
25 C
S
2
= 15V 1V p-p
= 25 C
NEGATIVE SUPPLY
1
LT1001
1k
–55 C
V
125 C
–55 C
25 C
125 C
S
25 C
1001 G12
3
= 15V
10k
2
1001 G18
1001 G15
80
100
120
140
160
180
200
220
1001fb
5
100k
4