LT6350HMS8#TRPBF Linear Technology, LT6350HMS8#TRPBF Datasheet - Page 6

• Manufacturer Part Number: LT6350HMS8#TRPBF
• Description: IC DIFF CONVERT/ADC DRIVER 8MSOP
• Manufacturer: Linear Technology
• Type: ADC Driverr

Specifications of LT6350HMS8#TRPBF

• Applications: Data Acquisition
• Mounting Type: Surface Mount
• Package / Case: 8-MSOP, Micro8™, 8-uMAX, 8-uSOP,
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

LT6350
TYPICAL PERFORMANCE CHARACTERISTICS
T
6
A
= 25°C, V
100
100
90
80
70
60
50
40
30
20
10
90
80
70
60
50
40
30
20
10
0
0
–0.3
–0.3
Differential V
Differential V
434 TYPICAL UNITS
V
T
V
MID-RAIL TO
V
434 TYPICAL UNITS
V
T
V
MID-RAIL
A
S
A
+IN1
S
+IN1
DIFFERENTIAL INPUT REFFERED V
+
CHANGE OF INPUT REFERRED V
= 25°C
= 5V
= 25°C
= 5V
–0.1V
+
= V2 =
= V2 =
–0.2
–0.2
= 5V, V
–0.1
–0.1
–0.1
–0.2
–0.3
–0.4
–0.5
0.5
0.4
0.3
0.2
0.1
0
OS
OS
–
0
Common Mode V
vs Input (+IN2) Voltage
V
V
(NOTE 6)
= 0V, V
+IN1
S
Delta Distribution
Distribution
0
0
= 5V
= V
1
0.1
0.1
+IN2
+IN1
+IN2 PIN VOLTAGE (V)
2
OS
= V2 = Mid-Supply, V
0.2
0.2
OS
(mV)
(mV)
6350 G02
6350 G01
T
T
T
3
OS
A
0.3
A
A
0.3
= 125°C
= –40°C
= 25°C
4
–0.1
–0.2
–0.3
–0.1
–0.2
–0.3
–0.4
–0.5
5
0.2
0.1
0.0
0.5
0.4
0.3
0.2
0.1
0
6350 G07
0
Offset Voltage vs Input Common
Mode Voltage, Op Amp 1
DC Linearity
0
V
TYPICAL UNIT
SHDN
S
6
0.5 1 1.5 2
= 5V
V
V
NO LOAD
TYPICAL UNIT
LINEAR FIT FOR 0.25V < V
INPUT COMMON MODE VOLTAGE (V)
+IN2
–
= 0V, V
1
= V
T
T
T
T
= 2.5V
A
A
A
A
= 125°C
= 85°C
= 25°C
= –40°C
+
+
, R
= 5V
2
V
F
+IN1
= SHORT, R
2.5 3 3.5 4 4.5
(V)
3
IN
< 4.75V
T
T
T
A
A
A
4
G
= 125°C
= 25°C
= –40°C
–0.2
–0.4
–0.6
0.6
0.4
0.2
= OPEN, R
6350 G10
6350 G21
0
–60
Differential V
5
5
–20
V1 = V2 = MID-RAIL
TYPICAL UNIT
L
= OPEN. See Figure 1.
TEMPERATURE (°C)
–0.1
–0.2
–0.3
–0.4
–0.5
–0.1
–0.2
–0.3
0.2
0.1
0.5
0.4
0.3
0.2
0.1
0
0
20
OS
–5
–6
Offset Voltage vs Input Common
Mode Voltage, Op Amp 1
DC Linearity
V
TYPICAL UNIT
+
vs Temperature
–5
–4
= 5V, V
INPUT COMMON MODE VOLTAGE (V)
–4 –3 –2 –1 0
60
–3 –2 –1
V
V
NO LOAD
TYPICAL UNIT
LINEAR FIT FOR
–4.75V < V
+IN2
–
–
V
V
S
= –5V, V
= –5V
S
= 10V
= 0V
= 5V
100
V
IN
+IN1
+
6350 G09
= 5V
< 4.75V
0
(V)
140
T
T
T
T
1
1
A
A
A
A
= 125°C
= 85°C
= 25°C
= –40°C
2
2
3
T
T
T
A
A
A
3
= 125°C
= 25°C
= –40°C
4
4
6350 G11
6350 G22
5
6350fb
5
6