LT1806CS6#PBF Linear Technology, LT1806CS6#PBF Datasheet - Page 12

• Manufacturer Part Number: LT1806CS6#PBF
• Description: IC, OP-AMP, 325MHZ, 125V/µs, TSOT-23-6
• Manufacturer: Linear Technology

Specifications of LT1806CS6#PBF

• Op Amp Type: Low Noise
• No. Of Amplifiers: 1
• Bandwidth: 325MHz
• Slew Rate: 125V/µs
• Supply Voltage Range: 2.5V To 12.6V
• Amplifier Case Style: TSOT-23
• No. Of Pins: 6
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

LT1806/LT1807
TYPICAL PERFORMANCE CHARACTERISTICS
12
–0.2
–0.4
–0.6
–1.0
–0.8
0.2
1.0
0.8
0.6
0.4
20
15
10
–1
–2
–3
–4
–5
–6
–7
–8
2
1
0
0
5
0
–50
1.0
0
Input Bias Current vs Temperature
Minimum Supply Voltage
Supply Current per Amp
vs Supply Voltage
T
1
–35
A
1.5
= 125°C
2
–20
TOTAL SUPPLY VOLTAGE (V)
TOTAL SUPPLY VOLTAGE (V)
T
2.0
A
3
= –55°C
TEMPERATURE (°C)
–5
4
T
2.5
A
NPN ACTIVE
PNP ACTIVE
V
V
= 25°C
5
V
V
S
S
T
10
CM
CM
A
= 5V, 0V
= 5V, 0V
3.0
= 125°C
6
= 5V
= 0V
25
7
3.5
T
T
A
40
A
8
= 25°C
= –55°C
4.0
9 10 11
55
4.5
18067 G04
70
18067 G07
18067 G10
12
85
5.0
0.001
–300
–400
–100
–200
–500
–100
0.01
500
400
300
200
–20
100
120
100
–40
–60
–80
0.1
80
60
40
20
10
0
1
0
0.01
1.5
0
Offset Voltage
vs Input Common Mode
Output Saturation Voltage
vs Load Current (Output Low)
Output Short-Circuit Current
vs Power Supply Voltage
V
T
V
TYPICAL PART
S
A
S
= 5V
= 25°C
= 5V, 0V
INPUT COMMON MODE VOLTAGE (V)
T
T
2.0
A
A
POWER SUPPLY VOLTAGE ( V)
= 125°C
= –55°C
T
1
A
0.1
= 25°C
2.5
LOAD CURRENT (mA)
T
T
T
A
T
A
A
A
= –55°C
2
= 125°C
= –55°C
3.0
= 25°C
“SOURCING”
“SINKING”
1
3.5
T
A
3
= 25°C
4.0
T
A
10
T
T
= 125°C
A
A
4
= –55°C
= 125°C
4.5
18067 G05
18067 G08
18067 G11
100
5.0
5
0.001
0.01
–10
0.1
18
16
14
12
10
–5
10
8
4
6
2
0
5
0
1
0.01
–1
0
Input Bias Current
vs Common Mode Voltage
Output Saturation Voltage
vs Load Current (Output High)
Supply Current
vs SHDN Pin Voltage
V
V
V
T
T
T
S
S
S
A
A
A
= 5V, 0V
= 5V
= 5V, 0V
= –55°C
= 125°C
= 25°C
0
COMMON MODE VOLTAGE (V)
T
1
A
T
0.1
A
T
SHDN PIN VOLTAGE (V)
= 125°C
A
LOAD CURRENT (mA)
= 25°C
1
= –55°C
2
2
1
T
A
T
A
= –55°C
3
T
= 25°C
3
A
T
A
= 25°C
= 125°C
4
10
T
T
A
A
4
= 125°C
= –55°C
5
18067 G12
18067 G06
18067 G09
18067fc
100
6
5