TS272IDT STMicroelectronics, TS272IDT Datasheet - Page 6

• Manufacturer Part Number: TS272IDT
• Description: IC AMP HI PERFORM DUAL OP 8 SOIC
• Manufacturer: STMicroelectronics

Specifications of TS272IDT

• Amplifier Type: General Purpose
• Number Of Circuits: 2
• Slew Rate: 5.5 V/µs
• Gain Bandwidth Product: 3.5MHz
• Current - Input Bias: 1pA
• Voltage - Input Offset: 1100µV
• Current - Supply: 1mA
• Current - Output / Channel: 45mA
• Voltage - Supply, Single/dual (±): 3 V ~ 16 V, ±1.5 V ~ 8 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (3.9mm Width)
• Number Of Channels: 2
• Voltage Gain Db: 83.52 dB
• Common Mode Rejection Ratio (min): 65 dB
• Input Offset Voltage: 10 mV
• Operating Supply Voltage: 5 V, 9 V, 12 V, 15 V
• Supply Current: 3 mA
• Maximum Operating Temperature: + 125 C
• Mounting Style: SMD/SMT
• Maximum Dual Supply Voltage: +/- 8 V
• Minimum Operating Temperature: - 40 C
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Output Type: -
• -3db Bandwidth: -
• Lead Free Status / Rohs Status: Details
• Other names: 497-2247-2

TS272C,I,M
Figure 5 : Open Loop Frequency Response and
Figure 6 : Gain Bandwidth Product versus Supply
Figure 7 : Phase Margin versus Supply Voltage
6/9
-1 0
5 0
4 0
3 0
2 0
1 0
4
3
5
2
0
48
44
40
36
32
28
1
0
1 0
2
Phase Shift
Voltage
0
T
V
R
C
A
a m b
C C
L
L
V C L
T
R = 10k
C = 1 0 0 p F
A = 1
+
= 10k
= 1 0 0 p F
1 0
a m b
L
L
V
S U P P L Y V O L T A G E , V
= 2 5 °C
S U P P L Y V O L T A G E , V
= 1 0 V
P H A S E
= 1 0 0
3
F R E Q U E N C Y , f (H z )
= 2 5 °C
4
4
1 0
4
G A IN
Gain
Bandwidth
Product
8
1 0
8
5
T
R = 10k
C = 1 0 0 p F
A = 1
a m b
V
L
L
1 0
Phase
Margin
6
1 2
= 2 5 °C
1 2
C C
C C
1 0
(V )
(V )
7
0
4 5
9 0
1 3 5
1 8 0
1 6
1 6
Figure 8 : Phase Margin versus Capacitive Load
Figure 9 : Slew Rate versus Supply Voltage
Figure 10 : Input Voltage Noise versus
7
6
4
3
5
2
7 0
6 0
5 0
4 0
3 0
3 0 0
2 0 0
1 0 0
4
0
0
Frequency
1
T
R = 10k
C = 1 0 0 p F
a m b
S U P P L Y V O L T A G E , V
L
L
6
C A P A C IT A N C E , C
2 0
= 2 5 °C
F R E Q U E N C Y (H z )
8
1 0
4 0
1 0
S R
6 0
V
T
T
R = 10k
A = 1
V
R
a m b
amb
C C
V
L
C C
1 0 0
S
1 2
L
C C
= 1 0 0
= 2 5 °C
= 1 0 V
(p F )
= 2 5 °C
= 1 0 V
8 0
S R
(V )
1 4
1 0 0 0
1 0 0
1 6