LT6600IS8-2.5#TRPBF Linear Technology, LT6600IS8-2.5#TRPBF Datasheet - Page 12

IC AMP DIFF LP FLTR 2.5MHZ 8SOIC

LT6600IS8-2.5#TRPBF

Manufacturer Part Number
LT6600IS8-2.5#TRPBF
Description
IC AMP DIFF LP FLTR 2.5MHZ 8SOIC
Manufacturer
Linear Technology
Datasheet

Specifications of LT6600IS8-2.5#TRPBF

Amplifier Type
Differential
Number Of Circuits
1
Output Type
Differential
Current - Input Bias
15µA
Voltage - Input Offset
5000µV
Current - Supply
28mA
Voltage - Supply, Single/dual (±)
3 V ~ 11 V, ±1.5 V ~ 5.5 V
Operating Temperature
-40°C ~ 85°C
Mounting Type
Surface Mount
Package / Case
8-SOIC (3.9mm Width)
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Current - Output / Channel
-
-3db Bandwidth
-
Slew Rate
-
Gain Bandwidth Product
-

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10 000
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APPLICATIONS INFORMATION
LT6600-2.5
the V
will be around 105°C/W. Table 3 can be used as a guide
when considering thermal resistance.
Table 3. LT6600-2.5 SO-8 Package Thermal Resistance
Junction temperature, T
temperature, T
dissipation is the product of supply voltage, V
supply current, I
is given by:
where the supply current, I
impedance, temperature and common mode voltages.
12
TOPSIDE
(mm
T
1100
330
J
35
35
COPPER AREA
0
= T
2
)
pin to provide a heat sink, the thermal resistance
A
+ (P
BACKSIDE
(mm
1100
330
35
D
0
0
A
, and power dissipation, P
2
• θ
)
S
. Therefore, the junction temperature
JA
) = T
BOARD AREA
J
(mm
2500
2500
2500
2500
2500
, is calculated from the ambient-
S
A
, is a function of signal level, load
+ (V
2
)
S
• I
(JUNCTION-TO-AMBIENT)
THERMAL RESISTANCE
S
• θ
JA
100°C/W
105°C/W
65°C/W
85°C/W
95°C/W
)
D
. The power
S
, and
For a given supply voltage, the worst-case power dissipation
occurs when the differential input signal is maximum, the
common mode currents are maximum (see Applications
Information regarding Common Mode DC Currents), the
load impedance is small and the ambient temperature is
maximum. To compute the junction temperature, measure
the supply current under these worst-case conditions, es-
timate the thermal resistance from Table 2, then apply the
equation for T
with DC differential input voltage of 1V, a differential
output voltage of 4V, no load resistance and an ambient
temperature of 85°C, the supply current (current into V
measures 37.6mA. Assuming a PC board layout with a
35mm
junction temperature is:
When using higher supply voltages or when driving small
impedances, more copper may be necessary to keep T
below 150°C.
T
J
= T
2
copper trace, the θ
A
+ (P
J
D
. For example, using the circuit in Figure 3
• θ
JA
) = 85 + (5 • 0.0376 • 100) = 104°C
JA
is 100°C/W. The resulting
660025fe
+
)
J

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