Datasheet

LT6600-2.5
12
660025fe
APPLICATIONS INFORMATION
the V
–
pin to provide a heat sink, the thermal resistance
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
COPPER AREA
TOPSIDE
(mm
2
)
BACKSIDE
(mm
2
)
BOARD AREA
(mm
2
)
THERMAL RESISTANCE
(JUNCTION-TO-AMBIENT)
1100 1100 2500 65°C/W
330 330 2500 85°C/W
35 35 2500 95°C/W
35 0 2500 100°C/W
0 0 2500 105°C/W
Junction temperature, T
J
, is calculated from the ambient-
temperature, T
A
, and power dissipation, P
D
. The power
dissipation is the product of supply voltage, V
S
, and
supply current, I
S
. Therefore, the junction temperature
is given by:
T
J
= T
A
+ (P
D
• θ
JA
) = T
A
+ (V
S
• I
S
• θ
JA
)
where the supply current, I
S
, is a function of signal level, load
impedance, temperature and common mode voltages.
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
J
. For example, using the circuit in Figure 3
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
2
copper trace, the θ
JA
is 100°C/W. The resulting
junction temperature is:
T
J
= T
A
+ (P
D
• θ
JA
) = 85 + (5 • 0.0376 • 100) = 104°C
When using higher supply voltages or when driving small
impedances, more copper may be necessary to keep T
J
below 150°C.