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
(mm2)
BACKSIDE
(mm2)
BOARD AREA
(mm2)
THERMAL RESISTANCE
(JUNCTION-TO-AMBIENT)
1100
2500
65°C/W
330
2500
85°C/W
35
2500
95°C/W
35
0
2500
100°C/W
0
2500
105°C/W
Junction temperature, TJ, is calculated from the ambient-
temperature, TA, and power dissipation, PD. The power
dissipation is the product of supply voltage, VS, and
supply current, IS. Therefore, the junction temperature
is given by:
TJ = TA + (PD θJA) = TA + (VS IS θJA)
where the supply current, IS,isafunctionofsignallevel,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 TJ. 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
35mm2 copper trace, the θJA is 100°C/W. The resulting
junction temperature is:
TJ = TA + (PD θJA) = 85 + (5 0.0376 100) = 104°C
When using higher supply voltages or when driving small
impedances, more copper may be necessary to keep TJ
below 150°C.