Datasheet
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Board thickness 62 mils(0.15748cm)
Boardsize 3.2in.x3.2in.
Boardmaterial FR4
Coppertrace/heatsink 1 oz
Exposedpadmounting 63/67tin/lead(Sn/Pb)solder
Heatsink Area
1ozCopper
P =
D(Max)
T T
R
-
J(Max) A
JA (System)q
(5)
P =(V V ) I +V I-
D(total) IN OUT OUT IN Q
´ ´
(6)
P =(V V ) I-
D(total) IN OUT OUT
´
(7)
P =
D(Max)
T T
R
-
J(Max) A
JA (System)q
= =1.4W
125 -
°C °
°
C
C/W
55
50
(8)
P =(V V ) I-
D(total) IN OUT OUT
´ =(5 3.3) 0.8=1.36W- ´
(9)
TPS752xxQ
TPS754xxQ
SLVS242C – MARCH 2000 – REVISED OCTOBER 2007
Figure 29 is an example of a thermally-enhanced PWB layout for use with the new PWP package. This board
configuration was used in the thermal experiments that generated the power ratings shown in Figure 27 and
Figure 28 . As discussed earlier, copper has been added on the PWB to conduct heat away from the device. R
θ JA
for this assembly is illustrated in Figure 27 as a function of heatsink area. A family of curves is included to
illustrate the effect of airflow introduced into the system.
Figure 29. PWB Layout (Including Copper Heatsink Area) for Thermally-Enhanced PWP Package
From Figure 27 , R
θ JA
for a PWB assembly can be determined and used to calculate the maximum
power-dissipation limit for the component/PWB assembly, with the equation:
Where T
Jmax
is the maximum specified junction temperature (+150 ° C absolute maximum limit, +125 ° C
recommended operating limit) and T
A
is the ambient temperature.
P
D(max)
should then be applied to the internal power dissipated by the TPS75433QPWP regulator. The equation
for calculating total internal power dissipation of the TPS75433QPWP is:
Because the quiescent current of the TPS75433QPWP is very low, the second term is negligible, further
simplifying the equation to:
For the case where T
A
= +55 ° C, airflow = 200 ft/min, copper heat-sink area = 4 cm
2
, the maximum
power-dissipation limit can be calculated. First, from Figure 27 , we find the system R
θ JA
is 50 ° C/W; therefore, the
maximum power-dissipation limit is:
If the system implements a TPS75433QPWP regulator, where V
IN
= 5 V and I
OUT
= 800 mA, the internal power
dissipation is:
Comparing P
D(total)
with P
D(max)
reveals that the power dissipation in this example does not exceed the calculated
limit. When it does, one of two corrective actions should be made: either raise the power-dissipation limit by
increasing the airflow or the heat-sink area, or loweri the internal power dissipation of the regulator by reducing
the input voltage or the load current. In either case, the above calculations should be repeated with the new
system parameters.
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