Philips Semiconductors
TDA1562Q application note
Koninklijke Philips Electronics N.V.
Application note Rev. 01.02 — 05 May 2006 36 of 62
Rth hs-a = = = 5.8 K/W
For the class AB amplifier the calculation is as follows:
Rth hs-a = = = 4.25 K/W
There is a difference of 1.65 K/W between the heatsink the TDA1562Q needs and the
heatsink the class AB amplifier needs.
Although 1.65 K/W does not seem to be a big difference, it usually means a considerable
increase in heatsink size, as is illustrated in figure 21.
Figure 21 shows a standard heatsink (Fischer Elektronik SK178) which can be used for
cooling an amplifier. To achieve a thermal resistance of 5.8K/W the heatsink needs to
have a length of approximately 50mm. To achieve a thermal resistance of 4.25K/W the
length of the heatsink should be approximately 100mm, so the class AB amplifier would
need a heatsink twice the size of the heatsink of the TDA1562.
5.8K/W : l=50mm , 4.25K/W : l=100mm
Fig 21. Example of a heatsink
This example shows that the TDA1562Q can be used with a considerably smaller
heatsink than a comparable class AB amplifier.
hsc
ac
Rth
T
−
−
−
0.1
65120
−
cjahs
aj
RthRth
T
−−
−
−−
05−10−
517
.