STK400-490 Sanyo Semiconductor Corporation, STK400-490 Datasheet - Page 5

• Manufacturer Part Number: STK400-490
• Manufacturer: Sanyo Semiconductor Corporation

Heatsink Design Considerations
The heatsink thermal resistance, c-a, required to dissipate
the STK400-490 device total power dissipation, Pd, is de-
termined as follows :
Condition 1: IC substrate temperature not to exceed 125 C
Where Ta is the guaranteed maximum ambient tempera-
ture, Pd (total) is the total power dissipation, Pd (L) is the
left-channel power dissipation, Pd (R) is the right-channel
power dissipation and Pd (C) is the center-channel power
dissipation.
Condition 2: Power transistor junction temperature, Tj, not
where N is the left and right-channel number of power tran-
sistors, N’ is the center-channel number of power transis-
tors, j-c is the left and right-channel power transistor ther-
mal resistance per transistor, and j-c’ is the center-chan-
nel power transistor thermal resistance per transisttor. Note
that the power dissipated per transistor is the total, Pd,
devided evenly among the N power transistors.
Expressions (1), (2) and (3) can be rewritten making c-a
the subject.
The heatsink required must have a thermal resistance that
simultaneously satisfies all three expressions.
The heatsink thermal resistance can be determined from
(1)’, (2)’ and (3)’ once the following parameters have been
defined.
• Supply voltage : V
• Load resistance : R
• Guaranteed maximum ambient temperature : Ta
The total device power dissipation when STK400-490
V
tinuous sine wave signal, is a maximum of 42.5W
(left+righ channels) and 34.3W (center channel), as
shown in the “Pd–P
CC
Pd (total)
Pd (total)=Pd (L)+Pd (R)+Pd (C)
Pd (total)
Pd (total)
c-a< (125–Ta)/Pd (total) ................................. (1)’
c-a< (150–Ta)/Pd (total)–[Pd (L)+Pd (R)]
c-a< (150–Ta)/Pd (total)–Pd (C)
j-c’/[Pd (total) N’] ......................................... (3)’
(1)= 25V, V
j-c+Ta<150 C ................................................. (2)
j-c’+Ta<150 C ................................................ (3)
j-c/[Pd (total) N] .......................................... (2)’
to exceed 150 C
c-a+Ta<125 C ................................. (1)
c-a+[Pd (L)+Pd (R)]/N
c-a+Pd (C)/N’
CC
O
CC
L
(2)= 32V and R
” characteristics graphs.
L
=6 , for a con-
STK400-490
When estimating the power dissipation for an actual audio
signal input, the rule of thumb is to select Pd correspond-
ing to (1/10) P
sine wave input. For example,
The STK400-490 has 4 left + right-channel power transis-
tors (N), 2 center-channel power transistors (N’), left + right-
channel thermal resistance per transistor
and center-channel thermal resistance per transistor
is 1.7 C/W. If the guaranteed maximum ambient tempera-
ture, Ta, is 50 C, then the required heatsink thermal resis-
tance, c-a, is :
From expression (1)’ :
From expression (2)’ :
From expression (3)’ :
Therefore, to satisfy all three expressions, the required
heatsink must have a thermal resistance less than 1.6 C/W.
Similarly, when STK400-490 V
(2)= 26V and R
From expression (1)’ :
From expression (2)’ :
From expression (3)’ :
Therefore, to satisfy all three expressions, the required
heatsink must have a thermal resistance less than 1.35 C/
W.
This heatsink design example is based on a constant-volt-
age supply, and should be verified within your specific set
environment.
Pd (L)+Pd (R)=25W [for (1/10)
Pd (C)=21.6W [for (1/10)
Pd (total)=Pd (L)+Pd (R)+Pd (C)=46.6W
Pd (L)+Pd (R)=30W [for (1/10)
Pd (C)=25.5W [for (1/10)
Pd (total)=Pd (L)+Pd (R)+Pd (C)=55.5W
c-a < (125–50)/55.5
c-a < (150–50)/55.5–30
c-a < (150–50)/55.5–25.5
c-a < (125–50)/46.6
c-a < (150–50)/46.6–25 2.1/(46.6
c-a < (150–50)/46.6–21.6 1.7/(46.6
< 1.35
< 1.51
< 1.41
< 1.86
< 1.75
< 1.60
O
L
=3 ,
max (within safe limits) for a continuous
P
P
/(55.5 4)
O
O
/(55.5 2)
max=5W]
max=5W]
CC
P
P
O
O
(1)= 21V, V
max=2.5W]
max=2.5W]
j-c) is 2.1 C/W,
No.5247–5/8
j-c’)
CC