Datasheet

LM4755
SNAS010E –FEBRUARY 1999–REVISED APRIL 2013 www.ti.com
HEATSINK DESIGN EXAMPLE
Determine the system parameters:
V
S
= 24V Operating Supply Voltage
R
L
= 4Ω Minimum Load Impedance
T
A
= 55°C Worst Case Ambient Temperature
Device parameters from the datasheet:
T
J
= 150°C Maximum Junction Temperature
θ
JC
= 2°C/W Junction-to-Case Thermal Resistance
Calculations:
2 • P
DMAX
= 2 • [V
S
2
/ 2 • π
2
• R
L
)] = (24V)
2
/ (2 • π
2
• 4Ω) = 14.6W
θ
SA
≤ [(T
J
-T
A
) / P
DMAX
] - θ
JC
–θ
CS
= [ (150°C - 55°C) / 14.6W] - 2°C/W–0.2°C/W = 4.3°C/W
Conclusion: Choose a heatsink with θ
SA
≤ 4.3°C/W.
DDPAK HEATSINK DESIGN EXAMPLES
Example 1:(Stereo Single-Ended Output)
Given: T
A
=30°C
T
J
=150°C
R
L
=4Ω
V
S
=12V
θ
JC
=2°C/W
P
DMAX
from P
D
vs P
O
Graph:
P
DMAX
≈ 3.7W (4)
Calculating P
DMAX
:
P
DMAX
= V
CC
2
/(π
2
R
L
) = (12V)
2
/π
2
(4Ω)) = 3.65W (5)
Calculating Heatsink Thermal Resistance:
θ
SA
< T
J
− T
A
/ P
DMAX
− θ
JC
− θ
CS
(6)
θ
SA
< 120°C/3.7W − 2.0°C/W − 0.2°C/W = 30.2°C/W (7)
Therefore the recommendation is to use 1.5 × 1.5 square inch of single-sided copper clad.
Example 2:(Stereo Single-Ended Output)
Given: T
A
=50°C
T
J
=150°C
R
L
=4Ω
V
S
=12V
θ
JC
=2°C/W
P
DMAX
from P
D
vs P
O
Graph:
P
DMAX
≈ 3.7W (8)
Calculating P
DMAX
:
P
DMAX
= V
CC
2
/(π
2
R
L
)= (12V)
2
/(π
2
(4Ω)) = 3.65W (9)
Calculating Heatsink Thermal Resistance:
θ
SA
< [(T
J
− T
A
) / P
DMAX
] − θ
JC
− θ
CS
(10)
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