LM4752

SNAS006E – FEBRUARY 1999 – REVISED APRIL 2013

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DDPAK HEATSINK DESIGN EXAMPLES

Example 1: (Stereo Single-Ended Output)

Given: TA=30°C

TJ=150°C

RL=4Ω

VS=12V

θJC=2°C/W

PDMAX from PD vs PO Graph:

PDMAX ≈ 3.7W

(10)

Calculating PDMAX:

PDMAX = VCC

2 / (π2R

L) = (12V)

2 / π2(4Ω)) = 3.65W

(11)

Calculating Heatsink Thermal Resistance:

θSA < [(T J − TA) / PDMAX] − θJC − θCS

(12)

θSA < 120°C / 3.7W − 2.0°C/W − 0.2°C/W = 30.2°C/W

(13)

Therefore the recommendation is to use 1.5 × 1.5 square inch of single-sided copper clad.

Example 2: (Stereo Single-Ended Output)

Given: TA=50°C

TJ=150°C

RL=4Ω

VS=12V

θJC=2°C/W

PDMAX from PD vs PO Graph:

PDMAX ≈ 3.7W

(14)

Calculating PDMAX:

PDMAX = VCC

2 / (π2R

L) = (12V)

2 / (π2(4Ω)) = 3.65W

(15)

Calculating Heatsink Thermal Resistance:

θSA < [(TJ − TA) / PDMAX] − θJC − θCS

(16)

θSA < 100°C / 3.7W − 2.0°C/W − 0.2°C/W = 24.8°C/W

(17)

Therefore the recommendation is to use 2.0 × 2.0 square inch of single-sided copper clad.

Example 3: (Bridged Output)

Given: TA=50°C

TJ=150°C

RL=8Ω

VS=12V

θJC=2°C/W

Calculating PDMAX:

PDMAX = 4[VCC

2 / (2π2R

L)] = 4(12V)

2 / (2π2(8Ω)) = 3.65W

(18)

Calculating Heatsink Thermal Resistance:

θSA < [(TJ − TA) / PDMAX] − θJC − θCS

(19)

θSA < 100°C / 3.7W − 2.0°C/W − 0.2°C/W = 24.8°C/W

(20)

Therefore the recommendation is to use 2.0 × 2.0 square inch of single-sided copper clad.

20

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