LM4755 datasheet(17/23 Pages) NSC

17 / 23 page

Application Information (Continued)

1.5 x 1.5 in. sq.

20–27˚C/W

A=28˚C, Sine wave

testing, 1 oz. Copper)

2 x 2 in. sq.

16–23˚C/W

The above values for

SA vary widely due to dimensional

proportions (i.e. variations in width and length will vary

SA).

For audio applications, where peak power levels are short in

duration, this part will perform satisfactory with less

heatsinking/copper clad area. As with any high power design

proper bench testing should be undertaken to assure the

design can dissipate the required power. Proper bench test-

ing requires attention to worst case ambient temperature

and air flow. At high power dissipation levels the part will

show a tendency to increase saturation voltages, thus limit-

ing the undistorted power levels.

DETERMINING MAXIMUM POWER DISSIPATION

For a single-ended class AB power amplifier, the theoretical

maximum power dissipation point is a function of the supply

voltage, V

S, and the load resistance, RL and is given by the

following equation:

(single channel)

DMAX (W)=[VS

2 /(2 •

π2 • R

L)]

The above equation is for a single channel class-AB power

amplifier. For dual amplifiers such as the LM4755, the equa-

tion for calculating the total maximum power dissipated is:

(dual channel)

DMAX (W)=2 • [VS

2 /(2 •

π2 • R

L)]

2 /(

π2 • R

(Bridged Outputs)

DMAX (W) = 4[VS

2 /(2

π2 • R

L)]

HEATSINK DESIGN EXAMPLE

Determine the system parameters:

S = 24V

Operating Supply Voltage

L =4

Ω

Minimum Load Impedance

A = 55˚C

Worst Case Ambient Temperature

Device parameters from the datasheet:

J = 150˚C

Maximum Junction Temperature

JC = 2˚C/W

Junction-to-Case Thermal Resistance

Calculations:

• P

DMAX =2 • [VS

2 /2 •

π2 • R

L)] = (24V)

2 /(2 •

π2 • 4Ω)

= 14.6W

≤ [(T

J-TA)/PDMAX]-

JC–

CS = [ (150˚C - 55˚C) / 14.6W]

- 2˚C/W–0.2˚C/W = 4.3˚C/W

Conclusion: Choose a heatsink with

≤ 4.3˚C/W.

TO-263 HEATSINK DESIGN EXAMPLES

Example 1: (Stereo Single-Ended Output)

Given:

A=30˚C

J=150˚C

L=4

Ω

S=12V

JC=2˚C/W

DMAX from PD vs PO Graph:

DMAX

≈ 3.7W

Calculating P

DMAX:

DMAX =VCC

2/(

π2R

L) = (12V)

π2(4Ω)) = 3.65W

Calculating Heatsink Thermal Resistance:

SA < TJ −TA /PDMAX −

JC −

SA < 120˚C/3.7W − 2.0˚C/W − 0.2˚C/W = 30.2˚C/W

Therefore the recommendation is to use 1.5 x 1.5 square

inch of single-sided copper clad.

Example 2: (Stereo Single-Ended Output)

Given:

A=50˚C

J=150˚C

L=4

Ω

S=12V

JC=2˚C/W

DMAX from PD vs PO Graph:

DMAX

≈ 3.7W

Calculating P

DMAX:

DMAX =VCC

2/(

π2R

L)= (12V)

2/(

π2(4Ω)) = 3.65W

Calculating Heatsink Thermal Resistance:

SA < [(TJ −TA)/PDMAX]−

JC −

SA < 100˚C/3.7W − 2.0˚C/W − 0.2˚C/W = 24.8˚C/W

Therefore the recommendation is to use 2.0 x 2.0 square

inch of single-sided copper clad.

Example 3: (Bridged Output)

Given:

A=50˚C

J=150˚C

L=8

Ω

S=12V

JC=2˚C/W

Calculating P

DMAX:

DMAX = 4[VCC

2/(2

π2R

L)] = 4(12V)

2/(2

π2(8Ω)) = 3.65W

Calculating Heatsink Thermal Resistance:

SA < [(TJ −TA)/PDMAX]−

JC −

SA < 100˚C / 3.7W − 2.0˚C/W − 0.2˚C/W = 24.8˚C/W

Therefore the recommendation is to use 2.0 x 2.0 square

inch of single-sided copper clad.

LAYOUT AND GROUND RETURNS

Proper PC board layout is essential for good circuit perfor-

mance. When laying out a PC board for an audio power

amplifier, particular attention must be paid to the routing of

the output signal ground returns relative to the input signal

and bias capacitor grounds. To prevent any ground loops,

the ground returns for the output signals should be routed

separately and brought together at the supply ground. The

input signal grounds and the bias capacitor ground line

should also be routed separately. The 0.1 µF high frequency

supply bypass capacitor should be placed as close as pos-

sible to the IC.

LM4755 Datasheet(PDF) 17 Page - National Semiconductor (TI)

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