STK433-040N-E

No.A2101-9/11

A Thermal Design Tip For STK433-040N-E Amplifier

[Thermal Design Conditions]

The thermal resistance (θc-a) of the heat-sink which manages the heat dissipation inside the Hybrid IC will be

determined as follow:

(Condition 1) The case temperature (Tc) of the Hybrid IC should not exceed 125°C

Pd

 c-a + Ta  125°C··································································· (1)

Where Ta : the ambient temperature for the system

(Condition 2) The junction temperature of each power transistor should not exceed 150°C

Pd

 c-a + Pd/N  j-c + Ta  150°C·················································· (2)

Where N : the number of transistors (two for 1 channel , ten for channel)

θj-c : the thermal resistance of each transistor (see specification)

Note that the power consumption of each power transistor is assumed to be equal to the total power dissipation (Pd)

divided by the number of transistors (N).

From the formula (1) and (2), we will obtain:

c-a  (125  Ta)/Pd ······································································ (1)’

c-a  (150  Ta)/Pd  j-c/N ··························································· (2)’

The value which satisfies above formula (1)’ and (2)’ will be the thermal resistance for a desired heat-sink.

Note that all of the component except power transistors employed in the Hybrid IC comply with above conditions.

[Example of Thermal Design]

Generally, the power consumption of actual music signals are being estimated by the continuous signal of

1/8 PO max. (Note that the value of 1/8 PO max may be varied from the country to country.)

(Sample of STK433-040N-E ; 25W×2ch)

If VCC is ±24V, and RL is 6, then the total power dissipation (Pd) of inside Hybrid IC is as follow;

Pd = 26W (at 3.13W output power,1/8 of PO max)

There are four (4) transistors in Audio Section of this Hybrid IC, and thermal resistance (θj-c) of each transistor is

4.2°C/W. If the ambient temperature (Ta) is guaranteed for 50°C, then the thermal resistance (θc-a) of a desired heat-

sink should be;

From (1)’

c-a  (125  50)/26

 2.88

From (2)’

c-a  (150  50)/26  4.2/4

 2.79

Therefore, in order to satisfy both (1)’ and (2)’, the thermal resistance of a desired Heat-sink will be 2.79°C/W.

[Note]

Above are reference only. The samples are operated with a constant power supply. Please verify the conditions when

your system is actually implemented.