STK403-430 데이터 시트보기 (PDF) - SANYO -> Panasonic
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STK403-430 Datasheet PDF : 8 Pages
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STK403-430
Thermal Design Example
The heat sink thermal resistance, θc-a, required to handle the total power dissipated within this hybrid IC is determined as
follows.
Condition 1: The IC substrate temperature Tc must not exceed 125°C.
Pd × θc – a + Ta < 125°C ... (1)
Ta: Guaranteed ambient temperature for the end product.
Condition 2: The junction temperature of each individual transistor must not exceed 150°C.
Pd × θc – a + Pd/N × θj – c + Ta < 150°C ... (2)
N: Number of power transistors
θj-c: Thermal resistance per power transistor
We take the power dissipation in the power transistors to be Pd evenly distributed across those N power transistors.
If we solve for θc-a in equations (1) and (2), we get the following inequalities.
θc – a < (125 – Ta)/Pd ... (1)’
θc – a < (150 – Ta)/Pd – θj-c/N ... (2)’
Values that satisfy both these inequalities at the same time are the required heat sink thermal resistance values.
Determining the following specifications allows us to determine the required heat sink thermal resistance from
inequalities (1)’ and (2)’.
• Supply voltage: VCC
• Load resistance: RL
• Guaranteed ambient temperature: Ta
Example:
Assume that the IC supply voltage, VCC, is ±23 V, RL is 6 Ω, and that the signal is a continuous sine wave. In this case,
from the Pd – PO characteristics, the maximum power will be 103 W for a signal with a frequency of 1 kHz.
For actual music signals, it is usual to use a Pd of 1/8 of POmax, which is the power estimated for continuous signals in
this manner. (Note that depending on the particular safety standard used, a value somewhat different from the value of
1/8 used here may be used.)
That is:
Pd = 65 W (when 1/8 POmax is 2.5 W)
The number, N, of power transistors in the hybrid IC's audio amplifier block is 12. Since the thermal resistance, θc-a, per
transistor is 3.6°C/W, the required heat sink thermal resistance, θc-a, for a guaranteed ambient temperature of 50°C will
be as follows.
From inequality (1)’: θc – a < (125 – 50)/65
< 1.15
From inequality (2)’: θc – a < (150 – 50)/65 – 3.6/12
< 1.23
Therefore, the thermal resistance that satisfies both these expressions at the same time is 1.15°C/W.
Note that this thermal design example assumes the use of a constant-voltage power supply, and is only provided as an
example for reference purposes. Thermal designs must be tested in an actual end product.
No. 7374-5/8
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