Default: 20W power transistor with T JMAX = 155oC, cooled by ‘top hat’ heat sink

Question

Default: 20W power transistor with TJMAX = 155oC, cooled by ‘top hat’ heat sink with CA = 9.5oC/W. Default ambient temperature TA = 25oC.

An ideal class-B output stage operates from power supply rails of ±12V. Assuming the default

power transistors and heat sinks, and assuming worst-case power dissipation determine:

(a) Smallest RL that the circuit can drive without exceeding the TJMAX limit of the transistors.

(b) If a junction safety factor TJ = 20oC is imposed, what is the PD(worst) for each transistor, VL (worst case) and the maximum allowed output voltage VL?

(c) For the VL (max)identified by part (b) determine the IS drawn from the power rails and case temperature TC of each transistor.

Explanation / Answer

a)The power dissipated within a semiconductor device be a consequence of Joule heating and will fall along its internal conducting path.

RL=V/i=20/12

=1.66 ohm

b)

The temperature difference is expected to be proportional to the dissipated power, for which we might expect

TJ - TA =JC PD

where PD = power dissipated in the device.

JC = 150 - 25/ 20W=6.25 C/W

The total thermal resistance is then JA = JC+CA = 6.25+9.5 =15.75

The maximum power that can be dissipated is then

PD =(TJmax - delta TJ -TA )/ JA =(155-25-20)/15.75 =6.98W

c)Vl =SQRT (2pd*rL)=7.64V

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.