5
\$\begingroup\$

I am brand new to heat calculations and trying to determine if I need a heat sink for the BJTs or not. Below is the circuit with voltages and current with the data sheet of the used BJTs. The maximum ambient temperature will be 25C.

enter image description here

enter image description here

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ What is the BJT you’re using here? \$\endgroup\$
    – Leoman12
    Commented Apr 22, 2020 at 2:51

2 Answers 2

6
\$\begingroup\$

Heatsinks are generally for power transistors or other power devices. But adding a heat sink Will help to keep the device cool. In your case, Q1 is dissipating only about 67mW which is safely below the max power rating of 625mW and no heatsinl required. The second transistor is dissipating about 200mW. This too is more than half lower than max power rating which is good an wont require heatsink. Note though that these are DC power though and the overall power (DC +AC) must be lower than max power. Take a look at peak power of Q2 when ac signal is applied at input of your circuit. It should be well below Pmax.

To determine if heatsink is require you need to check if Junction temperature of transistor is below a max of 150C . At this temperature the decide will be destroyed. Though a better temperature is 100C. If at your power dissipation the junction temperature reaches 100C then you require a heatsink. The formula for junction temperature is shown below.

enter image description here

\$\endgroup\$
3
\$\begingroup\$

Q2 is dissipating 350 milliWatts. Do you feel lucky?

WRONG !! The VDD is only 9 volts, not 12 volts.

Thus the power is somewhat lower. Thanks for catching that error on my part.

If you blow air on the leads, you may be OK.

To remove heat, the 3 leads are your friends. Particularly the collector lead can be the primary path, with the silicon die sitting on a small porch atop the collector lead, inside the plastic. So have 1cm by 1cm piece of metal, close to the collector-plastic interface.

Can much heat exit the base and emitter leads? No. Because of the gold bondwires hidden inside the plastic. Gold is a good heat conductor, but when used in 25 micron diameters (to save on the cost of gold), or about 1 mil (or 0.001 inch diameter), the thermal resistance of just the bondwire is about 10,000 degree Centigrade per watt. However, the transistor die atop collector mounting "porch" will dump heat into the plastic, and the plastic will dump heat into the base and emitter leads; plastic has Rthermal about 200X that of metal, but there are several cubic millimeters of plastic to offer heat transfer paths.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ How did you calculate Q2 to disspate 350mW? I thought the power dissipated across a BJT = Ic * Vce. Which would be 200mW. Im brand new to this and trying to learn thank you. \$\endgroup\$ Commented Apr 22, 2020 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.