Heat sink for tip36

Thread Starter


Joined Sep 26, 2014
good afternoon all
anyone can plz tell me what type of heat sink should I use to ic tip36 in the circuit
actually there are two rows of ic tip 36 in line. having 5 in each row.
Plz find the attachments for the circuit diagram.



Joined Apr 5, 2008

What will be the power dissipated in the transistors?
The power will determine the size of the heatsink.


Thread Starter


Joined Sep 26, 2014
hello sir,
this circuit is for 3kva inverter circuit .
3kva is the power dissipation,
each transistor is capable of handling 25 amp current
Thanks And Regards


Joined Nov 12, 2008
Maybe a rough idea, but not exact.
3000kva @ 220 volts = 13.6 amps.
If it is 100% efficient (it's not) it would take 4.6 time as much from 48 volts = 62 amps.
The voltage drop across the transistor when it is on is 1.8 volts, so 112 watts in the transistors.
Each set is on half the time so 56 watts in each set or about 11 watts each.
If each set was mounted to the same heat sink and that one had a thermal resistance of 1 degree C per watt the heat sink would be at about 81C.
A 1 degree C per watt heat sink needs about 2500 square cm of surface area. So a big one with lots of fins.


Joined Aug 23, 2012
The rating current of Tip36 is 25A, normally we just count it only I=25A*(1/5)A=5A for the power dissipation reason.

You need to in series with a 0.2Ω/20W in the c for each Tip36, that is to protecting the Tip36.

Eff = Efficiency.
Wo = Output Power.
Wi = Input Power.
Vi = Input Voltage
Ai = Input Current
P80 = 80%
P90 = 90%

Wo = Wi * Eff% = Vi * Ai * Eff% = 220V*Ai*Eff% = 3KW
Wo = 3kW/52V = 57.7A

If Converting Efficiency = 80%
Wi_P80 = Wo/Eff% = 3kW/80% = 3750W
Ai_P80 = Wi/Vi = 3750W/220V = 17A

If Converting Efficiency = 90%
Wi_P90 = Wo/Eff% = 3kW/90% = 3333W
Ai_P90 = Wi/Vi = 3333W/220V = 15.15A