Heat sink for tip36

Discussion in 'General Electronics Chat' started by nikhilsonar29, Feb 18, 2015.

  1. nikhilsonar29

    Thread Starter Member

    Sep 26, 2014
    68
    1
    hi,
    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.
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,347
    Hello,

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

    Bertus
     
  3. nikhilsonar29

    Thread Starter Member

    Sep 26, 2014
    68
    1
    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
    Nikhil
     
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,291
    1,255
    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.
     
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  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Or a fan...
     
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  6. nikhilsonar29

    Thread Starter Member

    Sep 26, 2014
    68
    1
    Thank you sir
    i will consider this
     
  7. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    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
     
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