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High Current Transistor TIP36 - Datasheet, Application Note

If you are looking for a power transistor able to support high current upto 25 amps and yet not incorporate the traditional cumbersome TO-3 package, then the TIP36 would certainly impress you.

High Current from Smaller Package


The package of TIP36 is TO-3P which means the accompanied heatsink would require just a single hole to be drilled and the device could be easily soldered over a PCB along with the heatsink, quite unlike the devices which are in TO-3 packages.

You would be surprised to know that this device is a lot advanced and powerful than the more popular MJ2955 (complementary pair of 2N3055) in terms of voltage and current ratings. Comparatively a TIP36 exhibits excellent high gain performance even at low or compressed saturation base voltages.



Let's proceed and learn the datasheet and the specifications of this outstanding high current transistor - the TIP36:

Typical Applications:


Audio Amplifier

Inverters

Motor Control

Solar Chargers.

Absolute Maximum Ratings:


The device will not tolerate anything above the following parameter magnitudes:

Collector-base Voltage (Vcbo) = 100 volts

Collector Emitter Voltage (Vceo) = 100 volts

Emitter Base Voltage (Vebo) = 5 volts

Collector Current (Ic) = 25 Amp continuous, peak 50 Amp for 5 ms only.

Base Current (Ib) = 5 Amps

Max Operating Junction Temperature = 150 degrees Celsius

Complementary Pair


TIP36 can be paired with TIP35, they both are ideally suited as complementary pairs.

Application Note:


One interesting and very useful application circuit using a couple of TIP36 transistor can be made for charging high current batteries in the order of 150 AH, directly from solar panels.



 

The circuit shown above may be used for charging 12 batteries having in excess of 100 AH capacities.

The circuit may be understood as follows:

How TIP36 Works to Produce High Current


The voltage from the solar panel enters the IC7812 first, the three diodes connected at the GND of the IC elevates the output voltage to about 14.26V which is the ideal voltage for charging 12V batteries.

However at this point the TIP36 stays inactive, because of the absence of any voltage across its base/emitter 1 Ohm resistor.

Once the output of the IC 7812 is loaded, sufficient voltage develops across the 1 Ohm resistor for saturating the TIP36 transistor.

The transistor conducts and shifts the required amount of current at the output.

However, in the course the transistor also tries to transfer the entire solar panel voltage to the output. As this tends to happen and the instant voltage at the output tends to go beyond the 14.26 mark, the IC7812 becomes reverse biased and stops conducting.

With 7812 not conducting, voltage across 1 Ohm resistor becomes zero and the transistor inhibits the flow of the voltage.
Due to the above action, the voltage tends to fall back below the 14.26 mark, which instantly prompts te IC to conduct and the cycle switches rapidly producing a voltage that's exactly 14.26 at the output.

Thus the IC is only responsible for holding the voltage to the set limit while the transistor TIP36 becomes responsible for delivering the required high currents.

Note: Use a common heatsink for the IC7812 and the transistor, this will ensure complete safety to the transistor from  thermal runaway situation. Be sure to use mica isolation kit for fixing the devices over the common heatsink.

 

Need Help? Please leave a comment, I'll get back soon with a reply!




Comments

  1. Hi..It's great to see this sort of resource freely available and supported..I've been searching unsuccessfully to find a 10 amp (or similar)constant current battery charger circuit for 12v lead acid batteries using an LM317 with power transistors inside the feedback loop. I have seen the many simple circuits for <1.5 amps but no application notes seem to cover that. Do you have one or can you link to one ? I realise that voltage sensing will have to added to prevent overcharging but I can cope with that. Thanks.

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  2. The above circuit can be utilized for the intended purpose but it will require an additional current limiting stage for the current limiting.

    I'll try to update the design hopefully soon.

    Alternatively there's one much easier option as shown in the following diagram, you can directly operate the circuit in conjunction with any suitably rated DC power supply for charging the battery. The output will be current controlled and safe.

    https://homemade-circuits.com/2011/12/make-hundred-watt-led-floodlight.html

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  3. Dear sir , what type of a diode is used before the 1 ohm 5 w resistor in the circuit. I don't know about such a diode which is rated 30 amp. Will you please help me sir?

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  4. DearArun,

    It's a rectifier diode, its current rating should be more than the maximum charging current or the load current at the output

    ReplyDelete
  5. hi swagatam
    this is srinivas
    thanks for sharing
    ihave made this circuit its working fine for some time After that 7812 regulator is became hot & blew
    please sugest me
    thanks & regards

    ReplyDelete
  6. hi srinivas,
    the 7812 IC will never blow as long as the input stays within 30V or the pinouts are correctly assigned.

    so please check the input voltage whether it's within 30V or not.

    ReplyDelete
  7. hi swagatam
    yes iam check it out input voltage is 18 volts
    here the main thing is iam not getting 18 volts from solar it is 18-0 volts transformer
    is it ok to check the circuit, and another thing is iam replace another 12 volts regulator after that it is working fine
    thanks & regards

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  8. Hi srinivas,

    the 7812 can tolerate upto 30V easily, so 18v is fine and it can be from any DC source, no issues.

    ReplyDelete
  9. Hi sir, can i use 18v 5amp transformer at the input with a 6amp diode to replace the 30amp diode to get 14v 5amp output or are there anymore changes i want to add a 741 automatic stage to the above design

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  10. Hi Victory, you can use this circuit for the said application, however for 5amps, TIP36 won't be required, you could use a TIP32

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  11. What if i use a 10amp transistor will it output 5a

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  12. excuse me sir ,

    If i want to use TIP36 and LM317 that i want current and voltage of charger about 5AH,12V.

    I have my schematic and i can send to your check ?

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  13. you won't need a TIP36 stage for a 5ah battery, you can use the LM317 regulator output directly for charging the battery

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  14. Goodevening sir,i have implemented the above circuit and i'm using a 12v 3ah transformer 6a10diode inplace of the 30amp one and tip42c inplace of tip 36 and 4 1n4007 diode at the adj of 7812 but i'm only getting 12.5volt and sometimes it increases to 13.8volt and back to 11.90-12.8v with current of 0.4amp could this be that the transformers voltage is too poor to provide steady 14v after rectifying? also(2)when i disconnect the collector of the tip42c from the positive line the output voltage becomes 0.7-1.2v and when i connect it back it increases to 11.9-13v is that ok?

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  15. without load the output from the 7812 must show 14V, if the input is also around 14 to 15V...all your results look incorrect there could be something wrong in your circuit.

    TIP42 won't be required for your application....it's only for charging high AH batteries in the range of 100AH

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  16. I only want 14v 6amp output that's why i use tip42 and i don't think there is something wrong with that since it's only there to produce 6amp current...what will happen if i remove the resistors at the base and enmiter of the transistor...i will try using 18v transformer as input

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  17. OK, that will do, I confused your comment with the comment previous to it. you can try removing the emitter.base resistors (vertical ones), these were included for extra safety to the transistor, anyway you can try removing them and see the response....

    ReplyDelete
  18. Hi, Thanks for the great circuit
    im just curious and want to ask about the power rating
    if this circuit outputs 14v @ 25A
    thats about 350W of power, but TIP36C is only rated for 125W
    how can it handle all of that power ?

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  19. Hi, according t me, the 125 watt is the power limit at which the device will generate sufficient heat to destroy itself...in short 125 watt is the max capacity of the device without a heatsink.

    with an optimal heatsink (cooling) this device would be able to provide you with the full 350 watts.

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  20. sir why we need mica isolation kit for fixing a voltage regulator ic to heat sinc ?

    ReplyDelete
  21. RT, when two devices are mounted over a common heatsink and their bodies are not supposed to come in contact with each other then mounting them through a mica isolator becomes mandatory...which prevents the device body from touching the heatsink and from shorting circuiting

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  22. so it is not necessary to use mica isolator to mount ics to heatsinks independently .. right ?

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  23. sir could you please tell me how to take high current from a -ve regulator ic in a way similiar to the one explained here. mine is a +18v 0V -18V dual supply capable of delivering maximum of 6A current.. i want operate 2 TDA2030 Ics in bridge mode as a 40W subwoofer amplifier. since the maximum supply voltage that this ic can handle is -18V and +18V as Vss and Vdd i need to reduce the voltage down to 15V . so i have used 7815 and 7915 for that.. now comes the problem of current delivery.. so please suggest the way by which i can efficiently connect one or two TIP36s to draw sufficient current for the load

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  24. yes, there's no need, in fact adding mica makes heat transfer much inefficient.

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  25. RT, an easier option would be to eliminate the LM7815/7915 and use series diodes for dropping the excess 3 volts

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  26. sorry sir i cant do that coz i need a stable 15 V supply which can be used for other purposes also.. and one more the circuiy given in this blog with TIP36C is not working.. the output current isnt increasing...

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  27. I have explained the working principle of the circuit in the article, you can refer to it and troubleshoot the fault in your design, or try a different transistor instead of TIP36

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  28. sir the current from the regulator increases if i increase the resistance value given from dc source to the regulator pin#1.. i have used one 100 ohm 1 W resistance instead of the 1 ohm 5W one and it results in an output current of 70 to 75 percentage of the maximum current that can be delivered directly from the source..

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  29. RT, increasing the resistor value will decrease the current and vice versa from the transistor....please make sure you are doing the procedures correctly.

    ReplyDelete
  30. Excuse me, i'm new at electronics. This IC 7812 is an L7812CV transistor? Thanks!

    ReplyDelete
  31. Hi, im a newbie at electronics. The L7812CV is an IC7812? Thanks!

    ReplyDelete
  32. Hello i want to charge lithium ion battery with high current with minimum input voltage of 5v and charging current should be 0.7A please suggest circuit with high charging current capability.

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  33. you can try the last circuit from this article

    https://homemade-circuits.com/2013/12/usb-automatic-li-ion-battery-charger.html

    ReplyDelete
  34. Its my first time here. I really appreciate for your job well done.

    Thanks for the everything it is very much interesting topics.

    Im Roger from the Philippines. May God bless you always.

    Best regards,

    ReplyDelete
  35. Thank you Roger, I am glad you liked my site,

    God bless you too!!

    ReplyDelete
  36. Hello! When I try this circuit with a power source at 20V I got 14.4V in output at 0.4 Amps when I plug in a 12V 7Ah battery. The 7812 gets hot but tip36 doesn't seem to be working. When i measure voltage in 1 ohm emitter resistance and in 2 ohm base resistance they are almost 0. Is it normal? Should I try to plug in the 150W solar panel and a 100Ah battery? Thanks!!!

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  37. Hello, the TIP36 will conduct only when the voltage across its base emitter resistor is around 1V.

    If you are getting 0.4amps at the output then the voltage across the TIP136 base emitter could be

    V = IR = 0.4 x 1 = 0.4V,

    so in that case the BJT will hardly conduct....you can try increasing the 1 ohm to 2 ohms or use a higher rated load that would consume at least 1 amp from the IC.

    ReplyDelete
  38. Are you sure tha emitter resistor, on schematic, is 1ohm/5w?

    ReplyDelete
  39. It's arbitrarily chosen, you will have to experiment the value with some trial and error, until the correct amount of current is measured at the output

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  40. I was referring to the vertical one, the horizontal one is reasonably correct.

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  41. Sir ,how much the ampire of IN 4007 diode?

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  42. sir i want to use the circuit with a24v batteries what are the changes i need to make thanks for always been there for me

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  43. You may have to replace the IC with a 7824 and configure it in a similar manner

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  44. sir does that means i will add more diod at the ground of the 7824 to obtain the 24v

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  45. 7824 will give you 24V directly, you may have to attach 7 diodes with a 7824 ground lead to get around 29V at the output

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  46. 29V is required for charging a 24V battery

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  47. thanks for your quick reply sir am very grateful

    ReplyDelete

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