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High Current Transistor TIP36 – Datasheet, Application Note

Last Updated on July 11, 2020 by Swagatam 90 Comments

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.

NOTE: Derate linearly to 150°C case temperature at the rate of 1 W/°C.
Derate linearly to 150°C free air temperature at the rate of 28 mW/°

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

To enhance the current gain of the circuit, you can add a TIP127 with the TIP36, as shown below. Remember both the transistors will require a large heatsink, possibly with a fan cooling.

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.

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Filed Under: Datasheets and Components Tagged With: Application, Current, Datasheet, High, TIP36, Transistor

About Swagatam

I am an electronics engineer and doing practical hands-on work from more than 15 years now. Building real circuits, testing them and also making PCB layouts by myself. I really love doing all these things like inventing something new, designing electronics and also helping other people like hobby guys who want to make their own cool circuits at home.

And that is the main reason why I started this website homemade-circuits.com, to share different types of circuit ideas..

If you are having any kind of doubt or question related to circuits then just write down your question in the comment box below, I am like always checking, so I guarantee I will reply you for sure!

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Reader Interactions

Comments

Charis Michalopoulos says:
July 6, 2017 at 10:59 pm

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

Reply
santiago troconis says:
November 16, 2016 at 4:07 pm

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!!!

Reply
Swagatam says:
November 17, 2016 at 5:00 am

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.

Reply
Swagatam says:
November 5, 2016 at 8:03 am

Thank you Roger, I am glad you liked my site,

God bless you too!!

Reply
Highlander says:
November 4, 2016 at 9:36 am

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,

Reply
Faizan Hamayun says:
October 22, 2016 at 1:09 pm

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.

Reply
Swagatam says:
October 23, 2016 at 3:07 am

you can try the last circuit from this article

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

Reply
Santiago_TP says:
October 2, 2016 at 12:33 pm

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

Reply
Swagatam says:
October 3, 2016 at 4:03 am

L7812 is IC 7812, so it's OK

Reply
Santiago_TP says:
October 2, 2016 at 12:32 pm

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

Reply
The Rocking Time says:
August 23, 2016 at 3:04 am

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..

Reply
Swagatam says:
August 23, 2016 at 7:03 am

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

Reply
The Rocking Time says:
August 22, 2016 at 6:10 pm

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…

Reply
Swagatam says:
August 23, 2016 at 2:01 am

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

Reply
The Rocking Time says:
August 21, 2016 at 5:27 pm

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

Reply
Swagatam says:
August 22, 2016 at 1:37 pm

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

Reply
The Rocking Time says:
August 21, 2016 at 2:56 pm

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

Reply
Swagatam says:
August 22, 2016 at 2:20 am

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

Reply
The Rocking Time says:
August 21, 2016 at 5:11 am

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

Reply
Swagatam says:
August 21, 2016 at 2:37 pm

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

Reply
PowerGuy says:
August 15, 2016 at 9:02 am

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 ?

Reply
Swagatam says:
August 15, 2016 at 2:12 pm

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.

Reply
victory says:
November 16, 2015 at 6:20 am

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

Reply
Swagatam says:
November 16, 2015 at 1:07 pm

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….

Reply
victory says:
November 15, 2015 at 4:29 pm

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?

Reply
Swagatam says:
November 16, 2015 at 4:16 am

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

Reply
Unknown says:
November 14, 2015 at 8:36 am

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 ?

Reply
Swagatam says:
November 14, 2015 at 1:21 pm

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

Reply
victory says:
November 1, 2015 at 1:31 am

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

Reply
Swagatam says:
November 1, 2015 at 4:10 pm

yes it will…

Reply
victory says:
October 24, 2015 at 11:33 am

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

Reply
Swagatam says:
October 25, 2015 at 5:16 am

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

Reply
network advise says:
March 1, 2014 at 12:25 pm

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

Reply
Swagatam says:
March 4, 2014 at 4:44 am

Hi srinivas,

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

Reply
network advise says:
March 3, 2014 at 7:34 am

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

Reply
Swagatam says:
March 1, 2014 at 3:03 pm

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.

Reply
Arun Dev says:
October 7, 2013 at 12:29 pm

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?

Reply
Swagatam says:
October 8, 2013 at 4:05 am

DearArun,

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

Reply
Roger says:
July 30, 2013 at 10:54 am

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.

Reply
Swagatam says:
July 31, 2013 at 2:18 pm

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

Reply
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