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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

Killy says:
March 8, 2022 at 5:24 am

Which car amp has the most tip36c transistors in it?
Like 24 32 + in one amp, is there any?

Reply
Bobtone says:
January 12, 2022 at 1:30 pm

Can this Transistor be configured with Lm317 to vary the voltage for a precise high current supply?

Reply
Swagatam says:
January 12, 2022 at 2:02 pm

yes that’s definitely possible…

Reply
Boolf Nasa says:
July 16, 2021 at 10:30 am

Hi,
I have a “JBL Sub200” audio amplifier that is operated based on a TIP35CW/TIP36CW circuit.
I changed many components on the circuit board including the TIP35CW/TIP36CW. However both TIP35CW/TIP36CW are the first to burn and become shorted upon turning ON the amplifier, even without getting hot.
What could be the cause for them to be the first to burn immediately knowing that they are 100V 25Amps, i.e. highest rated among all the other compnents.
They are powered by a AC transformer outputing 60V AC and 36V AC then rectifier bridge plus capacitors for the DC?
I already cooked 3 pairs of these TIP35CW/TIP36CW
Any hints?

Reply
Swagatam says:
July 16, 2021 at 11:57 am

Hi, did you adjust the quiescent current of the amplifier. If you don’t set the quiescent current, the transistor will unnecessarily draw huge current and get destroyed. Connect 100 mA bulbs in series with the +/- supply lines of the amplifier, then adjust the quiescent current preset so that the bulbs just shut off.

Before doing this make sure to keep the input terminal shorted to ground.

Reply
Tim Purvis says:
March 26, 2021 at 7:18 pm

Hi there,
I have a question regarding the TIP35-36 complementary pair:
I work on older GK 250ML amps quite often. These amps use the TIP33C-34C pair on the outputs. I have always been curious if I could sub the TIP35C-36C in place of the 33C-34C. These do have a thermal feedback loop A06 to prevent runaway. Did not know if the higher current capacity would alter the feedback and bias.
Just looking for an opinion, I have never tried it, but the 35-36 pairs are everywhere in the 247 case.
I would paste schematic, not sure if I can.
Best Regards,
thp

Reply
Swagatam says:
March 27, 2021 at 10:29 am

Hello, it is definitely possible to replace TIP33C/34C with TIP35C/36C as long as the supply voltage is below 100 V.

Sorry I am not sure about the feedback working of your amplifier, so I can’t suggest much about it.

Reply
john papa says:
April 7, 2020 at 9:00 pm

Hi my name is john and I liked the project solar battery charger using the tip 36 transistor I liked the way the diagram showed the orientation of the components it showed the acual pinout of the reg as is I have been studying electronics for 20 years and I think diagrams should show the pinouts for example every time I see a transistor in a diagram I have to refer to my manual for the correct pinout when it can be shown on the acual diagram at glance the orientation of each and every part. kind regards john

Reply
scott hetland says:
January 8, 2021 at 9:35 am

I am working on building this circuit but I am confused on the NPN and PNP power transistors .It calls for 0500 NPN
power transistors cant find that transistor. It also calls for tip 36 PNP power transistors. this is a strobe light i want to build any help would be much appreciated.
Thank You,
Scott Hetland

Reply
Scott Hetland says:
January 8, 2021 at 11:00 pm

what is an 0500 NPN power transistor

Reply
Swagatam says:
January 9, 2021 at 11:28 am

I am not sure which circuit and which part number you are referring to?

Reply
Swagatam says:
January 8, 2021 at 11:00 am

TIP127 and TIP36 are both PNP BJTs. If you cannot find them, you can try TIP2955 and 2N2907

Reply
Swagatam says:
April 8, 2020 at 12:12 pm

John papa, the actual pinout of 7812 is exactly same as shown in the square symbol. Holding the printed side towards you, left pin is input, center pin is ground, right side pin is out.

Reply
florin says:
September 5, 2019 at 5:24 pm

I used 2N6653 in a similar diagram and it became hot shortly at a current of 3 amps with cooler but no fan,and this one is rated at 20A and150w

Reply
Swagatam says:
September 5, 2019 at 7:53 pm

apply heatsink compound and a large finned heatsink and then check…the issue here is whether it can withstand the rated amount of current or not, and the datasheet can never be wrong, unless the device is duplicate or substandard

Reply
florin says:
September 5, 2019 at 11:51 pm

the data sheet is correct, the understanding is missing.I make everything right, I’m not a beginner.The transistor is original,what I want to tell you is that the reality in electronics is quite different, than in theory

Reply
Swagatam says:
September 6, 2019 at 7:44 am

Then probably you should tell this to the datasheet creator or the engineers who have written the datasheet without any special remarks.

Reply
florin says:
September 5, 2019 at 2:10 pm

TIP36 can only support 25 amps if the temperature of the PN junctions does not exceed 25 degrees Celsius, which is practically not possible.

Reply
Swagatam says:
September 5, 2019 at 7:43 pm

That’s not correct. TIP36 is rated to dissipate 125 watts without heatsink if the ambient temperature is at 25 degrees Celsius. It is definitely rated to work with 25 amps if adequate heatsinking is provided.

Reply
Swagatam says:
September 5, 2019 at 8:49 pm

This is my interpretation about the Pd, because I can’t figure out what the 125 watt indicates? I can’t find a clear definition for this anywhere on the internet

Reply
florin says:
September 6, 2019 at 3:54 pm

in completely open state (saturation) the transistor has Vce0= 4 volts at a current of 25 amps so a power dissipated 100 watts. The remaining 25 watts are dissipated by the emitter base junction, total 125watts.this at 25 degrees celsius.as the temperature of the junctions increases, the dissipated power decreases by 1wat per degree celsius.so at 50 degrees celsius it can dissipate only 100 watts and at 100 degrees celsius only 50 watts,and this only in the saturation state.in our diagram the transistor does not work in saturation mode so the dissipated power increases significantly up to 160 watts so for this reason we need more transistors in parallel,each with its 0.1 ohm emitter resistor.

Reply
florin says:
September 6, 2019 at 3:00 pm

125w in our case, represents the maximum power that can be dissipated by the PN junctions and is UxI, meaning the voltage difference between the collector and emitter multiplied by the current that flows.so,if we supply the diagram at 24 volts and at the output we have 14 volts the difference of 10 volts we find it on the transistor (resistor + diode).at 25 amps current at 0.1 ohm resistor will be 2.5 volts, at diode pins 1 volts total 3.5 volts.the transistor will have 6.5volt terminals which multiplied by 25 amps results 165wat which the transistor does not support. so I say that the diagram is not good. I hope you understood.

Reply
florin says:
September 5, 2019 at 8:44 pm

please check active region safe operating area and power derating rate depending on the capsule temperature rise for TIP36 and you will see why it is actually used at 50-60w dissipated powers and 6-8amps load

Reply
Swagatam says:
September 6, 2019 at 7:43 am

I know that, so please maintain 25 degree Celsius to prevent derating, if you can’t then its your problem not the transistors problem. I have only shown the maximum feasible limit in the diagram as per the datasheet.

Reply
florin says:
September 5, 2019 at 12:50 am

this schematic diagram is wrong

Reply
Swagatam says:
September 5, 2019 at 8:40 am

You only have to tweak the resistor values to get the right amount of current. The schematic is CORRECT!

Reply
florin says:
September 5, 2019 at 1:12 pm

if we mount 3 transistors in parallel then maybe we have 25 amps at the output

Reply
florin says:
September 5, 2019 at 1:52 pm

practically TIP36 is used at dissipated powers around 50w.so two TIP36 is not enough.

Reply
Swagatam says:
September 5, 2019 at 7:44 pm

Try it yourself, just make sure to add sufficient heatsink

Reply
florin says:
September 5, 2019 at 12:54 pm

ok, it is correct as a working principle, but with current values ​​it can only deliver 1 or 2 amps, not 25 amps as you say. even if you reduce the R value to 0.1 ohms it still cannot deliver 25 amp.TIP36 could not dissipate as much power.

Reply
Swagatam says:
September 5, 2019 at 1:27 pm

The TIP36 is rated at 25 amps, so it is designed to deliver 25 amps if sufficient heatsinking is provided.

Reply
florin says:
September 5, 2019 at 1:44 pm

I know that, but these are the absolute maximum values.It cannot be practically reached. Even if the diagram is powered by 24v the transistor must dissipate approximately 200w

Reply
florin says:
September 5, 2019 at 5:01 pm

what i mean is that,in the current configuration the diagram cannot deliver more than 5 or 6 amps with R0.1ohm without being destroyed by overheating

Reply
Swagatam says:
September 5, 2019 at 7:52 pm

As i said you have to apply a very good heatsink, after all the metal area of the transistor is so small, how can it dissipate so much heat so quickly.

The heatsink can be very large, that’s a different issue. Datasheet has clearly suggested that the device is designed to work with 25 amp continuous current, and I am sure they must have also applied some margin to it.

Reply
Swagatam says:
September 5, 2019 at 4:33 pm

Yes I agree, so here too it is assumed to be the absolute maximum value. If datasheet suggests that the device won’t burn until 25 amp is exceeded, the above article also refers to the same data. Moreover it is possible to include a 24 amp current limit by adjusting the various resistors

Reply
Favour says:
August 3, 2019 at 3:15 am

Well done sir, what is the efficiency of this circuit. Thanks teacher

Reply
Swagatam says:
August 3, 2019 at 9:34 am

Efficiency will be high when the input output difference is low, and vice versa.

Reply
abioye says:
February 25, 2018 at 1:17 am

thanks for your quick reply sir am very grateful

Reply
Swagatam says:
February 25, 2018 at 5:23 pm

you are welcome!

Reply
abioye says:
February 23, 2018 at 9:09 pm

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

Reply
Swagatam says:
February 24, 2018 at 8:35 am

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

Reply
Swagatam says:
February 24, 2018 at 8:36 am

29V is required for charging a 24V battery

Reply
abioye says:
February 23, 2018 at 4:39 am

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

Reply
Swagatam says:
February 23, 2018 at 5:06 pm

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

Reply
Robin says:
February 1, 2018 at 9:01 pm

Sir ,how much the ampire of IN 4007 diode?

Reply
Swagatam says:
February 1, 2018 at 9:06 pm

1 amp max limit

Reply
Swagatam says:
December 30, 2021 at 4:06 pm

You can replace the power transistor with an FET rated at the specified level.

Reply
Amoke says:
December 30, 2021 at 2:44 pm

Hi what if lwant to charge 400AH

Reply
Swagatam says:
July 7, 2017 at 3:19 am

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

Reply
Swagatam says:
July 7, 2017 at 3:17 am

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

Reply
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