TIP120 Transistor gets very hot...

Hi All,

I install the circuit as per attachment using TIP120 Transistor to deliver 2A current at the output, but the TIP120 get very hot after few seconds, is it something wrong with the circuit? Even with heat sink attach, the heat sink get very hot within 15 seconds. Am I using the wrong Transistor?





Thanks.

I didn't understood what is the significance of using a voltage regulator to drive the base of a transistor... Do you want to increase the current drawing capacity keeping voltage fixed to 12V???

The power dissipated by the transistor can be found by the voltage difference between collector & emmiter, multiplied by the current passing through it.... Use a multimeter to find them out...

The heating is normal in series pass transistor circuit. As Genovator has rightly pointed out the power wasted as heat can be found by the voltage difference between collector & emmiter, multiplied by the current passing through it. You can reduce the heat by reducing the input voltage a littile. Hope you understood the point if so dont forget to click the helpedme button below

Js_Ong said:

Hi All,
I install the circuit as per attachment using TIP120 Transistor to deliver 2A current at the output, but the TIP120 get very hot after few seconds, is it something wrong with the circuit? Even with heat sink attach, the heat sink get very hot within 15 seconds. Am I using the wrong Transistor?

Click to expand...

Usually if you needed to increase the current of a voltage regulator and maintain its output voltage you would use a power PNP transistor (mounted on a heatsink) installed on an “input” side of the voltage regulator IC.
In your design you decided to hook it up on the “output” side of the voltage regulator IC, and that means that the voltage regulation is compromised. The output voltage will vary with the load current, but if the regulation is not your goal and your application can tolerate small voltage drop that’s fine.
As far as the heat goes, the heat is just proportional to the product of (Vin-Vout) * Ic, so there is nothing wrong with the type of a transistor.
If the input voltage is, say, 24Vdc and the load requires, say, 2A, that thing (TIP120) will have to convert into heat (24V-12V-Ube-Ur) * 2A =>>> roughly 20W and that will heat it up rapidly ..

:wink:
IanP

The heat is normal and for the reasons already given but I'm at a loss to understand why the transistor is wired that way. The 7812 is rated at 1A but the resistor limits it's output to 0.012A and even under short circuit conditions, the resistor can only disssipate 0.12 Watts so why specify 5W?


Brian.

The TIP120 is a power Darlington with a beta of about 1000. So if you want to draw 2A from the output you must drive the base with at least 2mA. 2mA through the base means a 2V drop through the 1K resistor, so the base voltage is no more than 10V. And since the base-emitter voltage of a Darlington is about 1.2V, the output voltage is now 8.8V. Is that what you want? As for power dissipation at 2A, that depends on the supplied voltage. Assuming 24V input, the power dissipated by the transistor is 2A * (24V-8.8V) = 30.4W. That is a lot of power so of course it will get hot. So what is your input voltage? That is most relevant parameter at this point.

Since it is a linear regulator, heat will be there. You may need to mount the transistor on a larger heatsink and may need to use a fan if heat levels are too high. If heat becomes too large an issue, consider using a switching regulator.

Hope this helps.
Tahmid.