MJ2955 PNP Transistor heatsink

Please let me know that I connect MJ2955 PNP Transistor 10A Ic. But it becomes very hot at 1.2A

I am using 12VDC 21W bulb means 1.75A and 2955 gets very hot.

Hi,

A basic electronics formula: P = V x I
In your case this means:

P = power (generated heat) = voltage x current.

Current:
* There is current flow from base to emitter
* And there is current flow from collector to emitter

Voltage:
* bae current is from base to emitter, thus you need to calculate with V_BE
* collectir current is to emitter, too. Thus you need to calculate with V_CE.

Power:
* P_B = I_B x V_BE
* P_C = I_C x V_CE

Total power = P_B + P_C

Now you know the generated ammount of heat.

*****
This heat needs to be spread to the surrounding air.

Every electronic device comes with a datasheet. It contains the information about dissipated power and expectable rise of temperature.
This is true for a heatsink, too.
It's important to read both datasheets, of bjt and the heatsink.

R_th is the thermal resistance. It shows how much the temperature rises when it has to dissipate power. = delta_t / power.

*****
Example:
Maybe you calculate the power of your transistor to be 20W.
And maybe you want max temperature rise of 80°C which is 80K.

Then r_th needs to be 80K/20W = 4K/W or less. (From junction via heatsink to ambient)
This is a huge heatsink.

This is physics, you can't avoid this.

Klaus

Your heatsink is too small, a linear regulator from 20V to 12V is quite inefficient (lots of heat produced) look at a buck converter ...

You show an old little TIP2955 in a plastic TO-218 case, not an MJ2955 in a TO-3 metal case. You have it heating with (20V - 12V) x 10A= 80W but its datasheet shows its temperature is very hot with a huge heatsink and 80W. Your little heatsink is designed for about 5W.