Choosing a Transistor

Hi all,

I am designing a standalone DMX receiver for a light fixture, and for this I need a transistor that will survive switching the on and off currents of a 50watt LED. the led is rated at 16v 3200mA in normal operation. I have here some TIP122 NPN transistors, they are rated at 8A 100v, I was wondering if these are suitable, I am not sure whether the 8A is at a lover voltage thatn 100v or whether they are the maximum specs. Ie. will run fine switching 100v at 8A ,rather than being able to switch 100v at lets say 300mA and 8A at 12v.

It will take the current but will it take the heat. Look up the saturation voltage and multiply that by the current to give how much heat it will generate (in watts) then see how big a heat sink you need. http://www.thebox.myzen.co.uk/Tutorial/Power.html

http://itss.co.kr/product/pdf2/TIP120.pdf Thats the datasheet there, There are two values for the saturation voltage, which one do I use?

Power (watts) = Current x Voltage. Data Sheet says that with 3A going thru the Collector-Emitter voltage will be 2V. So, the part will be dissipating 3A x 2V = 6 watts. Its gonna be pretty warm!

The part wont be dissipating 6 watts, it will have 6 watts of power running through it. The Current for the LED i will be driving is 3.2 amps, the voltage on the other hand will be 16v, the power being run through the TIP122 would then be 51.5 watts. This is not the Dissipated Power otherwise the LED would not light.

The part wont be dissipating 6 watts

Yes it will.

This is the power that gets wasted and turned into heat by the transistor. It is not the power in the LED or in any other part of the circuit, it is the power you waste in doing the switching. This is determined by the saturation voltage and the current, nothing else.

So aside from that, any suggestions for reasonable alternatives for this task? I will have access to a 0.7 C/W heatsink as I am powering a 50w LED with it.

I would go instead with a MOSFET wth very low Rds, like 5mOhm (.005 ohm). Power (Watts) is also equal to Current*Current*Resistance, so 3*3*.005 = 0.045 Watt, much cooler running. And voltage drop across the part will be V=IR so 3A * .005 = 0.015V, so choose current limiting resistor appropriately: (Vsource - Vled)/3.2A = R Power rating needed for the resistor = I^2 * R, and I would double that for part longevity.