TomGeorge:
Hi
If we do the maths.PN2222 .. DC Gain == beta of 100 minimum.
beta = Ic/Ib
Ib = Ic/beta
Ic = LED current at collector = 1000mA
Ib = 1000mA/100 = 10mA base current.
3.3V from the arduino, 0.7 base-emitter volt drop.
voltage across base resistor = 3.3 - 0.7 = 2.6V
V = I x R
Voltage across base resistor = base current x base resistor
base resistor = Voltage across base resistor / base current = 2.6V / 0.01A = 260R
So 270R or 220R would be more suitable to get the transistor to saturate and pass full collector current.
Tom...
Guys, a 2222 will not be comfortable with a 1A collector current. End of story. And a DC gain of 100 applies only if the Vce is well above saturation. At saturation you'd be looking at more like Beta is 10 to 20 at best.
My advice is a FET, something with a Vgt (gate threshold voltage) around 2V and has its RdsON performance at less than 5V actually declared in the data sheet as FDD8447 does. When fets warm up the on resistance of the channel can double if it gets really warm so make an allowance for that as well.
If you are not looking to switch this load at any kind of frequency then any fet with the right on resistance specified in the data sheet with a 4.5V drive (assuming a 5V arduino?) could be fine without heatsink and without even getting warm itself.
A BJT as you were trying to use will drop a greater voltage than a MOSFET and either require a higher base current than you can provide or to be a darlington circuit arrangement with potentially slightly higher saturation voltage and definitely will not be a super fast switch and will likely require some heatsinking or at least surroundings that will tolerate some warmth.
Look at FDD8447 as a good example of what will do the trick if you have the drive voltage (5V) to use it.
PE