Your problem is not understanding the transistor datasheet for hfe vs Ic/Ib when saturated and you are operating in the linear region resulting in poor switch control or current control.
If this is a LED with a heatsink rated for 600mA then it could be rated for 2-3W and will have a Vf ~3V depending on power rating and temperature.
The TIP31A is a power NPN with a strong attenuated hFE as Vce drops below 2V where hFe~100.
For effective control, you should saturate the NPN hard to <<1.2V (pref0.4V) and not rely on the Rb value and uncontrolled Vce vs hFE drop.
Thus for Ic=600mA , choose Ic/Ib=50 thus Ib=12mA then
- R3=(5-1.4)/12mA= 300 Ohms
- next allow for a >= 0.8V drop (5-1.2-3.0) with a series R to LED or 1.25 Ohms rated >=1 Watt or use 8x 10 Ohm 1/4W R's in parallel.
If Vce <=0.6V then recalculate Rs and go from 1.25 towards 3 Ohms.
Then test and measure LED temperature with finger. If it is too hot to touch, you need a better heatsink.
This is a case where Rb and Rc are critical choices for driving a 3V diode from a 5 V supply, and frankly a Mosfet is easier with an RdsOn <<0.1 Ohm @1A.
This device also has an equivalent Rce that varies with Ic when Vce is saturated. Vbe must also be saturated and for most devices this ranges around Ic/Ib =20 but yet most devices , unlike this one, are rated for Vce(sat) @ Ic/Ib=10 and more expensive devices up to 50.
Here from the table, I compute Rce as follows:
- Vce(sat)=1.2V @ Ic=3A, Ib=375mA or Rce=1.2V/3A= 0.4 Ohms
- this uses an Ic/Ib = 8 , lower than most.
- if we design around a ratio of 20, Rce will rise as well as Vce
- if we design around the hFE=100, then Vce is no longer saturated and Vce=2V at 1A (typical not worst case) here Rce = 2V/1A = 2 Ohms
Thus we can conclude as a switch best case with higher drive current than you are using Rce starts around 2 Ohms and rises sharply by an order of magnitude with insufficient base current, making the LED dim as Vce rises.