5 ma to 2 amp variable current circuit using TIP120 BJT. Please analyze my circuit and comment
Start by doing your homework - looking at the transistor data sheet. Like this one. Pay attention to Figure 1 "DC Gain". Notice how the gain drops off at low base currents?

So, how much base current can you expect? Well, a Darlington base-emitter path has two junctions in series, so figure about 1.4 volts across it for a fairly wide range of base currents. This leaves 16.4 volts across your input resistor/pot combination. With the pot turned to zero, the base resistance is 120k, so your maximum base current will be on the order of $$i = \frac{V}{R} = \frac{16.4}{120k} = .14\text{mA} $$

What does this mean for this transistor? Look at Figure 1. At a collector current of 0.1 amps, a typical gain is about 400. Since gain is collector current divided by base current, $$Ib = \frac{Ic}{G} = \frac{100{\text{mA}}}{400} = .25\text{mA} $$

In other words, you would need twice as much base drive as you are providing just to get 100 mA of load current. And this is a conservative number, since figure 1 shows the gain drops with decreasing current, so I would guess that your circuit would produce a maximum of about 30 mA.

The only benefit I can see is that you won't need to worry much about heat-sinking the transistor. Of course, the bulb won't be glowing, either.