The TIP120 does not need 120mA at the base for normal operation, that's the absolute maximum rating, above which you don't want to go.
The spec you are mostly interested in is the hFE (current gain), which for a darlington is very high, since it's two transistors connected in a way so the current gains multiply.
For the TIP120 it's given as minimum 1,000 (compare with a typical 200 for a single bipolar transistor)
Also important are the max collector current (5A) and the collector emitter voltage (60V)
The main disadvantages are that the base-emitter voltage is doubled compared to a single transistor (~1.4V), and the saturation voltage is higher (typically ~0.8V compared to ~0.2V at low currents)
These points are rarely a problem for a simple switch driven from a micro pin. At higher collector-emitter currents though, the Vsat rises and can interfere with desired operation and cause problems with dissipation.
For example, in the TIP120 datasheet note that at 3A Ice, Vsat is given as 2V, but at 5A it has risen to 4V. That's 20W of dissipation, a lot to heat to try and get rid of to keep the temperature down. So when switching a large current you need to take these factors into account, and maybe decide to look at a more suitable part (e.g. logic level, low Rsdon power MOSFET)
Since we have a gain of 1000, we hardly have to draw anything from the micro pin. Let's say we want to switch 1 Amp:
1A / 1000 = 1mA into the base needed.
If we have a drive voltage of 5V, then we subtract the Vbe from the drive voltage and divide by the current:
(5V - 1.4V) / 1mA = 3.6k resistor. To give it a bit of leeway select something a bit smaller like 2.2k. This still only draws ~1.6mA.
I wouldn't read too much into the different prices - the price of components is often dictated by how popular they are, the more they sell the less they cost. If you see better specs at a cheaper price, go for it ;-)
You can some pretty odd prices when the component is scarce/new/obsolete - I saw a 10uF ceramic capacitor priced at £7.50 (qty 1) on Farnell the other week...
Dave is correct... I'll try to clarify some more.
In an NPN:
- The base-emitter voltage and the doping of the base determine the rate of emitter electron current injection into the base, which is swept into the collector due to the potential drop from base to collector and the narrowness of the base region.
- The base-emitter voltage and the doping level of the emitter also determine the rate of base hole injection into the emitter, which does reduce the collector current.
- The ratio of the dopant densities sets the current ratio between the base and collector (Beta).
BJTs are designed with light doping in the base and a very narrow base width to maximize the diffusion of the emitter current to the collector. As a result, the base current needed to develop the Vbe for a given rate of emitter current injection is very small compared to the emitter and collector current, and so BJTs have high current gain.
Here's an online reference that goes into some detail:
Modern Semiconductor Devices for Integrated Circuits, Ch. 8
Best Answer
A voltage doesn't pass through anything. But semantic details aside, yes, I suspect these numbers are the absolute maximum ratings for your transistor. You do not want to design with those alone.
Yes, at least approximately. The gain is usually used for small signals, so getting to the max. might require a (very) slightly different base current. Do read the fine print; the maximum continuous current likely requires keeping the ambient temperature below some value.