NPN Darlington circuit for 12V Diesel Glowplug
You are calculating this wrong.  You want 0.017 A going through a 1 Ohm device, so you must put 0.017 A * 1 Ohm = 0.017 V across it.  Creating that kind of current control with a non-saturated grounded emitter NPN topology is pretty difficult since the Hfe is not a tightly controlled parameter for BJTs.

The correct way to do this is to put a current limiter resistor inline with the 1 Ohm device.  Then when you saturate the device, according to the graph in the datasheet, the Vce(sat) at 0.1A will be about 0.6V, so you will have $$12 - 0.6V = 11.4V$$ across the pair. 

Ohms law says you need a resistance of $$\frac{11.4V}{0.017A} = 670 Ohms$$

680 Ohms is a standard resistor value in that range and should give about 0.0168A in current.

Now on to the calculation of the base resistor.  It is typical to ensure saturation by calculating as if the Hfe = 10.  This ensures that you are fully saturating the transistor.  So for a collector current of 0.017 A, you want a base current of 1.7 mA.  The TIP122 Vbe(sat) with a collector current of 0.1A is about 1.25V so the resistor will have 4V - 1.25V = 2.75V across it.  This results in:$$\frac {2.75V}{1.7mA} ~= 1.61 kOhms$$







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