Cheap, low-quality constant-current sink over wide voltage range
No, don't try to use the constancy of HFE to run your transistor, it varies with temperature, voltage, device, too many things.

As you have 3.3v available, use emitter degeneration like this. This is a standard technique for biassing transistors and creating current sources.







simulate this circuit &ndash; Schematic created using CircuitLab

The two diodes maintain around 1.4v on the base, or 0.7v across R1. With RE=130ohms, that's around 5mA. It will vary slightly with VBE and the diode drops.  Change the value of R1 for other sink currents.

R2 needs to be low enough to conduct at least Iout/HFE with 1.9v across it, ideally 5x that, so 5mA Iout, 100HFE, IR2 should be at least 250uA, R2=6.2k.

Depending on the VCEsat of the transistor, this will sink current down to around 1v. The transistor needs to be heatsunk to tolerate 16.3v * 5mA of power dissipation within it.

You can replace the two diodes with another resistor if you want, which gives more flexibility for Vre, at the cost of a little thermal and HFE stability. If you are happy to sink current only over the range 3v to 17v, then you can connect the base directly to the 3.3v, with 2.7v across R1, resized accordingly.