I implemented a breadboard prototype based on the circuit below (image created by a LED circuit calculator here).
The LEDs are SFH4556P. Voltage for the breadboard is provided by a wall wart adapter (phone charger) and supplies 5V (1A max) via a USB cable salvaged from a PC mouse (has 5 wires, two for data, shield ground, plus and minus - of which only the latter two are used). I also added a 500mA fuse to the circuit, the resistor is rated for 2 Watts.
I hooked everything up and started my measurements and immediately hit an issue. The current draw of the circuit was below 30mA meaning my LEDs were way too dimm.
I replaced the resistor with a 1.5 Ohm one (same wattage rating) and current draw rose up to around 40mA, which is still too low.
I then I removed the resistor and voila - around 50mA.
I also found the culprit - the salvaged USB cable. It seems that due to thin wires used in it, each of the two used cables cause a resistance of around 5 Ohm so it causes the circuit to already contain 10 Ohms of resistance.
How bad of an idea would it be to just leave the circuit as it is (without a resistor)? Can I calculate if the wires in the cable are capable of sustaining the circuit without any resistor in it?
The circuit is part of a DIY TrackIR project and that is also its intended usage.