I have a simple 5 volt DC circuit powered by a LiON battery. Running a fan that is rated at 5 volts 100mA-300mA. The battery has a 1000mA output max so the fan draws its full 300mA. I am looking for a way to lower the current available to the fan to 100mA so I can, in theory, triple my battery life. How do I accomplish this?
The fan will draw less current when it is given less volts. The current draw is roughly proportional to applied volts as is the speed. Consult the fan manufacturers specs and/or check on a variable lab supply to obtain a minimum voltage for reliable operation.
Lower fan voltages will mean that less air is shifted. Now make some PWM to control the fan and you will save battery power. If you run the fan at say 70% speed meaning about 70% voltage and about 70% current then power is halved if your proposed PWM system is perfect.
What you can do is connect your PWM control to a temp sensor so the fan speeds up when hot shifting its full specified amount of air and slows down when cold saving the battery and making less noise and increasing bearing life. Sure this is more complicated than a resistor but these days it is worth it when SMD is used.
A 50 ohm resistor would limit the max current to 100 mA. But that might be too much. How about a 50 ohm pot and adjust it till you get the current you want.