I'm want to use a TIP120 transistor (the only type I have laying around) to switch 5V (low amps) to my IC. This is for isolation from a higher voltage range (5-24V) which will be doing the switching. I realize that this is backwards from the way you normally use a transistor (normally to Amplify) but I want it to work anyways; please don't break my heart. I am trying to select a resistor for the base that will work with voltages as low as 5 volts and as high as 24V, but I really don't understand the logic behind the value of resistor. When I used a transistor before in the opposite configuration (5v to switch 24v) I was told to use a 1kohm (I did; it worked; but it was just PFM to me; didn't understand the purpose). I have been pouring over the data sheet and came to the conclusion that I need to keep the current across the base-emitter <120mA (IB Base Current (DC) 120 mA max), but >2mA (IEBO | Emitter Cut-off Current | VBE = 5V, IC = 0 | 2 mA) at a voltage of < 2.5v (VBE(on) * Base-Emitter ON Voltage VCE = 3V, IC = 3A | 2.5 V). I don't see how I can do any of the ohm's law equations for this unless I know the resistance of the Base-Emitter, which as I understand, varies depending on the current across the base-emitter (which seems quite circular to me). I am throughly confused. If someone just gave a resistance value to use, I would be grateful, but what I am really after is an explanation.
Thanks!
Thanks!