Holy shinola jeep head, is that a bomb! /wwwthreads_images/icons/smile.gif
Looks like a detailed project.
Save Jeepn's formulas, they will come in handy. Also, band's voltage divider is correct. IMPORTANT, whenever you put a resistor in series with a device (in your example the one that is requiring 3V) the same amout of current that the will flow through the device will flow through your resistor. If the resistor is not rated at the correct power lever (watts) the resistor will get hot and blow. Voltage dividers are very inefficient btw! They are quick and dirty. If I need to step something down I used a linear regulator with a variable output, a whole 'nother discussion.
Example,
say you have a light bulb that has R = 5 ohms but for some reason it can only operate at 5 volts, if you hooked a 5V battery to it you would have 1A flowing, I=V/R=5/5=1. You have a 10 volt battery, and want to step it down to 5V so you put a r = 5 ohms resistor in series with the light , you do indeed have 5V at the light now [using the voltage divider eqn, v = V*r(r+R)=10*5/(5+5)=5V, i.e. anytime the resistances are of the same value you cut the voltage in half]. Now lets see how much current is running in the system. I = V/R = 10 / (5+5) = 1A The current is the same but the voltage source has doubled! Where is the extra power being absorbed, the external resistor r! So the r has to be rated at greater than P=IV = 1*10 = 10W.
anyhow, hope I was somewhat clear, mostly just babbling though!!
big ed
/wwwthreads_images/icons/cool.gif Big Ed
'88 YJ, 4" susp,3" body,35's,283 Chevy V8,TH350,4.11's,D30,D35c