Carl H wrote:would you mind explaining how the resistor setup you posted works...im not gonna lie its been a long time since my physics class where we did power and reistance.
It's may day job... well sort of. I actually am a firmware guy, but have a pretty good working knowledge of hardware.
Not sure if you are familiar with a standard voltage divider, but that is effectively the same thing, just that typically they explain a voltage divider with one side grounded, the other at a voltage level and 2 resistors in the middle. In this case both sides have an arbitrary voltage level and 2 resistors in between.
But here for the sake of argument I will use the following explanation. The voltage values are simply to illustrate the concept. Lets say the top MAF in the picture is outputting 4v and the bottom MAF is outputting 2v.
V=IR
So in our case we know the combined R (1k + 1k = 2k), and we know the delta in voltage (4v-2v = 2v). Solve for the current thorough both resistors and you will get (I = V/R = 2v/2k = .001A . So because the 2 resistors are in series with each other this is the current through BOTH resistors.
So now that we know the current we can get the voltage at each of the nodes. So again knowing V=IR and now knowing the 1mA current lets figure out the voltage drop across the TOP resistor. V=IR = 1mA * 1k (remember just the top resistor) = 1V. Since there is a 1V drop that means the at the middle point in the drawing there is 4V-1V(drop across the top resistor) = 3V.
Now just to be sure lets verify the voltage drop across the bottom resistor. Again we have 1ma and again (because I told you to use 2 of the same sized resistors) we have 1k. V = IR = 1mA * 1k = 1V. So the voltage drop across the bottom resistor is also 1V. And that makes sense since the middle point on the drawing is 3V and the bottom MAF is outputting 2V there is a 1V drop (3V-1V = 2V).
Feel free to try it with any numbers on the MAFs. But as long as you use two of the same value resistor you and put them in series like that, the middle point where the 2 resistors connect will be the average of the 2 voltages being supplied.