I’m building an external controller for our driver station based on the EStop CCI. I want to use a couple of 6 position rotary switches for the analog inputs on the CCI and wire a resistor ladder for voltage dividing. Normally, for 5V I’d use something line multiple 20K ohm resistors that would give me roughly 1V per stage. At 3.3V, I think that the number should be roughly 13K ohms/stage. Am I doing that math right? My concern is that I don’t want to exceed the 100ma max current on the CCI.

Maybe I’ve been working on software too long to remember my basic physics. But, I have to admit that I’m getting confused dealing with voltage drops across the resistor ladder given that I want to minimize the current to keep it well below the 100ma. A quick clarification of the math would be greatly appreciated.

Without knowing how much current will be drawn away from a bridge point, you can’t compute the resulting voltage at that point. If you have data about the input impedance of the CCI port to which your bridge will be attached, you can factor that into your bridge calculation. The other option is to get the voltage divisions close, hitch up the interface and a voltmeter, and trim to get the proper division for each switch position. At least, voltmeters are designed with high input impedance, so when you take it away after trimming, the voltage will still be quite close to what you want.

P.S. I like the way you disguise the “math” in your post.

If I follow what you want to do correctly, it’s much simpler than you think.

All you are doing is making a voltage divider. If you use equal values of resistors, the voltage will be equally divided. Four resistors would give you voltages of 1/4, 1/2, 3/4 and the full source voltage. The actual value of the resistors is immaterial as long as they are equal.

Where it does matter is the total resistance determines the amount of current your voltages divider uses to create the division. You don’t want to use excessive current, but you want to use enough that the voltages are stiff enough to not be affected by the inputs they are connected to.

In other words, if you close a switch connected to a point that is 1.0v, you don’t want the act of closing the switch to change the current paths and end up with 0.8v In most cases, the inputs you are feeding are high impedance enough and draw so little current themselves, that it’s not a consideration.

Use what you have on hand. I’d grab a bunch of 100 ohm resistors and see what happens. At 3.3v, you’re not going to draw much current.

The input impedance of the analog channel is high enough that makes the current flowing through the analog input insignificant.
Rtotal = Vcc/Imax
R = Rtotal/(N - 1)
Round R up to the next one or more standard resistor value for good margin.