voltage divider 80V

60*1.5=90

54*1.5=81

You just keep cramming cells in till you get the voltage you want.

I'd power the arduino from it's own battery pack. I would use a photocouple to interface the 80v signal to the arduino digital input.

Connecting that directly to 80vdc regardless of the resistors is just asking for trouble.

Hi mrmeval. On other point of view. What do you mean about photocouple ? Do you have a specific component in list to do the job ?

'optocoupler' or 'optoisolator'

This is a good read:- http://en.wikipedia.org/wiki/Opto-isolator

OK/ I don't know if there's people around here, but i try .. I finaly did something like this = I use a voltage divier and a voltage regulator. I use a pull-down resistor, to get 0V on the arduino input when the switch is open. It's quite strange, because, the voltage divider works good, but the regulator gives me less than 5V : 3.6V The Arduino Digital Input understand this 3.6V as an High, so ... it's good for me ... but it's quite strange dont you think ? The regulator is made to give 5V, but not here. Is my circuit not good enought ? Did i forget something ?

Is my circuit not good enought ?

Not good enough you have no input / output capacitors.

The LM7805 draws 8ma of quiescent current (see manufacturer's data sheet) This has to be accounted for in your voltage divider circuit. The net effect is that the regulator is seeing too low an input voltage to permit it to regulate properly. Suggest you lower the main dropper resistor to 10k and see where you go from there.

As mike says, you also need a capacitor in both input and output legs. Suggest 0.1microfarad on output and 0.22 on input (again taken from data sheet)

jack

HI. Thank you for your help.. I didn't know about the capacitors, i saw the schematic on the datasheet but thought it would work without it. Richard, I am just trying to sense if the 80V is here or not. I thought i needed to have a precise 5V and a precise 0V on the Digital input of the Arduino, that's why i use a regulator, and a pull down resistor. Plus, the battery (80V) that we use to trig fireworks, which is the same for the arduino input, will never have 80V in it, but between 70V and 80V (we do a lot of shows before changing the battery) I thought the regulator was an acceptable solution to deal with the non-precise-battery. I thought that giving the regulator something like 11V, it would give me a regulated 5V. Richard, you tell me about a simple 100K resistor in serie. I don't understand what will be the result Voltage after the resistor, how could it be 5V ? How could it deal with the fact that the battery will never have a real 80V output (but something between 70 and 80) The pull down resistor is always usefull right ?

thx

OK. Now i understand , thank you very much for your explanation and the diagram. I tried it this morning , and it works good !!!! Arduino deals with the 80V perfectly. I use a 100KOhms Resistor. Many thanks. There's one more thing... The thing is,.. when the switch is closed the Arduino input gives me 1. But when it is opened, the input gives 10100011010001001... That's why i asked for the pull down resistor ... Do you think there is a way to have a real 0 when switch is opened ?

I just did = 80V (+) ----switch------ R 100K ----- Arduino Input 80V (-) ---------------------------------Arduino Ground

It seams like when the switch is opened, cables until the resistor act like an antenna, and gives me eratic result. I understand that a pull down resistor between the 100K and the arduino input will act like voltage divider ... ::) .. hooh electronic mysteries ...

Yes you need a pull down resistor, use 10K.

It matters not that it forms a voltage divider because that gets trumped by the protection diode which clamps it to +5V.