# Arduino minimum current

Hello from boat engine I get a signal of 12V and 2A. I want to drop the voltage to 5V. For that I will use the voltage divider. Less heat is better, that why I want large resistence resisiors, since the current is high.

So which resistors should I choose. Bigger is better? But not too big, since Arduino needs some current (digital in).

Maybe my post is not clear since english is not my native language. I appologize for that.

You should not use a voltage divider to power an Arduino.
You can connect the 12V direct to the power jack, or you can build or buy a 5V regulator.

Voltage dividers are not stable enough for the varying current demand of an Arduino and other electronic components you might want to add.

I will power it directly with 12V.

But I need the voltage divider to read the signal, that is 12 V and 2 A. Sure 2 A is not good for signal wire but this is not my decision but the engine manufacturer output. The signal will be 12V, 0V and alternating 0 V and 12 V.

Is the signal digital (on/off) or an analog voltage?
Okay looks digital.

Where comes the 2A from? You can measure a voltage or a current, but not both with a single input.

I'd dimension the signal voltage divider for 1mA, in order to make it somewhat insensitive to noise. Also add some safety precautions, because the 12V on a boat can mean anything from about 9V to 16V, with spikes of 100V and more. An opto coupler is fine for digital inputs.

iluvathar:
Sure 2A ... the engine manufacturer output.

I don't think that's what it means. It suspect it means maximum 2A. It makes no sense to say the output is always 2A, if you are also saying the voltage is fixed (12V). The current depends on the resistance. If you only want 1mA, you can have that by choosing the resistance you use.

Also not convinced that an opto-isolator is going to protect your circuit by itself. Sure, it will help prevent high voltage spikes reaching the Arduino. But won't the opto be damaged by these spikes? Whichever component fails, the circuit will still need repair. If this were a commercial product, the labour cost of the repair would dwarf the cost of any of the components. Surely its better to include some components to protect against those spikes? Like a zenner diode perhaps? I know ultimately nothing can protect 100%. But how to get from 99% to 99.99% protection...

iluvathar:
But I need the voltage divider to read the signal, that is 12 V and 2 A.

If you want to be able to detect the presence of a 12v signal with an Arduino I/O pin then a voltage divider will work. If (as would be normal) the voltage divider uses large resistance values the current will be tiny - just a few milliamps.

Just make sure that even if the voltage temporarily rises above 12v that the output of the divider can never exceed 5v. Using an opto-isolator in place of a voltage divider avoids that risk.

If you wish to measure the amount of current flowing in the 12v circuit then things are significantly more complex. But let's hear if that is what you want before providing unnecessary info.

...R

If you keep the current low enough (<0.5 mA) the clamping diodes of the Arduino pin will take care of it. So even the 100V spikes can be handled while normally 12V gives a 5V output. The problem is that the resistors are getting quite large so better use a separate clamping diode:

This circuit goes from 0-3.8V output for 0-12V input, so you can measure a bit over 12V as well.

Won't that send the 100V spikes to the 5V power supply and risk damaging that, and anything else connected to the 5V line. Ok, the resistors might reduce that 100V down to around 30~35V, but the 100V figure was just an example.

Depends on the internal resistance of the power supply.

At 100V peaks you’d get 10 mA of current. A power supply should be able to easily take care of that extra current.

In fact you could read a 100V signal with just a resistor on the Arduino pin, that’s what the data sheet suggests. A 220k or greater resistor is all you need. Current <0.5 mA and the clamping diodes take care of the rest.

Interesting, thanks for that.

and the clamping diodes take care of the rest.

Note the clamping diode does exactly the same thing, dumps 100V into the 5V supply. However as this is just a small spike it is absorbed by the decoupling capacitor on the 5V supply. The regulator also adjusts to compensate for any resulting slight over voltage.