# Arduino voltage misunderstanding

For arduino nano. I set digital pins 1,2,3,4,5 to HIGH (I do this using digitalWrite(pin,HIGH) for each pin). Does this mean that 5V is being sent through each of the pins so in total there is 25V going out from the arduino? How is this possible?

• think about this ... connect five wires to the positive terminal of a 9 V battery ... is there 45 V on the wires? Commented Jul 31, 2021 at 15:59
• i think that you are confusing voltage and current ... voltage is like the pressure in the water pipes in your house ... it is the same at all of the water taps ... if you open taps in the kitchen and in the bathroom, the pressure at the taps is the same, it does not increase, but the amount of water flowing is double ... the water flow is like electric current Commented Jul 31, 2021 at 18:22
• Voltage is a potential difference, like height. If you have five hills that are all 5m tall, you don't have a single 25m hill, you just have five places that are 5m tall. Commented Aug 2, 2021 at 1:10

Every pin gives you 5V on HIGH, yes, but these voltages don't add up.

Voltages are always measured between two points. In electronics we are defining one point in our circuit, that we use for all measurements, and call it ground. This is normally the negative pin of the power source.

Each pin has 5V in reference to the same ground, so they cannot add up (because that would mean the second 5V would be in reference to the 5V of the other pin, not in reference to ground). In the end they are connected in parallel to Vcc (where the power for the microcontroller is connected, in this case the 5V pin) of the Arduino (through the output drivers of the respective pin).

In the end this boils down to kirchoff's laws (if you want to research more about it). At a junction of connections, the voltage stays the same, but the current will be divided up between the connections.

Imagine that I have 5 flagpoles. I go to each of those flagpoles and raise a flag to a height of 2 meters. Since I have 5 flags, each at a height of 2 meters, does that mean that my flags have a total height of 10 meters?

Well, that question doesn't make a whole lot of sense, because "total height of the flags" isn't really a meaningful quantity. It's true that if you add up all the heights, you get 10 meters, but that number doesn't have any real meaning. Nothing in the system is 10 meters high.

Likewise, suppose you set 5 output pins all HIGH. Those output pins are just like flags on flagpoles, and you're raising each pin to a "height" of 5 volts. Nothing in the system is at a voltage of 25 volts; it doesn't make any sense to talk about a "total voltage."

• If a private were tasked with climbing or polishing every flagpole, then yes, he would have to climb or polish a total of 10 meters of flagpole. But that's just a nuance of the analogy, your comment about voltage is correct. Commented Aug 1, 2021 at 11:14

It's a fascinating question. The Nano has 22 pins, all potentially set to output. D0 to D13, and A0 to A7.

So we have a total of 22 outputs. Now if they did add up then we could connect up a 5V battery and get 110V out of it! That might kill you! So clearly something is amiss here.

It's like when an employee has "20 years experience". Do they really have 20 times the experience they had in the first year, or do they have the same experience 20 times over?

The fact is that all 22 pins are all referenced to the same ground pin, so we really have 22 x 5V but all from the same starting point. So we really have 5V but with maybe a bit more current.

If you look at the datasheet, you don't even have 22 x the current you might have from one pin. The Atmega328P is specified to have a total of 200 mA current output.

Don't worry, setting all pins to output won't kill you. :)