Hello there, I am new to Arduino and electronics in general, so I hope you can bear me :).
So, what happens if I sent a high voltage from an I/O pin directly to the ground pin? Could this damage the board even at 40mA? I understand the output of that pin is from the microprocessor … but who is in danger here? the microprocessor, the busses or what?
Also, why short circuit is dangerous? Is it because the current is too high and it should be consumed a bit, then travel back to the ground or I got the whole idea wrong.
Not sure. Never done that but it would be a fair assumption that the IO point could be damaged. On the other hand it may limit itself to 40mA but I don’t think so. It would probably also be governed by what is hosting the IO pin, Arduino, RPi etc and how rugged this device is. Keep in mind though that some damage can be accumulative and the point may not fail now but fail for no apparent reason in the future. For this reason it would be safest to assume damage if this situation should occur.
Not likely at this current level unless the actual track is extremely small.
The microprocessor or microcontroller.
Depends on what you mean by dangerous. Depends on the voltage involved. At 5V and decent current capability things might get a bit or very hot. At 240VAC mains it can literally blow a power point off the wall and probably start a fire, could also prove lethal if touched. Higher voltage like 11kV can blow the top off a power pole and once again can be lethal
Under normal operation this would be the case. The current would travel back to ground or the other side of a circuit doing some useful work or tasks along the way.
No not really. It is not a good thing under any circumstance and is not normal. You will be consuming a lot of power and generating heat and getting nothing for it.
In a nutshell, It is normally a fault condition and not wanted.
You may need to learn some basic electrical formulae. The most basic is V=IR where V is voltage (in Volts), I is current in Ampere, usually abbreviated to Amp, and R is resistance in Ohm. A rearrangement of this gives I=V/R or current is voltage divided by resistance. Or if we know the current and the voltage, we can work out the resistance R=V/I. In your case you talk of 5V and 40mA (0.040A) so we can work out R = 5/.04 = 125 Ohm. If the maximum current an I/O pin is rated for is 40mA, then there must be a load of at least 125 Ohm between that pin and ground.
A second formula worth thinking about is W=VI - heat generated in Watts is Voltage multiplied by current. So 5V and .04A generates 0.2W. This is not a lot, but enough to burn out a really small part.
A ‘short circuit’ is zero ohm. If we use the I=V/R then I=V/0 equals infinity. In reality, this can’t happen because there is always a bit of R. When you look at the I/O pin on a processor, there will be an R in the processor between the supply (5V) and the pin. Sometimes, that R is big enough to limit the current so if the pin is connected to ground it doesn’t supply enough current to burn out the processor. But that is not normally the case - the designers want external devices to see the maximum V (which is usually what external devices need to make them work) so they don’t deliberately put extra R on the pins. Shorting the pin to ground for any length of time generates enough heat in the processor to burn something out.
External devices vary in the Volt they require to work, and the amount of current they draw when that voltage is applied. Other integrated circuits like op amps, logic devices etc. usually have high resistance inputs (like millions of ohms) so draw tiny amounts of current (microamps) and are easily driven by an I/O pin. Other devices may have lower resistance, and may need extra components (like an external transistor) because they need more current than can be supplied by the I/O pin.
So when hooking things up to a microprocessor, it is useful to look at the specification of the device to find out what voltage it needs (or can handle) and what current it will draw. A rookie mistake is to hook a 3.3V part to a 5V part. The 3.3V part may not provide a big enough change in voltage for the 5V part to notice, and the 5V part may overload the 3.3V part. Sometimes it works, but not because it was supposed to do.
Hope that makes some sense