Reset
3V3
5V
Gnd
Vin
Ok, the "Vin" is where you put power in right ? That 9V battery thing ?
The Gnd is the ground, no problem here.
Ok the 5V, does it emit 5V or I could plug a 5V supply in here and do away with the Vin ? I am so confused.
3V3, now what is the deal with this 3V3 ? Can I supply 3V3 and it auto step up the voltage for the entire Arduino or Arduino emits 3.3 through this hole ? [3V3 = 3.3volts right ? Why can't they just say 3.3v ? so confusing!]
"Ok the 5V, does it emit 5V" Yes.
" or I could plug a 5V supply in here and do away with the Vin ?" Yes.
"Can I supply 3V3 " Yes
"and it auto step up the voltage for the entire Arduino" No.
" or Arduino emits 3.3 through this hole ? "Yes
"[3V3 = 3.3volts right ? Why can't they just say 3.3v ?" European thing I think. Like swapping . and , in numbers.
It's 3V3 because you couldn't see the decimal, the print's so small.
V_in is the "9V" (external) input.
The 5V will show up at 5V if V_in is used.
If you have a 5V source then you can connect that to "5V" instead of using V_in.
The 3V3 is only present when you're connected to USB.
Wooooooo
These are new info to me.
So basically, if I am not plugged into USB, the 3.3V pinout is USELESS ?!!
Wow...didn't see that coming....
Next question would be, if I supply 3.3v to the Pin and it doesn't step up for Arduino to use, and it doesn't emit 3.3v if not connected to USB, that pin is pretty useless isn't it ;p
Sorry, BBT, the 3V3 is present with or without the V_usb.
Yes, the 3.3V is the output of a linear regulator that is fed from 5V (from USB, from 5V regulator, from external). You'd a DC-DC converter (also called boost converter) to get a higher voltage.
Ok Ok
Let me try to Summarize this, I hope I get it right:
Vin is where you supply 6v to 20v.
Gnd - Meh
5v, if you supply power via Vin, this hole will "magically" become an output and give out 5v.
BUT
If you supply 5v here it will "magically" knows that it is the receiving end here and become a source in.
What if both Vin and 5v are supplied external power ?
No idea...
The 3.3v is a hole that keeps on emitting 3.3v, just think of it as a 3.3v output, it's not an input.
Is this correct ?
"If you supply 5v here it will "magically" knows that it is the receiving end here and become a source in."
No, the 5V regulator has no input voltage and does nothing. The rest of the board does not care where 5V comes from.
"What if both Vin and 5v are supplied external power ?" If the external source can supply enough current, the regulator will not see any voltge drop from the load and will not be doing much. The rest of the board does not care where 5V comes from.
"The 3.3v is a hole that keeps on emitting 3.3v, just think of it as a 3.3v output, it's not an input."
For practical purposes true. The 3.3V regulator can also be driven from offboard, but nothing on the board uses 3.3V, so might as well just power the external device directly.
For practical purposes true. The 3.3V regulator can also be driven from offboard, but nothing on the board uses 3.3V, so might as well just power the external device directly.
The auto-voltage switching comparator uses the 3.3vdc as a reference input to compare with Vin/2 to determine to switch off the USB +5vdc if external power is present.
Lefty
BlueBlondeTard:
If you supply 5v here it will "magically" knows that it is the receiving end here and become a source in.
The 3.3v is a hole that keeps on emitting 3.3v, just think of it as a 3.3v output, it's not an input.
There is no magic. Electrical circuits consist of nodes. A simplified view is: The node before the regulator is "Vin". The node after the regulator is named "5V". So if power is on "Vin", then the regulator brings the "5V" node to 5 volts. However, if you don't supply anything to "Vin" and instead put 5volts on "5V", then the node is already at 5volts.
(Note, I intentionally referred to the name "5V" to not mean voltage.)
Let me follow up on the previous post....
Electrical circuits consist of nodes. A simplified view is: The node before the regulator is "Vin". The node after the regulator is named "5V".
Electrical circuits consists of nodes and each node is given an identifying label (or name). The Arduino folks and many others label the nodes by the nature of the signal that is normally present at that node. The node before the regulator labeled VIN because that is where you normally apply your input voltage. The node after the regulator is labeled 5V because, if you apply the correct voltage to VIN, the regulator will provide +5 volts at this 5V node. This +5 volts powers the devices on the pc board and is also available, as an output at the 5V pin, for other devices.
If, on the other hand, you already have some external source of regulated 5 volts you can feed it in to the 5V pin and it will power the devices on the pc board.
It is not a good idea to supply power via both VIN and 5V although you will probably get away with it. The reason is that the 5 volts supplied by the regulator and the 5 volts supplied by your external source are not going to be exactly the same. There are no provisions on the pc board to account for this difference in voltage so something is likely being abused, however slightly.
Don