I know arduino can run safely off a 12V power supply which I have, but my project will be in an enclosure so to avoid long-term problems with heat from the regulator, I would like to use
Do you need 9V for other parts of the project or just to power the Arduino? If just to power the Arduino, you would be better served to use a buck converter that goes from 12V to 5V and connect directly to the 5V pin of the Arduino. That bypasses the weak and wasteful on board 5V regulator.
I am confused as to the pinout of the arduino. I thought it was:
input jack = 7~12V
Vin pin = 7~12V
V 5V pin is an output. Do not plug power input here.
V 3V pin is an output. Do not plug power input here.
I am using both an adafruit Metro 328 (pin compatible with arduino uno r3) and an adafruit 328 mini (also pin compatible with arduino uno r3).
"5V ... etc ... supplying voltage via the 5V or 3.3V pins bypasses the regulator and can damage your board. We don't advise it."
So you are saying if it is regulated by a buck converter it is safe or safe-er.
What if I just used my original idea. Anything wrong with that? It has the advantage that you can place the heat-consuming device (magnolian) outside the enclosure and get it wet.
Yup, that is what I am saying. Powering through Vin or the power jack means that the Arduino and all peripherals that are on the 5V rail are powered by the onboard 5V regulator. The on board 5V regulator is not heat sinked so will supply limited current before it overheats and shuts down. The amount of current depends on the voltage input to Vin or the power jack. The higher the voltage the less current can by supplied. I would use a buck converter to drop the 12V to 5V and connect that to the 5V on the Arduino, bypassing the, weak, 5V regulator. Then the rated current of the DC DC converter is available on the 5V line.
Do not power an Uno or Nano with USB and input to the 5V pin at the same time. Either one by itself, never both.
You can connect 9V to the Vin or power jack, but the above applies. It is more efficient to use the buck converter at 5V. And there is more current available for peripherals.
as you can tell i'm new here.
the original post was about a 12V power supply as input in parallel to
(a) other devices and
(b) the "magnolian" which is a particular brand of 12V in to 9V out converter for cars (link to amazon).
I don't know if it is regulated.
High efficiency: >96%; Reliable, low heat dissipation
With overload / over-current / over / low voltage protection, stable performance
Light compact, convenient to use and transport; Auto recovery
Widely used in Car device, such as LED Display, Hard Disk Player, MP3, DVD, GPS, etc
Then the 9V out from the magnolian is input to the barrel jack of the arduino.
If the magnolian is not regulated, the regulator on the arduino will do it.
So is this particular 12V to 9V converter suitable for this job?
Why wasn't the arduino just designed to accept 5V at the barrel jack if it wastes energy converting it to the 5V rail? Maybe so it could be powered by 9V batteries?
yeah i should have known that was a practical joke. you don't supply power to the output, although it IS possible it can fry the arduino. Look at the attached schematic. Note that the mosfet is a P-channel, so it is OFF for 5V on the gate and ON for GND on the gate. Note the Comparator Vin (+) is connected to the middle of an equal value voltage divider. Note that Vref(-) input is connected to 3.3V. Therefore the Barrel Jack Vin voltage must exceed 6V to turn OFF the mosfet which connects the USB 5V to the board's 5V power rail. Thus if nothing
is connected to the External DC Barrel Jack (or Vin pin) , then the comparator's Vin(+) < Vref and Vout= -Vcc (GND). Now suppose nothing is connected to the Barrel Jack or Vin pin, comparator output = GND (0V). The comparator is powered by the board's 5V so if there is no power on the Vin pin, the comparator cannot output 5V to turn OFF the mosfet and thus the mosfet is ON by default, allowing the USB 5V to connect to the board's 5V power rail. NOW suppose the USB 5V is CONNECTED, and there is 5V on the left side of the mosfet and THEN you connect an EXTERNAL 5V to the board's 5V pin, resulting in 5V on the RIGHT side of the mosfet. By Ohms Law, if there is no difference of potential between the LEFT side and the
RIGHT side of the mosfet, V=0V, .'. I = V/R = 0A | R = the mosfet RDs(ON). That being the case, is there any reason why you should not connect an external 5V power supply to the board's 5V OUTPUT pin ? Now consider the other schematic of the arduino 5V regulator circuit. Note the diode between the Barrel Jack output and the 5V regulator input. If nothing is connected to the barrel jack, the 5V regulator has no power, so if you connect 5V to the OUTPUT of an IC that has no INPUT power,
you cannot damage it. In short, as already stated, the 5V pin is ALSO an INPUT. It is in fact, both and INPUT OR an OUTPUT because with Vin power it is an OUTPUT, and with NO Vin power it is an External 5V INPUT.
Got it ?
Yes, you could use that to change 12V (or 9V) to 5V to power the Arduino through the 5V pin and power the peripherals that are on the 5V rail.
Probably because, at the time, there were few wall warts with 5V regulated outputs. 9V and 12V wall warts were easily sourced.
Maybe, but we now know that that is not a good idea. 9V smoke alarm batteries are expensive for the amount of current that they are capable of delivering. My opinion is that smoke alarm 9v batteries should come with the warning "Not for use on any Arduino project".
One part is almost correct. the 3.3 V pin is an output and in no way could be used to power any part of the Arduino - because no part uses 3.3 V.
Powering the 5 V pin with 5 V cannot damage the Arduino in any way at all. There is a concern.
And that is also wrong.
You see, it is rather more subtle than that.
The problem is that when you connect a PC to the USB port, if you supply a 5 V to the "5V" pin which is somewhat greater than the 5 V on the USB from the PC, it may feed current back into the PC's USB interface which may cause it to malfunction and possible even cause permanent damage not remediable by powering it down. Probably a greater risk for laptops where "powering it down" also means removing the battery.
I find this a trifle improbable considering most "powered USB hubs" do precisely this, connecting the auxiliary 5 V from their plug-pack directly to the USB input and we do not seem to hear about problems with this - or maybe it is just not publicised.
Anyway, the take-home message, power your Arduino with 5 V to the "5V" pin with the same "buck" regulator you use to power all the other 5 V devices in use. Forget that "Vin" or the "barrel jack" ever existed.
And that is precisely the reason. and no other
Mind you, those old 9 V supplies were not regulated and could rise to well over 12 V when lightly loaded due to their design, let alone a 12 V supply, so it would be rather risky particularly for "clones" using regulators with lesser ratings.
It is a sad "legacy" design. The more recent Arduino designs include proper switchmode regulators.
Oh yes, that is indeed totally wrong for the Nano. There is no problem whatsoever with powering the Nano while USB is connected to a PC or any other 5 V supply.
You could even - if suitably insane - connect a supply to "Vin" as well. Maybe a smoke alarm battery?
Based on taking physics digital labs a while back I understand your comments now. Thank you !!!! I am sorry I'm new and had not yet located the arduino circuit schematics you found for me kindly. My mistake.
I believe my comments also apply to the Nano. (again, V(potential difference)=(5V-5V)=0V)
Why wasn't the arduino just designed to accept 5V at the barrel jack if it wastes energy converting it to the 5V rail? Maybe so it could be powered by 9V batteries?
For starters, the Barrel Jack is by definition the "unregulated input" to the 5V regulator.
Secondly , It wastes energy because it is a linear regulator. If you are that much of an
energy conservationist, you can replace it with the Oki pin for pin compatible switching
dc to dc converter. Simply desolder it or cut the pins and solder the Oki to the same pins.
To be honest I didn't look at the Nano schematic but what I said about the 5V pin being both an INPUT OR an OUTPUTstill applies.
CASE 1: 9V connected to Vin => 5V pin =OUTPUT
CASE 2: Vin=NC, EXTERNAL 5V connected to 5V pin=> 5V pin = OUTPUT.
Everything I said about the barrel jack does not apply since there is none.
Hence, I did not need to mention "Nano" in my post since the relevant part
is still true.
"Word salad ?"
Really ?
Why ?
The onboard regulator has no power with no Vin . hence there is no output voltage. @LarryD,
if the USB is connected then there is 5V on the left side of the mosfet.
If you then connect external 5V to the 5V pin then the mosfet has 5V on BOTH
sides. the difference of potential across the mosfet = 5V-5V = 0V. No current can
flow from one side to the other. I have heard of many posters that have used external 5V connected to the 5V pin with no issues.
What then is the reason for saying this is not ok ?
If there is a valid reason then the external 5V could be left disconnected while
downloading sketches and then the USB disconnected before connecting the
external 5V but then this makes it impossible to use the Serial Monitor unless
you leave the external 5V disconnected when using it. I am just curious about
your reasons for stating that. I believe Paul_B agrees with me (as much as he
would hate to admit it) based on his statements in Reply #12:
Powering the 5 V pin with 5 V cannot damage the Arduino in any way at all. There is a concern.
I agree with him (as much as I hate to admit it) that the following is nonsense:
"5V ... etc ... supplying voltage via the 5V or 3.3V pins bypasses the regulator and can damage your board. We don't advise it."
