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Topic: Connecting 5V VCC to an Arduino Pro Mini (or to any Arduino) (Read 6639 times) previous topic - next topic

lkz7786

Hello, First time posting here and I am brand new to Arduinos!

I have a 5V Arduino Pro Mini and I see on the schematic that I can either supply a VIN to the regulator which will output 5V VCC to the board, or I can provide a regulated 5V VCC directly into the board. 

My question is, does it hurt the voltage regulator to connect an external VCC voltage to it when it is not powered at its input pin? In my limited electronics experience it seems that it would be bad to connect a voltage to an IC's output pin.  Does current flow into the regulator, and if so how much current?  Would it be better to disconnect the regulator if I never plan to use it?

Thanks for any help.  Many more questions to come...

Wawa

From the Products page here: The Arduino Pro Mini can be powered with an FTDI cable or breakout board connected to its six pin header...

5volt USB power connect directly to VCC this way.

The Pro-Mini pdf shows a solder jumper to remove regulator and onboard LED for battery powering.
https://www.arduino.cc/en/uploads/Main/Arduino-Pro-Mini-schematic.pdf
Leo..

OldSteve

That covers the Mini, I guess, but in the case of the UNO, it is recommended to not supply 5V power via the 5V pins because it can damage the board. I read that here, about half-way down the page:-
Arduino UNO & Genuino UNO

But looking at the schematic for an Arduino, that's exactly what powering the board from USB does, as mentioned by Wawa above. That's providing there is definitely no external supply connected to Vin, since it does bypass the isolating MOSFET.
In either case, the 5V is still applied to the output pin of the 5V regulator.

I guess if there's any chance at all that a voltage might be applied to Vin at the same time as 5V is supplied by an external source, the external 5V could be supplied via a USB plug, to the USB port, so that it is still isolated by the MOSFET.
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bwetman


   
I am planning to do the same thing, so I ask: in the end did you use the 5V pin or the USBVCC pin, or something else?

OldSteve

I am planning to do the same thing, so I ask: in the end did you use the 5V pin or the USBVCC pin, or something else?
Who are you asking, the OP or me?

If me, I'm supplying 5V directly to the 5V socket on the Arduino, but I never connect an external voltage to Vin.
I'm using a 7.2V NiMH battery, and using an external LM2940 low-dropout regulator to ensure that I can extract the maximum from the battery between charges without worrying about the 5V rail sagging. The LM2940 has a dropout voltage of only about 0.5V. The Arduino regulator has a dropout voltage of about 1V. (I'm drawing about 600mA, mainly powering two small continuous-rotation servos.)
Please do not PM me for help. I am not a personal consultant.
And others will benefit as well if you post your question publicly on the forums.

lkz7786

That covers the Mini, I guess, but in the case of the UNO, it is recommended to not supply 5V power via the 5V pins because it can damage the board. I read that here, about half-way down the page:-
Arduino UNO & Genuino UNO

Thanks for the link and here's what it says:

"5V.  This pin outputs a regulated 5V from the regulator on the board. The board can be supplied with power either from the DC power jack (7 - 12V), the USB connector (5V), or the VIN pin of the board (7-12V). Supplying voltage via the 5V or 3.3V pins bypasses the regulator, and can damage your board. We don't advise it."

That's pretty vague. They don't say how it could damage the board or what part of the board it could damage.   I wonder if they mean applying a voltage to the output pin of the regulator can damage the regulator.  I am still just as confused as before...

CrossRoads

Yes, the regulator is sensitive to being back driven. Add a diode from 5V (anode) to Vin (cathode) so the regulator is not 100% back driven.
The regulator on the Duemilanove was more rugged in this regard.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

raschemmel

The difference is the USB 5V is disconnected from the board when you plug an external supply to the barreljack. Connecting directly to the 5V pin bypasses that safeguard.
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

CrossRoads

Promini does not connect to USB directly unless one has an FTDI Basic or equivalent connected.
I have a dozen Promini's running at my Fencing club, powered with 5V on the 5V pin. Nothing else added, nothing connected to Raw.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

Paul__B

This (back-feeding the regulator) will only ever be a problem if a 5V supply with an output capacitor is powered up and then connected to the Arduino.  If the regulated supply is connected before it is powered on, its rise time will not be a problem.

Clearly, operation from USB always means that the regulator is being back-fed and is never a problem.

The Vin capacitor on the Pro Mini is of course, much smaller than on a UNO or Mega.


raschemmel

You're right, I was thinking of the UNO. My bad.
Quote
Nothing else added, nothing connected to Raw.
That means the onboard 5V regulator has no input voltage so there is no conflict since the onboard regulator is not running.

FYI, for newbies, "backfed" is an electronics term  for "connecting the input voltage to the output pin instead of the input pin of a regulator"

Adding the diode Crossroads mentioned means the diode will conduct with any voltage greater than 0.7V causing the input of the regulator to have the same voltage as the output - 0.7V.
This limits the backfed voltage to one diode drop.
Arduino UNOs, Pro-Minis, ATMega328, ATtiny85, LCDs, MCP4162, keypads,<br />DS18B20s,74c922,nRF24L01, RS232, SD card, RC fixed wing, quadcopter

Paul__B

You're right, I was thinking of the UNO. My bad. That means the on-board 5V regulator has no input voltage so there is no conflict since the on-board regulator is not running.
That is not the problem.

Wawa

Older 78xx/LM317 type regulators had an internal diode across in/out.
Not sure if newer 1117 type regulators have one. It does not show on diagrams I have seen in datasheets.
Maybe someone (not me) could ground Vin, and put a current controlled 5volt supply on the output of the regulator.
See if it releases the magic smoke.
Leo..

DrAzzy

As far as I can tell, some 1117 regulators don't mind it, while others will fail if you do that. They're not all the same (In working with some of the boards I sell on tindie, I discovered that one brand of regulators would get unhappy if I powered the output but not the input and started heating up. Other brands seem to work fine (I now use the ZLDO1117 series, which seems okay with it - and the datasheet indicates that under most conditions, this is supposed to work)
ATTinyCore for x4/x5/x61/x7/x8/x41/1634/828/x313 megaTinyCore for the megaavr ATtinies - Board Manager:
http://drazzy.com/package_drazzy.com_index.json
ATtiny breakouts, mosfets, awesome prototyping board in my store http://tindie.com/stores/DrAzzy

OldSteve

Older 78xx/LM317 type regulators had an internal diode across in/out.
How old? I've been using 7805s and LM317Ts for decades and in that time they've never had an internal diode. It's always been up to the circuit designer to add a diode if output capacitance is over about 10uF.
I have a Motorola data book, (yep, real paper), from 1993 with a block diagram in it, and there was no internal diode then.


Quote from:  DrAzzy
As far as I can tell, some 1117 regulators don't mind it, while others will fail if you do that
It depends on the output capacitance. For ≥1000uF, a diode is recommended. For lower capacitance values, the datasheet says that protection is not needed.

From the horse's mouth:-
Quote
PROTECTION DIODES
Under normal operation, the LM1117-N regulators do not need any protection diode. With the adjustable device,
the internal resistance between the adjust and output terminals limits the current. No diode is needed to divert
the current around the regulator even with capacitor on the adjust terminal. The adjust pin can take a transient
signal of ±25V with respect to the output voltage without damaging the device.
When a output capacitor is connected to a regulator and the input is shorted to ground, the output capacitor will
discharge into the output of the regulator. The discharge current depends on the value of the capacitor, the
output voltage of the regulator, and rate of decrease of VIN. In the LM1117-N regulators, the internal diode
between the output and input pins can withstand microsecond surge currents of 10A to 20A. With an extremely
large output capacitor (≥1000 μF), and with input instantaneously shorted to ground, the regulator could be
damaged.
Even without the above, the UNO does it all of the time when USB is connected and there is no voltage applied to Vin, and I don't see them dying all over the place from back-feeding the regulator.

So, it is perfectly acceptable to connect 5V to the 5V pins, as long as nothing is connected to Vin and output capacitance is comfortably below 1000uF. I'm doing nothing wrong. :)
Please do not PM me for help. I am not a personal consultant.
And others will benefit as well if you post your question publicly on the forums.

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