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Author Topic: External supply at 5v pin  (Read 2247 times)
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The first photo is the Leonarde R3 board. The soldering is too close to the via. I will redo that when I fix my other Arduino boards.

The diodes don't need to be fast switching, so I used normal 1N4007.


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« Last Edit: January 09, 2013, 02:42:17 pm by Krodal » Logged

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I went back thru my posts from Jan 2011 when I thought this came up (I have 948 pages of posts!) but could not find the original discussion.
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Designing & building electrical circuits for over 25 years. Check out the ATMega1284P based Bobuino and other '328P & '1284P creations & offerings at  www.crossroadsfencing.com/BobuinoRev17.
Arduino for Teens available at Amazon.com.

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I think that the diode throught the regulator is a good solution. Thanks CrossRoads.
I put a 1N4001 from 5Vpin to 12Vpin.
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Two more photos.

The first photo is the Arduino Nano V3.0.
A 1N4007 was too big for this, so I used a 1N4148. They are only 200mA, but can handle a repetitive peak of 450mA.
I glued the 1N4148 to the board, and after that I soldered it to the voltage regulator.

The second photo is the Arduino Mega.
Next to the ground connection is a smd capacitor. So the diode is up in the air. I used glue for mechanical strength.


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Good job Krodal.

Why you didn't use an 1N4001?
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Why you didn't use an 1N4001?
I just grabbed what was lying in my soldering place.
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I am wondering whether the caution added for the Uno R3 was really to protect the voltage regulator.  Here's what ON Semiconductor says about this:
Quote
Protection Diodes
The NCP1117 family has two internal low impedance diode paths that normally do not require protection when
used in the typical regulator applications. The first path connects between Vout and Vin, and it can withstand a peak
surge current of about 15 A. Normal cycling of Vin cannot generate a current surge of this magnitude. Only when Vin
is shorted or crowbarred to ground and Cout is greater than 50 F, it becomes possible for device damage to occur.
Under these conditions, diode D1 is required to protect the device.
Given that the VR input on the Arduino board should never go to ground, the built-in protection (15 AMP) seems to me more than adequate.

Perhaps the reason for not wanting you to attach power to the 5V pin is to protect the ATMEGA16 used in place of the USB-Serial converter used on older boards?  If that's the case, your bypass diode cure is worse than the disease!  Perhaps someone involved in Uno R3 development could chime in here.

The Nano doesn't use the Atmega16, and there's a blocking diode between the USB V+ and the 5V line.  That's why the Nano's USB converter can't be powered from Vin, but just from the USB jack, but it also means that externally-supplied power at the 5V pin can't in any way damage it.  TI's data sheet for the UA78M05 makes absolutely no mention of reverse current flow problems.  In other words, for the Nano while a bypass diode probably won't cause any harm, it just doesn't seem to serve any purpose either.

Ciao,
Lenny
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I also looked at the NCP1117 datasheet and came to the same conclusion.  With care it should be ok to use +5V pin on the Mega2560.  With care means look at 10 ways to Destroy an Arduino - http://ruggedcircuits.com/html/ancp01.html and dont try any of these! 
I also tried for a short period of time to use both external 5v regulator and the USBVCC power with no obvious problems.  It is clear that the USBVCC switch will be switched on because VIN will not have sufficient voltage to switch it off so both USBVCC and the external 5v regulator will be supplying the +5v rail and thats not good and you will need to protect your external regulator.  I would not run with both USBVCC and an external 5v regulator at the same time because you may not be lucky.  Also ensure that you use an isolated external supply because you dont want to have the external supply to have a different ground potential to the USB port.
« Last Edit: April 06, 2013, 03:57:08 pm by peteasa » Logged

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Quote
The Nano doesn't use the Atmega16, and there's a blocking diode between the USB V+ and the 5V line.  That's why the Nano's USB converter can't be powered from Vin, but just from the USB jack,

That is simply not the case. Look at the nano schematic ( http://arduino.cc/en/uploads/Main/ArduinoNano30Schematic.pdf )

The FTDI USB serial converter chip's Vcc pin is powered from the +5 bus, the same as the AVR 328P chip. So the FTDI chip is powered up from either the on-board +5vdc regulator or the VUSB the same as the AVR chip.

Lefty
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Ok, so everyone agree that it's a bad idea to connect a power source to the 5V connector.

But what about VIN? Does it have any more protection than the 5V pin?
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Hey guys,

I want to push this topic, because I'm in trouble with external 5V to 5V pin. I'm using an arduino mega.

So far I have tested the following: my pcb is consuming about 60mA. The Mega is consuming also about 60mA. Whenever I connect now the 5V Regulator on my pcb to the 5V of arduino both consume 250mA. Some times the regulator on my board is getting hot or sometimes the regulator on the mega is getting hot. So it means there is something wrong!
Same occurs when I'm connecting the 5V of my pcb to the arduino connector for external power, i.e supplying 5V from both sides to the arduino regulator.
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Here is a thread about this. Using the 5v pin to supply power and connecting a usb cable (powered) to the Mega bricks the usb IC.
http://forum.arduino.cc//index.php?topic=82046.0

To prevent the bricking, you should use a supply voltage higher than 6.6v into the Vin pin or power jack. That will disconnect the usb and 5v power buses.
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I only want to confirm the real solution is THE USB connector . I just did it a minute ago.
I divided a USB cable into two pieces.  Stripped the wires on both sides, because I liked to know which wires I needed for the 5 volt.
I my case, it was the red and the black wire. Checked the 5 V (4,81 V ) on the computer side (MAC) so I knew which wires I I had to connect at the USB connection,  direction Arduino .

The 5 Volt (5,64 V)  from an external pc power supply was enough, because on the USB entrance of the arduino there is a diode in serial . I did have the time to measure the 5v on the arduino, because I liked to tell you this.


Thanks for the idea. I works!
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