Powering Arduino from 5V source + usb

Hi,

I'd like to run a Duemilanove off an external regulated 5v supply but I'm worried that if I also plug the USB in (without disconnecting the 5v supply) I'll fry something...does anyone know if it's possible to do this?

I'm planning on plugging the 5v directly into the 5v jumper and not the Vin jumper to avoid the dropout from the regulator.

Cheers,

Chris

The Duemilanove is not prepared to run with external 5 Volt. The board can powered from the USB port or from an external power supply connected to the power jack (2.1mm) or the Vin-Pin.

The recommended range of the external power supply is 7 to 12 volts.

If you need a 5 volt version of Arduino then you can use the Arduino Pro.
A Solutions with 5 volt power supply for Duemilanove is only possible if you do some soldering. Only recommend for professionals!

You should use the power jack and then move the small jumper net to the USB jack.

The Duemilanove doesn't have the power jumper, it autoselects which power to use (DC power socket / USB).

Thanks, I guess I'll have to find some way of getting at the unregulated 9V then

The Duemilanove is not prepared to run with external 5 Volt.

Really? Is that documented/explained anywhere? I've been wondering how the new power circuit interacted with +5V via the power connector, but hadn't seen anything definitive.

I guess that technically, none of the USB arduinos were ever designed to be powered via the +5V power pins (and especially not with USB connected at the same time), but there were reports that it worked...

You can probably rig up something that feeds in +5V via the USB connector (disconnecting the USB power and using a common ground...)

UUUpps i was lookin at one of my way old boards

Really? Is that documented/explained anywhere? I've been wondering how the new power circuit interacted with +5V via the power connector

Technically it works with +5 Volts at the power jack or at the Vin. But the recommended range is from 7 to 12 Volts. The new arduino board uses a low drop voltage regulator. That means the input voltage must be only a little higher as the output voltage (5V).
If you run it with 5 volts it is not in a stable range and you can get troubles with your circuit.

I couldn't find any documentation but I checked the schema how the power supply is implemented.

We (or I, anyway) was talking about feeding a regulated 5V into the "5V" pins on the bottom "power" shield connector. This puts it in past the regulator circuit, similar to if it had come from the USB connector. In the old days, you could keep the USB power switched off manually and I think things would be OK. I'm not sure about the new scheme, though...

We (or I, anyway) was talking about feeding a regulated 5V into the "5V" pins on the bottom "power" shield connector

That's not possible because the Duemilanove has a power detection circuit that checks for power on pin Vin. Vin is high if power comes from power jack or from Vin pin.
Otherwise the +5 V is connected to the USBVCC-Signal and the board will be feeded from USB.

There is no more switch to manually switch between USB and external power supply.

I would not connect an external power supply directly to +5V while USB is connected. Rather I would look into what "westfw" suggested:

You can probably rig up something that feeds in +5V via the USB connector (disconnecting the USB power and using a common ground...)

There should be no issue however powering Duemillanove with a regulated 5V supply to +5V as long as USB is not connected.

The Duemillanove regulator drops only about 1V (Vin to +5V), but the switchover cicuit senses on Vin/2 versus 3.3V from USB. So as long as Vin is less than 6.6V, you should not connect the board to USB.

The power jack drops an additional 0.7V across the reverse polarity protection diode so this would add up to a minumum of 7.3V for a proper switchover between USB and onboard power. And as always, a safety margin wouldn't hurt.

The Duemillanove regulator drops only about 1V (Vin to +5V), but the switchover cicuit senses on Vin/2 versus 3.3V from USB. So as long as Vin is less than 6.6V, you should not connect the board to USB.

Don't forget that 'Vin' is defined after series diode 'D1' so we are talking about an input voltage 'PWRIN' of about 0.5 to 0.7 volts higher than the 6.6 V that is mentioned above.

Don

I really feel that if one wants to power the board with an external regulated +5vdc, the cleanest way is to use a spare hacked USB cable and wire the voltage to the two USB power wires and plug it into the board's USB connector.

Lefty

wire the voltage to the two USB power wires

The problem with this is that one of the reasons to use an external 5V supply in the first place is so that you can have more than the 500mA allowed through USB. If you feed in a supply via the USB power wires, you also go through the 500mA fuse...

I guess the other reason for the question involves a battery powered device that you want to connect to a computer "occasionally." It would be common and desirable to have the battery power the arduino via the +5V connection (run straight off batteries, and/or with a higher efficiency regulator than possible via the regulator and Vin.) The automatic switching makes this troublesome (probably you don't want USB feeding the battery supply backward! Could cause fire depending on exact circuit and battery technology!)

After reviewing the schematic of the Duemilanove (http://arduino.cc/en/uploads/Main/arduino-duemilanove-schematic.pdf), I see no reason there would be a problem running the board from a regulated 5V supply on the 5V pin at the same time as the USB. I've done it a few times myself, even before looking at the schematic.

If you try to plug a 5V supply into the Vin or the power connector, the board will run from the USB power instead because it auto-selects the best power source between USB and Vin/power connector. The 5V pin is not part of the auto-selection circuit though, so it will always be in use, regardless of whatever other power comes into the board.

There is a slight chance of back-flowing voltage into your USB port if the regulated 5V is slightly higher than the USB 5V. But I would expect the USB port to be protected from that.

Hi,

this can be a problem, because the +5V from USB is switched off when Vin is at 6V or higher. If you supply +5V directly to the +5V pin, the voltage regulator gets reverse voltage and also the +5V from the USB is connected to the +5V from the external supply. This can work but can also damage your board, so please be careful.

Mike

I sometimes think the original Arduino design of using jumper pins to select the voltage source was a cleaner method rather then the auto-voltage selection circuit. The Seeeduino current design is also more flexiable using a two position switch for voltage selection. In that case the switch can also be used as a power on/off switch if one of the two voltage sources are not plugged in.

So for example if one had an older Diecimila board and just removed the power selection jumper, then one could wire a regulated external +5vdc supply to the +5vdc connector pin and there would be no path back to the USB port +5vdc, no path back to the internal +5vdc regulator, no current limiting fuse, etc.

Lefty

I sometimes think the original Arduino design of using jumper pins to select the voltage source was a cleaner method

Agreed. If you want to see a versatile power supply configuration then take a look at the Bare Bones Board at www.moderndevice.com. Scroll down to the bottom of the page and download the Instructions for Bare-Bones Board Rev. E. The power supply configuration (done with two 3-pin jumpers) is shown in the schematics on pages 6 and 7. The earlier boards had the same configuration but the documentation was updated for this board revision.

Don

Hi, this is my first post to the forum, a great source for solutions !

I am finishing a robot powered by a 9.6v Nimh battery. a LM317 brings this source down to 5v (1.5A max), that I use to power both the duemilanove (directly through the 5v Pin), and the sensors and servos.

My problem is that when I plug the USB cable in, mainly for serial debugging, the Arduino draws power from the USB, not from the external 5v (which is normal, as I understand it after reading this post, since there is no Vin - thanks BTW).
This is not a problem per se, everything works (maybe I was lucky), but the 5v from my USB port is in fact 5.06V, and fluctuates much more than my external 5v. This prevents me from having reliable sensor values on the computer.

In other words, I want the USB cable to transfer data, but not power. What is the best way to do this ? cut the red wire in the usb cable ? or cut a trace on the duemilanove ?

Thanks in advance :slight_smile:

Manu (from France, sorry for my so-so english)

I would remove R10, and connect where R10 went to R11 to 5V. This will trick the circuit into thinking it is being powered by a large voltage and switch away from the USB power.
In fact thinking about it there would be no need to remove R10, just put in a link, then this is easy to remove to get it back to its original state.