Best way to provide 5v 2A battery power

Other Hardware General Electronics
21

afremont:
Interesting. I looked at the datasheet for the regulator and it does indeed have an internal protection diode connecting Vout to Vin. It is capable of handling a 50A surge which is far in excess of what might be encountered in normal cycling of the device. In case the Vin is shorted to ground and Vout contains more than about 50uF of capacitance could this limit be exceeded requiring another external diode. I think reports of connecting a truly regulated 5V supply to the +5V pin causing damage are highly suspect.

It is a puzzle, but a few posters have reported their boards damaged doing it soon after the rev3 boards were released, but who knows the true root cause?

It just got more interesting when out of the blue (or maybe based on analysis of returned damaged boards?) the Arduino folks added the 5V pin warning on several of their current board offerings.

Lefty

22

Sounds like more of a CYA thing than anything. It's obvious that you are right in that the USB power ends up on the +5V rail, but it appears that happens from within the mega32u usb chip? At any rate, +5V is being applied to the output of the regulator and not the input when powering from USB. The datasheet makes it clear that this shouldn't cause a problem with the regulator either. I suspect that maybe these "regulated" supplies provided an initial unregulated spike at power on.

23

afremont:
Sounds like more of a CYA thing than anything. It's obvious that you are right in that the USB power ends up on the +5V rail, but it appears that happens from within the mega32u usb chip?

No, the USB5v is routed to the AVR 5vdc bus via the FET isolation switch, which is on when no Vin is avalible. mega32U gets it's Vcc from same bus as the 328P does.

At any rate, +5V is being applied to the output of the regulator and not the input when powering from USB. The datasheet makes it clear that this shouldn't cause a problem with the regulator either. I suspect that maybe these "regulated" supplies provided an initial unregulated spike at power on.

I thought CrossRoads researched the datasheet for the regulator in question and it did show a recommendation of a 'bypass diode' wired across the regulator if there was a chance in the circuit for the output voltage to ever be higher then it's input voltage. I haven't the specific regulator part number so I've never looked over it's datasheet personally.

So what is your opinion on what to advice when people ask if they can do that? Should one mention of the 'official' arduino warning be given? I don't own a Uno, just a half dozen or so older arduino boards (mix of 328 and 1280 boards plus a few custom jobs). I have no qualm of doing it myself on my boards, but telling others to do it on a board model I haven't played with, plus with the official arduino stance on the issue, makes responding to this often asked question a little 'problematic' at least for me.

So I just tend to raise the question of why the arduino warning and sit back and watch the fun. :smiley:

24

That stuff I wrote about the internal protection diode, Vin shorted to ground, 15A of surge current and 50uF came straight from the datasheet for the part. It recommends an external diode if you are planning to exceed those limits. Since it has an internal diode, it's pretty much impossible to reverse bias the device.

Ok, I found the other "USBVCC" on the schematic that flows thru the P-channel FET. I really dislike disjointed schematics.

25

afremont:
That stuff I wrote about the internal protection diode, Vin shorted to ground, 15A of surge current and 50uF came straight from the datasheet for the part. It recommends an external diode if you are planning to exceed those limits. Since it has an internal diode, it's pretty much impossible to reverse bias the device.

Ok, I found the other "USBVCC" on the schematic that flows thru the P-channel FET. I really dislike disjointed schematics.

Got a link to the datasheet?

26

Let's see if this works. Look on page 10.

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&sqi=2&ved=0CE0QFjAB&url=http%3A%2F%2Fwww.onsemi.com%2Fpub%2FCollateral%2FNCP1117-D.PDF&ei=_AlqUYH_E8iIqQHe8IHgAg&usg=AFQjCNFNMljD1LWWuTQsGLIaVjTArGbNEA&sig2=r94Vns_KfZ8sjoaVOkQUFw&bvm=bv.45175338,d.b2I&cad=rja

27

Obviously the last diode drops worth of voltage won't be routed back, but will not cause a problem. If you measured +8V on the +5V pin with +12V connected to the barrel connector then something wasn't measured right. +8V would easily destroy the CPU.

The voltage was measured correctly. I measured the voltage when I noticed the LEDs on the arduino were noticeably brighter when connected to the +12v external supply. 9v on the external barrel connector produced ~7v on the arduino +5v pin. The arduino was not damaged, possibly due to quick detection of the difference. The original arduino external power supply setup seemed to me somewhat "tutti-frutti".

28

I believe that the MAX voltage is 6V, so I could see it scraping by at 7V or less, but that would really be tempting fate. I would really expect it to be all over, but the crying, at 8V.

29

afremont:
I believe that the MAX voltage is 6V, so I could see it scraping by at 7V or less, but that would really be tempting fate. I would really expect it to be all over, but the crying, at 8V.

It is what it is. Electronic devices usually "work the way they are wired" (both arduinos and multimeters) and usually ignore inexact thoughts of "I believe that the MAX voltage is 6V, so I could see it scraping by at 7V". If 6v is the "max", how would the arduino scrape by at 7v? Curious.

30

Well how about "the Absolute Maximum operating voltage is exactly 6.0V" as stated by the datasheet. But that will not immediately destroy the device. Anything above the 6V specified in the datasheet qualifies as tempting fate and scraping by in my book. 7V certainly fits that description so..... I assure you that the Arduino was never intended to run on anything that high. I don't know what kind of regulator you were using, but if it was letting 7 or 8V thru then it wasn't working as a "5V regulator", regardless of what you might believe.

EDIT: To try and better answer your question, there is no stated specification whereby the chip is guaranteed to be damaged or destroyed. There is only the Absolute Maximums which are not supposed to immediately kill the device, but are not guaranteed to keep doing so in the long run. Can you now see why I would consider 7V or 8V survival as scraping by or tempting fate? 8V would be exceeding the Absolute Maximum by 33%....

Also, when I say that "I could see" something, I'm generally saying that I might believe that it's true, but that I'm still somewhat skeptical about the point. So, yeah I could believe that it's possible that 7V didn't kill the board, but I wouldn't count on it to apply across the board to all devices manufactured. That's why I said it scraped by. I still would like to see a video of a chip running on 8V.

31

From my experience, I don't think the reverse-protection diode is much of an issue. I can't
remember seeing any actual boards with those diodes on there, but have seen dozens or 100s
of examples without them, and with no doubt 10s or 100s of 1000s of those boards in the field,
including my own.

The datasheets below indicate there is only a real problem if the input to the v.reg is shorted or
instantaneously crowbarred to ground, and you have a rather large cap on the output terminal.

Few of us ever use crowbar circuits, especially ones with 10-20A capability, and how often do
we short the battery/etc straight to ground to remove power [think: big spark, fried battery
wires, exploded battery], as opposed to simply unplugging it, and which isn't going to produce
an instantaneous 10A crowbar to ground. Also, I'd bet the vast majority of halfway decent
designs are going to have an electrolytic cap on the v.reg input terminal, which will limit the
Vin discharge rate right there.

ref LM7805 d/s:

reverse-bias protection
Occasionally, the input voltage to the regulator can collapse faster than the output voltage. This can occur, for example, when the input supply is crowbarred during an output overvoltage condition. If the output voltage is greater than approximately 7 V, the emitter-base junction of the series-pass element (internal or external) could break down and be damaged. To prevent this, a diode shunt can be used as shown in Figure 7.

ref LM1117 d/s:

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 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 ?F), and with input instantaneously shorted to ground, the regulator could be damaged. In this case, an external diode is recommended between the output and input pins to protect the regulator, as shown in Figure 4.

ref NCP1117 d/s:

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 uF, it becomes possible for device damage to occur. Under these conditions, diode D1 is required to protect the device.