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I want to bypass the onboard regulator and use the 5V pin to feed arduino with external, regulated source.

Everybody just keeps saying regulated 5V, but how precise does the regulation need to be exactly? Can you state a voltage range and what makes you think so?

10 ways to destroy arduino, Method #5: Apply >5V to the 5V Connector Pin mentions "applying more than 6V", can anyone elaborate how it can be calculated?

Is there a higher spike that arduino can handle for a short period of time? My concern is that the current draw from arduino might be under 60mA (right after switching power on and before the program initilizes and starts running), which is the lowest draw at which my source guarantees the 5V regulation --- when unloaded it gives out 6.5V. How could I resolve this better than putting a 60mA dummy load between the regulator and arduino itself?

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Unfortunately there is no one "clear cut" answer for all you ask.

There are some hard limits, though, which you can get from the main chip's datasheet:

Operating Voltage: 1.8 - 5.5V

But that's not the whole story. The minimum voltage depends on the clock speed of the chip, as shown in this graph:

enter image description here

But wait, there's more. The Arduino also contains a 3.3V regulator which is fed from the 5V supply. That will have a minimum voltage above 3.3V that it needs to maintain a 3.3V output. So if you want to use the 3.3V regulator then, according to the datasheet for the regulator (LP2985-33DVBR):

3 Description
The LP2985-N low noise linear (LDO) regulator delivers up to 150mA output current and only requires 300mV dropout voltage of input to output.

So a minimum of 3.3 + 0.3V, or 3.6V.

That gives you a usable range, clock frequency permitting, of 3.6V to 5.5V.

As to how to protect the Arduino from a power-on spike like you describe - you could:

  • Add your own extra regulation in the form of a 5.1V zener diode.
  • Add a dummy load which the Arduino can switch out of the circuit once booted up
  • Change your source to something with better regulation
  • Ignore it - the regulation with even a light load will be considerably better than open circuit. Add a resistor across your supply that would draw much less than 60mA and measure the voltage. See how bad the regulation at low load is.
  • I am more worried about the top limits rather than underpowering, but it's interesting thought, I didn't generally expect that low voltage could break stuff too. – jediz Jan 28 '17 at 15:38
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    According to the datasheet the absolute maximum is 6V - above that and you will kill it. The "safe" maximum though is 5.5V, above which they don't guarantee it will keep working reliably for any sustained period of time. – Majenko Jan 28 '17 at 15:39
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    It depends on what else they are wired to. In general you just get stability problems. In some situations, though, such as when you have a 5V signal feeding into a chip that is being under-powered, you may get physical damage from over-voltage on the input. – Majenko Jan 28 '17 at 15:44
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    Too low a voltage can make the micro controller brown-out and do funny things. It might jump to anywhere in the code. So it could theoretically set pins that are meant to be inputs, to output, and cause damage that way. Though the ATMega on the arduino has brown-out detection. So unless you specifically disable it, this isn't a problem. – Gerben Jan 28 '17 at 16:33
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    You want a zener reverse polarity across the supply with a small resistor in series between the supply and the zener (the Arduino is powered from the join between the two). The resistor is to limit the current through the zener and drop the voltage between the supply voltage and the zener voltage. electronics-tutorials.ws/diode/diode_7.html – Majenko Jan 28 '17 at 22:53
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The 6.5V is too much. I have had an ATtiny chip which seems to run fine after applying 8V to it. But how can I be sure that it is okay ? There are so many parts in the chip die that can be damaged.

There has been a few broken voltage regulators reported, when applying 5V to the 5V pin of an Arduino Uno. The reverse diode inside the voltage regulator might not keep up with a large current peak. In some of my projects I power an Arduino via the 5V pin. I have added an extra diode 1N4007 from 5V to VIN, to save the voltage regulator, and my 5V power is not very strong.

I also don't apply that 5V while it is on. It is from a DC/DC converter that is part of the project. So when I turn on the power, the 5V raises not too fast, and there is no high current peak.

When applying 5V to the 5V pin, the current can flow into the computer via the usb cable. The polyfuse of 500mA on the Arduino board will prevent most troubles, but it could damage the computer. When I turn off my computer with that project connected, I do indeed see an increase of the current to my project. My computer is luckely not broken (yet).

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Most decent 5V supplies (e.g. USB chargers) are within 5% of 5V. I use these all the time to power directly to the 5V pin. There has never been an issue with this. I don't use supplies outside 5% for this. Check with your multimeter first.

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You don't say why you want this information, but there are others who are trying to do similar. (Starting with NOT destroying their Arduino)

For instance, at the end of the article

10 Ways to Destroy An Arduino

is a reference (broken link in the article) to

The Ruggeduino

which might be more suitable to your requirements.

Here is an excerpt of its specs:

enter image description here

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