I am using the 3.7V Li Po battery with the charging and protection circuit for powering the Arduino via 5V PIN. My goal is to avoid the 5V PIN and use the VIN PIN.
I added the step-up convertor to reach stable 5V.
But I just found I have misread the Arduino specifications since the VIN PIN needs 7-12 V, while I have only 5V on output.
The Arduino specification is clear, but let me just double check that my layman conclusion, which is that 5V is not enough to power Arduino and I must use another step-up convertor to reach 7V is correct.
The VIN pin requires at least 7V as the input passes through a voltage regulator. If you want to power the board with a 5V supply then connect it to a 5V pin directly
I found this: 5PCS MT3608 DC-DC Step Up Converter Booster Power Supply Module Boost Step-up Board MAX output 28V 2A, less then $4.00 with shipping. The specs are: 1.Electrical characteristics
1). The maximum output current: 2 a
2). The input voltage: 2 v ~ 24 v
3). The maximum output voltage: 28 v
4). Efficiency: ＞93%
With parts like this why use two converters and burn the extra power and cost. This was Aliexpress.
The (largely useless) on-board regulator circuit includes a FET to disconnect the 5 V line from the USB port when powered by "Vin" or the "barrel jack" but this only happens when you provide (relatively minuscule) power this way, so in all other circumstances, the 5 V line is effectively connected to the USB port. This means that supplying a 5 V supply to the "5V" pin that is slightly greater that the USB voltage will tend to feed current back to the PC USB system and some PCs - notably laptops - do not deal with this and the USB system can shut down, possibly incurring permanent damage.
So you should not have the "5V" pin connected to a supply when connecting via USB to a PC with these specific Arduino versions. Others such as the Nano and Pro Mini, use a diode to prevent "backflow" in this manner and are entirely safe to connect both USB and a 5 V supply.
Do note that the only risk is to the PC, not the Arduino itself.
I also note that the majority of "powered USB hubs" also connect their 5 V supply directly into the source USB port, and I am not aware of reports of such a problem.
Why is that unwise? I would like to understand it please. Because of the power loss on convertor?
Quick rough calculation when 3.7V Li Po is converted to 7V considering 80mA current consumption for Arduino Mega and 90% efficiency gives me I = (0.087)/(3.70.9) = 168mA current drain from battery.
The reasons I used VIN PIN are:
Having the possibility to run Arduino on battery and USB at the same time during the development and preventing the risk that notebook USB / motherboard is damaged (as you also commented).
Having the VIN PIN connected I can use the 5V as analog PIN reference, otherwise I need to use the voltage dividers since for the analog reference will be used either the internal 1.1V or 2.56V voltage.
The possible solution I see is having two versions of circuit: One for development, which uses VIN PIN and second that uses 5V PIN and for analog PIN voltage reference use the same 5V input connected to AREF. Would that be a better solution?
Yes, at least the power loss of using a linear regulator to drop back from 7 V to 5 V and the limited current capability of the on-board regulator. If the Arduino is only drawing 80 mA and not providing any significant current to other devices connected to output pins and the "5V" pin, then it will work but when you decide to connect another device requiring significant 5 V current, you are likely to encounter trouble. This is independent of whether you are powering "Vin" from a boost regulator from a battery or any other source.
Boosting to 7 V and regulating back to 5 V just doesn't make sense.
It's dead simple. During development, you disconnect the "5V" pin when (before) you connect to the PC using USB. You have to plug things in and out, what's the difference?
No idea what you are trying to say here.
The (default) analog reference is whatever is on the "5V" pin. If you use the on-board regulator via "Vin", the reference will be the output of that regulator. If you use a boost converter to power the "5V" pin, the reference will be the output of that boost converter. Is it not a regulator?
