Arduino Nano Power and Programming - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 04:29:15 pm
content: I'm using an Arduino Nano Evo to read from some sensors, but it's also powering these sensors as well through its +5V pin. This can be circumnavigated by using an external DC voltage source and feeding it through the VIN pin (so, I don't have to worry about current limitations and whatnot from the Nano's pin). However, it says that doing so will cut off the USB power source. In this case, how do you program the Arduino Nano Every? Can you still program it via USB?EDIT: Initially, I wanted to ask about powering sensors through that +5V pin, but the FAQ on the Nano Every's page seemed to provide me a better way to power the Nano and my sensors, so I'm retooling my question to ask about another concern of mine. Though, the FAQ does say that you need to respect the voltage limits of 7-12 volts...makes me wonder if you couldn't just power the thing with a USB-5 volt wall plug.https://store.arduino.cc/usa/nano-every

author: JustMeHere
author_rank: Frequent Contributor
timestamp: March 30, 2021, 04:53:14 pm
content: The long USB cable will not provide good response to good responses to transients. I run my USB powered projects at 3.3V. This allows me to put a local 5V to 3.3V regulator on the device.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2021, 05:18:30 pm
content: Theschematictells all. Vusb (USB Vbus) feeds the +5V net via D2, aPMEG6020low Vfv Scottky diode. If its running from USB power, '+5V' will be approx 0.2V to 0.3V below Vbus at the device, which will almost invariably result in a voltage slightly under 5V. If present, Vin will be regulated down to 5V by IC1, theMPM3610synchronous buck converter module. Thus when Vin is present, D2 will either be reverse biased or only slightly forward biassed passing negligible current.The USB interface MCU IC3 is powered by the +3.3V rail, which is derived from the +5V rail, by U2, a linear LDO regulator. It gets power no matter the power source so USB should be fully functional when you power it via Vin or even if you feed an external 5V supply to the +5V pin.However, the buck converter's '+5V' output is much more capable than that of the regulator on previous Arduinos, and depending on cooling, should be good for hundreds of mA. As its got an on-board switching regulator, it wont have the overheating problems with 12V in and significant load on +5V that 'classic' Arduinos have. Due to the D2 voltage drop, powering it from a 'USB charger' wallwart would be a bad idea if you care about the accuracy of the 5V supply to your sensors. Use a 9V or 12V wallwart on Vin and you'll have no problems.N.B. Vin does *NOT* appear to be protected against reverse current flow when powered by USB Vbus. If you have other loads that operate from Vin (e.g. higher voltage motors, relays etc.), a Schottky diode in series with Vin would be advisable to prevent damage to the buck converter.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 05:26:58 pm
content: Thanks for your reply, but I'm a bit confused by your response. Do you mean powering the Arduino through its USB port, or converting the USB down to 3.3 V with a regulator like you said and powering the Arduino that way? Because the Arduino can't take in 3.3 V as an input to power it, unless I'm mistaken.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2021, 06:05:18 pm
content: Look at the schematic!IC3, the ATSAMD11D14A USB interface, (which presumably runs a variant of the usual Arduino USB CDC <=> UART bridge firmware) needs a 3.3V supply. Its serial I/O is level shifted up to 5V levels for U1, the main ATmega4809 MCU, by a bunch of BSS138 MOSFETs. You can be certain it will have enough voltage to work as long as the '+5V' rail is over 4V.However the ATmega4809 requires a *MINIMUM* of 4.5V to run at 20MHz (see p11 of its datasheet, and will get that as long as Vin is in the specified range, or USB Vbus>4.8V. It will be marginal if the +5V rail is heavily loaded and the upstream port Vbus is at its lower permitted limit for a 1 unit or greater load of 4.75V. It will fail abysmally if a non-compliant high resistance USB lead is used.TLDR: No, you cant power this Arduino at 3.3V, the absolute lower limit is 4.5V on its '+5V' pin.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 06:09:42 pm
content: Thanks for your reply. Looking at other forum posts, they don't really recommend powering the Arduino through their +5V or +3.3V pin, though they were referring to the Uno and not necessarily the Nano. The schematic shows that it may be possible; I see your points regarding accuracy and stability, but if we power it through that way using the +5V pin and not through USB (bypassing D2), that looks like it will work though yes, the buck converter is more stable.Adding an on-board regulator like the LM317 is going to be annoying, what with my limited space and all. Perhaps something like the 7805 might be enough,

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 06:25:29 pm
content: Quote from: Ian.M on March 30, 2021, 06:05:18 pmThanks for replying, but my first message was in response to JustMeHere. I was a bit confused by his post, so I replied before I read your first reply. Nevertheless, thank you for clarifying.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2021, 06:25:50 pm
content: How many amps do your sensors need at +5V? I strongly suspect that the MPM3610 on the Arduino will be good for several hundreds of mA,(as its a nominal 2A buck module, but there probably isn't enough copper area to heatsink it it get that much out of it), if you feed it 9V or 12V Vin, so you *SHOULDN'T* need an external regulator.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 07:00:02 pm
content: Quote from: Ian.M on March 30, 2021, 06:25:50 pmRight, sorry, that idea flew by me at first, but I thought about it more, and everything clicked. Powering via VIN appropriately, we use the +5V pin on the Arduino to power the rest of the circuit. We're not limited like the digital I/O pins, but we're only limited by the buck converter's limitations. Datasheet rates it for 1.2 amps of continuous load current.The eight I2C sensors (LSM9DS1) I'm using seem to take at best 4.3 mA normally (so from the sensors, that's 34.4 mA).The I2C switch (TCA9548A) doesn't seem to take up a lot of current for what it does.The big thing is the PCA9615 IC. The datasheet for it says that the input supply currents is 2 mA max for port A and port B, but the driving termination is typically 70 mA. Depending on that, the buck converter may or may not be enough to power the whole thing.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2021, 07:54:36 pm
content: It should have sufficient +5V rail current capability for the PCA9615 IC.Try it with your finger on IC1, the MPM3610 buck module (diagonally adjacent to the reset button). If it doesn't get too hot, it will be fine.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 08:14:48 pm
content: Understood. Thanks again for your help. Your comments have been very insightful.BTW, if you don't mind me asking, regarding the PCA9615, when it refers to the driving termination, how does that factor into the current consumption of the IC? Compared to the other parameters, that parameter indicates a lot of current, so how does that play in? I realized I neglected to mention how there's 16 of these ICs (2 per sensor), so understanding how the driving termination works may help reduce the amount of current drawn.

author: Ian.M
author_rank: Super Contributor
timestamp: March 30, 2021, 08:33:05 pm
content: 16 of them is probably too many for the arduino's 5V regulator. The differential termination resistors at each IC add about 7.5mA quiescent current for the idle state, and for each sensor, the end that's currently driving the cable will need another 70mA, so I'd expect a total draw up around 700-800mA.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: March 30, 2021, 09:10:45 pm
content: Thanks again for your reply. It just seems barely enough. For each sensor, only 1 PCA9615 IC will be driving the signals at a time, either transmitting from sensor->arduino or vice-versa, so worst case, 800 mA might just be fine. ...thinking about it, making a more powerful Arduino Nano (like a Powerduino) sounds like an interesting idea to pursue...seems like all it would take would to be replace the buck converter more or less.