Examining some of the Nano clone boards I have from various sources, I notice that some have an external 3.3v regulator (a 3 pin 662k) near the corner of the CH340 chip, and some do not.
Question 1: Is that a "normal" thing? From photos, I see the genuine Arduios do not have this regulator, so why would a knock-off add it?.
I have been making modifications to reduce power consumption for battery operation (replacing the main 1117 regulator, putting sleep modes into the code, disabling the on-board LEDs). The boards with the extra 3.3v regulator draw more current in sleep mode (about 3.5mA in my case) than the ones without the regulator (1.5mA). And removing that 662k does not help.
Question 2: Is there some other associated circuitry related to that extra regulator I should remove?
Thank you. Interesting you mention removing the CH340. I did try clipping the Vcc power lead to that chip, and it made no difference in the power consumption. Is that not sufficient? Do I really have to remove the whole chip? I was hoping to rig it up with a small slide switch, so I could turn it back on if I ever want to upload a code change via USB. Bear in mind, I'm not shooting for power consumption
down in the sub 100 uA range. Getting below 1 mA in sleep mode would be perfectly fine. So if removing that chip is only going to save, say, 100 uA, it's probably not worth it.
Yes, I've read that Gammon piece. I think I have done all I can from that write-up. But he does things I don't want to do for my application. For example, I need to power from a 9v battery, because I have other devices on the board that need 9v. So I removed the 1117 regulator, but replaced it with a switching regulator, which is more efficient, but not "zero" like you would get from just removing it. And Gammon's article does not talk about turning the power off to the CH340 vs fully removing it, or if that would even make a difference.
I tend to agree. Once you take off the USB, you may as well have a Pro Mini anyway. The Pro Mini has enough I/O pins, ADC, analog, and interrupt pins for anything I need to do. In the meantime... anybody really know how much saving there is by removing the CH340, and why I can't just disconnect its Vcc pin?
Just to add data to this, in case anyone ever wonders... removing the entire CH340 chip does nothing with regard to saving power. Maybe it shaves off on the order of 10-20 uA, if you're working at that level. But if you are hoping to knock off on the order of 1mA, don't bother.
The original Nano uses the 3.3V output of the FTDI chip as it's 3.3V. This has a capacity of about 50mA, which is pretty pathetic if you want to do much of anything with the 3.3V. Many clones use some version of the CH340 USB/Serial chip instead of the FTDI, and it isn't spec'ed to provide ANY 3.3V power (though it does contain an internal regulator.)
So pretty much any Nano derivative with an actual 3.3V regulator chip should be considered "improved."
(They still generally lack enough heat sinking for the regulators, so they're not good for very much power, but...)
It will probably get "phantom powered" through other pins, such as the Tx pin from the atmega328. Phantom powering is not good for any chip.
For low power circuits I prefer to use a bare atmega328 (28-pin DIP). I run them at 3.3V using the internal 8MHz clock and use a low drop-out regulator such as mcp1700-33 or ht7333.
One warning about dropping the clock frequency: current consumption of the CPU is linear with the frequency so you might think you are saving power.
Wrong! At least if your program is running on interrupts (which is logical to do).
When you drop from 8 MHz to 1 MHz current goes down 8 times but the time it takes for your interrupt routine multiplies by 8, so no net saving.
Of all the peripherials the ADC takes the most power (330-340µA, probably for the analog circuitry) so switching that one off (if possible for your application) is the first thing to do.
Here’s a list of current consumption of the 4809, the 328 will be more or less the same.
