Electronic – arduino – Power Consumption of Arduino Nano ( Current Draw Low Power )

arduinoatmegalow-power

anyone got measurements of Arduino Nanos power consumption ?

It guess it should be less than e.g. of the Uno or Mega as the USB Part only get powered when connected via usb and so does the 3.3 Regulator.

Also I should be able to easily disconnect the 5 V regulator and use a DC-to-DC converter instead with ~87% efficiency 34V -> 5V . More or less there should be not much more left than the LEDs (can be disabled by software to only blink every some seconds ?) and the Atmel 328P itself?

It will be able to sleep alot and beeing woken up by watchdog timer every 8s for some measurements and communicaton

Best Answer

According to the schematic, the FT232RL (which is the 3.3v regulator as well) on the Arduino Nano v3 is powered from the 5v line. Regardless if you use USB or VIN or a 5v in. As does the blue power led4 (680Ω resistor means 2.5mA). Led1 and LED2 are controlled by the FT232RL, so should only be on when there is usb/serial communication. Only LED3 is controlled by the arduino D13 pin.

According to this page, the Nano takes 20mA when running, 17mA with the blue power led removed. That is not standby/power down current, which should be a lot lower. The FT232RL has between 15mA full operational current, to at most 2.5mA or minimum 70µa (0.07mA) in USB suspend mode, or with reset pulled low. Since the reset pin is not used, I believe that it should be 2.5mA by itself without a usb cable plugged in, but can't be sure (the FTDI datasheet does not say). That is without any draw on the 3.3v regulator line.

You could remove the 5v regulator, or just ignore it and use the 5v pin to input the externally regulated 5v, which is almost exactly how the USB power is handled (that goes through a shockley diode to protect the USB when 5v is already present). But you should be aware of your regulator's actual efficiency. At a 29 volt difference, at say 5mA to 50mA draw, is it actually 87%? The datasheet's graphs will tell you.