I'm trying to power a Attiny2313A from 4V (which achieves the Voltage/Amperage ratio I want) but the power source I have to get 4V from, is 16V. So I'm wondering what some recommendations would be.

I have tried to do a voltage divider with ratio 0.25 (R1= 2k & R2= 6k) and this can power the device, but the current being used by the resistors clearly shows limitations and the output of the Attiny2313A isn't dependable.

I have also tried a buck converter I bought off eBay, which draws 5.5 mA without even connecting to Attiny2313A.

In case you wanna see the limitations from the voltage divider, here is two pictures, where the first is with a steady 4V from my Analog Discovery and the second is 4V from the voltage divider. Worth mentioning is that the signal you see is coming from the 16V and controlled by the Attiny via a transistor.

Analog discovery supply

Voltage divider supply

Any ideas or suggestions?

  • 2
    \$\begingroup\$ Don't buy undocumented products from peeBay. \$\endgroup\$
    – Andy aka
    Nov 29 '21 at 9:32
  • 3
    \$\begingroup\$ search for a switching regulator with low quiescent current. \$\endgroup\$
    – Mike
    Nov 29 '21 at 9:33

I checked the datasheet for your MCU. It looks like a sub-mA device. Buck converters might have trouble with that. I tried to search for some interesting buck converter that would fit your description on mouser, but they're all inadequately expensive, and there is no guarantee they are stable at 1mA output.

So I looked at the linear regulators. Haven't found too many, but there is this one: ADP1720 (Datasheet).

Micropower linear regulator with up to 50mA output and special capabilities for ultra low power, such as (from the first page of datasheet):

enter image description here

  • 1
    \$\begingroup\$ I will accept your answer, as this is good and thanks. But I chose to go ahead with a low power voltage regulator, as the lab at my school had some TS2950 laying around \$\endgroup\$
    – jonas
    Nov 29 '21 at 10:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.