3.3V voltage regulator for diy "bare-bone" arduino?

I would like to use whether six AA or one 9Volt external battery
for a diy "bare-bone" arduino*
(*just the 328p atmega, 8MHz oscilator, capacitors, resitor, LED)
tested LM317, LD1086, LM1117 voltage regulators
(it's a "budget" project – hence didn't try any step-up/step-down switching
regulators though they seem to be the best solution ...);


can i "simply connect" any of the regulators
(incl. the necessary resistors/capacitors according to their datasheets)
do i need to add resistors to reduce their current
in order to not "over-power"* them?

according to the values i measured
(i am a novice to arduino's power management
… could have made mistakes whilst measuring)

LM317  =  ca. 2A @ ca. 3.3V with 100 & 160 Ohm resistors
LD1086  =  ca. 1A @ ca. 3.3V with 100 & 160 Ohm resistors
LM1117  =  ca. 1A @ ca. 3.3V without resistors

according to dataheets/descriptions of several arduino boards
(but i might have misunderstood – this might not have any impact
on the power source …   )


LED to indicate that there is power flowing
… hoping that this reduces the power consumption:
- a 20 KOhm pullUp resistor ?

Protect from charge
… hoping that this protects the circuit from capacitors' stored charge:
- 2* 1N4001 on voltage regulator?

i look forward to your advice …

Picture of 3.3V voltage regulator for diy
power requirements.jpg
dot7242 years ago

Use a 5v regulator with a risistor in order to bring down the voltage from 5v to 3.3v

Use any regulator you like that you can set to 3.3V or that only outputs 3.3V. Just note that a 9V battery is a vary poor power source and won't last vary long. You would be better off using 4 AA batteries rather than 6. You don't need 9V for the regulator to work. 6V is more than enough. As for current. Keep in mind the device will draw the current it needs. Current is not pushed to the device.

Adding an LED power indicator isn't going to reduce power consumption. It will pull more power. But if you use an LED you will need a current limiting resistor (it's not called a pull up resistor in this case) so you don't burn out the LED. LEDs will continue to pull current until they blow which is why you need a resistor on them.

The diodes are not needed. The voltage drop across them means your output voltage will be reduced below the 3.3V you need.The capacitors are there to help maintain a nice clean input voltage without spikes and drops. They will not harm the circuit and no protecting between them and anything else is needed. Drop the diodes. Not sure why you have 2 capacitors on the voltage output from the regulator. You should have a ceramic cap on the voltage into the regulator and the electrolytic on the output to help maintain a clean power source. But since your using batteries you could do just as well without any caps.

marc_is_curious (author)  mpilchfamily2 years ago

Thanks a lot for your helpful reply!

good to know that i do not need any

additional components!


that i should

better use …

- 6V (4 aa(a)) batteries

think about …

- using a LED, diodes* or capacitors* at all


i looked at the example circuits**

within the voltage regulators' datasheets

but might have positioned them

in the wrong places …

** e.g.

http://www.ti.com/lit/ds/symlink/lm1117-n.pdf, page 9

http://www.ti.com/lit/ds/symlink/lm317.pdf, page 7 & 8

http://elcodis.com/parts/6259093/LD1086_p2.html#datasheet, page 2


thanks a lot, again:

am a novice to power regulation

& have been struggling with it

for quite some time!

Yes example circuits in the data sheets are going to call for filter caps. They are basic example circuits that asume your using a dirty power source such as mains power running through a transformer and bridge rectifier. But since your using batteries the filter caps are not that important.