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 ...);
SUITABLE POWER/CURRENT?
can i "simply connect" any of the regulators (incl. the necessary resistors/capacitors according to their datasheets)
or
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 … )
TRYING TO PROTECT & REDUCE POWER FOR ADDITIONAL FEATURES ...
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:
can i "simply connect" any of the regulators
(incl. the necessary resistors/capacitors according to their datasheets)
Yes
LED to indicate that there is power flowing
… hoping that this reduces the power consumption:
You don't reduce power consumption by lighting an LED, but it is not increased by much. However depending on the type of LED you might not see any light with a 20K resistor.
By the way they are called "low drop out regulators" not "drop out regulators not" . Drop out is the point when a regulator stops regulating, when the input voltage is too close to the output voltage. A low drop out regulators is one where this required excess voltage is small. A normal regulator drops out at about 1.5V difference with a low drop out one this can be 0.5 or even 0.25 Volts.
good to hear that i do not need additional components
& thanks for the low drop out regulator definition
(am still a novice to electronics, this forum & learning ... ) ;
i tried the 20K with the LED &, according to what you predicted, the light
is shallow but still bright enough to be recognised –
thought that the 20K might help to reduce the LED's power consumption ...
The extra power caused by dropping this resistor to say 1K is neither here nor there in the overall schematic of things. The processor is going to take about 30mA so what is another 0.5 mA going to matter.
The processor takes well below 30mA at 3.3V 8MHz...
For micro-power applications (that means what you are trying to do - use as little
power as possible to prolong battery life) you need to find a regulator with low
quiescent current drain (many take 5mA or more - some take a few uA only, look
for "micropower" in the description / datasheet. For 3.3V you would be crazy to
use 9V battery. 4.5V (3 AA's) would be much more performant.
@MarkT
... low quiescent ...
i will look for regulators with mA drops
or
might even spend the extra money & go for the
step-up/step-down switching regulators
(advertised to be the best solution within many blogs ...
there seem to be cheap options on eBay)
@MarkT & Riva
... 9V battery ...
i tried 3 AA(A)'s according to your advice:
they proofed to be far more efficient/long-lasting indeed
...
i still look for the best/most efficient power feed for
an 3.3V Arduino (probably LiPoly's are - i need to
get my head into charging circuits) ...
@ Mike
i did not measure the "bonus mA's" resulting from 20K yet,
saw them on other low-power circuits ...
but, true: i might as well use (much) less on that behalf
in order to have a brighter LED
The other factor you need to remember is design your DC circuit ahead of the regulator to provide the MINIMUM v-in-v-out (under all conditions). Consider the 30mA the processor draws at 3.3V, so it follows that if there is 3.3V of your 9V battery across the processor, the other 5.7V is across the regulator at 30mA as well. You will give away more energy to the regulator than the processor. Not a good idea.
Riva:
I'm not sure a 9V (PP3) battery will deliver the current you may need unless you have several in parallel.
Only some one who knew very little about electronics would wire batteries directly in parallel. They must go through a diode first to stop them cross charging.
Grumpy_Mike:
Only some one who knew very little about electronics would wire batteries directly in parallel. They must go through a diode first to stop them cross charging.
Bored and felt like bait fishing? Or trying to hone your !social skills.
pls. accept my apologies for replying that late ... (& many thanks again ... all of you help me "big style" to gather more inside-knowledge!)
@ rmetzner49
thanks for the input!
I originally thought of using a 9V (NiMh rechargeable) battery
(150 mAH) but learned-by-testing(-and-googling), that i get
much more lifetime (current/h ... mAH) with e.g.
4 AAA (NiMh rechargeable) batteries (350 mAH)
4 AA (NiMh rechargeable) batteries (1000 mAH)
...
on that behalf i might go for AA (NiMh rechargeable) batteries
using whether a
3 AA's & a step-up/step-down switching regulator
(not "loosing" any dropout rate ...
if i invest more money for the regulator)
... or ...
4 AA's & a linear, low-dropout regulator
("loosing" the dropout rate but saving money)
