Hi there, I'm trying to do a fun project that requires the battery to fit in a round tube with a diameter of 1 in (25.4mm) and a length of 4 in (10cm). I can fit a 9v battery in this space to power my arduino, but was advised 9v batteries are not great to use.
What are my other alternatives that powers arduino sufficiently within this space?
Thanks!!!
It depends on what else is in the tube and how long you want it to run for.
Weedpharma
weedpharma:
It depends on what else is in the tube and how long you want it to run for.Weedpharma
That's the spacing allowed for the battery only, nothing else other than the battery and wires from it are in that space.
As long as possible would be ideal, fully knowing that it will not be tooo long.
The length of the tube (10 cm) allows for 2 AA batteries. A lot of juice if these are 14500 Li-Ion 3.7V rechargeables. Parallel (3.7V/>4000mAh) or series (7.4V/2300mAh).
florinc:
The length of the tube (10 cm) allows for 2 AA batteries. A lot of juice if these are 14500 Li-Ion 3.7V rechargeables. Parallel (3.7V/>4000mAh) or series (7.4V/2300mAh).
Thank you, that sounds like a great way to go
Is your tube 1" internal diameter or 1" external diameter?
If it's 1" internal dia. you could fit two C size cells in it. They are exactly 25.4mm in diameter. An exact fit.
Another possibility (and/or for others) - Harbor Freight (and probably other discount chinese tool resellers) sells (and gives away sometimes with a coupon) the following flashlight:
Two dollars (USD) - or free if you have a coupon - nets you a set of 9 very bright white LEDs (mounted on a round PCB), a toggle rubber button switch - and as far as this thread is concerned - 3 AAA cells in a battery holder; the cells, IIRC, are arranged in series (so, 4.5 volts for alkaline, or 3.6 volts for NiMH). The flashlight is only 1 inch in diameter - each battery holder is 2 inches long.
Open them in the store if you can - I bought 3 one time, 1 was missing a battery contact spring.
cr0sh:
Another possibility (and/or for others) - Harbor Freight (and probably other discount chinese tool resellers) sells (and gives away sometimes with a coupon) the following flashlight:http://www.harborfreight.com/3-12-in-9-led-mini-flashlight-69052.html
Two dollars (USD) - or free if you have a coupon - nets you a set of 9 very bright white LEDs (mounted on a round PCB), a toggle rubber button switch - and as far as this thread is concerned - 3 AAA cells in a battery holder; the cells, IIRC, are arranged in series (so, 4.5 volts for alkaline, or 3.6 volts for NiMH). The flashlight is only 1 inch in diameter - each battery holder is 2 inches long.
That's a great idea for future use 
Thank you for that! I'm in Australia so no harbor freight here, but I have seen them at Bunnings or Woolworths here for under $5
Can 18500 batteries be safely used? At 3.7v and 3400mah, that sounds good and fits the profile?
Well now, a single 18650 lithium rechargeable cell will fit in there with considerable space to spare (it's only about 2½ inches long) - you can get them with solder tags which means you have no reliability problems with spring contacts. Nominal 3.6 V so essentially you can use 3.3 V componentry - but if something is rated no more than 3.3 V, you have to disconnect it while charging - the Arduino does not mind though as it is rated to 5 V.
Ever wondered why we specifically have 3.3 V devices? Well, for CPUs in PCs, it is fairly arbitrary, but it mostly is to suit the lithium-ion rechargeable battery. So what is important about those batteries? Mobile phones of course! And before Li-ions, they used three Ni-Cd or Ni-MH cells - which was also 3.6V.
All hail the mighty mobile phone!
If I used a single 18650 at 3.7v, or two 14500 at 7.4v, connect it to ground and vin, will I need some sort of step up/down regulator to use my arduino or will it work fine just as is by having a wire from 5v or 3v?
Sorry for my newb questions, I haven't used arduino with any power other than the computer or a crappy 9v.
If there is any particular link you think would be good for me, let me know.
Thank you all!
Well, a single 18650 at 3.7 V will feed your Vcc directly with no power loss, as long as the rest of the system can use 3.7 V (which means the mega328 runs with an 8 MHz crystal such as that version of Pro Mini).
If you use 7.4V, you have to feed it to Vin where the regulator drops it to 5V. Everything draws the same (or a little more) current, so you are essentially wasting the second battery power - the only advantage is if something must have 5 V to operate.
3.7v can be boosted to 5v and fed into the 5v, or, it can be boosted to 7v and fed into the barrel connector.
7.4v can be fed into the barrel connector.
Voltage fed into the barrel connector goes through an onboard regulator and is regulated to 5v.
Weedpharma