Need a little power design help:

I have a project I'm working on where I have measured that I will use approximately 14.4Ah over 24 hours. This is quite the load for a small timelapse setup where I need the camera clicking for 5 months. I need 24 hour power because once the camera turns off, it needs a human to press the power on button. I will have access to the site, but it won't be everyday access - most likely on the weekends.

I think I'm looking at needing to go with a deep cycle marine battery and a 25w solar panel to keep this system powered on and the battery charged.

As an example, if I have a 40Ah 12v car battery, and I know my draw is 14.4Ah a day, then this battery can give me 2 days of usage?

However my project requires only 5v, and my 14.4Ah is calculated from 5v.

This is where my math breaks up, but if I convert the car battery 12v to 5v output, and draw 14.4Ah a day from it, does this mean I'll get approximately 4 days of usage?

Ultimately I'd like to hook a 25w solar panel to the battery to try and keep it charged, too.

Thoughts? Thanks!

Edit: adding a little info about my project. It may help with the design.

I'm an electronics rookie, I know enough to get me in trouble, but not enough to help fix it :)

I'm looking to setup a long-term timelapse camera. The enclosure will consist of 1 camera and 1 mobile broadband device. I'd like to use solar power to try and keep the unit operational.

Some brief details on my setup: The camera is a Canon ELPH 130 point and shoot which has firmware on it to take 1 photo every 10 minutes, between 6am and 6pm. When a photo is taken, it'll save it to an Eye-Fi SD Card. When the Eye-Fi receives a new photo, it'll connect to the MiFi and automatically upload it to the internet.

I've connected my Multimeter in-line with these devices to gauge their amp usage.

Here's my findings:

  • Camera: average current of 0.25a over 2 hours.
  • Mobile Broadband: average current of 0.35a over 2 hours.

Both devices showed various spikes - mainly when a photo was taken and uploaded. Then the current seemed to "settle down" back into the averages above.

The downside: I need these devices up 24 hours a day because if they lose power it requires human intervention to push a button to power them back on.

So here's my math:

  • Camera: 250mA x 24 hours a day = 6000mAh
  • Mobile Broadband: 350mA x 24 hours a day = 8400mAh
  • Totaling: 14400mAh battery required to keep this thing going for 24 hours.

One thought on the camera and current usage: The camera does not have a sleep function, so it's powered on all the time. It does have a power-reducing mode (LCD off), but that's about it. Perhaps I can connect some jumper wires to the power button to be able to control on/off without a human needing to press it?

  • \$\begingroup\$ For calculations close to reality, do not forget to include battery C-rate and Peukert factor. \$\endgroup\$
    – GR Tech
    Commented Apr 17, 2014 at 17:18
  • \$\begingroup\$ @Pat you need to look at the comments I provided on PV panel sizing and how much energy you are liable to get out of a battery for a given PV panel size and real-world sunshine. Look at the gaisma site I referenced to see how much solar energy you can expect daily on average. Use worst case month if you want this to work in winter. Note the relationship that I suggested between panel "Wmp"rating and actual effective wattage and energy you can expect to get. Note my comments on the effect of an MPPT controller versus any other. You may come to some sobering conclusions re panel size needed. ... \$\endgroup\$
    – Russell McMahon
    Commented Apr 18, 2014 at 21:21
  • \$\begingroup\$ Your revised power requirement estimates will allow a smaller panel size than I originally suggested. \$\endgroup\$
    – Russell McMahon
    Commented Apr 18, 2014 at 21:21

2 Answers 2


To do these calculations, it's easiest to work in terms of power instead of current. For example, if your camera is averaging 250mA at 5V, then that is 5V x 0.25A = 1.25W. If you are running it for 24 hours, then that's 1.25W x 24h = 30 Watt-hours ("Wh") per day.

For comparison, your 40 Amp-hour ("Ah") battery, at 12V nominal, has 12V x 40Ah = 480 Wh of capacity.

With no other considerations, this would mean that the camera (by itself) would operate for 480Wh / (30Wh/day) = 16 days on the battery. Of course, there are many other considerations, including the inefficiency of converting your 12V to 5V...

Please look at this answer I made for a different application. It details the considerations and calculations for a battery powered system with possible solar input.

Good luck!

  • \$\begingroup\$ Thank you! Very clearly worded in a way a layman like me can understand. Your other answer you linked is great too. Can you spot check my math to spec out a solar panel for this setup? You calculated 30Wh a day and my solar insolation says I get 4 hours of good sun a day. 30Wh / 4 hours = 7.5W panel, and if I use a charge controller, it's still about 7.5w? So I would be OK with a 10w panel? \$\endgroup\$
    – Pat
    Commented Apr 17, 2014 at 17:31
  • 1
    \$\begingroup\$ @Pat That's the right idea, but I wouldn't ignore the 10% controller loss. Also, this is only for the camera, not the broadband. There will be still more power loss when you convert the 12V to 5V, which depends on the type of converter you choose. And, finally, you might get cloudy days! People are often surprised how much panel they need :) \$\endgroup\$
    – bitsmack
    Commented Apr 17, 2014 at 18:03
  • \$\begingroup\$ Oh bummer, I thought this was for both devices. If only I can save power by being able to power off/on the camera remotely somehow... Okay so it looks like both devices combined are 72Wh a day. (480Wh battery) / (72Wh device) = 6 days. So also for the panel: 72Wh / 4 hours of sun = 18W panel. Plus 10% controller loss, 20W panel. Plus 20% overage = 25W panel to charge the battery. I think? \$\endgroup\$
    – Pat
    Commented Apr 17, 2014 at 18:19
  • \$\begingroup\$ @Pat Yep! Sounds right. Is there a way for your modified ELPH firmware to go into low-power mode between pictures? That could help your power balance. \$\endgroup\$
    – bitsmack
    Commented Apr 17, 2014 at 18:56
  • \$\begingroup\$ Yeah, I'm using the CHDK firmware which has a power-reduction mode. That's when the camera is mainly at 250mA. It only spikes into the 400mA's briefly for the photo, then drops back down. \$\endgroup\$
    – Pat
    Commented Apr 17, 2014 at 19:46

You want to work in W.h OR convert the A.h with voltage change.

Below Z12_v = conversion efficiency from 12V to 5V. Use say 80% if not known.

Note that for a Pv panel rated at X Watt_mp you get about 50% of this out of battery after storage IF you are not using MPPT.

Get hours/day avg of sun from http://www.gaisma.com

For 2 hours of equiv full sun / day x 25W panel you get about 2h x 25W x 50% = 25 Wh at 12V.
At say 80% Z12_5 you get about 20Wh out of system at 5V.

At 14.4 Ah/day x 5V you need 72 Wh at 5V so panel is low by about 3 or 4 x with 2h sun.

Note that equiv full sun hrs/day is much lower than daylight hours. See Gaisma kwH/m^2/day table for equiv sun hours/day.

Ah at 5V = Ah at 12v x Z12_5 x 12/5
Ah at 12V = Ah at 5V x 5/12 / Z12_5

Wh at 5V = Wh at 12v x Z12_5 <- no factor of 12/5
rearranging Wh at 12V Wh at 5V x / Z12_5

What sort of camera? What mean I draw? How many exposures /day at what resolution?
14.4 Ah/day sounds VERY high.
14.4/24 = 600 mA average continuous.
A camera should shut down to FAR less than that when not taking photos. Far less.

  • 4
    \$\begingroup\$ Your answer is so full of undefined acronyms, abbreviations and jargon that it's practically gibberish. Please take the time to write full sentences, define the terms you use and organize your thoughts into a logical sequence. \$\endgroup\$
    – Dave Tweed
    Commented Apr 17, 2014 at 15:52
  • 1
    \$\begingroup\$ Thanks @DaveTweed, glad to see I wasn't the only one confused by it all. \$\endgroup\$
    – Pat
    Commented Apr 17, 2014 at 15:56
  • 1
    \$\begingroup\$ @Pat Answer doesn't need tailoring - assuming no technical error or typo (never certain when I come home at 3am+ and have a 'quick look' at SE to see if I can be helpful). If your answer is not provided by others (and it may be) then this largely tells you what you asked about. Most of the terms in there should be fully familiar to you OR understandable with a very small amount of effort. Pv should be PV. MPPT - easy search. Watt_mp = Wmp = max power from PV panel under optimum conditions. ALL other terms are standard and non-jargon, despite assertions to contrary 'by others'. \$\endgroup\$
    – Russell McMahon
    Commented Apr 18, 2014 at 1:34
  • 1
    \$\begingroup\$ ... I intially defined Z12..., clarified a typo in Pv, pointed out how MPPT may be elucidated if elusive, and explained Wmp or similar. Apart from that the perceived undefined acronyms, & 'jargon' and 'gibberish' seems to amount to use of W.h, A.h, gaisma.com, Wh , Ah, sun hours/day. | If people reading this cannot deal with such then they are not overly interested in the answer. | \$\endgroup\$
    – Russell McMahon
    Commented Apr 18, 2014 at 21:12
  • 1
    \$\begingroup\$ ... Anyone who is interested in an answer of technical quality as opposed to one of profound phraseology, enlightening erudition or prolific prosody would be well enough served by taking the time to chug through what was offered. Those who care enough about trappings or trivia or tidiness are welcome to mellifluorise \$\endgroup\$
    – Russell McMahon
    Commented Apr 18, 2014 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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