Amp/Watt Hours and battery capacity, will raising voltage raise capacity? - Page 1
Forum section: electronics / beginners

author: rwgast_lowlevellogicdesin
author_rank: Frequent Contributor
timestamp: January 07, 2013, 10:25:49 pm
content: Now I stumbled upon an article the other day on Hack a Day where the guy was building a power supply for his project and he wrote something I had never heard said before, but always wondered about.In simplest terms lets say you have a battery that is 5v with 1000maH, this means you can pull 1 amp at 5v for an hour, theoretically (yes I have watched all Daves videos about batteries and discharge curves and such). Well in this article the guy basically says if I were to throw two of my 5v batteries in series I would have a 10v 1000maH pack, so far so good. Then he claims that if I were connecting this battery pack to a device that uses 5v I would roughly get 2000maH out of my 10v pack, because I was only using half the voltage. To me this sounds reasonable and logical but Im not sure if its acually true, and it may bot be practical as a linear regulator will just burn off the extra 5v so you will have no capacity gains but what about a switching regulator?Im curious because I have a 19.2v pack Ive built and I use switching regulation for all my digital logic. I only need an actual 18v to briefly supply to motors in sticky situations, I plan on feeding them a PWM signal of tat averages out to about 12v on a regular basis. So according to this guys logic I should be getting alot more than 2.5amp hours out of my battery pack if Im mostly going to draw 12v/5v and 3.3v from the battery.Is this actually true...

author: EEVblog
author_rank: Unknown
timestamp: January 07, 2013, 10:56:25 pm
content: Yes, you need a switching regular, otherwise with a linear regular you are still drawing 1A form the pack, so the AmpHour capacity remains the same.Dave.

author: Psi
author_rank: Super Contributor
timestamp: January 08, 2013, 12:43:15 am
content: Quote from: rwgast_lowlevellogicdesin on January 07, 2013, 10:25:49 pmIt's kinda true, but as Dave said, this "device" would need to have a switchmode powersupply.If you connect two 5V 1000mAh packs in series you now have one 10V 1000mAh pack.If you connect that 10V pack to a switchmode regulator which outputs 5V you will end up with the capacity of 2000mAh (at 5V).(assuming 100% efficiency)It's easy to imagine when you think about a switchmode system and how the current changes. eg..Input= 5V @ 2A ---->switchmode boost converter---> output = 10V @ 1Aor the other way.Input= 10V @ 1A ---->switchmode buck converter---> output = 5V @ 2A

author: amyk
author_rank: Super Contributor
timestamp: January 08, 2013, 11:56:54 am
content: Or you could just connect them in parallel to get the same voltage and increased capacity.

author: ptricks
author_rank: Frequent Contributor
timestamp: January 08, 2013, 01:48:14 pm
content: The easiest way to understand serial battery calculations like this is to just convert the sources to total watts , watts = volts x amps .5 watts per battery , total of 10 watts to use however you want, 1V for 10 hours, 3V for 3.3 hours, etc.If you are doing parallel batteries just add the amps together and assume the voltage is the same as the battery volt. Note: make sure parallel batteries are at the same charge level or one battery will be the supply while the others do basically nothing.