Calc battery life - Page 1
Forum section: electronics / beginners

author: bootloader9800
author_rank: Contributor
timestamp: July 08, 2022, 06:30:54 pm
content: Hey guys. So I was looking to find the correct capacity battery for my circuit. The circuit is pulling 1.2 Amps @ 5 Volts.So that means it is using 6 Watts of Power ( P = I X V = 1.2 X 5 = 6W)----- (1)I figured that I will use a regular 2S Lipo battery that is rated at 7.4 V and 6500 mAh. I figured that the battery can provide me with 48.1Wh of energy.7.4V x 6500mA = 48,100mWh = 48.1Wh ------- (2)Later I figured that if this battery can give me 48.1Wh, that means it can run my circuit for 8.01 hoursI got 8.01 hours using equation (2) / equation (1)48.1Wh / 6 W = 8.01 hours.I was all happy telling myself that the battery can give me 8.01 hours power.But this freaking Digikey website keeps telling me that my Battery will last only 5 hours and 25 minutes.https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-battery-lifeThe Digikey website's calculations are below:Battery Life = Battery Capacity ( in mAh) / Load Current (in mAh)Battery Life = 6500 / 1200 = 5.41 = 5 hrs 25 minutesCan some one tell me whats going on here? Why did I get 8.01 hours while Digikey got 5 hours and 25 minutes?Thanks folks!!

author: Uunoctium
author_rank: Frequent Contributor
timestamp: July 08, 2022, 07:20:16 pm
content: the mistake happens by typing in the same current as at your "secondary" 5V side. At 100% converter efficiency your primary source (7.4V) is only loaded with 5V/7.4V = 0,67567 x Isek. -> 810,81mA.If we type that in, the result is 8h1min

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 08, 2022, 07:46:09 pm
content: The important thing is how do you convert 7.4V to 5V with little loss? If you power the device with 7.4V which for many devices rated at 5V would work just fine but the power consumption goes up because both the supply voltage is higher and the current draw is higher too.

author: Uunoctium
author_rank: Frequent Contributor
timestamp: July 08, 2022, 08:11:42 pm
content: Yep, that' s clear. A converter with 100% efficiency is unrealistic. Just trying to figure out what the bug is.The mysterious disappearance of 2.4V!In other words:Digikey's calculator is based on the same voltage at source and sink side. So you can easily shorten voltages out on both sides of equation.But in OP's math suddenly 2.4V are missing.And 2,4V x 1200mA = 2,88W are also missing.Added to 6W result is 8,88W.48,1Wh/8,88W = 5,41hIf we calculate right with 7.4V on both sides and a pure resistive load, the overall balance looks much more worse:5V/1,2A = 4,166Ohm.7,4V/4,166Ohm = 1,776A=13,142W48,1Wh/13,142W = 3,659h

author: Benta
author_rank: Super Contributor
timestamp: July 08, 2022, 08:58:24 pm
content: The mistake lies in you not understanding the voltage characteristics of batteries.You can't discharge a battery to 0 V (well, you can, but it's useless, and with a Li-cell it's also dangerous).A LiPo cell has a fully charged voltage of 4.2 V. The minimum allowable discharged voltage is 3.0 V (or 3.2 V, depending on manufacturer).The "3.7" or "7.4" V is a mean value computed from the discharge curve from 4.2...3.0 V or 8.4...6.0 V defined by the battery manufacturers.That was just to get you prepared for your switching regulator/converter that will extend operating time by lowering battery current draw.A 90% efficient buck converter (not unrealistic) will drop current draw to 900 mA at 5 V output.Now, your maths mistake is taking the roundabout way over power.The correct calculation is capacity / current draw.6500/1200 = 5.4 hrs or 5 hrs, 25 minutes.With buck regulator:6500/900 = 7.22 hrs or 7 hrs 10 minutes.It's really that simple.Cheers.EDIT: LiPo capacity is defined at 20 hrs discharge. Your actual capacity will be less at 900 mA due to internal cell resistance.

author: Kasper
author_rank: Frequent Contributor
timestamp: July 09, 2022, 09:37:38 pm
content: Are you using a regulator? What kind?Digikey equation assumes you have no regulator or a linear regulator aka LDO.Your equation assumes you use a buck regulator that has efficiency of 100%

author: Siwastaja
author_rank: Super Contributor
timestamp: July 10, 2022, 05:41:21 am
content: Quote from: Kasper on July 09, 2022, 09:37:38 pmSorry to derail butLDO = Low Drop-OutNot equal to linear regulator. Some linear regulators are Low Drop Out type, some are not. LDO can also used to describe non-linear regulators.I'm going to correct this every time I see this from now on because correct terminology is important.

author: Kasper
author_rank: Frequent Contributor
timestamp: July 10, 2022, 06:25:02 am
content: Quote from: Siwastaja on July 10, 2022, 05:41:21 amCorrect terminology is good. For example, I said 'aka' which means 'also known as'. And you changed 'aka' to 'equals', which is not good.Aka != equalsLinear regulator != LDOLinear regulators are commonly called LDOs. If you want to search digikey for linear regulators, save some time by searching 'ldo' and selecting the top result.I probably should have said linear regulator AIKA LDO.Aika = as incorrectly known asPs great xkcd pic. I was just telling my wife about it the other day and I am repeating it to her now.