August 9, 2016  
How to calculate battery runtime 

For a Javascript calculator that gives reasonable estimated of battery runtime click here. Notes for Design Engineers: How to calculate how much battery capacity you need. I know, I feel your pain. The marketing department gave you a specification and all it says is “maximize run time, minimize the battery size and cost." But they won't tell you much run time is acceptable, how much size and weight will the market put up with, what cost is acceptable? Hey, the reason that they aren't more specific is they are hoping for a miracle and don't want overspecify in case they don't get the miracle. The miracle you were hoping for was a complete specification, but let's get real here. Your revenge is to wait 2 weeks and come back with “Good news, I fit it in a fountain pen for a BOM of only $5000 and by trimming the power budget (i.e. eliminating all but one of the features) we got it to run for over 5.5 seconds before recharging.”And then sit back and hope for better guidance from marketing! You already knew that I couldn’t help you with your specification, but at least you can use the following design estimation tools to give the marketing department a matrix of choices. How much battery capacity do you need to run your device? Here is how you estimate it.Step 0. A little tutorial on measurements of electronic charge. After all, it is electrons (really ions) that are stored in the battery. In phreshman fisicks we all learned that the measure of charge is the coulomb and that a single electron has 1.602e19 coulombs of charge. One amp flowing in a wire for one second will use one coulomb of charge, which is 6.24 x 10^18 electrons,. Q = I*t where Q is the charge in coulombs, I is the current in amps and t is the time in seconds. The amount of charge passing through that wire (conducting 1.0 amps) in 60 seconds is 60 coulombs, and in one hour you would have had said “hello” and “goodbye” to 3600 coulombs of charge. Batteries were evidently developed by engineers who subscribed to the “whatever’s easiest” system of measurement. They got tired of pulling out their slide rules to divide by 3600 every time they wanted to know how long 24000 coulombs would last them and came up with the unauthorized unit of amphours. Later, when smaller batteries were used they came up with milliamphours. Don’t be confused by the hyphen. Amphours means amps times hours. Divide by amps and you get hours, divide by hours and you get amps. So it isn’t amps, and it isn’t amps per hour, it is amphours. And, by the way, I have even used the term ampseconds because when you say “coulombs” everybody goes glassyeyed on you. Don’t get me wrong, I love amphours for units, it is a handy rule of thumb. Amp hours is how much charge is stored in the battery. Since a battery changes voltage during the discharge, it isn’t a perfect measure of how much energy is stored, for this you would need watthours. Multiplying the average or nominal battery voltage times the battery capacity in amphours gives you an estimate of how many watthours the battery contains. E = C*Vavg Where E is the energy stored in watthours, C is the capacity in amphours, and Vavg is the average voltage during discharge. Yes, watthours is a measure of energy, just like kilowatthours. Multiply by 3600 and you get wattseconds, which is also known as Joules. As long as we are in the prelude, I might also mention that since the charge in a capacitor is Q=CV that a battery can be rated in farads as well. A 1.5 volt AA alkaline battery that stores 2 amp hours of charge (that’s 7200 coulombs) has the equivalent capacitance of 4800 Farads. Of course a battery makes an awfully weird capacitor because the voltage doesn’t drop proportionally to the stored charge, it has a high equivalent resistance, and etc. Also, I should mention that you don't always get all the amphours you expect out of a battery. This is explained in Part 3 below as the Peukart effect. This is why I called it a ruleofthumb rather than a theorem. The biggest errors come when you discharge batteries fast. Some batteries, such as CarbonZinc, Alkaline, or Lead Acid become less efficient when you discharge quickly. A typical sealed lead acid battery will give only half of its rated capacity when discharged at the C/1 rate compared with the C/20 rate. The following method assumes that you know how many amps you need for the gadget under power. If you know the watts go to Step A below.Step 1. Back of the envelope If the current drawn is x amps, the time is T hours then the capacity C in amphours is C = xT For example, if your pump is drawing 120 mA and you want it to run for 24 hours C = 0.12 Amps * 24 hours = 2.88 amp hours Step 2. Cycle life considerations It isn’t good to run a battery all the way down to zero during each charge cycle. For example, if you want to use a lead acid battery for many cycles you shouldn’t run it past 80% of its charge, leaving 20% left in the battery. This not only extends the number of cycles you get, but lets the battery degrade by 20% before you start getting less run time than the design calls for C’ = C/0.8 For the example above C’ = 2.88 AH / 0.8 = 3.6 AH Step 3: Rate of discharge considerations Some battery chemistries give much fewer amp hours if you discharge them fast. This is called the Peukart effect. This is a big effect in alkaline, carbon zinc, zincair and lead acid batteries. For example if you draw at 1C on a lead acid battery you will only get half of the capacity that you would have if you had drawn at 0.05C. It is a small effect in NiCad, Lithium Ion, Lithium Polymer, and NiMH batteries. For lead acid batteries the rated capacity (i.e. the number of AH stamped on the side of the battery) is typically given for a 20 hour discharge rate. If you are discharging at a slow rate you will get the rated number of amphours out of them. However, at high discharge rates the capacity falls steeply. A rule of thumb is that for a 1 hour discharge rate (i.e. drawing 10 amps from a 10 amp hour battery, or 1C) you will only get half of the rated capacity (or 5 amphours from a 10 amphour battery). Charts that detail this effect for different discharge rate can be used for greater accuracy. For example the data sheets listed in http://www.powerstream.com/BB.htm For example, if your portable guitar amplifier is drawing a steady 20 amps and you want it to last 1 hour you would start out with Step 1: C=20 amps * 1 hour = 20 AH Then proceed to Step 2 C’ = 20 AH / 0.8 = 25 AH Then take the high rate into account C’‘=25 /.5 = 50 AH Thus you would need a 50 amp hour sealed lead acid battery to run the amplifier for 1 hour at 20 amps average draw. Step 4. What if you don’t have a constant load? The obvious thing to do is the thing to do. Figure out an average power drawn. Consider a repetitive cycle where each cycle is 1 hour. It consists of 20 amps for 1 second followed by 0.1 amps for the rest of the hour. The average current would be calculated as follows. 20*1/3600 + 0.1(3599)/3600 = 0.1044 amps average current. (3600 is the number of seconds in an hour).In other words, figure out how many amps is drawn on average and use steps 1 and 2. Step 3 is very difficult to predict in the case where you have small periods of high current. The news is good, a steady draw of 1C will lower the capacity much more than short 1C pulses followed by a rest period. So if the average current drawn is about a 20 hour rate, then you will get closer to the capacity predicted by a 20 hour rate, even though you are drawing it in high current pulses. Actual test data is hard to come by without doing the test yourself. If you know the watts instead of amps, follow the following procedureStep A: Convert watts to amps Actually, watts is the fundamental unit of power and watthours is the energy stored. The key is to use the watts you know to calculate the amps at the battery voltage . For example, say you want to run a 250 watt 110VAC light bulb from an inverter for 5 hours. Watthours = watts * hours = 250 watts * 5 hours = 1250 watt hours Account for the efficiency of the inverter, say 85% Watthours = watts * hours / efficiency = 1250 / 0.85 = 1470 watthours Since watts = amps * volts divide the watt hours by the voltage of the battery to get amphours of battery storage Amphours (at 12 volts) = watthours / 12 volts = 1470 / 12 = 122.5 amphours. If you are using a different voltage battery the amphours will change by dividing it by the battery voltage you are using. Now go back to Steps 24 above to refine your calculation. 

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