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Custom design and manufacture of state-of-the-art battery chargers, DC/DC Converters, and power
supplies
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 do you need to run your device? Here is how you
estimate it.
Step 0. A little tutorial on
measurements of charge. In phreshman fisicks we all learned that the measure of charge is the coulomb and that a single
electron has 1.602e-19 coulombs of charge. One amp flowing in a wire for one second will use one coulomb of charge.
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 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
amp-hours. Later, when smaller batteries were used they came up with milliamp-hours.
Don’t be confused by the hyphen. Amp-hours 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 amp-hours. And, by the way, I have even used the term amp-seconds because when you say “coulombs”
everybody goes glassy-eyed on you.
Don’t get me wrong, I love amp-hours 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 watt-hours. Multiplying the
average or nominal battery voltage times the battery capacity in amp-hours gives you an estimate of how many watt-hours the
battery contains.
E = C*Vavg
Where E is the energy stored in watt-hours, C is
the capacity in amp-hours, and Vavg is the average voltage during discharge. Yes, watt-hours is a measure of
energy. Multiply by 3600 and you get watt-seconds, 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 alkaline battery that stores
2 amp hours of charge (that’s 7200 coulombs) has the 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.
Step 1. Back of the
envelope
If the current drawn is x amps, the time is T
hours then the capacity C in amp-hours 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 a big effect in alkaline, carbon zinc, zinc-air and lead acid batteries. It is a small effect in
NiCad, Lithium Ion, Lithium Polymer, and NiMH batteries.
For lead acid batteries the rated capacity is typically given
for a 20 hour discharge rate. If you are discharging at a slow rate you will get the rated number of amp-hours 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 amp-hours from a 10 amp-hour
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(3559)/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. |
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