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The X110 netbook from LG is what I purchased to advance my many computer-related projects during vacations, and I consider it a good buy. What I like about it:
My X110 has the small battery, with a nominal 2200 mAh. The manufacturer claims a battery life of 2 hours 20 minutes with that battery. Curious as I am, I wanted to know
To measure the battery lifetime, I let the netbook run on battery until it was nearly empty and extrapolated to get the time it would have taken to run empty completely (see below for details). I did this for several scenarios:
Here are the results:
| Scenario | Battery life | Average current |
| Minimum load | 2:42:49 | 813 mA |
| Moderate load | 2:18:06 | 954 mA |
| CPU 100% | 2:01:43 | 1075 mA |
| Brightness 100% | 1:54:21 | 1145 mA |
| Maximum load | 1:27:29 | 1502 mA |
Going by the data, turning the display brightness up takes even more additional power than number crunching. So a good way to prolong your battery life is to keep brightness down – sit in the shadow, not in the sun when using it outdoors, and take care to look at the display perpendicularly, which gives you maximum brightness; set a low default and turn it up only when needed, not the other way round.
The CPU and the display between them seem to account for most of the power consumption of a fully loaded system. When one takes the difference between minimum consumption and 100% CPU and brightness respectively and adds these three components up, the result is 1407 mA, only about 100 mA less than the maximum load result. So the additional power required for increased hard disk activity and an external mouse and network is comparatively low.
What I did not test: None of the scenarios included an external monitor, audio
output, and USB devices other than a small laptop mouse. These ports may
consume additional power when active. A USB device may consume up to 100 mA at
5 V according to the USB specification (see Wikipedia
), which amounts to 45.5 mA for a 11.1 V (nominal) battery
(which the average currents above imply).
Considering you may use your netbook primarily when travelling (I do), how best
to store it between trips deserves some mention. Litium ion batteries should
best be stored half empty and must not be discharged completely (see
Wikipedia
). So it is of some interest how fast
the battery discharges on its own.
Here I used a simpler approach than for the battery lifetimes of the running netbook. I just left the X110 switched off for some time and checked the remaining charge after the next reboot. The average discharging currents computed from the results were:
| Situation | Average current |
| Battery present | 14 mA |
| Battery removed | 0.2 mA |
These numbers seem low compared to the discharge current when the netbook is running, and indeed they are. But the first row of the table implies that the netbook's battery will be completely discharged after one week unless you remove the battery. The second row says that the battery will discharge on its own within 1½ years. (I had put up a much larger average current earlier, but it was probably due to measuring only a short period after the battery had been fully loaded, when it presumably discharges fastest.) So, to prevent degradation of your battery, you should remove it from the netbook. (If you have the space to store it in your fridge, better still, for at fridge temperatures lithium ion batteryies degrade most slowly.)
The time it takes to charge the battery fully seems to be independent of the system load; presumably the current from the AC/DC converter increases as needed. Charging increases the reported battery fill level linearly most of the time, only when the battery is very nearly full, the charge current is reduced, and the charging process is completed more slowly, almost asymptotically.
I measured this in the same way as the discharging when the netbook was running, by logging the battery fill level over time. The total charging time including the asymptotic part was 2:28:05, the end of the linear charging (which makes the battery almost full) was at 2 hours, and the interpolated end of the charging process at the linear charging current (974 mA) was at 2:12:03. Translation: if you charge the battery for a good 2 hours, you're good. Only if you want to squeeze the very last electron inside before your trip to the jungle, you have to charge it for 2 1/2 hours.
As inspiration for those who want to measure their battery lifetime themselves, here's how I did it. All the following requires that you are running Linux; as usual non-UNIX operating systems are unsuitable for anything other than flashing pretty pictures at stupids.
First, I opened a log of the terminal session with the script command to record the output of the following commands. Then I started the following infinite loop:
while true ; do date ; cat /proc/acpi/battery/BAT1/state ; sleep 60 ; done
The commands inside the loop print out the current time and the state (including charge level) of the battery, respectively. The latter uses Linux's all-you-can-eat kernel information shop, the proc file system. After the first output, I disconnected the external power supply. The loop causes the battery state to be printed every minute, allowing to track the charge level. (In those scenarios where the display was blanked, I displayed the screen briefly at interals.)
I then waited until the battery was nearly empty (around 50 mAh), reconnected
the power supply, interrupted the loop (Ctrl-C) and terminated the
session log. The log now contained times and battery state information at
approximately minute intervals. I wrote a small Perl script (download) to convert the log to
a tab-separated value table with the times in seconds and shifted so that the
first time was zero. (The script uses the Date::Parse module to extract the
time; I could have made this easier by using date +%s instead of just
date.) I then performed a linear fit to those data using gnuplot
.
The abscissa (x-axis) offset is the total battery lifetime in seconds, and the
slope is the (negative) discarge current in the unit mAh/s = 3.6 A. It
did not work for me to fit the x-axis offset directly; this turned out to be
numerically ill-behaved, so I computed it from the y-axis offset and the slope.