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Message-Id: <D659872E-E276-4E00-BAFC-1E461EDA1FA7@goldelico.com>
Date: Wed, 1 Apr 2015 22:39:17 +0200
From: "Dr. H. Nikolaus Schaller" <hns@...delico.com>
To: Pavel Machek <pavel@....cz>
Cc: Tony Lindgren <tony@...mide.com>,
Marek Belisko <marek@...delico.com>,
Benoit Cousson <bcousson@...libre.com>,
Sebastian Reichel <sre@...nel.org>,
Dmitry Eremin-Solenikov <dbaryshkov@...il.com>,
David Woodhouse <dwmw2@...radead.org>,
Rob Herring <robh+dt@...nel.org>,
Pawel Moll <pawel.moll@....com>,
Mark Rutland <mark.rutland@....com>,
Ian Campbell <ijc+devicetree@...lion.org.uk>,
Kumar Gala <galak@...eaurora.org>, devicetree@...r.kernel.org,
LKML <linux-kernel@...r.kernel.org>, linux-omap@...r.kernel.org,
linux-arm-kernel <linux-arm-kernel@...ts.infradead.org>,
linux-pm@...r.kernel.org
Subject: Re: [PATCH v4 3/6] Documentation: DT: Document twl4030-madc-battery bindings
Hi,
Am 01.04.2015 um 22:16 schrieb Pavel Machek <pavel@....cz>:
> Hi!
>
>>> As I explained in some other mail, those tables should not be
>>> neccessary at all. They can be computed from li-ion characteristics
>>> and internal resistance, and assumed current during charge and
>>> discharge.
>>
>> I already explained that we do not know the charging and discharging
>> current well enough for such a calculation.
>>
>> And I explained that the “internal resistance” is a system (battery + cables +
>> connectors + other circuits) parameter that is not easy to derive or measure
>> and type into the .dts source code.
>>
>> At least I have no idea how I should find it out for my boards. While I can
>> easily determine the curves (and we already have them for the platform_data
>> driver).
>>
>> Please propose your own code doing that so that we can test if it is
>> better.
>
> So, how does this look?
>
> It looks to me like you have cca 0.1 Ohm resistance in your system,
This is completely unknown.
> and are using cca 75mA while discharging, and charge by cca 1.4A.
Where do you get these figures from?
A GTA04 board discharges usually between 50 and 400 mA (depending on activities)
and if you turn on WiFi, it will be up to 600 mA. If you use 3G it can draw even more
during operation.
Total current going in is 500-800 mA (depending on USB negotiations) and this is
split into system operation and charging current. So 1.4A charging current is not
possible. Rather 200-500 mA.
So it might be that the battery is discharged although the system is connected
to a charger.
> (And
> these are all the coefficients the code should need; rest is battery
> characteristics -- common to all li-ions, and charger characteristics
> -- that will be common to all cellphones. If current can be measured,
> this code should go more precise answers).
>
> pavel@amd:~/g/tui/ofone$ ./liion_maps
> Charging
> Voltage 4.2 V ; table 100 % internal voltage 4.18 V current 0.233 A computed 97 %
> Voltage 4.1 V ; table 75 % internal voltage 4.08 V current 0.233 A computed 87 %
> Voltage 4.0 V ; table 55 % internal voltage 3.93 V current 0.700 A computed 69 %
> Voltage 3.9 V ; table 25 % internal voltage 3.76 V current 1.400 A computed 26 %
> Voltage 3.8 V ; table 5 % internal voltage 3.66 V current 1.400 A computed 3 %
> Voltage 3.7 V ; table 2 % internal voltage 3.56 V current 1.400 A computed 2 %
> Voltage 3.6 V ; table 1 % internal voltage 3.46 V current 1.400 A computed 1 %
> Voltage 3.3 V ; table 0 % internal voltage 3.16 V current 1.400 A computed -1 %
> Badness 395.4861761427434
> Discharging
> Voltage 4.2 V ; table 100 % internal voltage 4.21 V current -0.075 A computed 100 %
> Voltage 4.1 V ; table 95 % internal voltage 4.11 V current -0.075 A computed 91 %
> Voltage 4.0 V ; table 70 % internal voltage 4.01 V current -0.075 A computed 79 %
> Voltage 3.8 V ; table 50 % internal voltage 3.81 V current -0.075 A computed 46 %
> Voltage 3.7 V ; table 10 % internal voltage 3.71 V current -0.075 A computed 3 %
> Voltage 3.6 V ; table 5 % internal voltage 3.61 V current -0.075 A computed 2 %
> Voltage 3.3 V ; table 0 % internal voltage 3.31 V current -0.075 A computed 0 %
> Badness 171.69576218433212
>
>>> Running below 3.3V.. not really. At that point, the battery is really
>>> _empty_, and voltage is going down really really fast.
>>
>> It is the diffference between 2% and 0% where a fuel indication might
>> be most important…
>
>>> Plus, you are damaging the battery at that point.
>>
>> The power controller will shut down - but the driver should report
>> reasonable (but IMHO not necessarily perfect) values until the last
>> moment.
>
> It is tricky to do a good job near 0%... or anywhere else. See for
> example
>
> http://cseweb.ucsd.edu/~trosing/lectures/battery.pdf
>
> You start a call, percentage goes down, end a call, it will go
> back up. I'm pretty sure you will not be able to make a call with "5%"
> indication from your code at low battery temperature (say -10C).
The whole system is heating up a little, so that you never have -10C for the
battery.
I think you are trying to solve situations that don’t exist in practice.
And AFAIK, the GTA04 board is the only user of this driver in case the battery
has no built-in fuel gauge. So why improve it beyond what the GTA04 users need?
I repeat myself: this driver is not built for highest precision because there are
better methods (fuel gauge chip). These might not be available so this fall-back
driver has been built.
It is definitively better than no driver and worse than the optimum.
>
> Anyway, see above, I think I provide reasonable values even in that range.
>
> Signed-off-by: Pavel Machek <pavel@....cz>
> Pavel
>
> #!/usr/bin/python3
> import math
>
> def percent_internal(v):
> u = 0.0387-(1.4523*(3.7835-v))
> if u < 0:
> # Formula above does gives 19.66% for 3.756, and refuses to
> # work below that. Assume 3.3V is empty battery, and provide
> # linear dependency below that.
> u = (v - 3.3) * ((3.756 - 3.3) * 19.66)
> return u
> return (0.1966+math.sqrt(u))*100
>
> charging = [ [4200, 100], [4100, 75], [4000, 55], [3900, 25], [3800, 5], [3700, 2], [3600, 1], [3300, 0] ]
>
> discharging = [ [4200, 100], [4100, 95], [4000, 70], [3800, 50], [3700, 10], [3600, 5], [3300, 0] ]
>
> # current > 0: charging
> def percent_ohm(v, current):
> v_int = v - current * 0.1
> print(" internal voltage %1.2f V current %.3f A " % (v_int, current), end='')
> return percent_internal(v_int)
>
> def percent(v, charging):
> if charging:
> # Charger model. Chargers will do constant current then
> # constant voltage, so current will go down as voltage
> # approaches 4.2V
> i = 2.8
> if v >= 4.:
> i = i/2
> if v >= 4.05:
> i = i/3
> else:
> i = -0.15
>
> # With current forced to 0, we get badness 4014 and 258
> # 2.5A, sloped: badness 576
> # +4A -> badness 1293
> # +3A -> badness 890
> # +2.5A -> 339
> # +2.4A -> badness 389
> # +2.3A -> badness 444
> # +2.2A -> badness 504
> # +2A -> badness 634
> # +1A -> badness 1450
> # +0.5A -> badness 2865
> # -0.2A -> badness 252
> # -0.15A -> badness 251.37
> # -0.1A -> badness 252
> # -0.05A -> badness 254
> i/=2
> return percent_ohm(v, i)
>
> def compute(map, charging):
> diff = 0
> if charging:
> print("Charging")
> else:
> print("Discharging")
> for v, p in map:
> v /= 1000.
> print("Voltage ", v, "V ; table ", p, "% ", end='')
> p2 = percent(v, charging)
> print("computed ", int(p2), "%")
> diff += (p-p2)*(p-p2)
> print("Badness ", diff)
>
>
> #perc = percent(volt)
> compute(charging, 1)
> compute(discharging, 0)
Please explain what you calculate here. Especially what “Badness” is?
BR,
Nikolaus
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