[Gta04-owner] 4.3-rc6 available
Jens Seidel
jensseidel at users.sf.net
Wed Oct 21 13:54:01 CEST 2015
Hi Nikolaus and others,
2015-10-19 8:39 GMT+02:00 H. Nikolaus Schaller <hns at goldelico.com>:
> And, I have done some simple suspend tests
> and came to an interesting conclusion:
>
> root at letux:~# ./measure-suspend 10
> suspending for 10 seconds
> did wake up
> 192960 uA over 10 seconds
> root at letux:~# ./measure-suspend 20
> suspending for 20 seconds
> did wake up
> 128520 uA over 20 seconds
> root at letux:~# ./measure-suspend 30
> suspending for 30 seconds
> did wake up
> 107040 uA over 30 seconds
> root at letux:~# ./measure-suspend 40
> suspending for 40 seconds
> did wake up
> 80280 uA over 40 seconds
> root at letux:~#
>
>
> It is really unexpected that the average is going down such
> significantly.
>
> This indicates that the system is drawing a lot of energy,
> for some initial seconds and then goes down.
>
> Let's try to model it by
>
> Iavg = (Iinitial * initial + Isuspend * (total - initial)) / total
>
> where initial is the initial time with high current and total the
> total time.
>
> There are 3 unknown variable (Iinitial, Isuspend, initial)
>
> So we have to solve 3 equations, e.g.
>
> 192 * 10 = Iinitial * initial + Isuspend * (10 - initial)
> 128 * 20 = Iinitial * initial + Isuspend * (20 - initial)
> 80 * 40 = Iinitial * initial + Isuspend * (40 - initial)
let's not forget the fourth equation (for 30 s result):
107 * 30 = Iinitial * initial + Isuspend * (30 - initial)
You avoided it probably to not get an overdetermined system but even
without this the given three equations have no solution at all (even
if very small measure errors are respected).
> An approximate solution is (haven't tried to exactly solve the equation system):
>
> Iinitial = 4 s
> Isuspend = 50 mA
> initial = 400 mA
You swapped Iinitial and initial:
Iinitial = 400 mA
Isuspend = 50 mA
initial = 4 s
> So this means that it takes 4 seconds after suspend
> until power is really going down. So a suspend with
> less than 4 seconds has no effect. Interesting.
I was thinking about this a little bit more (from a mathematical
context) and tried to solve the system consisting of all 4 data sets
you have in a least squares approach:
Find a residual r so that the system
192 * 10 + r(1) = Iinitial * initial + Isuspend * (10 - initial)
128 * 20 + r(2) = Iinitial * initial + Isuspend * (20 - initial)
107 * 30 + r(3) = Iinitial * initial + Isuspend * (30 - initial)
80 * 40 + r(4) = Iinitial * initial + Isuspend * (40 - initial)
has a solution with minimal (Eucledian) norm of r: r(1)^2 + r(2)^2 +
r(3)^2 + r(4)^2 -> min.
I found r to be
128.1599
-64.0800
-256.3200
192.2399
What really surprised me is that all components of r are small
multiplies of 64. So I think that your model is 100% correct and the
accuracy of the measurement is a multiple of 64 mAs (here 256). Is
this possible?
The solution Isuspend =44.856 mA (instead of 50 mA) follows now easily
and uniquely by subtracting the equations from each other whereas
Iinitial and initial cannot determined (there are infinity solutions
all fulfilling (Iinitial - Isuspend)*initial = 1609.2 as only the
power consumption during the startup phase is unique, not the
splitting in startup time and startup consumption).
Is this useful to you?
Jens
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