Op 03-02-11 20:53, Werner Almesberger schreef: > Bas Wijnen wrote: >> If anyone has info about how to do this (or pointers to Linux source >> files which read the capacity), I'm interested. :-) > > I'd look at the schematics [1]:
The citation is missing, but your explanation is good enough. :-) > the red LED is connected to the > charger chip U5 and - with diode D4 preventing high VUSB flowing into > the CPU - to the CPU. > > You could try to force it off by outputing a "high" level on CHARGE_N, > "fighting" U5's FET pulling down, but you probably won't reach a > sufficient voltage difference to accomplish much. Not sure if this is useful anyway, but I looked in the Jz4720 datasheet and can't find a CHARGE_N pin. The schematics I found so far weren't readable enough (for me anyway) to conclude what was connected where, so I have the question: is CHARGE_N just a gpio which is named because of its use (by the kernel), or is it a missing pin on the Jz4720? > Because of D4, you can't force it on. Which makes it not very useful anyway... > The only feedback on battery status you have is +VBAT on the ADC > (ADIN0). And now I see that this is located in the "touch screen" section of the programmer's manual, which explains why I never saw it before. :-) > A USB current < 400 mA would suggest that the PW_ON_N switch (Q3) > isn't acticated. In the case, the nominal charge current (which > includes overall system consumption as well) should be 1100V/R15 = > 110 mA. If PW_ON_N is set, it should raise to approximately > 1100V/(R14 || R15) = 550 mA. So you're saying that the battery always charges with (at most) 110 mA? That's quite a lot less than the maximum USB current of 500 mA. > Note that this is the current that leaves U5 and is shared by the > battery and the rest of the system. > > Disclaimer: all this is according to schematics and data sheets. > Reality may choose to differ :) Of course. :-) The 180 mA I measured was for an empty battery with the device turned off. It does still supply the rtc though (but I suppose that's what you mean by the overall system consumption). When I remove the battery, it takes 4.4 mA, so that should be subtracted. I suppose 4.4 mA is also the amount by which it drains the battery when it is turned off, leading to an empty battery in: capacity: 850 mAh * 3.7 V = 11.3 MJ Draining with 4.4 mA * 5 V = 22 W leads to 11.3e6 / 22 = 515e3 s = 143 h = 6 days. When switched on, it drains with values depending on its use, but the maximum is around 500 mA. In that case, the result is multiplied by 4.4/500, leading to 75 minutes. Thanks, Bas
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