battery charge time and tests

Werner Almesberger werner at openmoko.org
Thu Feb 3 17:30:14 EST 2011


Bas Wijnen wrote:
> The citation is missing, but your explanation is good enough. :-)

Oops, sorry ! This one:
http://en.qi-hardware.com/w/images/9/9c/Lb60_schematic.pdf

The core of the charger circuit is in the box on the first page.

> 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?

CHARGE_N is what the signal is called in the Ben. The Jz4720 pin is
PC27. See page 5 of the schematics.

>> Because of D4, you can't force it on.
> 
> Which makes it not very useful anyway...

I think you could just short D4 without suffering too many ill
consequences (you drop up to ~2 V in the LED and then another ~0.3 V
in the clamp diode of PC27, which should always yield less than 3.3
V), but it's probably not worth the risk.

> 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.

No, it only does this if PW_ON_N (CPU signal PWRON_, in the schematics
called "POWERON"_) is low. If PW_ON_N is high, the charger should pass
~550 mA to the battery.

I think I figured out the logic now. I had it wrong before:

- if we're on USB power and the CPU is switched off, PW_ON_N is high
  and Q3 does conduct. Thus, we charge the battery with 550 mA.

  USB current is these 550 mA plus at least 10 mA on LED and R33
  (see below.)

- if we're on USB power and the CPU is switched on, the CPU pulls
  PW_ON_N low, Q3 stops conducting, and we charge the battery with
  only 110 mA.

  Furthermore, +VBUS drives the gate of Q1 high and Q1 doesn't conduct.
  This means that no current flows between the battery and the rest
  of the system, and the system is supplied via D2. Hence, total USB
  current is 110 mA plus whatever the rest of the system demands.

- if we're not on USB power, the gate of Q1 is low, Q1 conducts, and
  thus supplies the system from the battery.

> When I remove
> the battery, it takes 4.4 mA, so that should be subtracted.

Whee ! That's the R33-(R35 || Q5) path on page 2 ;-)

With a battery present and being charged, you'd also burn about 6 mA
in the R12-D3-U5 path.

> 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:

R33-... is on +VUSB, so most of these 4.4 mA should disappear when
running from battery alone.

- Werner




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