>
>
>Dan Michaels wrote:
>>
>> Brian Gregory wrote:
>> >Simon Nield <
TakeThisOuTsimon.nieldEraseME
spam_OUTQUANTEL.COM> wrote:
>> >> if the pic you are using has an adc you could use that.
>> >
>> >I would need some kind of voltage reference or some resistors to divide the
>> >battery voltage if I used the supply (from an LDO regulator I guess) as the
>> >reference. That's run the battery down.
>> >
>>
>> Yes, it would seem that just about anything you connect to the
>> battery itself is likely to pull it down - except possibly a zener
>> with high enough voltage and hopefully low leakage [??????]
>>
>> Here is a variation on some of the other - diode+resistor schemes.
>> Assuming you aren't running straight off the battery - but are using
>> an LDO to power the bus - eg, LM2936-5.0 [15 uA]:
>>
>> battery---+-----LM2936--+----> +5v
>> | | |
>> | gnd |
>> zener |
>> | +-----+---+
>> | | Vdd |
>> +--10K--|RB7 |
>> | | |
>> +-------|RA0 |
>> Vsense | |
>> +---------+
>>
>> Assume battery = 9v, then use 5.1v zener. RB7 = 1 while in sleep
>> mode. Wake up --> pull RB7 = 0 --> measure voltage on RA0 -->
>> Vsense = Vbattery-5.1v --> accurate reading, since the A/D Vref=5v
>> from the LM2936.
>>
>> At full charge, Vsense ~ 9-5.1v = 3.9v. At low-charge, when
>> Vsense ~ 0.1v, then Vbatt has dropped to 5.2v, and the LDO is about
>> to fall out.
>>
>> If the zener has too much leakage, then assign RB7 as input
>> during sleep, instead of holding it as output = high.
>>
>> Who knows, this thing might even work,
>> - Dan Michaels
>> Oricom Technologies
>>
http://www.oricomtech.com
>> =========================
>>
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