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Author Topic: Can a pin handle a voltage greater than Vcc if it is just sinking to ground?  (Read 739 times)
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It will read 255 (or whatever the max value for 10-bit resolution is) until the voltage falls below 3.3V.

1023.

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The internal clamping diodes will protect the pin as long as I supply enough current limiting to it.  Is this right?

That is my understanding.  It is also my understanding that protection is obtianed by the diode conducting.  Which would be a constant drain on the battery just like the voltage divider (until the battery drops below 3.3V).  Which brings us full circle to my earlier question.  Is the voltage divider that much of a drain to be a problem?


I don't really know.  It's a 400mAh battery.  I hate the idea of it just draining down until the battery protection circuits kick in even when the device isn't being used.
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It will read 255 (or whatever the max value for 10-bit resolution is) until the voltage falls below 3.3V.

1023.

Quote
The internal clamping diodes will protect the pin as long as I supply enough current limiting to it.  Is this right?

That is my understanding.  It is also my understanding that protection is obtianed by the diode conducting.  Which would be a constant drain on the battery just like the voltage divider (until the battery drops below 3.3V).  Which brings us full circle to my earlier question.  Is the voltage divider that much of a drain to be a problem?


And I might add that as long as your battery cut-off device to disengage the battery once it's below your low point cut-off setpoint (most use a 3.0vdc min value not 2.7) is upstream of the voltage divider tap off point why be overly concerned about it's current draw? Or are you not planning on actual battery cut-off, just some kind of a minimum current draw mode?

Lefty
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 Or are you not planning on actual battery cut-off, just some kind of a minimum current draw mode?

Lefty


That's the idea.....put everything into deep sleep mode and have a pin change interrupt that will wake everything back up again when it is time to turn back on.  Its just a small GPS logger that my wife can wear on a lanyard around her neck when she runs.  I'm trying to make it really small (my goal is for it to all be on one board that will fit inside of an "altoids smalls" tin along with the battery.

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I hate the idea of it just draining down until the battery protection circuits kick in even when the device isn't being used.

What does "isn't being used" mean?  Processor asleep?  In Power Down?  On / off switch set to off?

How much wall clock time do you expect to get out of the 400mAh battery?  (How many minutes, hours, days before recharging?)

If you make the total resistance across the voltage divider ~10K ohms, the maximum current is 0.42 milliamperes.  In theory, draining to half the battery's capacity, that's 19 days eaten by the voltage divider.  Too much?

I assume you could add a transistor (MOSFET) to the circuit to separate the processor from the battery and eliminate almost all of the drain.  (A "hardware head" will have to advise you.)
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that's 19 days eaten by the voltage divider.  Too much?


Okay perhaps I'm over thinking this.  I was under the impression that it would drain much quicker.  Voltage divider should be fine then.
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Thanks everyone for helping me work through this!
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