P8X32A QuickStart (#4000) Input Pin max voltage pressure?

garyg

Hi
I wish to detect a 7.2 volt battery, if it is connected or not.
I've been looking for a specification of what the maximum voltage pressure that the
internal diodes that limit voltage connected  to an input pin can handle.
What I would like to do is use this formula 7.2v - 3.3v / 0.0005A = 7800 ohms.
It seems like I should be able to use a 7800 ohm or more likely 10K ohm connected
directly from 7.2V+ to my input pin.
Question is, will the Voltage of 7.2 or higher damage the input diode?
I realize this is a very basic question.  I just can't seem to find the answer.
Thanks for any help you can provide on this issue.
Garyg

Duane Degn

You should probably use 8.5V as the high voltage since 7.2V is likely the average voltage from the battery. Two cell LiPo batteries have a voltage of over 8.4V when freshly charged.
Still your 10K ohm plan should still be okay. You could certainly use a higher resistance. I'd think 100K ohm would work fine.

garyg

Thanks Duane
I will try using the 100K resistor as you suggest.
I think that keeping the diode current as low as is possible will help with my
overall Quickstart board current draw.
I'm still wondering though, how high of a dc input voltage can be when just using a resistor
to limit the current through the Diode.
Thanks again.
gg

Duane Degn

I hope someone corrects me if I'm wrong, but I don't think there is an upper (or lower) voltage limit. From what I've read it's the current that's important. As long as an appropriate resistor is used you can hook the Propeller up to high voltage power lines.
I recently read an old forum post where Chip Gracey mentioned one can monitor RS-232 with just a resistor on the line. Even the negative voltage swing of RS-232 is okay with the appropriate resistor.
I wish I had known this when I was first starting with the Propeller because I ended up using special chips to receive RS-232 signals.
One still needs these special chips to send RS-232 but not to receive it.

Phil Pilgrim (PhiPi)

It's the current that's important, not the voltage. The series resistor is there to limit the current. You will also need a resistor from the pin to ground. The reason is that, if the input is left floating, you could still see a high reading. The "pull-down" to ground will solve that problem. It you use, say, a 22K series resistor, a 100K pull-down will suffice.
-Phil

Ariba

A simple resistor will not be sufficient to detect if the battery is connected or not.The problem is that you have a floating pin if no battery is connected and you read a high or a low in this case. You need some kind of pulldown resistor for a defined "no battery" state.
This results anyway in a voltage divider. Such a voltage divider allows to detect not only if the battery is connected, but also if the battery has still enough voltage.
This circuit detects if the battery is connected and if the voltage is higher then ~6V :
Andy
Edit: I forget: There will be absolutely no current flow through the protection diode up to a Bat voltage of 13V.

garyg

Thanks to all.
I understand now.
I'll use a voltage divider.
I never thought about the pull down resistor.
gary