[sdiy] Protecting microcontroller inputs from - voltage?

[harrybissell]

Hi Bert...

yes precisely. Good news... if you choose a zener (my preferred solution)
you don't need another diode... the reverse voltage of the zener is a normal
diode drop, so you get both polarity protection in one component

The only thing you need to watch out for is zener tolerance... the 5.1V
zener might actually be higher.  I might use a lower voltage zener (like a 4.8V)

and have it almost in conduction already.  Or use a tight tolerance part...

This solution has the big advantage that it protects from power supply sequence
issues as well... the internal protection diodes of the chip (or the diode clamp
you
add from the pin to the POSITIVE supply...can allow the circuit to power up
just on the current from the input pin... this can be fatal in some circuits.
The Zener
cannot do this because it is not connected to the positive supply... it will
just return the
current to the driving source...

H^) harry

Bret Truchan wrote:

> Hi Harry,
>
> Do you mean something like this?
>
>                 R1       R2
> I/O pin] -----/\/\--+--/\/\----------< to whatever
>                     _|_
>                     /_\
>                      | anode
>                      |
>                     gnd
>
> R1, R2 = 200 ohm ~ 500 ohm
>
> My current design already looks similar to that, except that I use a
> 5.1v zener diode between the two resistors.  How could I incorporate
> both diodes into my design?  Would this work?
>
>                 R1                   R2
> I/O pin] -----/\/\--+-----------+--/\/\------< to whatever
>                     _|_          |
>                     /_\         _|_
>                      | anode    /_\  = 5.1 zener
>                      |           |
>                     gnd         gnd
>
> >>Hah... a trick question
> >>
> >>Putting the diode right AT the chip will probably not help you, unless you
> >>use shottky diodes.  A diode at the chip is a race... will the external
> >>diode
> >>conduct before the internal diode ?
> >>
> >>I'd even suggest using two resistors in series... with the diodes
> >>connected at
> >>the
> >>center of the resistors.. It would protect in both directions and not
> >>bother
> >>either
> >>the chip or the external input/output with too much current draw in a
> >>fault.
> >>
> >>Chip protection should be in proportion to the chances that someone will
> >>connect
> >>
> >>something unusual / stupid to it... almost a sure bet in a modular :^P
> >>
> >>H^) harry
> >>
> >>Ken Stone wrote:
> >>
> >> > >>   Use a diode clamp.  That is, a diode from the I/O pin to ground,
> >>with
> >> > >> the cathode on the I/O pin.  Check the microcontroller electrical
> >>spec,
> >> > >> but most will tolerate -1V or so.  If even less drop is needed, use
> >>a
> >> > >> Schottky diode.  In practice, a 1N4148 will work fine.  Caveat: put
> >>a
> >> > >> small resistor in series with the I/O pin and whatever it is
> >>sensing.
> >> > >
> >> > >On which side of the resistor would you put the diode ?
> >> >
> >> > The diode goes directly to the IC pin. The resistor goes between the
> >>pin and
> >> > whatever you are connecting to. The purpose of the resistor is to limit
> >>the
> >> > current flow through the diode in the "protected" condition. If you put
> >>the
> >> > resistor on the other side, you'll probably blow the diode up when the
> >>input
> >> > goes negative, or at very least, you will be sending whatever is doing
> >>the
> >> > driver to it's current limit.
> >> > _______________________________________________________________________
> >> > Ken Stone   sasami at hotkey.net.au
> >> > Modular Synth PCBs for sale <http://www.blaze.net.au/~sasami/synth/>
> >> > Australian Miniature Horses & Ponies <http://www.blaze.net.au/~sasami/>
> >>
> >
>
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