AVR-Chat

Sealed metal box - check.  :-)

It's going into an automotive environment, which is inherently noisy.  

Now just so I can make sure I'm not confusing myself, take a look at
this diagram I whipped up:

http://i6.tinypic.com/1zf3b6b.jpg

The 1k resistor keeps any current through the diodes to a minimum if
the user accidentally connects the pin to power.

The diodes protect the mcu from any static by shorting anything below
0.6V to ground, or above 5.6V to 5V.  

The 0.01uF cap filters out any AC noise.

Whaddaya think?  It's a relatively low frequency 0-5V DC signal (the
input).  I doubt it could switch from 0 to 5V in less than 50mS.

btw, thanks for pointing out the polyswitches.  I knew they existed, I
just didn't know what they were called.  I'll have to look into those
for current limiting the outputs.  I'll have to see if they make them
in 5 amp'ers.  


--- In AVR-Chat@yahoogroups.com, "Larry Barello" <yahoo@...> wrote:
>
> Well, if you are military and making something to survive a nuclear EMP
> event, you can do it: sealed metal box, with EMI filters on all
lines going
> into/out of the box (choke, capacitor, clamp diodes, resistors). 
That is
> about as idiot proof as possible (and very expensive).
> 
> For me, like Roy, below, 1-100k resisters and letting the input
protection
> diodes handle the load works well.  For analog inputs it is a bit
trickier.
> You need to understand how they work and the signal you are dealing
with,
> but again, for me the 1-10k + .01uf works great.
> 
> For digital signals, the input impedance is very high (100's megohms) so
> 100k is not that serious.  If you are using pull-up resisters, of
course,
> they have to be external (can't use the internal 50-100k resistors
in the
> AVR).  Even a 10kv discharge will only dump 100ma max through the input
> protection diodes and I believe that most modern CMOS inputs are
latch-up
> proof.  In any case, a 1kv discharge is only 10ma and that is safe
for any
> but the most archaic semiconductors.  The story gets better if there
is a
> bit of capacitance on the input as the usual source of static
discharge is
> pretty low capacitance (1000pf?) so the spike gets smoothed out and the
> current spread out for a lower overall peak value.
> 
> For analog inputs, Atmel recommends < 5k source impedance for accurate
> results.  So, if you use > 5k input protection, then you need some
> capacitance to lower the effective impedance at the pin.  That bandwidth
> limits the signal.  For anything I do that is never an issue as my
sample
> rates are rarely over 1-4 kHz.
> 
> Cheers!
> 
> -----------
> Larry Barello
> www.barello.net
> 
> 
> | -----Original Message-----
> | From: AVR-Chat@yahoogroups.com [mailto:AVR-Chat@yahoogroups.com]
On Behalf
> | Of Roy E. Burrage
> | Sent: Saturday, July 08, 2006 10:45 PM
> | To: AVR-Chat@yahoogroups.com
> | Subject: Re: [AVR-Chat] Re: Circuit for protecting analog inputs?
> | 
> | You might use a diode in series with the power input, Ryan, to protect
> | from polarity inversion.  Better yet, use a bridge rectifier and
you can
> | use either polarity DC as well as AC input from a wall wart.  That way
> | all you do is connect power...no problems with polarity or type as
long
> | as the voltage limits aren't exceeded.
> | 
> | For inputs, we use a 10K resistor in series with the input and
depend on
> | the internal diodes for clamping.  If speed isn't a factor, we
might use
> | 2 resistors and a 0.1uf or a 0.01uf cap between them to common for a
> | noise filter...which isn't a bad idea any time you can do it.
> | 
> | You can only go so far in making things idiot proof, you know.
> | 
> | 
> | REB
> | 
> | 
> | 
> | Ryan wrote:
> | 
> | >Thanks, I know exactly what you're talking about.  However, for
> | >someone who might inadvertantly connect a +12V supply to the
pin... It
> | >would obviously try to sink lots of current through the diode, likely
> | >blowing it up, then blowing up the MCU.  Would the effectiveness be
> | >reduced by putting series resistors (say, 1k?) with the two diodes?
> | >
> | >Or if the supply was connected backwards, all the current would
try to
> | >go through both diodes....  Trying to create an idiot proof box, but
> | >you know how that goes.  ;)
> | >
> | >Ryan
> | >
> | >
> | >
> | >--- In AVR-Chat@yahoogroups.com, Ken Arck <ah6le@> wrote:
> | >
> | >
> | >>At 10:38 PM 7/8/2006 -0000, you wrote:
> | >>
> | >>I need to protect an analog input for an AVR.  Protect it from ESD,
> | >>users connecting it wrong (i.e. to 12V power), etc.  Any suggestions
> | >>on what to use?  Unity gain opamp?  "transorb" or similar transient
> | >>device???  Discussion welcome.  Thanks,
> | >>
> | >><---Some expouse using zeners and currentl limiting resistors
but I've
> | >>always found the leakage current to be an issue. So now I use
regular
> | >>diodes to ground and the Vcc rail. See the attached schematic. The
> | >>
> | >>
> | >diodes
> | >
> | >
> | >>are general purpose silicon like 1N914 or 1N4148.
> | >>
> | >>Ken
> | >>
> | >>  ----------
> | >>
> | >>
> | >>
> | >>
> | >>
> |
>-------------------------------------------------------------------------
> | -----
> | >
> | >
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