Now, I'm no electrical engineer, but it seems to me that you have two paths
for the discharge of this capacitor, one is through the pullup to VCC, and
the other is through a protection diode to VCC (MCLR line).  As VCC is
dropping won't the resistor start discharging first because of the step
function nature of the forward bias voltage on the diode (i.e. until there
is at least a .6V drop on the resistor ...60uA current... the diode does
nothing)?

So, in other words, as long as VCC is decaying at a rate that is equal to
or slower than the time constant of your MCLR RC combination you should be
ok...  Even a little faster might be ok because then you have two parallel
paths (as well as the .6v 'hysteresis' forward voltage of the MCLR diode).
For me this time constant is .1uF * 10k Ohm = 1ms.  I've got decouplers on
my VCC line on the order of 22uF +.  If for example, my micro typically
burns 1mA or less that gives an equivalent load at 3.6V of 3.6KOhm giving a
time constant for VCC decay of : 3.6k * 22uF = 79mS.  Seems like it should
never be a problem unless some yahoo figures out how to yank out a battery
and short VCC to GND in less than a couple of mS.  I'm not saying its
impossible, I'd just like to see it!

Well, its all academic as next spin of board gets a schottky diode to VCC
anyhow.

Let me know if there is something wrong with my logic...  In practice, my
EE skills stop around the V=IR stage.

-E

>At 11:53 PM 02/23/2000 -0700, you wrote:
>>At 03:28 PM 2/23/00 -0700, Dan Michaels wrote:
>>
>>>Dwayne,
>
>[snip]
>>
>> >I agree with one of Dan's comments - a large cap tied directly to MCLR
>>> >might cause damage if VCC is ever lowered suddenly (shorted to gnd).  This
>>> >is not a normal situation and I am not aware of any failures in my
products
>>> >that can be attributed to this.
>>>
>>>Personally, I don't think you need the "sudden" qualifier, but it's
>>>a potential problem every time you power down the system.
>>
>>I don't see how!  If VDD collapses slowly (ie - you remove AC), the cap
>>will begin to discharge either thru the substrate or the pullup resistor or
>>both.  If the power supply can supply all the current the circuit needs,
>>the cap on MCLR will supply only enough current to discharge - usually a
>>very low current.
>>
>>dwayne
>>

Erik Reikes
Software Engineer
Xsilogy, Inc.

ereikes@xsilogy.com
ph : (858) 535-5113
fax : (858) 535-5163
cell : (858) 663-1206