Note that if you accidentally short the power supply by dropping a
screwdriver on it, the dV/dt will be very high.

Sherpa Doug

> -----Original Message-----
> From: Olin Lathrop [mailto:olin_piclist@EMBEDINC.COM]
> Sent: Thursday, August 30, 2001 8:07 AM
> To: PICLIST@MITVMA.MIT.EDU
> Subject: Re: [PIC]: PIC reseting due to relays
>
>
> > Is 1uF really that big ?
>
> Yes, I definitely wouldn't do this.  When power is shut off
> and the power
> supply drops, the cap will discharge thru the PIC.  The cap
> current is I = C
> * dV/dt = 1uF * 5V / Td,
> where Td is the power supply discharge time in seconds.  If
> your supply
> discharges in 5mS then the average cap current will be 1mA.
> If the PIC
> doesn't allow that current, then you enter unspecified
> territory with MCLR
> at 5V and everything else at 0.
>
> > I recall others on the list who routinely
> > use 10uF or more with no apparent problem.
>
> Now you get 1mA average at 50mS discharge time.
>
>
> ********************************************************************
> Olin Lathrop, embedded systems consultant in Littleton Massachusetts
> (978) 742-9014, olin@embedinc.com, http://www.embedinc.com
>
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>

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