Hi Folks,

I want to develop my own Programmer (Everybody else is doing it), but I want
to do a couple of things differently.

A big one is, I want to use a AC/DC Wall Adapter (aka "Wall Wart") for power
and another PIC (probably a 16C54) to control and communicate with the host
PC.  Now to simplify it, I want to use the incoming DC (15V) as the Vpp
reference.

Now, when I started looking at controlling the Vpp, I wondered about the
following circuit for Vpp:


  +12V ---------+
                |
                <
                > 10K Current Limiting
                <  Resistor
                >
                |            1401
 PIC Output ----+-------------|>|-------- To Programmed Part _MCLR


For Programming, the PIC Output Bit would be set to Input Mode (the
corresponding TRIS Bit = 1).  Otherwise (ie when the PIC is being pulled out
of the socket), TRIS=Bit=0 to drop the line down low and isolated it using
the diode.

I don't think there will be any problems.  According to the Programming
Spec, a max of 200 uAmps is drawn through the _MCLR Bit (which would cause a
2 Volt Drop across the 10K Resistor) and there is a 0.7Volt drop through the
1401 diode.  When the PIC pulls the Line down, there will only be 1.5 mAmp
being sunk.

If I start with 15 Volts, I will loose a max of 2 Volts through the 10K
resistor and 0.7 Volts thru the diode.  This leaves 12.3 Volts for
programming (enough to trip the PIC into programming mode).


Does anybody see any problems with this/reasons why I wouldn't want to
program a PIC using this method?

myke

Today, the commercial sector is advancing computer and communication
technology at a breakneck pace.  In 1992, optical fiber was being installed
within the continental U.S. at rates approaching the speed of sound (if
computed as total miles of fiber divided by the number of seconds in the year).

Aviation Week and Space Technology, October 28, 1996