2006/10/4, Shawn Wilton wrote:

> Yeah, I've seen that worthless document.  I have the 9V adapter plugged in
> to the ICD,

This is fine.

> so I don't think I need to apply an external VDD so long as I
> set the ICD to power the target (which is unconnected).

Then you have been applying Vdd to the header.

>  Should still go
> ahead and connect.

I would suggest you just connect a 10K resistor and do the test
without plug it into your target.  That's the way I tested it this
evening.  I don't have any target circuit at the moment.  In this way
it would at least eliminate other possibilities.




> Iff VDD were not powered wouldn't the ICD fail to connect and fail the
> self-tests as well?

No Shawn.  The Vdd exist on the header.  You power it from your ICD2
with the 9V external power supply.  This is fine.  But the problem is
you'll still need to pull the 'actual MCRL' to Vdd.  Let me explain
this in deep.

Think of the header as 2 part.  Part A: the extra pins that designed
for ICD2 communication and setting up some options (like jumper, some
other type of headers don't have jumper).  This part also include the
~ICDMCLR, which has taken some reposibility (Vpp) off the usual
MCLR/Vpp pins as you will find in usual ICD capable device, like the
high pin count PIC.

PART B: The 8 pins that are mapped to emulate the actual F675.  For
this part to work, you have only one option regarding the
configuration of the MCLR.  It must be configured as MCLR. Not a true
MCLR as some of its responsibility (Vpp) has been taken by PART A
(~ICDMCLR),    BUT it must still be pull-up to Vdd, as you did with a
normal PIC's MCLR.  Without that the PIC won't run, and the ICD2
communication can't be established. Can you just find a 10k resistor
and connect it to the header to perform the test.  It sould be joint
to pin 3 and pin 6 of the -ICD.  You can find the PCB trace and join
it at the top side of the PCB.  Or connect the 10k resistor at your
target and plug the header in.


Best regards,

-- 
WH Tan
-- 
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