Aren't the connection between the programmer/debugger used
and the -ICD version of the PIC correct simply by using
the supplied cable between the programmer/debugger and
the header ? The header itself would internaly connect
to the correct pins, not ?

The fact that the standard chip and the debug-enabled chip
has different pin-out is something internal to the header.

And what is connected to the *breadboard* would be the
correct pinout of a real 627A/628/648A. And the crystal
would have to be connected to the target-connector of the
header just as if the header *was* a real processor.

I also saw the schematic, but it didn't help me understand
how things are setup. The "unusal" pin numbers of the debug-
enabled PIC is of no concern for the target circuit. If not,
it wouldn't be usable for in circuit deebugging.

Just connect the header to the programmer/debugger as the
docs says and connect the header to the target circuit just
as *if* it was a standard 628A/648A.

Or have I completely missunderstod how the ICD-header works ?

Jan-Erik.







jim wrote 2011-11-05 19:16:
> PGC and PGD are RB6 and RB7 respectively.  In your schematic you show the
> PGC and OGD pins going to the OSC1 and OSC2 pins.
> This is why there is no ICD functionality with your circuit.
>
> If you connect the ICD pins to OSC1 and 2, you effectively disable the
> oscillator.
>
> Try moving the ICD PGC and PGD pins to RB^ and RB&.  You should be able t=
o
> use the ICD then.
>
> Regards,
>
> Jim
>
> -----Original Message-----
> From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf =
Of
> jana1972@centrum.cz
> Sent: Saturday, November 05, 2011 9:45 AM
> To: Microcontroller discussion list - Public.
> Subject: Re: [PIC] IS this faulty?
>
> Thanks ALL who replied.
> I do use capacitors in the schematic.
> Please see the schematic here
> http://www.dilynamobily.cz/Media/Uploaded/Normal/Problem.jpg
> ( the link is case sensitive)
>
> My problem is that I can not debug any program using PIC16f648A - ICD
>
> Please check and let me  know if you can see any problem
> Thanks
> L.
>
>
>> OK, whatever.
>>
>> We don't know for sure if there are any capacitors or not.
>> And if not, isn't it a possible "problem" ?
>> I think it is.
>>
>>
>> jim wrote 2011-11-05 14:53:
>>>
>>> The main reason for the capacitors in a crystal circuit is to provide a=
n
>>> intentional imbalance between the input of the oscillator and output pi=
n
> of
>>> the oscillator so that when power is applied, the imbalance causess
> current
>>> to flow, and thereby guarantees the oscillator will start oscillating.
>>>
>>> Jim
>>>
>>> -----Original Message-----
>>> From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behal=
f
> Of
>>> Isaac Marino Bavaresco
>>> Sent: Saturday, November 05, 2011 7:45 AM
>>> To: Microcontroller discussion list - Public.
>>> Subject: Re: [PIC] IS this faulty?
>>>
>>> Em 5/11/2011 10:57, Jan-Erik Soderholm escreveu:
>>>>>> I *think* that the header is useless without a target circuit.
>>>>> But what is the smallest target circuit?I thought adding only crystal
> is
>>>>> OK for starting debugging
>>>> Right, but a crystal is useless without capacitors. And it is
>>>> a bit unclear how the header is powered.
>>>
>>>
>>> Not exactly. It may become unstable, or not to oscillate at all, but it
> may.
>>>
>>>
>>>>> Sorry, I have just only  started with microchips  - so a lot  of new
>>>>> knowledge for me to learn.
>>>> It's not about learing PICs, it's about telling everything and
>>>> not hiding details. :-)
>>>
>>>
>>> By my experience with teaching newbies in electronics and
>>> microcontrollers, it takes time to notice that there's missing
> information.
>>> Because of this, I generally take it easy with them.
>>>
>>>
>>> Isaac
>>>
>> --
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>
>
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