On Mar 27, 8:08 am, vasile surducan <[email protected]> wrote:
> On Wed, Mar 24, 2010 at 9:06 AM, m...@watty <[email protected]> wrote:
>
> > see alsohttp://www.embedinc.com/picprg/icsp.htm
> > Quotes
> > "Beware that some linear regulators, like common 7805 for example, can
> > be damaged by raising their outputs above their inputs.  A diode from
> > the regulator output to its input my be required.  "
>
> All 7805 I have seen  has that diode internally, check the manufacturer
> specifications before trust in everything Olin says.
>
> > Professional ICSP Programmer will control VDD. The main PSU must be
> > somehow isolated or off and only the PIC vdd connected to ICSP.
>
> No the main PSU does not need to be isolated nor off. Pickit2 is sensing the
> PSU voltage and is programming using that voltage and not the internal
> voltage. Did you measured the current flow value from Vpp to Vdd when
> programming ? It's a simple application of  Thevenin law there.
>
> Vasile
>
> P.S. If you want to discover new things, never believe entirely what is
> written in a datasheet.

Indeed. Test. maybe the Datasheet was written by a Student on Work
Experience :-)

Older 78xx seem to be more easily broken, esp. if the input of
regulator is large capacitors and no supply and you put +5V (or what
ever) on output pin, hence many circuits have diode From output to
input to shunt to input capacitor (avoids the 0.6V drop of series
diode on output).

However on my boards I may have circuits that can't be powered from
ICSP programmer VDD, or that I don't want powered. Some programmers
can also vary VDD. So I isolate the USB slave socket power (as the
spec says you must not feed +5V OUT on a USB slave socket) and also
the main PSU.

A new problem is the 18FxxJxx series which runs on 3.3V. If you want
more than 32K Flash and USB, this seems only solution,

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