During programming, you have a 8V drop (13V-5V) across a 27k  
resistor, effectively pumping 0.3mA into your VDD rail.  As long as  
your circuit pulls more current than that from VDD (and your chip  
does), you'll be okay.

They also connect VDD to VCC to allow the chip to be powered by the  
programmer, with a diode protecting the VCC supply.  This means that  
you *MUST* turn off the VCC supply before programming.  Personally, I  
don't use the VDD connection from my PICkit2 programmer at all since  
the VDD on it is tied low when not actively programming.  I just let  
the VCC rail power the chip during programming.  This *will* be an  
issue if you have a VPP-before-VCC chip, but I just know that there  
would be some point where I'd either connect the programmer before  
turning off VCC or turn on VCC before removing the programmer  --  
both of which would short VCC to ground.  (Is there a list of those  
sneaky VPP-before-VCC chips?)

I hope this helps,
Matt

On May 4, 2007, at 1:17 PM, Kevin wrote:

> Hi All,
>
> Can someone explain why Vpp is connected to Vdd as shown in the ICSP
> diagram at http://kitsrus.com/gif/icsp.gif
>
> The associated page at http://kitsrus.com/icsp.html mentions "The  
> 27K is
> recommended to prevent VPP current from raising VCC" but I don't  
> see how
> this works.
>
> What I figure is that during ICSP when Vpp is applied (13V) shouldn't
> Vcc (5V) be (lightly) pulled up to 13V because of that 27K resistor?
> Wouldn't this then exceed the "Max voltage on Vdd with respect to Vss"
> spec and damage the PIC?
>
> I'm sure I'm missing something here (because I have used this ICSP
> circuit successfully), but I just can't see it ;-/
>
> BTW a related question - is the MCLR pin more susceptible to ESD  
> because
> it doesn't have an internal clamp diode to Vcc?
>
> Thanks in advance,
> Kevin
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