RussellMc wrote: > His crucial assumption, that made his surprise at least reasonable > (and in no ways in conflict with the laws of physics) was - > > " ...all inputs are protected by static protection diodes which can > usually handle > several mA which is below the current limit of most programmers." I saw that, but I still don't see what that's got to do with anything unles= s you assume Vdd is tied to a fixed voltage source that can *sink* current. We were also talking about programming, so a sink on Vdd honestly didn't occur to me. To make his statement reasonable you have to assume: 1 - Vdd is being held solidly, and can sink substantial current. 2 - The programmer can only produce "a little" Vpp current, which certainly must be below the Vdd sink capability. The USBProg, for example, can produce about 10mA at 13V, and a lot more at 3.9V (one diode drop above 3.3V). Early PIC programmers were specified by Microchip to require 50mA Vpp drive capability. I don't consider these values to be "a little". 3 - The MCLR protection diode can handle said current without damage. Even if you believe all of 1-3 should have been clear (on reading again I agree 2-3 was stated well enough), that still leaves the glaring problem of attempting to violate the datasheet by putting way too much current thru th= e protection diode. Or you can think of it in voltage terms. To get 10mA thru the diode will require MCLR well more than the maximum drop above Vdd, and in reality we're talking about 10s of mA. Even if the programmer only puts out 10mA, the diode can handle it, and Vdd can sink it (not at all given, especially in a programming situation), thos= e 10mA can still end up running thru bad places in the chip. Whether that causes damage I don't know, but I'm pretty sure this setup would violate at least one datasheet spec although I didn't look it up this time. Given all that, it should be no surprise that something bad, possibly permanent happened. Perhaps there was honest confusion on the part of the OP, but I took it to meean yet another was trying to get away with willful datasheet violation, as you well know we see way too often here. I= n this case he'd actually tried it, and then was "surprised" that it didn't work. He made faulty assumptions about the internal workings of the PIC, decided therefore which datasheet limits he could ignore, then was surprise= d when the PIC blew up. I can see your interpretation might be correct if you try really hard to apologize for him as best as possible. Maybe you're right, maybe not. I don't think so, but since there isn't enough information your interpretatio= n must remain a possibility which didn't occur to me at the time. > He is saying that the protection diodes will USUALLY sink more current > than MOST programmers will source But how does he know that? Where is basis for this wild guess? > AND IF SO, the pin will be clamped > to a diode drop above Vdd Only if Vdd is a good sink and the current itself inside the PIC does no other harm. > AND SO the IC would not be expected to die > BECAUSE the absolute rating of voltage and current has not been > exceeded. But it has. In this senario the Vdd max rating is not exceeded, but the MCLR to Vdd maximum probably has (if it's the usual 300mV, I didn't check). Some PIC datasheets also specify a maximum protection diode current, although later datasheets seem to deal with that by specifying such a low voltage on the diode that it can't possibly conduct any "real" amount of current. Those parts with 5V tolerant inputs get more complicated as the high side protection is much more than a simple diode to Vdd. In that case his other assumptions would be wrong and violate his whole chain of logic. In short, I don't see how you can dump 10s of mA onto MCLR and stay within spec, regarless how well any other pins are held to any other voltage. > To me it seems that his key assumption (which he stated) is that the > programmer max current is < the protection diode max current. But there is no basis for that assumption, and it was apparently shown wron= g by experience as you stated. This is a especially bad assumption because h= e apparently doesn't know the range of Vpp currents programmers could provide= , nor the current the protection diode can tolerate, nor whether there even i= s anything as simple as a protection diode there. > If you had "addressed" him on the validity of that assumption we may > have had a reasonable discussion. Maybe :-). Maybe. I took his assumption to be so obviously absurd and without basis that it really seemed to me he was trying to justify getting away with violating the specs. I therefore addressed what I thought the real problem was. Perhaps my assumption was wrong, but you may not have liked my debunking of his assumption any better. ******************************************************************** Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products (978) 742-9014. Gold level PIC consultants since 2000. --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .