On Fri, 18 Oct 2002, Wouter van Ooijen wrote:

*>> The positive pulses can cause reverse breakdown of the BE
*>> junction in the multiemitter input transistor.
*>
*>I fail to see how that would be possible when the pull-up is taken from
*>the Vcc pin of the chip...

The pullup and the input pin capacitance form a lowpass filter but another
lowpass filter is formed inside by the base pullup resistor of the
multiemitter transistor and the base proper vs. substrate. Consider the
case of a NAND gate with one input driven low from outside, and the other
input tied high. The base of the first transistor will be at ~0.6V (less
than 1V probably). A positive spike enters through the Vcc line.

If the emitter is tied hard high it will rise with the Vcc line but the
base will follow later or not at all (it is kept low by the other input).
Meanwhile BE breaks down at ~5V and a 0.5-1V spike should be enough to put
the BE junction tied hard high into reverse breakdown mode and probably
cause a logic glitch in the gate output.

If the pullup is a resistor then it forms a filter with the pin input
capacitance and the tied high input will not follow the spike.  Also there
will not be enough injected current in case of reverse breakdown to cause
the transistor to glitch. Resistor values of 1k to 10k come to my mind.

Peter

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