>> Bipolar transistors have negative temperature coefficient,
>> i.e. resistance goes down as they heat up.

>        I've been thinking about this for a week now and I'm not sure I
>understand.  I thought resistance always increased with increasing
>temperature.

The problem is that the forward-biased voltage drop decreases with
increasing temperature.

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Consider the simplified version (above) of two bipolar transistors in
parallel. Essentially you're dealing a pair of diodes in parallel.

Nominally, a conventional silicon diode has a forward voltage drop of
600mV. If the diodes are absolutely identical, no problem.

If, for argument's sake, the upper diode in the diagram has a voltage
drop of 600mV and the lower one a voltage drop of 590mV, then,
theoretically, the bottom diode will conduct all of the current and the
upper diode will conduct nothing.

As the bottom diode heats up, its voltage drop decreases, ensuring that
the voltage drop difference becomes even greater and so on.

In reality, both components would conduct some current, with the bottom
one conducting more than the top one and with this difference increasing
as the bottom one heated up faster due to its conducting a higher
current.

Although the voltage drop in a bipolar transistor is lower than a simple
diode the same principle applies.

Tim. 


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