Adrian (a@acm.org) on 6/24/11 wrote:
---------------------------
>David Hess (davidwhess@gmail.com) on 6/22/11 wrote:
>---------------------------
>>
>>Now I am confused. How does that explain the negative temperature coefficient
>>for MOSFET channel resistance? It can not be because of threshold voltage change
>>because it occurs in saturation. It does however become a positive temperature coefficient at very high voltages.
>>
>>Or does the above only apply to power MOSFETs and not to the submicron ones used in logic processes?
>>
>>How does this relate to gain increasing in bipolar transistors with increased temperature?
>>Secondary breakdown is a very real problem in bipolar power transistors.
>>
>
>The temperature coefficient of the channel resistance of MOSFETs is *positive*, not negative.
>
>For power applications this is a major advantage of MOSFETs, because it allows
>their connection in parallel without the danger that the current will pass mostly through a single device.

Sorry about that. I was thinking the right thing but somehow reversed it when writing it down. :) I was up all night diagnosing a Tektronix 2230.

>For logic circuits this leads to the familiar behavior that they are faster at
>low temperatures, where the channel resistance is small, so the RC products are
>also small, thus the switching times become smaller.
>
>Nevertheless, there are certain operating points (e.g. close to breakdown or in
>subthreshold mode) where the channel resistance may *seem* to have a negative temperature
>coefficient. However, in those cases, what you see is not the current that corresponds
>to the drain-source voltage divided by the channel resistance, but a current whose
>largest part is due to other effects, e.g. due to the parasitic bipolar transistor
>that exists in parallel with any MOSFET. Whenever the leakage current through the
>parasitic bipolar transistor is not negligible compared to the drain current, you
>will see a negative temperature coefficient.

Unfortunately my two Siliconix books are in a box somewhere or I would have looked it up. I just remembered that the current sharing fails at high voltages and something similar to secondary breakdown becomes possible.