[sdiy] Temperature compensation results

Ian Fritz ijfritz at earthlink.net
Sun Jun 15 16:50:27 CEST 2003


At 10:32 AM 6/14/2003, René Schmitz wrote:

>I would want to know what makes you think that these nonideal effects 
>aren't present in the transistors which are considered "matched"?
>Or put differently, a mismatch of 50uV would be caused by a difference in 
>saturation current and a 1mV mismatch won't?  Where would the boundary be?

I won't pretend to have any expertise on what kinds of sources of 
non-idealities occur in Si devices, since most of my work has been in 
compound semiconductors.

In an extreme case, I would imagine that one transistor might have a large 
defect, such as a dislocation pileup originating from a substrate 
imperfection, and that the other transistor would not.  Then the Is factors 
would be *qualitatively* different for the two devices, since one would 
have to include an extra term to describe the large current from the 
defect-induced leakage path.

More generally, I believe a good model would include leakage from a 
distribution of defects.  Mismatch would then depend on what the actual 
distribution in each device is.  So I would guess that the boundary you are 
asking about isn't sharp.

Interestingly, I think this discussion has a connection to some work I did 
a couple of years ago to select a transistor noise source.  I went through 
about 50 2N3904 devices looking for the ones with the lowest 1/f noise.  I 
only found one with really low 1/f noise, and that was the one with the 
lowest overall noise level.  This suggests to me that most of the devices 
have significantly more leakage than is ideal.  So I agree that there are 
non-ideal leakage processes in the pairs we choose as matched.

Perhaps the devices we pick as being matched have simply have a low level 
of similar defect distributions.  This suggests a new way to pick matched 
pairs: compare reverse leakage currents and pick pairs with matched and 
relatively low leakage.  Of course no one will agree to doing it this way, 
because it's not how Moog did it.  :-(

Best regards,

   Ian




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