Actually no where near so much is required. All you need is twice enough to counteract the temperature coefficient over a useful temperature range (say up to 150C)
The drop involved should be on the order of 0.12 to 0.25 V at operating current, usable down to about 5 V, maybe 3.3 V. At higher voltages and modest currents use just enough to swamp the temperature coefficient possibly dropping so much a 0.5 V).
-- JosephKK Gegen dummheit kampfen Die Gotter Selbst, vergebens. --Shiller
Yes, but the V(f) is pretty high (1.5 to 2.2 V). Interesting device though.
-- JosephKK Gegen dummheit kampfen Die Gotter Selbst, vergebens. --Shiller
Gee, here I was thinking the important part was the fundamental principles, with which any such problem can be solved. silly me.
Cheers Terry
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Don't you know him? :-)
-- Thanks, Fred.
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