As junction temp increases, r goes up and I goes down, right?
This must be a misprint:
(This is IRFM150/2N7224)
(see fig. 1 & 2 on p. 4). I think these 2 graphs are transposed. The higher condition should result in less current, yes?
Just testing my grasp of the knowledge...
Thanks,
DNS error:
Page 4 of a GIF??????
NOT FOUND in DigiKey; either 2N7224 or IRFM150.
Fantasy?
Sorry:
Click on the data sheet image.
Thanks,
The graphs look correct to me.
BTW there is no I vs. T curve, it's R vs. T on fig 4, page 4, and none of the other graphs are even close to I vs. T.
In the graphs you're looking at, you can hold Vds constant (say one volt) and compare Id at the two temperatures. It is clear that r decreases with increases in temperature. More carriers or something. Isn't that a characteristic of all junctions?
You will find a similar result for the IRF510, for example.
Chuck
Look carefully; the figures are correct; it takes a lower Vgs for a given Id at 500V (linear current sink region) and one gets a lower Rds(on) at a given gate drive (look at Vds=1V, Vgs=7.0V).
Gate threshold voltage has a negative coefficient, so falls with temperature (causing the Vgs = 4.5V curve to move up from about 1.1 to 4.5A). This can cause runaway conditions in paralleled linear circutis.
Rds(on) rises, as can be seen by looking at the saturated region of the, say, Vg = 15V curve: at 1V, Ids falls from about 30 to 17A. This causes stability in paralleled switching circuits (unlike bare silicon junctions in diodes and BJTs, and to a lesser extent, IGBTs).
It can be seen in Fig.4 that Rds rises about by a factor of 2 from 25 to
150°C, so my rough reading of the log plot is within reason.Tim
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