That's fun. Now we need all the divider combos for an eight-pack. :-)
Here's a start: 'n' in series feeding 'n' in parallel yields a (n^2+1) : 1 divider ratio.
Cheers, James
That's fun. Now we need all the divider combos for an eight-pack. :-)
Here's a start: 'n' in series feeding 'n' in parallel yields a (n^2+1) : 1 divider ratio.
Cheers, James
Right. Most surface mount resistors are made on 20 mil alumina, with about the same aspect ratio, so their capacitance (and their thermal resistance, and power dissipation capability [1]) is independent of size.
[1] assuming you heat sink the end caps ============================================================For completeness, was the mistake made when doing the calculation, or when typing up the post? Is the estimate for a 1206 SMD resistor 30 fF and you just happened to type SiO2 instead of Al2O3, or really 9.7/3.9 times that,
75 fF? Interesting that one is about 50% below the measured value, and the other about 50% high; I would predict that the estimate would be low due to field leakage around the sides. Anyway, just pedantically curious.----- Regards, Carl Ijames
I meant Al2O3, honest. The capacitance values are right. I measured several resistor values. Details if how I went about it are at: . Gosh, that's five years ago, already.
Jeroen Belleman
Don't stop now; you're on a roll.
Good for 30kV then! So your 1200 volts is no problem :)
Of course. However, as you know, I build a lot of low-level stuff where the noise of high- value resistors is as inconvenient as their slowness. Finding ways round it is a fun problem, even better than Fields's 555s. ;)
Cheers
Phil Hobbs
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