Hiya, I read a while ago that 100k draws less current than 470k, is this true. As i’m building a new lunetta type machine I have mostly 470k that I bought from Maplin,when they were having their closing down sale. I’m using a 9v battery to power this going into a 5v regulator, just worried about using up the battery power with all those 470k pots (10 in total). Will that be an issue? I have a few 100k so may use those too. Many thanks.
No. Ohm’s law: I = V/R. Larger resistance, smaller current.
If you’re dropping 5 V across a 470k pot, I = 5/470k = 11 µA. 10 such pots is 110 µA = 0.11 mA. Typical 9 V alkaline batteries have a capacity of about 550 mAh, so if the pots were the only thing drawing current:
t = 550 / 0.11 = 5000 hours
That assumes you’re using the pots as attenuators and that current out of the wiper is negligible. If they’re used as variable resistors, R will be smaller (down to 0 Ω, though there should be a fixed resistor in series to limit the current) and the current will be higher.
But even then, if R is typically ~ 100k, I = 0.5 mA and 550 / .5 = 1100 hours.
Presumably your device’s active components (ICs, transistors) will be drawing significantly more current than that and the pots will not be significant in how long the battery lasts.
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Again you’ve explained it very well, thanks. Yes I’m using the pots for both attenuators and variable resistors. I thought the same when you said the IC’s, transistors will be drawing more current than the pots. It’s just nice to have some backed up info from a well learned source as yourself, thanks again.