Timeline for how to make a W potentiometer from a log or linear pot?
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:32 | history | edited | CommunityBot |
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| Jul 19, 2013 at 5:57 | history | edited | Metalhead1247 | CC BY-SA 3.0 |
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| Jul 15, 2013 at 12:57 | answer | added | Vovanium | timeline score: 0 | |
| Jul 15, 2013 at 6:42 | vote | accept | Metalhead1247 | ||
| Jul 15, 2013 at 4:34 | comment | added | Kaz | Tick mark label values are easy to calculate if a pot is linear. Just make sure the first and last tick line up accurately with the extreme rotations of the knob. Then linearly interpolate the resistance in between and plug into the circuit's formulas. | |
| Jul 15, 2013 at 4:32 | comment | added | Kaz | If this is for guitar, it is a waste of effort. You just need a scale around the pot to remember the knob positions. Put some equally spaced ticks around it and for the technically minded player you can put the actual calculated values next to these like kHz or dB or whatever. | |
| Jul 14, 2013 at 23:04 | history | tweeted | twitter.com/#!/StackElectronix/status/356549919722586112 | ||
| Jul 14, 2013 at 21:12 | comment | added | Dor | What about using a transistor-based resistor? With 3 Op-amps, each one for each curve, complementing a s-curve. | |
| Jul 14, 2013 at 17:07 | answer | added | Olin Lathrop | timeline score: 2 | |
| Jul 14, 2013 at 16:31 | answer | added | Andy aka | timeline score: 0 | |
| Jul 14, 2013 at 16:22 | history | edited | RedGrittyBrick | CC BY-SA 3.0 |
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| Jul 14, 2013 at 16:09 | history | asked | Metalhead1247 | CC BY-SA 3.0 |