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zorgrian
modified 4 years ago

TweakingLinearPotsForAudioToMakeAudioTaperOrLlogPot.png

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Search tags: log, log pot, logarithmic potentiometer, audio pot, audio taper 1V p-p input There are two reasons for this little trick: 1) EC has EC doesn't have audio taper or log pots or any for of taper that's non linear. 2) Audio pots are becoming more difficult to buy The second reason is the most important. Audio pots are necessary as our hearing sensitivity in many ways, including volume, is not linear in response. This means that if you use a linear pot for audio, you get something approximating an on/off switch. The region at the lower end will appear to have some control but it wont seem very responsive, while the region at the top or quiet end of the pot will seem to have almost no function at all. So, to solve this problem you can put a resistor across the wiper and earthy end of the pot. The value should be about 20% of the pot value. Here we have a pot value of 100K We have a resistance of 20K This presents a 100K (approx) impedance to the input side, while presenting a 20K impedance to the output side. This set-up will work for microphones. Inside a guitar the story is different. you would have to change the pot value to 1Mohm and use a 200K resistor across the wiper and earthy end. Like this you will not load-up the pickups and will be presenting a reasonable impedance to the amplifier. ENJOY
published 9 years ago
thebugger
9 years ago
Yeah it adds up, but the resistor should be 1/4 the potentiometer value.
zorgrian
9 years ago
Why do you say a quarter of the value. Ive been doing this for years. I found that between 10 and 20% works well. 25% is too much. Test it out yourself

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