I'm following this here:
http://www.ladyada.net/make/midisense/sensormodel.html
So I have some 50k log pots I'm trying to use with the midisense....I think I understand the principle behind choosing a feedback resistor value, but I can't figure out one that works with a 50k log pot. If I understand what's going on correctly, the voltage can vary from .5 to 5V. If I choose a feedback resistor that's say 10k...that means at the lowest resistance of my pot (which is zero but for the sake of argument let's say) is 500 that would give me a voltage of 10v and at the highest resistance of 50k I get .1V. OK so that means when the pot is at 10k the voltage has already dropped to .5 volts so the top 40% or so of the pot's travel is not registered since the voltage has bottomed out. You can adjust for this by increasing the resistance of the feedback resistor but then you have the exact same problem at the bottom end of the pot.....
I'm really BANNED confused...it seems like a linear pot would make the most sense...but limor clearly said to use a log pot, so somebody please tell me what the hell I'm doing wrong here.
-thanks
jason
feedback resistor values for log pots
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http://www.ladyada.net/make/midisense/sensormodel.html
the first case, 'minimum' resistance is 0 so it will pin (any feedback resistor will work)
the second case, 'maximum' is 50K, Vout = .5V * Rf / 50,000 and you want Vout to be as small as possible but not smaller than 0.5, so make Rf = 50K and you're done
short version: feed back resistor == max resistance of pot
dont forget there's two ways to hook up the pot, try both.
that help?
the first case, 'minimum' resistance is 0 so it will pin (any feedback resistor will work)
the second case, 'maximum' is 50K, Vout = .5V * Rf / 50,000 and you want Vout to be as small as possible but not smaller than 0.5, so make Rf = 50K and you're done
short version: feed back resistor == max resistance of pot
dont forget there's two ways to hook up the pot, try both.
that help?
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yeah that makes sense but the problem is the response of the pot is not linear and doesn't utilize the full range of the pot. Basically the midi output of the pot's motion is only changed when the voltage is between 5 and .5, correct? If I turn the pot up (clockwise) the voltage goes to infinity as R goes to zero, but the midi output maxes out once the voltage goes over 5V which happens when the 50K pot reaches 5K ohms, basically all of the pot's range between 5k and 0k is useless "dead space" on the pot...should I just put a 5k resitor inline with the pot?
I guess I'm confused, it seems like a linear pot works better....I hooked up a linear pot and it gave me a much nicer response...it seems from the math that the best pot would be a 5k->50k ohm pot coupled with a 50k ohm resistor...that would give me a linear response from .5 to 5 volts. Why am I wrong?
also the midisense is constantly sending out signals for midi channel 2 cc 33 which I don't have specified for anything...any ideas what that could be?
Help me Limor Fried, you're my only hope....
-jason
I guess I'm confused, it seems like a linear pot works better....I hooked up a linear pot and it gave me a much nicer response...it seems from the math that the best pot would be a 5k->50k ohm pot coupled with a 50k ohm resistor...that would give me a linear response from .5 to 5 volts. Why am I wrong?
also the midisense is constantly sending out signals for midi channel 2 cc 33 which I don't have specified for anything...any ideas what that could be?
Help me Limor Fried, you're my only hope....
-jason
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ok I got some thing that works but maybe you can explain why...i put a 10k ohm resistor inline with my 50k ohm log pot. Then I wired the pot backwards, so the calibration slider goes up(right) as I turn the pot down(counterclockwise) this setup gives me a nice clean linear looking response in ableton live....albeit backwards...but that's easy to fix in live
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the reason a logarithmic is better is that Vo = Vout = .5V * 50K / R where R is the potentiometer resistance. therefore Vo is inversely proportional to R ( V = 1 / R * const) if you graph 1/R you get a logarithmic curve. But if R is not linear on the scale, but inverse log...you'll cancel out the logs and get a linear.
soundsl ike its working now, though?
soundsl ike its working now, though?
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i cant understand your comments because you're trying 10491843 things at once and asking why each one may or may not work.evilxsystems wrote:yeah that makes sense but the problem is the response of the pot is not linear and doesn't utilize the full range of the pot. Basically the midi output of the pot's motion is only changed when the voltage is between 5 and .5, correct? If I turn the pot up (clockwise) the voltage goes to infinity as R goes to zero, but the midi output maxes out once the voltage goes over 5V which happens when the 50K pot reaches 5K ohms, basically all of the pot's range between 5k and 0k is useless "dead space" on the pot...should I just put a 5k resitor inline with the pot?
I guess I'm confused, it seems like a linear pot works better....I hooked up a linear pot and it gave me a much nicer response...it seems from the math that the best pot would be a 5k->50k ohm pot coupled with a 50k ohm resistor...that would give me a linear response from .5 to 5 volts. Why am I wrong
-jason
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Please be positive and constructive with your questions and comments.