offsetting a modwheel...

[bosone]

maybe not midibox related... but midi-related... :)

my masterkeyboard has a modwheel (CC1) that has a "center-stop", like a detent, in the middle, so that when the CC value is 64 there is a "dead band" around 64. this is not so easy to manage.

i opened the box and found that the pot is 10K, but it is smaller than the normal pots i can buy in my store, so there is no way to change it.

moreover, it doesnt' work at full-range, but onlt say, between 4 and 6 Kohm (can remember well...) the "detent" or the center-stop is at mid-range, so i think is at 5Kohm.

my idea to overcome this problem is to "rotate" the modwheel pot, so that it will work between (let's say) 1 and 3 kohm instead that between 4 and 6 (to avoid the center-stop) and add a simple resistor (in the form of a variable resistor) to add the offset required. if i'd add a constant 3 Kohm the min-max resistance of the modwhell will be preserved, and it will work like before, but without the center-stop.

shuold it work, isn't it?  ???

i hope i was clear...  :P

[stryd_one]

Yep, it'll work fine.

Resistors in series:

R_t = R₁ + R₂ + R₃..... + R_n

Resistors in parallel:

¹/R_t = ¹/R₁ + ¹/R₂ + ¹/R₃ ... + ¹/R_n

[QBAS]

hey

last days ago I prepared in asm solution for pitchbander (with spring and center position - short: regular one).

It is the same like in your setup: 10 kohm pot, but only some range is used.

I have ready and working code, also with option for setting offset.

Of course since using smaller range give smaller resolution (less than 10 bit), but still enough for midi events.

My pitch bender send only 128 values , so on channel 1 is from E0 00 00 to E0 00 7f. The middle value is always 00 on my pitchbender.

I put additional space at center position for "catch" unpitched note. The margin of this center pole is definable.

this work without additional resistors, but who knows, maybe with resistors is better...

[stryd_one]

Hmmm.... QBAS I was thinking...

Let's say you have a pitchbender pot, and it only travels 90 degrees out of a possible 270... that means it will only use 1/3 of it's throw, so if you want to get better resolution, you just need to use a 30K pot :)

So:

Required pot value = 10000 / (Throw used/Maximum throw)

[QBAS]

... if somebody really need full 10 bits. Personally I will try to play with this 7 bits only (at pitch), but if it is not good then I will change pot like you said.

[doc]

...ähmmm.. @stryd_one

Resistors in parallel:

Rt = 1/R1 + 1/R2 + 1/R3 ... + 1/Rn

Shouldn't it be:

1/Rt = 1/R1 + 1/R2 + 1/R3 ... + 1/Rn  ?!

(I know, australia has is own laws ... but this one may be the same all over the world  8) ;D )

greets

Doc

[stryd_one]

Heh sorry I got lost in cut'npaste land, thanks for spotting it doc!

[sup]1[/sup]/R[sub]t [/sub]= [sup]1[/sup]/R[sub]1[/sub] + [sup]1[/sup]/R[sub]2[/sub] + [sup]1[/sup]/R[sub]3[/sub] ... + [sup]1[/sup]/R[sub]n[/sub]

Hehehehe