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 Forum index » DIY Hardware and Software » Klee sequencer
Using 100K programming pots???
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marvkaye



Joined: Mar 14, 2011
Posts: 225
Location: Fla

PostPosted: Thu Feb 07, 2013 9:43 am    Post subject: Using 100K programming pots??? Reply with quote  Mark this post and the followings unread

I got to thinking how neat it would be to use lighted sliders for the programming pots and use the slider LEDs in place of the free-standing LEDs more commonly used. So I searched around, found some sliders that would do the job and picked up a batch... unfortunately I dummied out and ordered them in 100K instead of 50K. Is there a way I can use them, perhaps by changing values of other components elsewhere in the circuit, or do I have to go back to square one, take the lump on these and see about getting 50K's if I'm to go ahead with this plan?

I hate when I do stupid shit like this........ Rolling Eyes Rolling Eyes Rolling Eyes

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Osal



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PostPosted: Sat Feb 09, 2013 7:12 am    Post subject: Re: Using 100K programming pots??? Reply with quote  Mark this post and the followings unread

Hey marv,

When more small is the resistance of the input attenuator potentiometer in relation to the input resistor of an inverting amplifier, more linear is its response.

For this potentiometers the linearity could be desirable.
For example, if the "range select switch" is set to "one octave", if the potentiometers response is linear, when the pot is in its center, the output would be 6 semitones.

It can be calculated, but I guess that the difference is not very big. However if you want the same response as with the 50K potentiometers using the 100K pots, change to 200K the values of R1 to R16(page 3 of 4), R17 and R10 (page 4 of 4)
The pages are from the schematics of the analogue board on "em_klee_PCB_schematics_122"

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marvkaye



Joined: Mar 14, 2011
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PostPosted: Sat Feb 09, 2013 8:52 am    Post subject: Re: Using 100K programming pots??? Reply with quote  Mark this post and the followings unread

Osal wrote:
Hey marv,

When more small is the resistance of the input attenuator potentiometer in relation to the input resistor of an inverting amplifier, more linear is its response.

For this potentiometers the linearity could be desirable.
For example, if the "range select switch" is set to "one octave", if the potentiometers response is linear, when the pot is in its center, the output would be 6 semitones.

It can be calculated, but I guess that the difference is not very big. However if you want the same response as with the 50K potentiometers using the 100K pots, change to 200K the values of R1 to R16(page 3 of 4), R17 and R10 (page 4 of 4)
The pages are from the schematics of the analogue board on "em_klee_PCB_schematics_122"


Thanks for the detailed response, Osal, it's much appreciated.

Unfortunately I have already populated my analog board as per the build docs and am less than enthusiastic about desoldering all those components. So now the question is what happens to the range if I simply use the 100K pots without changing the values of R1-R16 & R17-R18 (R18 can be mistaken for R10 unless you enlarrge the image... I had mine printed on 11x17 so they're fold outs... lots easier on these old eyes Cool ) Would that double the range or cut it in half? Or is that really undesirable because the linearity gets screwed up? My sliders are center detented, so if the range doubles in a reasonably linear fashion I could use that to my advantage and simply modify the legend on the range switch accordingly. It also looks like adjusting that range can be accomplished with the single Variable Range Resistor (R33)... maybe that's the answer??? What do you think?

<marv>
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Osal



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PostPosted: Sat Feb 09, 2013 1:56 pm    Post subject: Reply with quote  Mark this post and the followings unread

marv

The range is not affected, it is just affected the linear response.
This is because the voltage divider formed by the potentiometer interacts with the input resistor of the inverting amplifier.
For-example, imagine that the potentiometer is in its center. So, a 100K potentiometer will form a voltage divider with two 50K resistors. If we apply, say, 10V in one end of this voltage divider and the other end is 0V, apparently, the node between the two resistors would be 5V. However the inverting input of the inverting amplifier is also 0V. So, actually, you can see it as the input resistor is in parallel with the second resistor of the voltage divider. Don't know if I explain it clearly.

Anyway, if you calculate it, you will see that for a 100K inverting amplifier's input resistor and a 100K potentiometer, when the pot is in its center the voltage output is about 4V instead of the 5V expected.
With 50K potentiometer the voltage output is about 4.4V instead of the 5V expected.
This is not that important, I have my klee with 50K rotary potentiometers and more or less when the range switch is at one octave, you listen a halve octave when they are about its center. I think that you could use 100K potentiometers if you are not very concerned about it. Maybe you would find that the halve of octave is little more to the right or left than its center.

But I think that detented center potentiometers are not appropriate here. The center will not give you any relevant value in some positions of the range switch and neither will be accurate.

Edited for clarity.

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marvkaye



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PostPosted: Sat Feb 09, 2013 5:59 pm    Post subject: Reply with quote  Mark this post and the followings unread

Hey Oscar,

Thanks for the clarification. Actually, I'm not that concerned about the linearity, as there's going to be some variation due to the pot tolerances anyway... I think I'm just going to go ahead with them and see what happens. I'm designing a PCB with connection headers to mount them on so if they don't work out I can do another board with 50K pots and swap them out. Should be interesting.

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