will a basic buffer be sufficient?

CDRIVE

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highest and lowest input voltage?
9V highest ofcourse but im not sure what the lowest would be.
I feel this is turning out to be much more complicated than it should be and its my fault
heres is the schematic of the device i hope to control: http://www.musicfromouterspace.com/analogsynth_new/NOISETOASTER/images/noisetoaster_page_1.gif

I'm forced to concur with you. You are guilty of a practice (repeated here constantly) that drives us nuts! This topic should have been posted like this...

Hello, I have one of these white noise synthesizers..
http://www.musicfromouterspace.com/...MAINTAB=SYNTHDIY&SONGID=NONE&VPW=1356&VPH=513
That page contains a product description and links to two schematics but I'll also post the links here.
Main schematic 1:
http://www.musicfromouterspace.com/analogsynth_new/NOISETOASTER/NOISETOASTER.php#SCHEMPG1
Schematic 2:
http://www.musicfromouterspace.com/analogsynth_new/NOISETOASTER/NOISETOASTER.php#SCHEMPG1

Here's what I'd like to do........XYZ goes here.

Why forum nubes insist on feeding us vital information in microbits escapes me. Especially when they're in a hurry to complete their project. Why the hell make us chase this information down?

You'll have to excuse me for a while, as I'm going to read through the description and study the schematics. When I feel I've got a good understanding of the circuit I'll post again.

Chris
 

Arouse1973

Adam
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Yes I agree Chris. I could specify the op-amp in less than 5 mins if we had all the information. On the flip side to this you get people apologising for giving too much information, makes me laugh.
Adam
 

AFex54

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As I stated earlier,... that is an AC coupled amplifier. It can't be used for DC applications. The first OpAmp circuit that you posted is the correct configuration.

Chris
this one the im talking about:http://www.muzique.com/images/buff7.gif

edit: oh sorry that was very stupid of me...
dont know how i managed to find the same circuit 3 times and still think it was a different one.
 
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AFex54

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Yes I agree Chris. I could specify the op-amp in less than 5 mins if we had all the information. On the flip side to this you get people apologising for giving too much information, makes me laugh.
Adam
yeah sorry, I felt giving a pile of info would
Yes I agree Chris. I could specify the op-amp in less than 5 mins if we had all the information. On the flip side to this you get people apologising for giving too much information, makes me laugh.
Adam
yeah sorry, I will know for again, I thought the solution would be as simple as ''you need a unity gain buffer''. felt a wealth of info would be bit rude but now i know it only makes your job easier.
 

Arouse1973

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Ok so when you have drawn your circuit let us have a look and we can sort it out for you.
Cheers
Adam
 

AFex54

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Ok so when you have drawn your circuit let us have a look and we can sort it out for you.
Cheers
Adam
I dont know what circuit you mean?
the noise toaster schematic has everything , its just 30 or so duplications of the frequency control pot**.
one for each keyboard key,
you select the pot with a keypress which closes a switch underneath sending 9V through it.
the more I read about this the more I think maybe it doesnt even need a buffer,
the inverting summer after R1 would have high impedance input anyway and what other reason would i need one?
the noise toaster low fidelity so the utmost precision its not important/necessary
**(R1, in the very top left, http://www.musicfromouterspace.com/analogsynth_new/NOISETOASTER/NOISETOASTER.php#SCHEMPG1 )
 
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Arouse1973

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I thought you were going to draw your own circuit that you could follow? I misunderstood
Adam
 

Arouse1973

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Ok I see now I think. You are basically switching in different voltages instead of you adjusting the original POT each time?
 

Arouse1973

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yeah exactly,

So I think the answer is no, you are just replacing the current POT with different ones with a different setting. As far as the circuit is concerned all you have done is adjust the POT.
Adam
 

CDRIVE

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If you tell us how many keys you have I think we can draw a schematic for you.

Chris
 

CDRIVE

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OK, after much consternation I'm reasonably certain what you want to do. I agree that I don't think you need a buffer because you will be connecting the keyboard (resistive divider) to pin2 of R2 which is 75KΩ. In short I think this schematic will work for you. The resistive divider is comprised 4.7KΩ resistors. I chose this value because they're common and total 94KΩ, which is close to the value of your 100KΩ Pot. 5.1KΩ's would work too but they're not as common.KeybdVdivider.JPG

Chris
 
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AFex54

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So I think the answer is no, you are just replacing the current POT with different ones with a different setting. As far as the circuit is concerned all you have done is adjust the POT.
Adam
right well that was a load of fuss over nothing haha.
I made a thread a few weeks ago for advice before I built the keyboard and they recommended a buffer at the output bus of the pots so it stuck with me
 

AFex54

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OK, after much consternation I'm reasonably certain what you want to do. I agree that I don't think you need a buffer because you will be connecting the keyboard (resistive divider) to pin2 of R2 which is 75KΩ. In short I think this schematic will work for you. The resistive divider is comprised 4.7KΩ resistors. I chose this value because they're common and total 94KΩ, which is close to the value of your 100KΩ Pot. 5.1KΩ's would work too but they're not as common.View attachment 20719

Chris
and using potentiometers instead of resistors wouldnt be problem?
I just assume you used the resistors conform to a more traditional musical keyboard.
my keyboard is rather unconventional, its more like a hybrid between a keyboard and an analog sequencer
 

CDRIVE

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and using potentiometers instead of resistors wouldnt be problem?
I just assume you used the resistors conform to a more traditional musical keyboard.
my keyboard is rather unconventional, its more like a hybrid between a keyboard and an analog sequencer
If your keyboard has 20 pots with each wiper connected to a keyboard key then yes. What values are the pots?

Chris
 

CDRIVE

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Ouch! That may put you back in the Buffer category because 20 of them will total 2MΩ. The input resistance of the Noise Toaster will certainly load it. Are they wired as potentiometers or rheostats? If you don't know what the difference is you're not alone. On the other hand, all pots can be rewired as a rheostat but adjusting each one to ~ 5KΩ can be hairy on a 100KΩ pot. Not impossible though.

The biggest problem will be maintaining resistance setting because just rapping the keyboard could change the resistance settings. When multiplied by 20 rheostats (if they 'all' shifted)..well you get the picture. When wired as Pots this problem is reduced by 19 because the resistance of the fixed terminals are just that..fixed. The total resistance of the divider is fixed as the sum of all the pot's fixed terminal resistance. Not so when wired as rheostats. When wired as a rheostat any resistance change in any single rheostat will change the divider's total resistance, thus changing the voltage output on all of the keys instead of just that one.

Chris
 

AFex54

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Ouch! That may put you back in the Buffer category because 20 of them will total 2MΩ. The input resistance of the Noise Toaster will certainly load it. Are they wired as potentiometers or rheostats? If you don't know what the difference is you're not alone. On the other hand, all pots can be rewired as a rheostat but adjusting each one to ~ 5KΩ can be hairy on a 100KΩ pot. Not impossible though.

The biggest problem will be maintaining resistance setting because just rapping the keyboard could change the resistance settings. When multiplied by 20 rheostats (if they 'all' shifted)..well you get the picture. When wired as Pots this problem is reduced by 19 because the resistance of the fixed terminals are just that..fixed. The total resistance of the divider is fixed as the sum of all the pot's fixed terminal resistance. Not so when wired as rheostats. When wired as a rheostat any resistance change in any single rheostat will change the divider's total resistance, thus changing the voltage output on all of the keys instead of just that one.

Chris
heres a diagram of how mine are wired,
hope it helps: http://postimg.org/image/n3w1j5f0d/

edit: and of course I made a mistake... it should be R2 not R1
 
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CDRIVE

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heres a diagram of how mine are wired,
hope it helps: http://postimg.org/image/n3w1j5f0d/

edit: and of course I made a mistake... it should be R2 not R1
If that's the way your pots & keys are wired you have no problem at all because that's not a voltage divider in any sense of the term. All your pots are wired in parallel across the 9V source. Each pot is totally independent of the other. Actually each pot constitutes a totally independent voltage divider unto itself.

If that schematic is accurate we've been spinning our wheels here. o_O As far as I can see you're good to go! So get to it! ....I'm hitting the sack. ;)

Chris
 
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AFex54

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If that's the way your pots & keys are wired you have no problem at all because that's not a voltage divider in any sense of the term. All your pots are wired in parallel across the 9V source. Each pot is totally independent of the other. Actually each pot constitutes a totally independent voltage divider unto itself.

If that schematic is accurate we've been spinning our wheels here. o_O As far as I can see you're good to go! So get to it! ....I'm hitting the sack. ;)

Chris
thanks a lot and good night!
 
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