When should you buffer a signal?

[AFex54]

I dont know where or when my signals need current buffering,
I feel buffering randomly could cause a circuit to malfunction,
but I do know I need it.
Is someone more knowledgeable could shed some light on this that would be great!

 

[poor mystic]

Hi
In general, signals need buffering when the output impedance of the signal source is too high for the load. This can result in the signal being lost.
However that might be cryptic advice. You're welcome to be a more specific about your problem, and then you might get a more specific answer.
Mark

 

[AFex54]

Hi , Im building a simple CV keyboard for a synthesizer,
This is the idea but my design is different:
http://www.musicfromouterspace.com/analogsynth_new/SINGLEBUSSKEYBOARD2007/kbbus.gif

the string of resistors in the image will be a bare 300mmx2mm copper wire busbar connected to a 9V 2.5A supply,
when a key (only one key will be pressed at a time) is pressed it will send 9v straight to a 2nd busbar an also to a third but through a potentiometer and diode.

the potentiometer busbar will connect to this devices potentiometer which you can see in the top right connected the battery and the bare bus bar will connect to bottom right switch:
http://www.musicfromouterspace.com/analogsynth_new/NOISETOASTER/images/noisetoasterwiring3.gif

 

[poor mystic]

I see. Where did you suspect a need for the buffer?

 

[AFex54]

I see. Where did you suspect a need for the buffer?

The 'Bus buffer' in the CV Keyboard diagram is making me think I need one

 

[BobK]

An op amp buffer as in the circuit you showed is a good idea here. The impedance of your divider network varies quite a bit depending on which key is pressed. The buffer would lower the impedance and make it the same for all keys.

Bob

 

[AFex54]

An op amp buffer as in the circuit you showed is a good idea here. The impedance of your divider network varies quite a bit depending on which key is pressed. The buffer would lower the impedance and make it the same for all keys.

Bob

sorry only seeing your reply now,
Would that not remove the specific impedance of the potentiometer/resistor for each key aswell?

 

[BobK]

It is not the impedance that matters, it is the voltage across the divider. The buffer ensures that the input impedance of the next stage does not effect those voltages.

Bob

 

[poor mystic]

sorry only seeing your reply now,
Would that not remove the specific impedance of the potentiometer/resistor for each key as well?

Yes, that's the idea. The next stage is a VCO - a voltage controlled oscillator, which is controlled by voltage. Let's call the voltage Vc.
Ideally, a voltage-controlled circuit samples a voltage without changing it. This is difficult with transistor technology, but easy with op-amps.
The op-amp puts very little load on the circuit which provides the control voltage Vc, yet provides a low-impedance output which follows Vc exactly.

 

[AFex54]

now i understand thanks