[sdiy] Leaving off a pot?

[Michael Bacich]

On Oct 8, 2006, at 9:31 AM, Sam Ecoff wrote:

> I'm a firm believer that pots go on inputs but not outputs on  
> modules, so I've decided I want to leave off the output pot of his  
> design. So, my question is, is it just a simple matter of leaving  
> it out and keeping the 1k resistor between the pcb and the output  
> jack, or is there something more involved?


Yes, just connect the output of U5-B to a 1K resistor, and then  
straight to the output jack.  That's exactly the way to do it.

I'm assuming you want your status LED to get brighter and dimmer as  
the ADSR level changes?  You could try connecting the output of U5-B  
directly to an LED (through a resistor first, of course), but the LED  
may load down the output of that opamp, and cause the ADSR output to  
not be consistent, depending on what destinations you connect the  
ADSR to.  By the same token, this loading might also cause your LED  
brightness, relative to ADSR level, not to be consistent.

It would be best if you were to make some kind of active buffer  
circuit for your LED, and drive that buffer with the output of U5-B.   
It could be something as simple as one unity gain non-inverting opamp  
stage.  Basically, connect the ADSR signal to the + input of an  
opamp, and connect the output of the opamp back to the - input;  
connect the output to the anode of your LED through a resistor,  
anywhere from 1K to 2K; connect the cathode of the LED to Ground.

If you don't want to (or can't) waste an opamp doing this, then you  
could use a single NPN transistor and a two or three resistors to  
make an LED driver stage that will act as a buffer.

Michael Bacich