Question regarding starving circuit

[ketem13]

I want to add a starving circuit to 40106 osc circuit.

I have the following schematics:
Some questions regarding it -
1 do I need to add a small value resistor in series with the button and the capacitor lug? if so, why?
2 the capacitor creates an effect of the voltage rise up slowly before fiding the pot lug 1 (and the rest of the circuit), right?
Will it work without it?
3 why the capacitor prevent from the charge to arrive straight to that node? I understand it need to charge up but why the current can't go to that place in addition to the fact the capacitor is charging?

4 the diode is to block current to flow to ground, so how the capacitor can have a path to ground in order to discharge?

[EATyourGUITAR]

Delete the cap and the diode. The bulk capacitance should be on the same supply as the CD40106. Change the pot to 500K. I have built something similar. I think I used 2200uF 16v.

The CD40106 is a load. That will drain the cap over time.

Putting the cap after starve changes the attack portion of the sound when the cap is charging. What you have here will work but it puts unlimited current through the battery for a short time.

You have a lot of excellent questions. It would be the first few lectures of a physics class or circuit analysis class.

There are a lot of resources in the sticky threads for learning basic circuit analysis. We can do it here but it isn't really efficient to rewrite the book. I highly encourage you to go look that up since you are at that stage where you have questions and enthusiasm in learning.

[Electric Druid]

Is VCC the supply to the 40106? In which case, it's all back-to-front. What's the diode for? It looks like a "parallel" protection diode in a stompbox, but if you've got a battery wired in, you don't have to worry about incorrectly wired power supplies. I suppose it would protect the circuit if you connected the *battery* the wrong way around, but it'll toast your battery if it does, since it shorts it out.

Also if you're charging a cap through a pot like that, you should have a small resistor in series with the pot to provide a minimum resistance. Otherwise, when the cap is empty it's effectively a dead short, and if the pot is turned close to zero a large current can flow. That will burn a dead patch in the end of your pot track. The small resistor limits the current to a safe level for the pot (how much depends on the power rating of the pot).

[EATyourGUITAR]

That is an excellent point about burning up a potentiometer. Put a 500 Ohm resistor in series to the pot so you have a minimum resistance. Move the cap to the other side of the starve pot so the cap is in parallel to the power pins of the 40106.