Small Clone Depth Question

[Unlikekurt]

I recently built a small clone chorus using tonepad pcb and included the depth pot in place of the switch.
For the life of me I can't seem to figure out where the stock positions would be in relation to the pot (not necessarily by sound, but by looking at the schematics and playing with numbers).
http://tonepad.com/getFile.asp?id=97

Switch
LM358 Pin 1 ---- 47K Resistor ---- 2K7 Resistor ---- SPST ---- Ground

Pot
LM358 Pin 1 ---- Lug3 ---- Wiper ---- Lug1 ---- Ground

For the Switch, the junction of 4k7 and 2k7 go through a 39K Resistor and onto CD4047
With the pot, the wiper takes the place of that junction.

Can someone possibly help explain what's going on in both instances?
Thanks

[Mark Hammer]

In the stock unit, the 4k7 and 2k7 act like a voltage divider.  Imagine a 7k4 depth pot turned down about 2/3 of the way, and that's what the switched-closed position is.  When the switch is open, there is no attenuation of the depth, but there is another interesting effect that kicks in.

The 4k7 resistor and 10uf cap form a single-pole lowpass filter, with a rolloff starting around 3.4hz.  So, at slow speeds, it has no impact, but as you start to turn the speed up, the cap starts to "soften" the sweep, such that the triangular LFO has a different "turnaround" at positive peaks.

In the modified SC with a 10k pot, the 10uf cap interacts with both the speed of the LFO and the depth pot setting.  The rolloff is a function of the resistance in series interacting with the cap to ground.  With the cap value and LFO rate held constant, the rolloff kicks in at progressively lower and lower frequencies the more you turn the depth down.  In a sense, for faster speeds, the modified SC has a different sweep form at higher depth than lower depth.  Kinda neat.

[Pablo1234]

With the switch -
its a 4.7k and 2.7k resistor , if you take the sum 7.4k total resistance then divide the lower resistor by the total resistance you get 34% voltage roughly of the total voltage dropwith the switch closed. With the switch open you get a current limiting resistor that may have a division with down stream components but probably not so you get 100% voltage. So you get 36% of the depth with switch closed and 100% with it open.
With the Pot -
you get a variable voltage division from 0-100% for depth

[Unlikekurt]

Thanks for the replies.
I was thinking pretty much the same about the voltage divider.
Mark: Thanks for going into how replacing the resistors with the pot will alter the design of the low pass network and providing details about that.