I have read the previous post, but I still don't understand totally...
What happen when the switch is set to full power ? How can it bypass the resistor on the switch since that resistor is between the pin that have the bias tap coming from the PT and the bias tap going to the board?
Here's what I mean :
Habitually, when you don't have a low / high option, and using only a 2 position switch, the bias tap goes directly from the PT to the board 27 k resistor
This is that same tap that we remove from the board, to connect it to the switch, is that correct ?
But the connexion must still go from the PT to the board (via the switch) even in high mode, right ?
So how can it go from A: the PT to the switch and then B: the switch to the board without being affected by the extra resistor is that resistor is set exactly between A and B ?
I don't know if it's clear, or if you understand what I'm trying to say because I'm not that good in english, but visually, I don't understand it.
Here's what I see in a graphic
Let say that the bias tap coming from the PT is attached to pin 6 of the switch, and that the switch bias tap is going to the board from Pin 3 (don't know how they are named, so we will use number...
PT bias TAP -> Switch Pin 6 -> Resistor -> Switch Pin 3 -> on Board 27k resistor
But on high power it should go from :
PT bias TAP -> Switch Pin 6 -> Switch Pin 3 -> on Board 27k resistor
is that right ? But how can it bypass the resistor on the switch if Pin 6 is link to Pin 3 with that resistor ?
I hope you will understand what I'm trying to say
