
Originally Posted by
buffo
I wonder what that current control circuit will do when the resistance goes to infinity for the short period of time it takes you to throw the switch? The tube might wink out...
I think it would be better to set R2 to the high current position (presumably this would be a low resistance on the pot), then add a second resistor in series with R2 that is large enough to bring the current down to idle.
Then use the switch to short around that other resistor. When the switch is open, the current control circuit sees the full resistance of both R2 and the fixed resistor you added, and thus the tube is idling. When the switch is closed, it shorts around the second resistor and thus the current controller only sees the resistance of R2, so the current ramps up.
Adam