Bias resistors

[DELAMBG]

Hi all, I'm trying to get the current draw down a little more in an amp running 4 6L6's. The drop resistor in series with the pot is a 3 watt 100k

you think it's safe to put a 1 watt 10K in series with it to bring it down a little more?

Thanks!

[MWJB]

What is the load resistor value?

[DELAMBG]

Do you mean the resistor feeding the pot for the bias adjustment?

[Jazz P Bass]

You have not described the circuit!
If it is a grounded cathode output stage, than making the bias voltage (6L6 pin #5) more negative will lower the idle current.

[DELAMBG]

sorry it's a super lead running 6L6's

[MWJB]

Is there not a resistor running to ground in series with the pot? If so, making this larger will reduce plate current. The bias feed typically comes from the junction of the supply "dropping" resistor(s) & the "load" resistor (the load resistor can be either a fixed resistor, a trim pot, or a trim pot in series with a fixed resistor). Increasing the value of the dropping resistors (between voltage supply & bias feed) pushes current up. I'd leave the 100K 3W as it is and look at the other resistors in the voltage divider.

To work out what you need to do measure the -dc voltage off the bias diode. Then apply the following formula:
Load resistor/(load resistor+dropping resistor) = x
x multiplied by the voltage at the bias diode = your supply voltage.

E.G your load resistor might be a 22K pot in series with a 47K resistor, fed by a 15K dropping resistor (after the diode), with perhaps -68v at the bias diode, so with the pot set to maximum resistance (coldest bias)...
69K/(69K+15K) = 0.82
0.82*-68v = -55v

To make the plate current higher you need to reduce negative voltage (maybe to -50vdc?) to make the plate current lower you need to increase negative voltage (say to -60vdc).