Resistor Values; best practices

[Tim Hall]

Hi all,

I recently took some measurements on my amp, and found that the cathode-biased power tube is running way too hot. I need to bump up Rk from 330Ω to 435Ω, by my math.

This is my first time to make any value changes to an amp, and I hope someone can help me determine the best way to achieve this value. Should I use two larger value resistors in parallel, two smaller value resistors in series, or go with the closest value single resistor I can find?

Thanks,

Tim

[Jag]

Go with the closest standard value. I'd err on the side of caution and go with 470, better to be a little high. Is this a 6L6 output tube(s)?

[R.G.]

Put in a 470, and parallel it with another resistor to bring it down to 435.

How? Rp = (R1*R2)/R1+R2), so if Rp=435 and R1=470, then
435=(470*R2)/(470+R2) and solving for R2,
R2=5841

if you use a 5.6K, the resistance is Rp = 5600*470/(5600+470) = 433 ohms.

The power rating needed is a function of the voltage. Since P = V^2/R,
the power in the paralleling resistor is inversely related to the first resistor, and the power in the parallel resistor is P = P1* (470/Rp), so the paralleling resistor dissipates 470/5600 =0.084 times the power of the main resistor.

If the 470 is a 5W, you need a paralleling resistor of at least 5*(470/5600) = 0.419W to have them burn up at the same time. A 1/2W resistor would make the 5W 470 burn out first. You could put in a 10K 1/2W pot and tweak it in.

[Tim Hall]

Thanks for the replies. I wasn't sure how the power rating worked out...I'm going to copy-paste that into my notes for future reference.