Why is there a cathode to ground resistor on a KT66 power tube in an amp kit?

[Planobilly]

There is a cathode to ground resistor on a KT66 power tube on a amp kit. The value is 1 ohm. I ASSUME the only reason for it being there would be to be able to measure the current flow. I ASSUME 1 ohm is not going to change the cathode bias. Is my assumption correct?

The KT66 is a beam power tetrode and this amp kit is based on a Marshal JTM 45 schematic. The original schematic has the cathode going directly to ground.

Thanks,

Billy

 

[Svein]

The cathode to ground resistor introduces a slight negative feedback. At a cathode current of 100mA, it will change the grid bias by 100mV, which is not very much. The famous Williamson amplifier (http://www.preservationsound.com/?p=3823) uses a complex cathode resistor network, but assume an equivalent of 120Ω per tube, giving a cathode bias (at 80mA) of ≈ 10V.

 

[jim hardy]

Do i perhaps remember discussing that amp ?
If i recall it was for measuring cathode current when you adjust bias ?
While amp is off , make sure it's still one ohm.

 

[Planobilly]

No Jim, not the same amp. This is a new kit I am try fix for a kid to keep him from killing himself. You would not believe how bad he had it messed up. Solder running everywhere. Really dangerous stuff. I had to re-do everything.

I am now looking it over to see if the instructions are correct. I really dislike these kit projects. No bleeder resistors. No cover on the AC line. On and On...not good.

 

[jim hardy]

I ASSUME 1 ohm is not going to change the cathode bias. Is my assumption correct?

I agree with your assumption. See curves at http://www.drtube.com/datasheets/kt66-jj2007.pdf
100 millivolts isn't much bias at all.

 

[Planobilly]

Thanks Jim

Billy

 

[Bandit127]

I am learning this at the moment but can I assume that a 1 Ohm resistor in the cathode circuit allows the plate current to be measured directly as a voltage because a 1 Ohm resistor gives V=A. Once I have my plate voltage I can work out my plate dissipation and then adjust bias accordingly.

So if my plate voltage is 400 V, the voltage drop on the 1 Ohm resistor is 25 mV (0.025 V = 0.025 A) then my plate dissipation is 10 W.

By measuring the current in this way I guess I am ignoring the small current added to the plate by the input signal on the grid. Or am I?

 

[jim hardy]

By measuring the current in this way I guess I am ignoring the small current added to the plate by the input signal on the grid. Or am I?

grid current is microamps
and the electrons came from the cathode so plate current will be a teeny bit less than cathode, negligible when using analog meters of tube days.