5F1 Circuit Question - V1 Cathode Resistors

First off.... Im not a theory expert and hope to not add confusion...

The resistor holds the cathode above ground potential and sets the bias for the triode. If it were not there the cathode would be at zero. The grid is at zero therefore negative in respect to the cathode. This alows us to control the flow of electrons. As you apply signal to the grid positive flows more electrons and negative suppresses electron flow. The more positive the grid to cathode runs the tube into saturation. The more negative the grid to cathode runs the tube into cutoff.

These old military videos are some of my favorites. You can imagine the alternator being guitar signal on the control grid.

 

With one milliamp of current through the tube, there is a voltage drop across the cathode resistor. E=I*R.

So 1 ma. through 1.5K ohms is 1.5 volts.

 

GREAT videos and I enjoy those, as well. Simple enough for my lizard brain.

 

Oh sure, thats the easy answer

 

Me too.

They had to teach young impressionable teenagers to be radio warriors!

 

No. It's the law.

 

The resistor brings the cathode voltage (zero potential at ground) to 1.5DC at the cathode pins.

Valve Wizard to the rescue. Chaper 1.9 on biasing.

http://www.valvewizard.co.uk/Common_Gain_Stage.pdf

Cathode is ground potential, so zero. It can be negative with respect to the grid, but it is zero. A cathode resistor (for cathode biasing) makes the cathode voltage stay within a range positive with respect to grid, as opposed to 'fixing' the grid negative with respect to cathode (fixed bias). Putting resistance in series with the cathode allows for the tube to self-bias with anode current changes.

At least this is how I understand it. Someone far more knowledgeable please correct me if I ain't got it right.

 

The absolute basics with another short video thrown in:

https://robrobinette.com/How_Tubes_Work.htm

 

pretty much right

And then not so right

The filament heats up the cathode material which causes it to "boil" electrons off into the space around it. They form a cloud around it called the space charge. Electrons have a a negative charge. Since they have a negative charge they only ever want to flow towards a more positively charged point. They came from the ground, attracted by the positive voltage on the plate. They can't flow back the way they came because ground isn't more positive than the cathode, so no, the cathode resistor isn't "acting like a valve to dump charge to ground". It's a one way street in the opposite direction. The negatively charged electrons come from ground, travel up through the cathode resistor to the cathode, and then want to cross the vacuum to the plate because it's strongly positive. The grid is put in between them to throttle that flow.

The purpose of the cathode resistor is to raise the cathode voltage above the grid voltage so the grid will be an effective throttle. The grid is at ground potential, so the bigger you make the cathode resistor the higher the cathode voltage climbs and the more negative the grid looks to the cathode. Which makes the grid a more effective throttle. At some point you will make the throttle effective enough to cut off current flow completely.

 

I like your way of explaining it. Adds a nice mental picture to the process.