Can someone explain why the anode output is out of phase?

[egellings]

As grid goes higher in voltage, anode goes lower in voltage because the tube conducts harder, pulling the anode down, if there is an anode resistor. That's inversion.

 

[Windcrest77]

Voltage at grid goes more negative, voltage at anode becomes more positive. Simple

I'm a newbie, probably not qualified to answer but here goes, shoot me down please...

I think of the cathode like a fire hose spraying a constant stream of water at the plate, and the plate is a "always dry" sponge that forever "soaks up" the stream, doesn't reflect it.

The grid is a gate valve in the middle of this fire hose stream, when you make it more positive you divert more water out of the stream, when you make it less positive you divert less water out of the stream. So the stream is thick or thin at any position of the gate valve.

At the same time that mechanism is happening, "conventional" electrical current is trying to flow from your B+ supply from the positive plate to the cathode nearer ground. That current flows stronger when the fire hose "stream" is "thicker" (grid more negative) and that conventional current flows weaker when the fire hose stream is "thinner" (grid more positive).

So I guess it inverts because the grid moves more negative to make the stream "thicker" (less resistance) for the conventional current to conduct more from plate to cathode. So they move opposite.

Am I close, or way off?

 

[nigelwright7557]

Thanks for the reply.

I understand that as grid voltage increases, more electrons will be attracted to anode, so the current through the tube will increase, and the increasing current will increase the voltage across Ra. But why is it inverted? It seems to me that they move in the same direction, so they should be in phase?

Because the anode resistor is tied to B+ and the voltage across Ra increases so the anode voltage will go down.

 

[DIYish]

I really appreciate all the inputs here, my friends, especially the detailed illustration from @JMFahey and @NickKUK

My initial thought that Anode voltage = (B+) - (voltage drop across Ra) was correct, but then I overthought it using the formulae. I think I got it now 🙂

 

[hbc]

It could be easier to understand transistor first?
Simplified:
The base is 0.6v above the emitter (cathode).
The current flowing through the collector (anode) and emitter are the same.
Therefore using ohms law, the current flowing through Remitter is (Vbase (grid) - 0.6)/ Remitter.
The voltage across Rcollector is then Iemitter * Rcollector.
Understanding this simple relationship is about the only useful thing i learned at university and has resulted in 30 years of paid employment (so far)!
Valves / tubes are pretty similar.