Triode - how come grid is "more negative" than cathode?
There are several types of biasing in tube amplifiers.  Your schematic shows one particular type, grid-leak biasing.  Grid-leak biasing is by far the hardest to understand, and many erstwhile explanations expect you to push the "I Believe" button and go on.  At first blush, it does seem magical:  How can the grid get negative with respect to the cathode, there being no negative biasing voltage supply?  It's even harder to understand if you were trained that current flows from positive to negative.  It’s easier if you were military trained or your training is in physics (both of which acknowledge that current flow is negative to positive).

The key to understanding how grid-leaking biasing works is to dig deep into the physics of how a tube works.  The heater heats the cathode.  The cathode has a coating that, when heated, creates a cloud of electrons around it.  Some of these electrons fall back into the cathode, some escape, and some bang into the nearby grid wire - not many, because, after all, nothing is attracting them (compared to plate B+).  But there are some with high enough energy and direction to hit the grid wires and stay there.  So now we have a grid with more electrons than it started with.  If it were not for the resistor between grid and ground(cathode) the electrons would keep accumulating.  But with this resistor the extra electrons have somewhere to go -- through the resistor and back to the cathode, so they are constantly bleeding off.  Electrons flowing from the grid to the cathode (through the resistor) make the grid negative with respect to the cathode.  Making the grid negative is biasing the tube.

Just a few, possibly helpful, comments...


Understanding how transistors work does not help you to understand how a tube works.
Thinking of a tube as a resistor network (e.g., the sum of currents into and out of a node equals zero, that kind of thing) also does not help here.
Thinking of how a thermocouple works DOES help, i.e. you get current flow without an external battery.  Replace, in your mind, the cathode and grid with two dissimilar metals in contact (a thermocouple) and heat them.  The process (though different) produces the same effect – a small current flow with no external voltage source.


I hope you find this explanation helpful.