This question is about the voltage drop across a vacuum tube diode.
Consider a circuit that consists of a battery, a vacuum tube diode, and a resistor all wired in series.
My understanding is that, at high-enough voltages, the current through a tube diode is approximately constant. Let me call this \$I_\text{out}\$.
This current must run through the resistor and back to the battery. By Ohm's law, the voltage drop across the resistor should be \$\Delta V_\text{resistor} = I_\text{out} R\$.
My question is: what happens when \$\Delta V_\text{resistor} < V_\text{battery}\$. Where does the remaining voltage go? Presumably it is lost at the diode, but I don't understand the model for voltage drop across a diode.
Best Answer
First, if you want to know this in detail, there are some excellent textbooks on this subject. They mostly stopped getting published some time in the 1970's, but they are there, and they treat the subject in great detail.
Second, in a vacuum tube diode, the thing that pulls electrons from the cathode to the anode is the electric field at the cathode, that's set up between the anode and cathode. More field = more current, and IIRC it's exponential with voltage -- just like a semiconductor diode, except for entirely different reasons.
Using your terminology, \$\Delta V_{resistor} + \Delta V_{diode} = V_{battery}\$. That \$\Delta V_{diode}\$ is the diode voltage drop. The voltage drop in a vacuum tube diode comes about because you need a certain field strength at the cathode for a certain current to flow, but after the electrons are off the cathode and traveling toward the anode, they have to move through the vacuum, accelerating and gathering energy all the way until they smack into the anode (and giving up that kinetic energy as heat). Basically, you need some energy to pull the electrons away from the cathode and to the anode, and that comes from the diode's voltage drop.
In a typical hot-cathode vacuum tube diode when the anode is heated, it forms a cloud of electrons around it. These electrons form what's called a "space charge" around the cathode. This leaves the anode just a bit positively charged, which holds the space charge in place around the cathode*. In order for current to flow, there needs to be some electric field, as mentioned above.
* Unless the anode -- or something -- is very close to the cathode. In some triodes, you can bias the grid by using a very high "grid leak" resistance (\$10\mathrm{M}\Omega\$ is typical) to ground; electrons in the space charge will collect on the grid, bringing its voltage negative; typically to negative one or two volts. Very little current flows -- which is why grid leak resistors are such high resistance.
(BTW: the reason that a triode works is because of the whole "current is a function of electric field" thing -- screen the cathode from the anode with a grid, and when you lower the grid voltage it reduces the electric field at the cathode, and less current flows.)