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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.

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  • \$\begingroup\$ Ah, I always thought that you need at least 100V to conduct a vacuum tube. Or do you have a high voltage battery? \$\endgroup\$ – tlfong01 Sep 19 '20 at 2:22
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    \$\begingroup\$ Ah, I'm asking as a theoretical question. It could be a battery or any other source of DC voltage. :-) \$\endgroup\$ – Ned Ruggeri Sep 19 '20 at 2:38
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    \$\begingroup\$ Voltage is a measure of energy of a "quantity" of electrons. 1 Volt is 1 Joule per coulomb (6.24 × 10^18) electrons. In order for the vacuum tube diode to work there is a heater element creating a "cloud" of electrons local to the cathode. These electrons are then accelerated in an electric field created by the potential difference between the anode and cathode. This corresponds to a change in the energy of those electrons and therefore a change in voltage. This voltage difference corresponds to the difference between the battery voltage and that across the load (resistor). \$\endgroup\$ – mhaselup Sep 19 '20 at 5:53
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    \$\begingroup\$ Per @tlfong01 the operating voltage of a vacuum diode is going to be over 100 volts. \$\endgroup\$ – mhaselup Sep 19 '20 at 5:55
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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.)

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    \$\begingroup\$ Good stuff. I think you forgot to address the question in the last paragraph - when \$ \Delta V_{resistor} < V_{battery} \$ which it can't be. \$\endgroup\$ – Transistor Sep 18 '20 at 23:13
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    \$\begingroup\$ Thanks. Addressed in edit, hopefully. \$\endgroup\$ – TimWescott Sep 18 '20 at 23:19
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    \$\begingroup\$ Thanks Tim. I thought the filament provided the energy to free electrons from the cathode, and that even a very a small electric field would want to pull electrons toward the anode. Since electrons are traveling from a higher potential to a lower potential, I would have thought that it doesn't require energy input to move the emitted electrons to the cathode. But perhaps, as you indicate, the work done by moving through the electric field is stored as kinetic energy of the electron. When this "smacks" the anode, perhaps the kinetic energy is lost as waste heat? \$\endgroup\$ – Ned Ruggeri Sep 18 '20 at 23:36
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    \$\begingroup\$ I expanded the answer a bit. And yes -- it's electrons hitting the anode that heats the anodes. In amplifying tubes with considerable currents and voltage drops, the limiting factor on the power of the tube is often the ability to draw heat away from the anode -- it's not uncommon for serious power tubes to have anodes that glow red, or even white, in normal operation. \$\endgroup\$ – TimWescott Sep 19 '20 at 0:00

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