This must be a rater basic question about terminology.

I frequently come across the following statement: when vacuum diode is working in space charge mode (as opposed to saturation mode), the cathode emission current is larger than anode current

Iemission > Ianode

This is is supposedly caused by the potential barrier created by the space charge around cathode, which prevents some emitted electrons from reaching the anode.

I would immediately understand that inequality if we were talking about the initial stages of vacuum diode operation, when the space charge is still being accumulated, i.e. when emitted electrons leave the cathode and just stay in space charge area around it.

However, it seems that the above inequality is also widely applied to vacuum diodes working in well-established mode, when the space charge is already fully formed, i.e. the number of electrons that enter the space charge equals the number of electrons that leave it.

In that case, how it is possible to have such current inequality? Where does the extra emission current "disappear", if it does not reach the anode?

I suspect that the answer is simple: Iemission, by definition, is intended to include only the current that leaves the cathode, but it does not include the current that returns to the cathode from the space charge. If we designate that return current as Ireturn and assume that it has negative value, then the following equality will hold

Iemission + Ireturn = Ianode, where Ireturn < 0

That would explain the inequality when Ireturn is left out of the picture.

Is this the proper understanding of what is meant by the inequality in question? Or am I missing something else?

  • \$\begingroup\$ not all the thermal emission current reaches the anode, because the emittance goes in all directions from the surface of the cathode. \$\endgroup\$ Commented Dec 9, 2016 at 18:17
  • \$\begingroup\$ @Tony Stewart. EE since '75: That's understood, but that cannot override fundamental laws of physics, like Charge Conservation Law. All charge (all current) that goes into the tube must come out of the tube, period. That's is what this question is about: how can cathode emission current (in-current) be larger than out-current (anode current). \$\endgroup\$ Commented Dec 9, 2016 at 18:34
  • \$\begingroup\$ Most of the flow of charge per second goes into heating the cathode heater and escapes the tube as heat. \$\endgroup\$ Commented Dec 9, 2016 at 18:42
  • \$\begingroup\$ @Tony Stewart. EE since '75: Um.. Sorry, but that just does not make sense. Firstly, electric current (and charge) does not "disappear" into heat or "escape" as heat. Secondly, we can limit out consideration to indirectly heated cathodes, in which case heater current is not even part of our picture. The question is purely about "signal" current in vacuum diode. \$\endgroup\$ Commented Dec 9, 2016 at 18:50
  • \$\begingroup\$ When you apply a signal to the Anode opposing the acceptance of -ve charges from thermal emission , the current reduces from the Cathode slows down until it stops the flow. The relationship between potential , U and current , I is defined by Child's Law with the Perveance ,P (somewhat like conductance) which includes space charge effects which limit current. \$ U= (\frac {I}{P})^{\frac {2}{3}}\$ \$\endgroup\$ Commented Dec 9, 2016 at 20:21

1 Answer 1


Think of the device with zero anode current and both plates at 0V and the the device has been allowed to run a while:

If the inequality holds:

\$ \textrm{If }I_{anode} = 0 \textrm{ then } I_{emission}+I_{return} =0 \$

this only makes sense if you put a direction on the emission and return currents so one opposes the other. This represents the electrons jumping off the hot cathode, hanging around in between the plates (or bouncing off of one and back to the other), but the net charge on the cathode is going to be zero.

In circuit you have to have define a current with a direction and a polarity. So in this case I'm going to define all currents as cathode to anode as positive and use the actual electron flow moving in the direction as positive (which is reversed from circuit current nomenclature). But this means that in this operation \$ I_{emission} \$ and \$I_{return} \$ are have opposing signs on their values

\$I_{emission} + I_{return} =0 \$

Now lets suppose that we increase the voltage on the anode to a positive value: (Upper diagram). Electrons are heated off the plate (\$ I_{emission} \$) but not all of them come back because some of them hit the anodes metal (its now positive) and flow out of the diode.

\$I_{emission}+I_{return} = I_{anode} \$

Since there are always electrons bouncing back to the cathode \$ I_{return} \$ is always negative \$ I_{return}<0\$ (in the way I've defined things). With the author including the inequality \$ I_{return}<0\$, it means that the return current is defined as going from cathode to anode and is always negative.

The important takeaway is that with a negative return current you can't have more anode current than emission current. \$I_{emission}> I_{anode} \$ And that means you'll have to keep the cathode sufficiently hot.

diagrams from http://www.electrical4u.com/vacuum-diode-history-working-principle-and-types-of-vacuum-diode/

Diagrams from and further reading here.


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