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Thermionic Emission Question

  1. Mar 27, 2010 #1
    I have a question regarding Child's Law with thermionic emission. I understand that due to space-charge effects, that the emission current from a cathode reaches a saturation current at a certain temperature, explained mathematically by Richardson's Law. Utilizing an accelerating positive potential will raise this emitted saturation current. What does Child's Law exactly say? I know it relates a current density to a voltage raised to the 3/2. I've been looking into it, and I can't seem to make it clear to myself whether it's one of these two distinctions. In terms of a emission current vs. anode plate voltage curve, does Child's Law just show the increasing emission current as a function of the plate voltage disregarding the saturation current? Or does it show the saturation current as a function of the voltage across the cathode emitting the electrons? Thank you for any clarification!
     
  2. jcsd
  3. Mar 27, 2010 #2
    The Childs Law is also known as the Langmuir Law or the Childs-Langmuir Law or equation.

    It can appear in more than one form, perhaps this is your difficulty.

    [tex]{I_a} = GV_a^{\frac{3}{2}}[/tex]

    Where [tex]V_a[/tex] is the anode voltage, [tex]I_a[/tex] is the anode current and G is a constant, called the perveance.

    G is dependant on the geometry and material of the diode.

    It is possible to derive the current density version, I assume you have from first principles (Poissons equations and KE of and electron) in terms of the current density of a flowing space charge (electron beam), J.
    To get to my equation above you must then also take into account the electron time of flight and the distance between the anode and cathode.

    To correct a couple of your statements.

    The emission is given by Richardsons equation, not Childs-Langmuir.
    Voltage 'across the cathode' has no meaning. In the derivation the cathode voltage is reckoned as zero and the origin of the space coordinate system is based at the cathode.
     
  4. Mar 27, 2010 #3
    Okay, thanks! That makes sense. That just leaves me with one (kinda silly) question. So when measuring this effect in the lab, we were able to apply voltage across the anode and measuring the current through the anode. What is the difference between the anode current and the emission current? To me, it doesn't seem that Child's Law takes into effect the saturation current that is reached when increasing the anode voltage?
     
  5. Mar 28, 2010 #4
    I have atached some sketches. Let's have a quick look at your lab.

    The first sketch is what you would have seen plotting anode current against external anode voltage ( what you get connected to a variable power supply). This is with the vacuum diode fully warmed up.

    The current increases with increasing anode voltage to a limiting value, beyond which the current remains sensibly constant. The rising part of the curve is concave down (below a straight line).

    The second sketch is what happens when you fix the anode voltage and increase the cathode temperature. The curve looks similar to the first one, there being a max temperature beyond which we cannot raise the current.

    The third curve is a composite of the first two.

    I stress that the currents and voltages are what you would measure on external meters connected into the circuit.

    The rising section of the graphs are modelled by the Richardson law (temperature) and Childs law (voltage).

    We have two laws because the actual measured current is less than would be expected by the Richardson law.
    This is because of the space charge that builds up between the electrodes. If you like, the space charge reduces the accelerating voltage seen by transiting electrons within the body of the diode.

    Unless you need to go into the details of the space charge I would not delve further as it gets quite complicated quite quickly.
     

    Attached Files:

  6. Mar 29, 2010 #5
    Wow, thank you! This makes sense to me now. Thanks so much!
     
  7. Oct 10, 2010 #6
    I noticed in one of your sketches that the current is slightly above zero when the external anode voltage is zero. Wouldn't the current simply be zero since there is no driving force for the electrons to leave the cathode?
     
  8. Oct 10, 2010 #7
    Yes indeed. This thread is about thermionic emission. So electrons are driven off the cathode by heat, regardless of the anode voltage. So if you think about it the anode, even at zero volts, will be one of the most 'positive' places in the tube. Some current flows at negative anode voltages, the amount depending upon the cathode temperature. So there is practically zero at room temps, but measureable amounts at normal heater temps.
     
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