# emission current and electron energy

Discussion in 'General Electronics Discussion' started by amphx, Sep 4, 2015.

1. ### amphx

2
0
Sep 4, 2015
Hi all,

I'm building an electron gun using a light bulb tungsten filament and aluminium foils for the Wehnelt and Anode. I'm trying to measure the emission current of the gun by placing a positively charged plate (collector) biased with respect to the filament in front of the gun. I'm also measuring the emission current by placing a amp meter along the connection from collector to the filament. through several experiments I have realized that the emission current is the same for the electrons emitted from the gun or when testing the filament only.

1)
My question is assuming the gun is working and anode accelerates the electron for high energy around 2000 eV in my case, must this affect the emission current or does it only increase the energy of the electrons i.e. J/C and not their current i.e. C/s ?

if so, is it not true that the kinetic energy of the electrons is increased and thus their velocity?
therefore, they move the distance in filament-collector electrostatic field much in short amount of time which in turns means C/s is increased. ergo, emission current must increase with the gun?

2) also I'd like to know if there is something I can buy to make the effect of electrons visible?
something similar to phosphor sheet but something I can get from your average electrical or hardware store.

Thank you,

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2. ### davennModerator

13,786
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Sep 5, 2009
hi, welcome

you realise your filament and anode have to both be inside a vacuum tube ?

again the phosphor sheet/screen needs to be inside a vacuum tube
along with the filament and anode
get an old basic oscilloscope tube and use it to do all your experiments.
with its phosphor screen you will see the effect of the electrons hitting the screen

3. ### Ratch

1,093
334
Mar 10, 2013
I would think that the emission current is dependent on the ability of the filament to supply electrons, and that in turn is dependent on the temperature of the filament and the space charge around it. If you crank up the anode voltage, each electron will reach the anode faster, but without more coulombs/time from the filament, the emission current will be the same. The higher speed of the electrons from the higher anode voltage assures that each electron will be spaced farther from its neighbor along the beam in order to keep the same current. If the voltage is high enough, x-rays can be generated when the electrons strike the anode. By the way, J/C is a unit of voltage, not energy.

Yes, kinetic energy increases with velocity.

Everything depends on the filament's ability to supply electrons. The space charge around the filament inhibits the emission from the filament, so a higher voltage from the anode that draws the space charge away from the filament will facilitate the emission discharge. There are a few factors working here.

I don't think so. Electrons are so small you cannot see them individually, only their effects when they strike something. Why not buy a small CRT that has already been shaped, wired, and coated with a phosphorescent material?

amphx likes this.
4. ### davennModerator

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Sep 5, 2009
thanks for the back up

5. ### amphx

2
0
Sep 4, 2015
Hi,

So, to sum up, since the filament is limited in its electron generation, the amount of coulombs per second it can produce remains constant but the kinetics energy of each emitted electron grows larger due to the presence of the anode plate.

so, we can say that if originally filament is producing 0.7mA of emission current at some input voltage. then addition of an anode of 2kV, provides each of those electrons with 2 keV of kinetic energy.

therefore, in my circuit schematic above where my amp meter is connected to the collector plate to measure the emission current, the meter must still read 0.7mA, assuming all the electrons can pass through the Wehnelt and anode holes. otherwise, i need to adjust the distance of Wehnelt to anode, and the hole diameter to reach my 0.7mA discharge.

is my analysis correct?

Also, regarding the energy of the electrons, since the anode is biased 2kV w.r.t the filament, theoretically, the electrons must obtain 2 keV of energy. But, is there a way for me to measure this or make certain of ?

Thank you

6. ### Ratch

1,093
334
Mar 10, 2013
That would appear to be the case.

Don't forget that if the anode is close enough to the filament, it might reduce the effect of the space charge around the filament, and contribute to increased emission current. I am only listing factors that might affect filament current. I don't have the expertise in tubes to know for sure.

I know of no way to do this. That does not mean that there is no way.

Ratch

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Sep 5, 2009