# Power consumed by electron gun

• A
• rsc9421
In summary, the power consumption due to accelerating electrons is very small and can be replaced by a DC motor running at a low speed.
rsc9421
First Assume the following basic circuit:

I read in many textbooks that the electrons in the circuit are accelerated by the positive voltage and decelerated by the collisions, so the speed is constant.

We also know that the circuit current is I = 10A so the power consumed is P = V * I = 100W

Now assume an electron gun:

Here we have the same situation, electrons are accelerated (by high voltage) but without collisions

So we can conclude that the power consumption by accelerating electrons is P = V * I ?

Welcome to PF.

rsc9421 said:
So we can conclude that the power consumption by accelerating electrons is P = V * I ?
Power consumption of what? The heated element that provides the thermionic emission of the electrons? Or just the power consumption due to accelerating the electron beam (and dissipated in heat and radiation when the electrons hit the target)?

rsc9421
The power consumption due to accelerating the electron beam, I am not assuming that the electrons will hit a target, I am only interested in the power consumption of acceleration part

rsc9421 said:
The power consumption due to accelerating the electron beam, I am not assuming that the electrons will hit a target, I am only interested in the power consumption of acceleration part
So you have the acceleration voltage multiplied by the beam current to give the ##VI## beam power. Adjusted by the efficiency of generating that high voltage (typically about 90% or so, depending on the application)...

rsc9421
Thanks a lot! one more question:
What happens if instead of a high voltage source we put a positively charged metal plate as anode

The electron beam will be accelerated but there is no power consumption ? (assuming that the beam is perfectly focused and no electron will hits the anode)
Who is doing the work in this case?

rsc9421 said:
Who is doing the work in this case?
The anode charging circuit.

rsc9421
Ok but there is no power consumption to accelerate the electron beam?

rsc9421 said:
Ok but there is no power consumption to accelerate the electron beam?
Yes there is. If you didn't charge up the anode with respect to the cathode, there is no reason for those pesky little electrons to want to leave their warm cathode home and go out crashing into the anode...

https://en.wikipedia.org/wiki/Cathode-ray_tube

sophiecentaur, Klystron and rsc9421
I had a hard time understanding how the consumption of hundreds of watts using an accelerator voltage source can be replaced by a simple 12v x 0.5A dc motor running a van der graaf generator

Thanks a lot again!

rsc9421 said:
I read in many textbooks that the electrons in the circuit are accelerated by the positive voltage and decelerated by the collisions, so the speed is constant.
The total Kinetic Energy of all the electrons is tiny and makes only a minuscule contribution to the power delivered to any electrical load.
rsc9421 said:
I had a hard time understanding how the consumption of hundreds of watts using an accelerator voltage source can be replaced by a simple 12v x 0.5A dc motor running a van der graaf generator

Thanks a lot again!
Engineering comes in here; the actual numbers count if you want to make a valid comparison. A VDGG produces a very tiny current. VI will still be the Power dissipated. In your example, the maximum rated motor Power is 6W but this will not be the actual power (less than a Watt) a lot of this will be used in the mechanism (belt friction and slippage) If the High Voltage is, perhaps 200kV, the current that's produced by the movement of the charges carried by the belt will be picoAmps average. So the actual 'mean spark power' will be probably microWatts. The VDGG spark is never continuous.

Otoh, the power in the continuous beam of a TV cathode ray tube will be several W (to make it bright enough to see). Fewer kV but a significant current.

vanhees71
sophiecentaur said:
The total Kinetic Energy of all the electrons is tiny and makes only a minuscule contribution to the power delivered to any electrical load.

Engineering comes in here; the actual numbers count if you want to make a valid comparison. A VDGG produces a very tiny current. VI will still be the Power dissipated. In your example, the maximum rated motor Power is 6W but this will not be the actual power (less than a Watt) a lot of this will be used in the mechanism (belt friction and slippage) If the High Voltage is, perhaps 200kV, the current that's produced by the movement of the charges carried by the belt will be picoAmps average. So the actual 'mean spark power' will be probably microWatts. The VDGG spark is never continuous.

Otoh, the power in the continuous beam of a TV cathode ray tube will be several W (to make it bright enough to see). Fewer kV but a significant current.

thanks for your response!
ok, let's forget about VDGG for a moment and go back to the electron gun.

What happens if previously before accelerating the electrons we first positively charge the anode by some known method (friction, conduction or induction) so that the anode acquires a high voltage, the power consumed to accelerate the electrons is equivalent to the power consumed to positively charge the anode, right?

If I am correct, with this method, the power consumed could be reduced to accelerate the electrons.

The beam current will discharge the anode unless you maintain its +charge by taking the electrons away. Another power supply needed.

davenn and vanhees71
The energy supplied to a charge moving across a voltage qV and consequently the power supplied is IV. In the situatons you are considering the external energy used by the system is to maintain the potential difference. The methods used differ in their efficiency.

sophiecentaur and vanhees71
sophiecentaur said:
The beam current will discharge the anode unless you maintain its +charge by taking the electrons away. Another power supply needed.
but we can add a passive focusing for the beam, I would love to do a simulation about it

hutchphd said:
The energy supplied to a charge moving across a voltage qV and consequently the power supplied is IV. In the situatons you are considering the external energy used by the system is to maintain the potential difference. The methods used differ in their efficiency.
is there any documentation or paper about that?

About what particularly? This is basic EM theory in any introductory book on circuits or fields.

hutchphd said:
About what particularly? This is basic EM theory in any introductory book on circuits or fields.
you mentioned that the proposed methods differ in their efficiency, I wanted to know if there is any paper that reports on the efficiency of the method of positively charging the anode to reduce the power consumed

There are many kinds of electron beams, produced by a variety of methods. If you are interested in a specific system, you (with help from PF here) can work it through. Of course energy is always conserved and you will need to actively maintain any potential difference no matter what. I don't know any generic answer book.

You did not consider that the filament becomes positive by emitting electrons. The missing electrons have to be provided by the accelerating power supply . In the setup with electrostatically charged anode and no accelerating power supply the filament will emit until it will reach the same positive potential as the anode and the beam will stop, after being accelerated les and less. All the initial potential energy was converted into KE of the electrons.

artis and hutchphd
nasu said:
You did not consider that the filament becomes positive by emitting electrons. The missing electrons have to be provided by the accelerating power supply . In the setup with electrostatically charged anode and no accelerating power supply the filament will emit until it will reach the same positive potential as the anode and the beam will stop, after being accelerated les and less. All the initial potential energy was converted into KE of the electrons.
nice observation, I will keep it in mind
I think I have an interesting project in mind

hutchphd said:
In the situatons you are considering the external energy used by the system is to maintain the potential difference.
A regular CRT TV tube uses a high positive voltage on the (behind) face of the tube where it is in a 'safe place'. You can see a well insulated red cable feeding the front / side of the tube from the EHT unit.

This allows the rest of the circuitry to operate at relatively safe voltages because 'somewhere' the electrons have to be produced, launched and controlled from regular circuit components. It keeps the rear area of the TV safer for measuring and maintenance.

Any issue of efficiency will be just a practical one because the same total potential difference is needed and the current will also be the same.

Alternatively, when you need to have a low (safe) voltage where the electron beam arrives, the source will be at a high negative potential - for instance when electron beam welding or in an electron microscope.

It's all down to the total PD.

PS Focussing makes no difference to the energy consumption or efficiency as long as all the electrons get to the target. The focussing is a conservative field.

Last edited:
hutchphd
I think for focusing and for modulating the beam current with a "grid", these electrodes are held negative to avoid current being drawn. Control can still be had at positive potentials but then there is current being drawn. The screen-end of the tube is coated with graphite, and the screen is usually aluminium coated, so a current will flow from there as well as from the anode(s). As mentioned, it is safer to ground the positive side of the supply, but the tube does not actually care.

tech99 said:
The screen-end of the tube is coated with graphite, and the screen is usually aluminium coated, so a current will flow from there as well as from the anode(s).
The Anode(s) in a CRT do not take any appreciable amount of current. Their purpose is to accelerate and focus the electrons. The fact that they are flimsy and do not get hot is testimony to that. As with many electron beam tubes (as opposed to conventional 'tubes' with sweaty anodes, the (narrow) beam needs to be carrying the Energy, which is released on contact with either the phosphors or the 'Collector' which absorbs the energy that remains in the beam. See Klystron and Travelling Wave Tube. Badly focussed beams can heat the electron gun and also the helix in a traveling wave tube; the electron optics in a dark art, IMO.

PS if the tube were very long, the trajectory of the electrons would bring them back to the Anode but, first time through, they just rush past.

tech99
@rsc9421 The reason you need a power supply for the anode even if you charge it once and the electrons don't strike it is because electrons are charged and negative while the anode is also charged and positive.
As the electrons move past that charged anode , even if they don't strike it they still have an effect on the anode charge. It is like a capacitor where both plates don't touch each other but charging one with excess of electrons or lack of them influences the other plate (the charge on the other plate)If you could just charge an anode and then leave it alone and make it constantly accelerate electrons then you could get free energy, because the accelerated electrons gain kinetic energy which is released as heat/radiation upon hitting a target which is a form of energy. This of course cannot happen! without energy input and the energy input goes to maintain the charge on the anode and the potential difference between cathode-anode.

artis said:
@rsc9421 The reason you need a power supply for the anode even if you charge it once and the electrons don't strike it is because electrons are charged and negative while the anode is also charged and positive.
You need to be careful here; what you say doesn't apply to the situation with all 'Anodes'. It's correct for any tube that can be considered basically as a Diode but the electron gun in a CRT supplies approximately Zero power to the anodes (the tubular structures in the gun structure that provide the accelerating field and the focussing). The power that's supplied to the CRT consists of the EHT volts X the current arriving at the screen. That can be several tens of Watts. No current flows through the Anodes so there is no power to it or from it.
There is an equivalent situation in orbital mechanics; an asteroid arriving near the Earth will (at some distant point) have a certain amount of Kinetic Energy in the Earth's reference frame and a certain amount of Potential Energy. It sweeps past the Earth in a near-miss trajectory and arrives at a point at the same initial distance (different direction, of course) with the same initial KE and PE situation. This assumes that the asteroid is small compared with the Earth Mass. The Earth's energy is the same as before the encounter. The Anode in the gun is in the same situation; no more or less energy after the encounter with each passing electron.

sophiecentaur said:
You need to be careful here; what you say doesn't apply to the situation with all 'Anodes'. It's correct for any tube that can be considered basically as a Diode but the electron gun in a CRT supplies approximately Zero power to the anodes (the tubular structures in the gun structure that provide the accelerating field and the focussing). The power that's supplied to the CRT consists of the EHT volts X the current arriving at the screen. That can be several tens of Watts. No current flows through the Anodes so there is no power to it or from it.
There is an equivalent situation in orbital mechanics; an asteroid arriving near the Earth will (at some distant point) have a certain amount of Kinetic Energy in the Earth's reference frame and a certain amount of Potential Energy. It sweeps past the Earth in a near-miss trajectory and arrives at a point at the same initial distance (different direction, of course) with the same initial KE and PE situation. This assumes that the asteroid is small compared with the Earth Mass. The Earth's energy is the same as before the encounter. The Anode in the gun is in the same situation; no more or less energy after the encounter with each passing electron.
Ok, but isn't this situation only possible because in the CRT, the screen is the final anode into which the beam energy is also dumped?
If one takes away the HV cable attached to the upper side of the CRT and only leaves the gun intact with it's thermionic cathode heating wires and small focusing anodes, do then we still have a beam exiting the tube neck? You know basically just taking the electron gun alone.

Because the way I understand it is like this. The filament only produces electrons in vacuum, aka the energy supplied is bit larger than the work function of that particular metal of the filament but the electrons that come off of it have no large KE only some random thermal KE that allows them to be free so all KE given to them comes from the E field produced between the cathode-anode, if this is so, then where would the energy come to continually accelerate them without the screen anode?

I suppose you can have no power to the small focusing anodes only if you have the final larger screen anode right? Because then as the electrons exit the smaller tube anodes they are pulled away by the screen anode and therefore don't get attached to the smaller anodes which they otherwise would be ?

I was being pedantic. The thread title is about the gun. The screen is not part of the gun. Without the screen, to collect the electrons with its high + potential, they would return to the anode(s) in the gun, after a long trajectory. Energy has to come from somewhere and I have just been identifying where.

## 1. What is an electron gun?

An electron gun is a device that generates and accelerates a beam of electrons. It is commonly used in cathode ray tubes (CRTs) and electron microscopes.

## 2. How does an electron gun work?

An electron gun works by using a high voltage source to create an electric field, which accelerates the electrons towards a target. The electrons are emitted from a cathode and focused into a narrow beam by a series of electrodes.

## 3. What factors affect the power consumed by an electron gun?

The power consumed by an electron gun is influenced by several factors, including the accelerating voltage, beam current, and efficiency of the gun. The type and design of the electron gun also play a role in determining its power consumption.

## 4. How is the power consumption of an electron gun measured?

The power consumed by an electron gun is typically measured in watts (W). This can be calculated by multiplying the voltage applied to the gun by the current of the electron beam. The power consumption may also be listed in the specifications of the gun.

## 5. Can the power consumption of an electron gun be reduced?

Yes, the power consumption of an electron gun can be reduced by optimizing the design and operating conditions of the gun. This may include using lower accelerating voltages, reducing the beam current, and improving the efficiency of the gun. Additionally, using alternative electron gun technologies, such as field emission guns, can also result in lower power consumption.

Replies
13
Views
1K
Replies
16
Views
2K
Replies
8
Views
437
Replies
7
Views
1K
Replies
7
Views
1K
Replies
2
Views
1K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
10
Views
2K