Kilovoltage Peak role for an X ray tube

  • Context: Undergrad 
  • Thread starter Thread starter superduke1200
  • Start date Start date
  • Tags Tags
    Peak Ray Tube
Click For Summary

Discussion Overview

The discussion revolves around the role of kilovoltage peak (kVp) in X-ray tubes, specifically focusing on its relationship with radiation dose and the underlying physics. Participants explore the implications of kVp on the intensity of X-ray production, the effects of filtration, and the dependency of dose on kVp squared.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the radiation dose to the patient is directly proportional to the square of kVp, suggesting that higher kVp results in more photons produced during electron interactions with the anode.
  • Others explain that while the intensity of the beam is proportional to the power, which is proportional to kVp squared, the actual dependence may be greater due to the effects of filtration that eliminate lower energy X-rays.
  • A participant questions the meaning of kVp squared and seeks clarification on how power relates to the X-ray tube's operation.
  • Some participants discuss the relationship between tube current and kVp, with one asserting that tube current is dependent on the voltage applied to the cathode, while another mentions that Ohm's law applies in this context.
  • There are differing views on the applicability of Ohm's law in vacuum tubes, with one participant arguing that the tube current should not change with varying kVp since the electrons are already emitted.
  • Another participant introduces the concept of dynamic resistance in diodes and vacuum tubes, suggesting that resistance changes with voltage, affecting current in a non-linear manner.
  • One participant offers an alternative explanation linking dose to the fluence and energy of electrons, stating that the product of these factors is proportional to kVp squared.
  • There is a request for clarification on the relationship between fluence and Bremsstrahlung yield, specifically whether it refers to photons or electrons.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the relationship between kVp, tube current, and radiation dose, with no consensus reached on the exact nature of these dependencies or the applicability of certain physical laws in this context.

Contextual Notes

Some discussions involve assumptions about constant mAs and the effects of filtration on exposure rates, which may not be universally accepted or fully resolved.

superduke1200
Messages
57
Reaction score
1
Hello everyone,

while reading about the X ray tube, I came across this:

https://radiopaedia.org/articles/kilovoltage-peak

In the second paragraph, it is stated that: "The radiation dose to the patient is directly proportional to the square of kV."

I understand that by accelerating the electrons with a higher voltage, more photons will be produced while they interact with the anode. Thus, the dose to the patient will be greater

But I fail to figure out why there is a dependency on the square of KV.

Any contribution would be highly appreciated
 
Physics news on Phys.org
The simplest explanation is that the intensity of the beam is proportional to the power provided which is proportional to the KVp2. For a real x ray tube though the dependence is greater than the square because of the filtration that is used to eliminate lower energy x rays. This is because as you increase the KVp the penetration of the filters increases faster than the inherent intensity increase due to the change in KVp.
 
  • Like
Likes   Reactions: geoelectronics
gleem said:
The simplest explanation is that the intensity of the beam is proportional to the power provided which is proportional to the KVp2. For a real x ray tube though the dependence is greater than the square because of the filtration that is used to eliminate lower energy x rays. This is because as you increase the KVp the penetration of the filters increases faster than the inherent intensity increase due to the change in KVp.
KVp2? I don't understand the meaning of this. Could you be more specific?
 
KVp is peak KV, the KV that is set on the x ray machine.
 
Yes, my bad. Something unknown happened and I couldn't read your reply correctly. I'm familiar with what KVp is.

In which part of the X ray tube does the power you describe refer to?

I forgot to mention that my question is made under the assumption that the mAS remains the same. And we just duplicate the KVp for example
 
The KVp times the tube current (not the filament current) is the power delivered to the electron beam. This is mostly deposited as heat in the anode with about 0.1% converted into x rays.
 
gleem said:
The KVp times the tube current (not the filament current) is the power delivered to the electron beam. This is mostly deposited as heat in the anode with about 0.1% converted into x rays.
I couldn't agree more. But I fail to determine the squared dependency of the KVp.

Power = KVp * Tube current

Since we have the same mAS, hence the same tube current why the squared dependency?
 
The tube current is also dependent on the tube voltage according to Ohms law. V = constant x I where the constant depends on the filament current. So the power is proportional to the voltage squared.
 
If I'm not wrong Ohm law is not applicable in vacuum tubes.

The tube current I.e. the stream of electrons that travel from the cathode to the anode, is dependent on the the voltage applied to the cathode. In other words only the cathode voltage will determine the number of electrons that will be emitted and make the tube current. Since they are already emitted and we have a vacuum tube, why would the tube current change with a change of KVp.

If the filament emits from example 100 electrons why would we have a change in current with different KVp since we have vacuum?

There is no doubt that my stream of thoughts is somewhere wrong.

The question is, where am I wrong..
 
  • #10
Finally, I came across with the following fact:

In diodes , transistors and vacuum tubes it is the resistance that changes with voltage which changes the current in the circuit.

So that seems to explain the reason why Ohm's law is applicable.

The thing is that in these cases the relationship between voltage and current is not linear.
 
  • Like
Likes   Reactions: gleem
  • #11
Sorry I took so long to reply but I had something to do. In circuit theory the current is taken proportional to the voltage. In the case where the relation is not linear the the factor is called the dynamic resistance. For a diode the dynamic resistance in relatively constant in most of a diodes normal operating range

For an x ray tube the tube current depend on the filament temperature (current in the filament circuit) as well as the potential difference between the filament (cathode) and the target (anode).
 
  • #12
I must admit it is far better now.

One last question based on something that you wrote down earlier:

"This is because as you increase the KVp the penetration of the filters increases faster than the inherent intensity increase due to the change in KVp".

I would appreciate if you could explain this a bit more thoroughly
 
  • #13
superduke1200 said:
One last question based on something that you wrote down earlier:

"This is because as you increase the KVp the penetration of the filters increases faster than the inherent intensity increase due to the change in KVp".

This statement was made with regard to the change in the increase in the exposure rate with KVp as more filtration is added. I.e. exp. rate ∝ KVp2 for low filtration but changes to KVp(>2) as the filtration is increased although the absolute exposure rate will decrease. Am I clear?
 
  • #14
Perfectly clear!
 
  • #15
Hi
I have another explanation. the dose (in fact the kerma) is proportional to the fluence and energy of the electrons. the fluence is proportional to the Bremsstrahlung yield itself proportional to the energy. the product of the 2 is proportional to E2 (kVp2)
I read that in : http://www.springer.com/gp/book/9783319486581
 
  • Like
Likes   Reactions: mfb
  • #16
PSRB191921 said:
Hi
I have another explanation. the dose (in fact the kerma) is proportional to the fluence and energy of the electrons. the fluence is proportional to the Bremsstrahlung yield itself proportional to the energy. the product of the 2 is proportional to E2 (kVp2)
I read that in : http://www.springer.com/gp/book/9783319486581
I'm afraid that the link that you provided is not open source.

Just one question: When you write " the fluence is proportional to the bremsstrahlung yield"

Which fluence do you refer to? Photons or electrons?
 
  • #17
The photons (X-rays) fluence
 

Similar threads

Graduate Larger X-ray dose in traditional X-ray of head vs. Head CT?
  • · Replies 7 ·
Replies
7
Views
4K
Undergrad Why Bremsstrahlung <20keV for Tungsten?
  • · Replies 17 ·
Replies
17
Views
4K
Undergrad X-ray tube to gamma tube question
  • · Replies 7 ·
Replies
7
Views
2K
Fluoroscopic optimal conditions
  • · Replies 4 ·
Replies
4
Views
1K
Graduate X-ray production - maximum pressure
  • · Replies 4 ·
Replies
4
Views
2K
X-ray production and ionization energy
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Question about X-ray power emitted by the Coolidge tube
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad The intensity of x-rays in a Coolidge tube
  • · Replies 3 ·
Replies
3
Views
3K
X-ray Tube Output: 80μGy/mAs at 1m, 900µGy/s @1m
  • · Replies 4 ·
Replies
4
Views
3K
Calculating Roentgens from an X-Ray source
  • · Replies 7 ·
Replies
7
Views
3K