Kilovoltage Peak role for an X ray tube

In summary: The filtration on an x ray tube is designed to reduce the number of lower energy x-rays that reach the patient. So as you increase the KVp the number of x-rays that make it to the patient is less proportional to the intensity of the beam.
  • #1
superduke1200
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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
 
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  • #2
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.
 
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  • #3
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?
 
  • #4
KVp is peak KV, the KV that is set on the x ray machine.
 
  • #5
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
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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.
 
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  • #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
 
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  • #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
 

What is the purpose of the kilovoltage peak (kVp) in an X-ray tube?

The kVp in an X-ray tube controls the energy of the electrons produced by the cathode and accelerates them towards the anode. This results in the production of X-rays with a specific energy level, which is important for obtaining clear images and minimizing patient exposure.

How does the kVp affect image contrast and brightness?

The kVp has a direct impact on image contrast and brightness. Higher kVp values result in an increase in both contrast and brightness, while lower kVp values decrease contrast and brightness. This is because higher kVp produces X-rays with higher energy levels, which can pass through denser tissues and create a higher contrast image.

What is the optimal kVp for different types of imaging?

The optimal kVp for imaging depends on the type of tissue being imaged. For soft tissue imaging, a lower kVp (around 70-80 kVp) is ideal, as soft tissues have a lower density and do not require as much energy to produce clear images. For bone imaging, a higher kVp (around 100-120 kVp) is needed to penetrate through the dense bone and produce a clear image.

How does the kVp affect patient radiation dose?

The kVp is directly proportional to the patient radiation dose. Higher kVp values result in higher radiation doses, while lower kVp values result in lower radiation doses. This is because higher kVp produces X-rays with higher energy levels, which can penetrate deeper into the body and expose more tissues to radiation.

What are the potential risks of using high kVp values?

High kVp values can increase patient radiation dose, which can have negative effects on health, especially with repeated exposure. Additionally, using high kVp values can result in overexposed images, which can decrease image quality and make it difficult to interpret the images accurately. It is important for radiographers to carefully select the appropriate kVp values for each imaging procedure to minimize risks and ensure high-quality images.

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