# I Kilovoltage Peak role for an X ray tube

1. Mar 5, 2018

### superduke1200

Hello everyone,

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

2. Mar 5, 2018

### gleem

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.

3. Mar 5, 2018

### superduke1200

KVp2? I don't understand the meaning of this. Could you be more specific?

4. Mar 5, 2018

### gleem

KVp is peak KV, the KV that is set on the x ray machine.

5. Mar 5, 2018

### superduke1200

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. Mar 5, 2018

### gleem

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. Mar 5, 2018

### superduke1200

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. Mar 5, 2018

### gleem

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. Mar 5, 2018

### superduke1200

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. Mar 5, 2018

### superduke1200

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.

11. Mar 5, 2018

### gleem

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. Mar 5, 2018

### superduke1200

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. Mar 5, 2018

### gleem

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. Mar 5, 2018

### superduke1200

Perfectly clear!

15. Mar 9, 2018

### PSRB191921

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

16. Mar 12, 2018

### superduke1200

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. Mar 12, 2018

### PSRB191921

The photons (X-rays) fluence