# Dose calculation of X-Ray tubes

I am working on a X ray tube project and I want to calculate approximate dose. Do you know any equation ?

I have found one on internet but I think this is wrong:

$$D=g\cdot kV^2\cdot \dfrac{mAs}{d^2}= \left[\dfrac{Sv}{h}\right]$$

So here is $$g$$-factor a constant and how can I find it (I mean if it is what is it for this equation, if it is not then how can I calculate it, because as well as I know $$~g~$$-factor depends on the angle of the anode and these things). And here the $$~d^2~$$ stands for distance but this the distance of what ? The distance between the tube and the object or the distance between cathode (filament) and anode (tungsten) ?

And can I calculate $$~mAs (I\cdot t)~$$ by $$~\dfrac{kV}{R}~$$ of the filament ?

• Proportional to voltage squared? What is your source? (Not the x-ray source, I mean where did you get this from?) – Pieter Feb 19 at 9:32
• I am trying to understand the expression. I suppose "mA" means milliamps? And "k" is not some constant but also a prefix, so the voltage in kilovolt? Units do not seem to match: the left hand side multiplies by time (in seconds), the right hand side has a dose rate in Sv/h? No, the current of a vacuum tube does not follow Ohm's law. – Pieter Feb 19 at 9:44
• @Pieter I found it on researchgate.net, but it seemed me wrong so I wanted to verify it. mAs is miliampere second and kV is kilovolt yes. Do you know any other equation for my purpose sir ? – fissile_uranium Feb 19 at 11:06
• @nmasanta It's great that you've converted the formulae to MathJax, but I think you may have changed the meaning slightly. It's hard to tell because the OP hasn't responded to Pieter's requests for clarification. But I think $m$ is supposed to be the milli- prefix, not a separate quantity (eg mass). Also, $mAs (I\cdot t)$ is (probably) supposed to be 2 separate expressions, with the $(I\cdot t)$ being a synonym or explanation of $mAs$. Ah, fissile_uranium just responded as I was writing this comment. ;) – PM 2Ring Feb 19 at 11:09
• @fissile_uranium Can you give us a link for the source of that expression from researchgate.net? – PM 2Ring Feb 19 at 11:11

X-ray flux is proportional to current $$I$$ (which is normally of the order of milli-ampères for x-ray tubes) and inversely proportional to $$d^2$$ when the distance $$d$$ is large compared to source size. Dose is also proportional to time $$t$$ (which can be measured in seconds or hours - unclear what here).
Electrical power is $$P = IV$$ and when voltage is in kV and the current in milliamps you get the product in watt.
Only a very small part of this power is emitted as x-rays. That fraction is increasing with voltage. A proportional increase would result in that $$V^2$$ factor, but there is no theoretical justification for proportionality.
So the $$g$$ in this parametrisation will depend on lots of things: anode material, for example. And angle. And on voltage.