How to calculate X ray dose using mAs and kV without using any detector?

Dear Sandeep,

Marius is right; for calulation of the organ dose, the dose value depends on the patient geometry.

But I think in your case I think you are interested in the exposure in the sense of KERMA.

Therefore please have a look at my answers of a recent very similar question here on RG.

It is not a trivial task.

You may also use 'Caldose', please see the second  and third links. But I havn't tested the program.

for all the above answers . The dose value depend on the patient geometry  

To clarify the answers given above, when people say  the dose depends on 'patient geometry', I assume what they mean is that patient dose depends on where the beam is pointing at, i.e. the body part being irradiated. The same combination of kV and mAs would produce very different values of patient dose if the beam was pointed at the foot than if it was pointed at the head.

Dear all,

I think it is quite clear to Sandeep that the dose of organs in addition to the mAs and kVp setting depends on the patient geometry.

But making the ouptut of x-ray tubes comparable with respect to dose the air KERMA is often taken as reference.

In addition to the mAs and kVp setting the anode material, the anode angle, inherent and additional filtration, the distance and often also the voltage ripple/modulation are used as imput parameters for calculation of the x-ray flux or air KERMA respectively.

It is quite sure that any modell behind the x-ray flux calculation only fits the ground thruth to a certain extent and that other parameters such as the anode roughness and window contamination, which both depend on life time and  history of heat dissipation of the tube, do not really enter current calculations. Therefore it could be advised to measure the dose; such  advices very often pop up here rightly in RG in relation to dose issues. 

This advice should be adhered when you will have a certain degree of accuracy because you know it from the data sheet of your dosemeter.  

@ Sandeep. By the way: please tell me your anode material, anode angle and filtration of your tube, so I will give you  the expected air KERMA values for the above mentioned kVp settings.

An additional info, we need the filtering.

Mr Sandeep Mittal

I am working out below two independent methods to arrive at the output of x-rays for a given combination of kV and mA. Please note that both these are only approximate but they do indicate the different approaches to the answers. Let us take the distance from the x-ray tube target to the point at which the output is required as 100 cm.

Method 1:

Let us assume kV = 60 and mA = 10; then power dissipated on the x-ray target is:

60 x 10 watts or 600 joules/sec = 600 x 10⁷ ergs/sec ……….(1)

The fraction of electron energy converted to bremmstralung (f) is given approximately by:

f = E_e x Z x 10^-6 (Z is the atomic number of target = 74 for tungsten; E_e is the electron energy in keV)(from medical physics literature).

Since the average energy of the unfiltered x-ray photon energy is one-third of the electron energy from Kramer’s equation, the photon energy available as bremmstralung therefore is:

Hence f = (60/3) x 74 x 10^-6 = 0.148% …………………………..(2)

Therefore the x-ray energy output at the target = (1) x (2) above = 0.888  x 10⁷ ergs/sec …… (3)

Assuming the x-ray distribution from the target is isotropic (valid for a thick target), the energy output at 100 cm is: [(3) above] / 4π (100²) = 0.0705 x 10³  ergs/sec/cm2 ………………(4)

Assuming the total (inherent + added) filtration in the tube is 3 mm:

For 60kV, 3mm filtration offers 2HVLs, since the first HVL is 1mm and the homogeneity coefficient is 0.5 (obtained from literature); Hence the attenuation offered by 3 mm Al = ¼ (0.25)

This leads to an output after filtration = [(4) above]/4  = 0.0176 x 10³ ergs/sec/cm² ……….(5)

To get the output air in R/sec at 1 meter, we have to multiply (5) above by the mass energy absorption coefficient of air for 60 kV with 3 mm filtration. For an unfiltered x-ray beam, Kramer’s Law gives an average energy of one-third kV; for a filtered(hardened) beam, let us assume the average energy as 30 keV (this is a valid approximation). The mass energy absorption coefficient µen /ρ for 30 keV x-rays = 0.1501 cm² /gm [From Hubbel’s 1982 revised 2012 data]…………….(6)

Therefore the output at 1m = [(5) x (6)]/ 87.7 [1 R = 87.7 ergs/gm of air] = 0.0301 R/sec  = 1.81 R/min…..(7)

This can be converted to air kerma in rads/sec by multiplying by 0.877

It may be noted the above derivation involves a number of approximations. Moreover, the values of  (µen /ρ) vary widely over the low keV region.

 Method II

This  method uses the RadPro calculator available free on-line. Here you have to enter the values of kV (60), mA (10), filtration(3 mm Al) and distance(100 cm) from the target to the point of measurement. This calculation then yields a value of 1.8719 x 10⁸ µR/hr = 187.19 R/hr = 187.19/3600 R/sec = 0.052 R/sec = 3.12 R/min ………………………….(8)

It may be noted that these values (7) and (8) above are obtained by altogether two different approaches and both involve approximations. Both these denote typical outputs of a fluoroscopic unit.

Please consult any radiation physics text or website for references cited above.

Hope this helps.Good Luck.

I wish to add to my above. Both the methods show the x-ray  outputs in air in roentgen or air kerma. These values can be converted to the tissue dose by multiplying  with the ratio of energy absorption coefficients of tissue and air at the stated energy, available from Hubbel's data. This completes the answer to your query.

Sankaran sir excellent  and comprehensive  reply. I learnt so much while reading. Thanks for detailed reply. 

Thank u, Munish Kumar. I put out above what I remembered while teaching Dip.R.P trainees many years ago. X-Ray technology was one of several subjects I was lecturing on. Perhaps trainees may still be using my lecture materials. Best wishes.

An excellent explanation by Sankaran Sir.

http://iopscience.iop.org/article/10.1088/0031-9155/31/4/010/pdf

The formula for radiation dose of an X-ray unit

D = g*kV*mAs/d^2

where g is constant (factor) would be dependent on anode composition (generally Tungsten), anode angle, and inherent and added filtration and d is distance. The units would usually be in mGy (milliGray). Entrance air kerma can be converted to an entrance skin dose by first multiplying by the appropriate 'f' factor (for diagnostic energy ranges this is a small adjustment, usually about 1.07 to account for the differences in atomic number of air and soft tissue) and then multiplying by a backscatter factor to account for x-ray photons scattered back to the surface from deeper tissue. Backscatter factors depend on beam energy and field size and are in the neighborhood of 1.3 or 1.4.

PCXMC should be also helpful.

A bit late, but nevertheless a remark:

x-ray dose is a very good approximation proportional to the square of the tube voltage;

so we have: D=g*kV²*mAs/d² , not linear with the voltage as Hany has answered above.

You can calculate effective atomic number (Zeff) for tissues with Phy-X/ZeXTRa.

Hi,

For my x-ray tube (YXLON.TU 160-D03 ), we have derived experimentally the following equation for the dose-rate:

where: D is the delivered dose rate in air in units [Gy/min]

I is the current of the x-ray tube in units [mA]

V is the voltage of the x-ray tube in units [kV]

R is the distance between the x-ray anode and the object in units [cm]

It gives similar results as the online tool at http://www.radprocalculator.com/XRay.aspx

Regards

Krassimir

Phy-X Team : The question is on dose calculation and not on effective atomic number for tissues; so this answer has no relevance here. moreover, when we click on either of the references cited below, we get a dim map of the world only. what is this?

Krassimir's Answer:

The formula is not clear mathematically. What is V.V and R/R? Please give step by step calculation of the dose in Gy/min for 60kV using your formula. Take I = 10mA and R = 100cm

Dear Sankaran Ananthanarayanan ,

we all know (at least from the statements above) that the dose is roughly proportional to the square of tube voltage V and inverse proportional to the square of distance R.

So we have: V.V = V * V = V² and 1/R/R = 1/(R*R) = 1/R²

I must admitt that the notation of Krassimir Stoev 's formula above is a bit curious.

In addition I also wonder why the Phy-X Team answers with an effective atomic number statement on a question related to dose.

Suggest you download "RADCOMP" software available for free from my website--

Please see my earlier answers. For accurate dosimetry, purchase a Farmer dosimeter. Pl see net for details.

@ Respected Sankaran sir

Thanks for your kind reply on the similar question asked by me again.

could you please guide me, about how could I find out the relation between thickness of filtration, HVL and kV? is there any specific literature or physical rule? Sankaran Ananthanarayanan

I parameters for kVe, mAs and ffd x-ray examination

Nice to share.

How to calculate X ray dose using mAs and kV without using any detector:

Dose rate (mGy/hr) = (kV)^2 * mA * exposure time / 10^8

Plugging in the values from your question, we get:

Dose rate (mGy/hr) = (400)^2 * 100 * 100 / 10^8 = 1600 mGy/hr

This means that the x-ray beam would be delivering a dose of 1600 mGy per hour to the object being radiographed.

The mass of the object does not affect the dose rate. The dose rate is only affected by the kVp, the mA, the exposure time, and the distance from the source.

Dear Professor Gad. Could you provide a reference for the equation you mention? Also the exposure time in the equation is in hours, am I correct ?

calculate the dose rate from a 50 kV X-ray machine, with a tube current (mA) of 10, an exposure time of 10 seconds, and a target distance of 10 cm from a 50 kg human body? The conversion factor for X-rays at 50 kV is 0.175 Gy/(s*mA). using formula: Dose Rate (Gy/s) = (mA * Exposure Time * Conversion Factor) / (Distance^2)

Dose Rate (Gy/s) = (mA * Exposure Time * Conversion Factor) / (Distance^2) Given: mA = 10 Exposure Time = 10 seconds Conversion Factor = 0.175 Gy/(s*mA) Distance = 10 cm First, let's convert the distance from centimeters to meters: Distance = 10 cm = 0.1 m Now, we can substitute the values into the formula: Dose Rate (Gy/s) = (10 * 10 * 0.175) / (0.1^2) Dose Rate (Gy/s) = (100 * 0.175) / 0.01 Dose Rate (Gy/s) = 17.5 / 0.01 Dose Rate (Gy/s) = 1750 Gy/s. The dose rate from a 50 kV X-ray machine, with a tube current of 10 mA, an exposure time of 10 seconds, and a target distance of 10 cm from a human body is calculated to be approximately 1750 Gy/s.

Ashraf Gad Dear Professor,

Thank you for your time and the answers. It would be helpful if you could provide the references for the same.

thanks

Could you please calculate for ''40 KV'' with ''40 mA'' dose rate for operating unprotected X-ray tube?

Dear Muhammad khalid Hussain ,

you know, that, apart from 'kV' and 'mA', the output of an x-ray tube depends on various other parameters such as filtration, take off angle and ageing, etc.

So any number given here will only be a very rough estimation.

For such numbers of please have a look at the diagram of our colleague Ulrich Neitzel, which is presented in his answer of the following question:

So for 40kV you can expect an air KERMA in the order of about 10µG/mAs^*) in 1m distance from the x-ray tube.

*) I have mutually rounded the given 11µGy/mAs to 10µGy/mAs because of the large uncertainty range...

Best regards

G.M.

Thank You So Much Dear Prof.