The Revision Guide for Student Nurses (Part I)

Radiographic Settings Intro - Key Notes

The exposures required to take good quality, diagnostic radiographs depend upon many factors including the type of x-ray machine, the processor and of course the patient. It is for this reason that veterinary practices should keep a log of all radiographs taken comprising as much information as possible and accurate reports of the quality of the resultant exposed films. This will aid personnel in that a useful exposure chart can be created to allow quick reference to the settings required for any part of the anatomy of as many species and breeds as possible. In addition, film speed is relevant; medium speed is the best compromise. Slow film/screen combinations produce the highest definition suitable for small areas such as the limbs and skull. A fast film-screen combination requires lower exposures than a slow film-screen combination, but produces less definition than the latter.

EXPOSURE LOG
The following should be noted:

  • Date.
  • Owner's name and patient identification.
  • Species & breed.
  • Size & weight (note whether the animal is obese, under-weight, juvenile, pregnant etc).
  • Part of the patient's anatomy radiographed.
  • Position of the patient.
  • Film/screen combination.
  • Film focal distance.
  • Type of grid used if applicable (used for tissue depths of 10cm or greater).
  • Exposure settings used (kV, mA, mAs).
  • Notes, eg. quality of resultant film, plaster cast present, contrast agents used.

LINE VOLTAGE
A line voltage compensator may be found in lower output x-ray machines. This is used to control fluctuations in the mains electricity that may result in inconsistent output of x-rays despite normal exposure factors. Line voltage should only be adjusted if indicated, and must never be used to directly alter other exposure factors.
Larger x-ray machines are likely to possess an autotransformer which automatically controls mains voltage fluctuations.

KILOVOLTAGE (kV)
Kilovoltage is commonly shortened to kV. It is the kV control that affects the energy of the x-ray photons and thus affects both contrast and blackening of the image. kV is defined as the potential difference applied between the filament (cathode) and target (anode) of the x-ray tube head. The subsequent speed and energy with which the electrons bombard the target affects the penetrating power of the primary beam. The kV control may be directly linked to the milliamperage (mA) control in smaller machines. This means that mA is lowered as kV is raised, which is not ideal in circumstances where both a high mA and kV are required (such as in large patients) since longer exposure times are therefore necessary.
If the kV is increased by 10, the mAs should be halved to achieve the same exposure and if kV is decreased by 10, the mAs should be doubled.

MILLIAMPERAGE (mA)
Milliamperage is commonly shortened to mA. The mA setting will affect the quantity of electrons passing across the x-ray tube head from the filament to the target and has an effect only upon the blackening of the image. mA is also referred to as the tube current. Since the current is so small, it is measured in milliamperes rather than amperes. Raising the mA causes the filament to increase in heat subsequently causing the release of more electrons.

MILLIAMPERE SECONDS (mAs)
mAs is the measurement of the length of exposure time. The longer the exposure time, the more likely you are to encounter movement blur. If the mA (tube current) is increased, the exposure time is lowered. It is sensible therefore to use the highest mA possible. A higher kV also allows the mAs to be kept small.
mAs should be doubled to penetrate a dry plaster cast, and doubled again to penetrate a wet plaster cast.
Some useful x-rays are absorbed by grids, which means that if a grid is used, mAs must be increased; multiply the mAs required without a grid by the grid factor.

THE EXAMINATION
When calculating exposures, always remember that:

EXPOSURE TIME (in seconds) = mAs divided by mA.
NB. 15 mAs = 150 mA for 0.1 sec OR 15 mA for 1.0 sec.

Since all x-ray machines require different exposures, it is unlikely that you will be required to state settings in the practical examination. You will however encounter written questions necessitating a good working knowledge of the formula involved.

The example questions below are similar to those that have appeared in the written papers and should be of help to you during revision:

1) Which of the following radiographic settings will produce an image with the greatest density?

a) 100 mA, 0.1 sec.
b) 150 mA, 0.05 sec.
c) 200 mA, 0.05 sec.
d) 200mA, 0.1 sec.
The answer is d) since these settings produce the greatest tube current for the longest time period.

2) A radiograph is to be taken of a patient post casting (the cast is still wet). How would you adjust normal exposure factors in order to obtain a good quality image?

a) Increase kV by 10 and double the mAs.
b) Decrease kV by 10 and double the mAs.
c) Halve the mAs only.
d) Double the mAs only.

The answer is a) since increasing the kV by 10 means that an equivalent exposure would be achieved by reducing the mAs by half; we require a greater exposure and therefore instead of reducing the mAs we double it. Thus the effect of twice doubling the mAs is achieved.

3) If a radiographic exposure is given as 60 kV and 15 mAs, what would the exposure time be if the output of the machine is 10 mA?

a) 15 secs.
b) 0.15 sec.
c) 1.5 secs.
d) 2 secs.

The answer is c). kV is not relevant to this question; exposure time = mAs divided by mA (15 divided by 10 is 1.5).

4) If a radiographic exposure is given as 70 kV and 10 mAs and the output of the machine is 10 mA, what would the exposure be if a grid with a grid factor of 2 were used?

a) 20 mAs.
b) 70 mAs.
c) 15 mAs.
d) 20 mAs.

The answer is a) since the original mAs is multiplied by the grid factor (10 x 2 =20).

5) The settings are as per Q4; what would the exposure be if the kV was raised to 80 and no grid was used?

a) 20 mAs.
b) 12 mAs.
c) 2 mAs.
d) 5 mAs.

The answer is d) since mAs is halved for every 10 kV increase (10 divided by 2 = 5).

6) Again, the settings are as per Q4; what would the exposure be if the kV was lowered to 60 and no grid was used?

a) 5 mAs.
b) 20 mAs.
c) 15 mAs.
d) 2.0 mAs.

The answer is b), mAs is doubled for every 10 kV decrease (10 x 2 = 20)