RAD 350 Chapter 8 X-Ray Emission

RAD 350 Chapter 8 X-Ray Emission

<p>RAD 350 Chapter 8 X-ray emission</p><p>X-ray QUANTITY is directly controlled by mA and time (mAs). The relationship of mAs and quantity of radiation is directly proportional – i.e. – if you double the mAs, the amount of radiation is doubled. X-ray QUALITY (intensity/hardness/penetrability of the beam) is controlled by kVp. As kVp increases, the intensity increases (shorter wavelength photons). X-ray quality/penetrability may also be expressed as HALF VALUE LAYER (HVL). As kVp increases, the tube becomes SLIGHTLY more efficient in producing x-rays.</p><p>Filtration – two types of filtration are usually associated with the x-ray tube/unit. Inherent (the tube itself – exit window. etc.) and is about .5 mm al equivalent. ADDED filtration is based upon the kVp of the unit (about 2 mm al equivalent at 70 kVp). The SUM of the two is called TOTAL FILTRATION.</p><p>COMPENSATING FILTERS may also be used to attenuate/absorb a portion or all of the beam at certain areas to more evenly apply the radiation. A WEDGE filter can absorb radiation at the toe end of the foot on an AP foot, but allow more radiation to penetrate the heel end thus giving a nice exposure instead of “burning out” the toes or having the heel end be light. A TROUGH filter is most often used in dedicated chest x-ray units/rooms. A “BOW TIE” filter is used in some CT scanners. Conic filter – usually for digital fluoro units.</p><p>X-ray tube : 2 primary components – Filament (thoriated tungsten) and anode. A dual filament = dual F.S. (controlled by mA station L or S); the filament can vaporize somewhat with use and gradually increase the inherent filtration over the .5mm al equivalent. Anode rotates 3400 – 10,000 rpm LOWER anode angle = HIGHER heel effect</p><p>Protective tube housing = insulating oil and shielding Leakage radiation = less than 100 mR/hr at 1 m 1/2 and full wave rectified = 100% ripple 3-phase power 6 or 12 pulse = 13% or less ripple Hi frequency = 1 % ripple Heat units (HU’s) = mA X time X kVp 3-phase = additional factor to multiply</p><p>Tube rating charts MUST match the anode angle/speed to get the correct chart!</p><p>Housing cooling chart</p><p>Tube cooling chart</p><p>How many HU’s if full wave rectified unit uses 5 consecutive exposures of 200 mA, ½ sec, and 80 kVp?</p><p>200 X ½ X 80 = 800 X 5 = 4000 HU</p>

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    2 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us