Fluke Biomedical 35080B - Manual (Page 38)

35080B
Operators Manual
7-2
7.3 Theory of Operation
The Model 35080B kVp Divider contains a pair of solid-state detectors which when exposed to x-rays,
send signals to a ratio module. The resulting ratio depends on the kV of the CT beam and the difference
in thickness of filters placed between the detectors and the beam source.
The filters are carefully selected to give the best response curve possible over the kV range of CT interest
(see Specifications).
The narrow dimensions of the CT Filter Pack's (33551) detector area are suited to the beam area
restrictions of CT applications.
7.4 Operating Instructions
1. Select 2.5 mm Al filtration and remove beam-restricting devices from the CT beam path on the CT
machine. Open the beam width to 10 mm to cover the sensor area on the CT Filter Pack (33551).
Use the patient-positioning laser to align the Model 35080B kVp Divider within the beam.
2. Connect the kVp Divider output to a storage oscilloscope.
3. Set the oscilloscope sensitivity to 0.1 V/div, DC Coupled.
4. Put the kVp Divider into the ‘‘CAL’’ mode and verify its output is 0.644 volt ± 1%.
5. Turn the kVp Divider OFF and position the oscilloscope trace to read 60 kV (the reference baseline
on the oscilloscope which is added to the measured value).
6. Again, put the kVp Divider into the ‘‘CAL’’ mode and verify that the oscilloscope trace corresponds
to 124.4 kV (± 1 kV).
7. Place the tube head in its top position.
8. Switch the kVp Divider into the ‘‘RUN’’ mode and make the exposure.
7.5 Correction of Readings
Filtration
No correction is needed for 2.5 mm Al (Refer to Figure 7-1.) . For a given kV reading: if the effective
filtration of the CT beam is greater than 2.5 mm Al, subtract the difference (in mm) times the correction
factor from the apparent reading; if the effective filtration is less than 2.5 mm Al, add the difference
times the correction factor to the apparent reading.
Example 1
The kV reading is 100, and the total filtration of the beam is 3.5 mm Al. The difference between the
effective and calibration filtration is 1 mm. This difference times the correction factor for 100 kV is 1 x
0.35 = 0.35 kV. Therefore the corrected kVp is calculated as 100kV - 0.35 kV = 99.65 kV.
Example 2
The kV reading is 80, and the total filtration is 2.0 mm Al. The difference between the effective and
calibration filtration is 0.5 mm. The difference times the correction factor for 80 kV is 0.5 x 0.2 = 0.10
kV. Therefore, 80 kV is 0.5 x 0.2 = 0.10 kV. Therefore, 80 kV + 0.10 kV = 80.1 kV.