PHYS C7

X-ray Energy

kV_p (highest voltage applied across tube) determines maximum energy of electrons and thus x-ray photons (E_max = kV_p)

Average photon energy in clinical x-ray beams is ~1/3 to 1/2 of maximum energy

X-ray Emission Spectrum

total number of x-rays emitted is equivalent to area under the curve of x-ray emission spectrum

always bell-shaped

position along the energy (x) axis can change

larger area under the curve = higher x-ray intensity (quantity)

farther to the right a spectrum is, the higher the effective energy (quality) of x-ray beam

Beam Parameters

X-ray tube output controls beam’s quantity and quality

Rule: When intensity/quality (amplitude) is effected it is always for bremsstrahlung and characteristic x-rays

Factors Effects on Emission Spectra:

Tube Current effects intensity

Tube voltage effects intensity and energy of bremsstrahlung x-rays

Filtration effects intensity and energy of bremsstrahlung x-rays; most effective at low energy

Target material effects intensity and energy of characteristic x-rays peaks

Tube Current

Number of electrons passing from cathode to anode, mA

Doubling mA = twice as many electrons flow = twice as many x-rays at every energy (amplitude of each point on emission spectra curve is doubled)

Increasing tube current = better image quality (because image noise decreased) but increased dose

X-ray Quantity (Intensity) - mGy(a)

Number of x-rays in beam

Measured as area under spectrum curve

Measured in units of air kerma (energy transferred to electrons in air), mGy_a which is proportional to number of ion pairs produced in air by a quantity of x-rays

mAs

determines total radiation quantity used during an exam

= tube current(mA) x time (sec)

Changes x-ray quantity (mGy_a) and image receptor exposure linearly - Doubling mAs doubles number of electrons (thus number of photons) and thus number of photons hitting image receptor

Tube Output

mGy(a)/mAs

how much radiation the tube produces per mAs at a specific kVp and distance

Signal-to-Noise Ratio (SNR)

Ratio of mean photon intensity (signal) to the random fluctuation in number of detected photons (noise)

SNR =N / √N

Noise = √counts because discrete photons follow Poisson statistics

High SNR = high resolution because signal (N) increases faster than noise (√N)

Image Noise

Mostly caused by a lack of photons (low SNR) reaching receptor

Noise is reduced by increaisng mAs, which increases x-ray quantity to provide more signal to receptor

However caution is needed as increasing mAs will increase dose

Although increasing kVp (thus photon energy) can reduce noise it also increases scatter and reduces overall subject contrast

X-ray Quality (Energy)

Increasing energy increases penetrability

High penetrability = high quality/harder beam

Half Value Layer (HVL)

measures beam quality

the thickness of an absorber needed to reduce beam intensity by half

Image Receptor Exposure

Number of x-rays that reach image receptor

X-rays diverge, speading over larger areas, as they move away from focal spot (source)

Intensity (x-ray quantity) and image receptor exposure = 1/d² or changes by (d₁/d₂)²

where d = distance from focal spot

Increasing Tube voltage (kV_p) Effects On Quality and Quantity

Increases Quality (position) - faster acceleration, electrons hit target with higher kinetic energy, higher photon mean and max energy

Increases Quantity (amplitude) - higher energy electrons have more interactions with target before losing energy

*No effect on position of characteristic x-rays

Other effects of Increasing Tube voltage (kV_p)

Increases HVL

Decreases image noise

Increased forward direction of Bremsstrahlung photons relative to incident electron beam

Increases relative dose because increase in photon number outweighs increase in pentrability (which woud usually decrease dose as more photons pass unabsorbed) due to secondary ionisations depositing more energy in body

Tube voltage (kV_p) Equations

X-ray quantity proportional to (kVp)²/changes by (kVp₂/kVp₁)²

Relative dose proportional to (kVp)³/changes by (kVp₂/kVp₁)³

Image receptor exposure proportional to (kV_p)⁵/changes by (kVp₂/kVp₁)⁵

15% kV_p Rule

If you increase kVp by 15% (x1.15), you must halve mAs (and vice versa) to keep image receptor exposure the same

If you decrease kVp by 15% (x0.85), you must double mAs (and vice versa) to keep image receptor exposure the same

Because (1.15)⁵ ≈ 2, (0.85)⁵ ≈ 0.5

15% kV_p Rule - Application

used to change image contrast or reduce patient dose while maintaining consistent image density (degree of blackening)

15% kV_p Rule - Image Contrast

Increasing kVp by 15% produces longer scale of contrast (more shades of grey, lower contrast, more penetration power to visualise deeper structures)

Decreasing kVp by 15% produces shorter scale of contrast (higher contrast between air, tissue, and bone)

15% kV_p Rule - Dose

Increasing kVp by 15% and halving mAs reduces patient dose, as higher-energy photons are more likely to pass through patient without being absorbed

What does 15% Rule Not Account For?

Beam hardening

Detector response

Scattering

15% kV_p Rule - Digital Radiography Limitations

rule was developed for film–screen radiography

• not needed to maintain image brightness because of wide detector dynamic range (ability to produce images over large exposure range with less obvious exposure errors) and automatic brightness adjustment by processing algorithms

• radiographers increase exposure factors to reduce noise which increases patient dose, but over-exposure is now automatically adjusted/hidden to look acceptable

• standard image protocols and automatic exposure control (AEC) terminates exposure when sufficient detector signal reached