Week 9 - X-Ray Tubes (Part 1)

Question 1

How to Form an X-Ray Image

Answer 1

To form an image:
o Some x-ray must be absorbed in the patient (more in some areas than in others)
o Some must reach the x-ray film
o Uniform Beams of x-rays (100’s of billions of them

Question 2

Effect of Number of X-Rays

Answer 2

• Too few leads to an underexposed image or noisy image
• Too many x-rays leads to an underexposed image or too high a radiation dose to the patient
• In both cases we need to be able to control the quantity and quality of the x-ray beam
  o Quality refers to the energy of the x-ray photons

Question 3

X-Ray Tube

Answer 3

• X-rays are produced by an x-ray tube
• An x-ray tube converts electrical energy to electromagnetic energy (x-ray photons).
• The x-ray beam is controlled by controlling the electrical energy delivered to the x-ray tube.
• The operator (you) chooses the appropriate electric potential (kilovolts), electric current (milliamperes) and the time period (seconds) for which the current flows.

Question 4

The Electron Volt

Answer 4

The energy required (work done) to move a single electron through a potential difference of 1V is very small and known as the electron volt
o 1 eV = 1.602 x 10-19 J

If an electron is accelerated through a potential difference of 1000 V it will gain 1000 eV of kinetic energy
o 1000 eV = 1 keV
o 1000000 eV = 1 MeV

The electron volt is a useful derived unit of energy when dealing with x-ray units

Question 5

Main Components of an X-Ray Imaging System

Answer 5

The x-ray tube
 Accelerates electrons and generates x-rays

User controlled parameters
 current (mA), voltage (V) and exposure time (s)

Question 6

Direct Current

Answer 6

Simple circuit
o Battery supplying electric potential V (voltage)
o Resistor R

Current flowing through the circuit
o I = V / R
o Ohm’s Law

Battery supplies power to the circuit
o P = I x V = I2 R

This power I x V will be dissipated in the resistor
o Resistor heats up
o A lot of the power lost –>due to wire heating up

Question 7

Problems with DC Power

Answer 7

• Transporting electrical power from where it is generated to where it is needed results in significant losses (heating of cables)
  o (P = I2 R)
• The loss is proportional to the length of cables/wires (resistance) and the current squared
• Desirable to reduce the current to minimise losses
• Increase the voltage to maintain the required electrical power (P = I V)
• Increasing the voltage for DC circuits is not simple

Question 8

Alternating Current

Answer 8

• Electricity is generated by converting mechanical work into electrical energy (opposite of a motor)
• Usually a turbine – steam used to rotate a shaft attached to a large coil
• AC current induced in the coil i.e. direction of current flow changes or alternates at a rate proportional to the rate of rotation
  o Coil rotates around –> rotates through opposing magnetic fields
• Sinusoidal

Question 9

Electrical Supply in Australia

Answer 9

• Electrical Potential is 240V and the frequency is 50 Hz
• Supply voltage is measured in Vrms
  o Measure of the average magnitude of the voltage
• Peak Voltage (Vpeak)
  o Root 2 times the rms voltage

Voltage of alternating current can more easily be increased using devices known as transformers

Used in the national grid

Question 10

Transformers

Answer 10

A transformer is a device for increasing or decreasing an ac voltage
o e.g. Mobile phone battery may require 3-4 V to be recharged. Mains electricity is 240 V so voltage needs to be reduced using a transformer
o e.g A TV might require significantly higher voltages than 240 V so the voltage would need to be increased.

• AC Current in primary coil induces a changing magnetic field in the iron core (Amperes Law)
• Changing magnetic field induces a changing current in secondary coil (Faraday’s Law)

Question 11

Transformer Equations

Answer 11

• Voltages are proportional to the number of turns in the windings
• The currents are inversely proportional to the number of turns

Question 12

Types of Transformers

Answer 12

Isolating:
o Has separate primary and secondary windings. The two windings are isolated from each other as far as DC voltages and currents are concerned.

Autotransformer
o A single winding with many tapping points (secondary winding is not isolated form the primary).

Step-up: Ns > Np, i.e., the secondary voltage is higher than the primary voltage.

Step-down: Ns < Np, i.e., the secondary voltage is less than the primary voltage

Question 13

X-Ray Tubes AC Current: How does this effect X-Ray Production?

Answer 13

• If the current and voltage are rapidly changing direction 50 times per second

How will this effect x-ray production?
o Output would be reduced
o The x-ray beam would be pulsed
 Only produced when the target is positive voltage and target is negative voltage (50% of time)

o Ideally want a nice steady output of x-rays (DC)

50% of the time anode will be positive
o Therefore, 50% of the time electrons will be accelerated to anode

Maximum X-ray energy is equal to the accelerating voltage across anode-cathode

Question 14

Rectification

Answer 14

• Can convert AC to DC through a process known as rectification
• A device known as a diode will only allow current to flow in one direction
  o Can minimise/remove the oscillations

Question 15

Half-Wave Rectification

Answer 15

• 2 diodes
• No current flows for half the time

In secondary circuit, negative voltage has been removed by the diodes
o Waste half the power
o Double the exposure times

Question 16

Full-Wave Rectification

Answer 16

Output is still pulsed
o Changing from 0 to a maximum

One pulse every 1/100s
o In Australia

Voltage is always positive
o Voltage is being oriented in the correct way

• Half the exposure time of half-wave rectification
• The voltage on an x-ray tube at any moment determines the maximum energy of the accelerated electrons at that moment striking the anode
• The maximum energy of the electrons determines the maximum x-ray energy –this will be changing with time!

Zero or low energy x-rays are of little use for imaging
o No penetration in patient (stopped at skin)

Question 17

Smoothing

Answer 17

• Can be achieved with a capacitor
  o It stores and release charge in a similar way to tank stores and releases water
• Capacitor charges when voltage increases
• Capacitor discharges when voltage is decreasing
• Ripple (closer to DC current)

Question 18

Three Phase Power Supplies

Answer 18

• Most of the power generated in the world is what is called 3 phase
• Three-phase power is the most efficient way that electricity can be produced, transmitted, and consumed

How to produce 3 phase power?
o 3 coils at 120 degrees to each other in the turbine/generator (independent of each other)
o A current will be induced in each of them but with a phase difference of 120

Question 19

Rectifying 3 Phase Power

Answer 19

• Rectifying circuits can be used with the 3 phase power supply
• There are now 6 pulses per cycle and variation in amplitude is much less.

Question 20

Main Components Controlling the X-Ray Tube Output

Answer 20

Three parameters to control
o	Voltage (kV) – Maximum energy of x-rays produced by Bremsstrahlung target 

o Current (mA) – Number of x-rays produced (actually number of electrons generated by cathode)

o Time (s) – How long to leave electron source (cathode) on for.

Question 21

Filament Circuit

Answer 21

• The filament circuit controls the x-ray tube current.
• The resistors control the current in the primary side of the filament transformer, which in turn controls the current in the secondary side, which in turn controls the heating of the filament.
• The filament temperature controls the rate of thermionic emission and hence the tube current.
• Transformer is located in the high voltage section
  o It isolates the high voltage at the tube from the operators console
• Controls the number of electrons passing from the cathode to the anode to the filament
• Number of electrons emitted by the filament is determined by the temperature of the filament which depends on the current supplied to it
• As the current supplied to the filament increases it becomes hotter and more electrons are released by thermionic emission
• Filament current normally around 3 – 6 A
  o Very high!
• Connections from the autotransformer provide the voltage to the filament circuit.
• The voltage from the mA selector is then delivered to the filament transformer

Question 22

Filament Circuit: Ohms Law

Answer 22

Ohms’ Law (V = IR)
o We can adjust the voltage with precision resistors to provide the selected mA
o The x-ray tube current typically has preset values e.g. 100, 200, 300 mA

Question 23

Step Down Transformer

Answer 23

Step-down transformer
o Decrease in voltage supplied to the filament
 Proportional to the number of turns

o Increase in current supplied
 Inversely proportional to the number of turns

o IP / IS = NS / NP

Question 24

Thermionic Emission

Answer 24

Thermionic emission is the release of electrons from a heated filament

Question 25

kVp Adjustment

Answer 25

Minor kVp
- Fine tunes the voltage

Major kVp
- Coarse tuning of the voltage

• Voltage is set in the low voltage part of circuit (before stepped up to high voltage)

Question 26

High Voltage Section

Answer 26

• Select the voltage as precisely as possible at low voltage THEN increase it to high voltage using the step-up transformer

Question 27

High Voltage Step-Up Transformer

Answer 27

• The high voltage (or high tension) transformer is a step-up isolating transformer.

Ns /Np –> 500 to 1000
o e.g., Ns 1/50 s :Np = 500:1

Voltage induced in the secondary winding of a high-voltage step-up transformer is alternating like the primary voltage but has a higher value

Question 28

Types of Timers

Answer 28

Time selector
o The operator chooses a specific exposure time.

mAs timer
o Monitors the tube current & terminates the exposure when the desired quantity of charge (mAs = mC) has passed from the cathode to the anode.

Automatic Exposure Control (autotimer)
o Terminates the exposure once the required amount of radiation has been incident on the image receptor.

Question 29

Auto-Timer / Automatic Exposure Control (AEC)

Answer 29

The current for charging the timing capacitor comes from a radiation detector placed between the grid and the cassette. The radiation detector measures the radiation dose received by the image receptor.

The exposure is terminated when the correct amount of radiation has reached the image receptor. The auto-timer therefore will automatically compensate for: 
o	FFD 
o	mA 
o	kV 
o	Patient thickness, and 
o	Anti-scatter grids 

The detectors are radiolucent and can’t be observed in the image.

Question 30

Selection of Monitoring Area

Answer 30

• Usually, there are 3 sensitive areas (not all have to be used).
• Satisfactory results require careful patient and x-ray beam alignment with the monitoring area(s).

Question 31

Pulse Counting Timers

Answer 31

• Mains Frequency
• Digital logic circuits count the number of half cycles of the mains voltage waveform.
• Exposure times are multiples of 10 ms.
• The minimum exposure time is 10 ms.