G Practical MRI 1 – 26 th March 2015 G Practical MRI 1 Review of Circuits and Electronics

G Practical MRI 1 – 26 th March 2015 Current Current is the flow of electrical charge through an electronic circuit – The direction of a current is opposite to the direction of electron flow Current is measured in Amperes (amps) – 1 A = 1 C/s French physicist and mathematician 20 th January th June 1836 André-Marie Ampère

G Practical MRI 1 – 26 th March 2015 Voltage Voltage, or electric potential difference, is the electrical force that causes current to flow in a circuit Voltage is measured in Volts (V) – One volt is the difference in electric potential across a wire when an electric current of one ampere dissipates one watt of power: 1 V = 1 W/A Italian physicist, inventor of the battery 28 th February th March 1827 Alessandro Volta

G Practical MRI 1 – 26 th March 2015 Resistance The electric resistance is the opposition to the passage of an electric current through an element Resistance is measured in Ohms (Ω) – One ohm is the resistance between two points of a conductor when a constant potential difference of one volt produces a current of 1 ampere: 1 Ω = 1 V/A German physicist 26 th March th July 1854 Georg Simon Ohm

G Practical MRI 1 – 26 th March 2015 Ohm’s Law Defined the relationship between voltage, current and resistance in an electric circuit It states that the current in a resistor varies in direct proportion to the voltage applied and it is inversely proportional to the resistor’s value V IR

G Practical MRI 1 – 26 th March 2015 Kirchhoff’s Laws Kirchhoff’s voltage law (KVL) – The algebraic sum of the voltages around any closed path (electric circuit) equal to zero Kirchhoff’s current law (KCL) – The algebraic sum of the currents entering a node equal to zero German physicist 12 th March th October 1887 Gustav Kirchhoff

G Practical MRI 1 – 26 th March 2015 Kirchhoff’s Voltage Law (KVL) _ _ _ _ v1v1 v2v2 v4v4 v3v3 v 3 + v 4 – v 2 – v 1 = 0

G Practical MRI 1 – 26 th March 2015 Kirchhoff’s Current Law (KCL) i1i1 i2i2 i3i3 i 2 + i 3 – i 1 = 0

G Practical MRI 1 – 26 th March 2015 Problem Use Kirchhoff's Voltage Law to calculate the magnitude and polarity of the voltage across resistor R4 in this resistor network

G Practical MRI 1 – 26 th March 2015 Problem Use Kirchhoff's Voltage Law to calculate the magnitudes and directions of currents through all resistors in this circuit

G Practical MRI 1 – 26 th March 2015 Inductor The energy stored in magnetic fields has effects on voltage and current. We use the inductor component to model these effects An inductor is a passive element designed to store energy in the magnetic field

G Practical MRI 1 – 26 th March 2015 Physical Meaning When the current through an inductor is a constant, then the voltage across the inductor is zero, same as a short circuit No abrupt change of the current through an inductor is possible except an infinite voltage across the inductor is applied The inductor can be used to generate a high voltage, for example, used as an igniting element

G Practical MRI 1 – 26 th March 2015 Inductance The ability of an inductor to store energy in a magnetic field Inductance is measured in Henries (H) – If the rate of change of current in a circuit is one ampere per second and the resulting electromotive force is one volt, then the inductance of the circuit is one henry: 1 H = 1 V  s/A American scientist, first secretary of the Smithsonian Institution 17 th December th May 1878 Joseph Henry

G Practical MRI 1 – 26 th March 2015 How Inductors Are Made An inductor is made of a coil of conducting wire μ = μ r  μ 0 μ 0 = 4π × (H/m)

G Practical MRI 1 – 26 th March 2015 Energy Stored in an Inductor power Energy stored in an inductor

G Practical MRI 1 – 26 th March 2015 Capacitor The energy stored in electric fields has effects on voltage and current. We use the capacitor component to model these effects A capacitor is a passive element designed to store energy in the electric field

G Practical MRI 1 – 26 th March 2015 Physical Meaning A constant voltage across a capacitor creates no current through the capacitor, the capacitor in this case is the same as an open circuit If the voltage is abruptly changed, then the current will have an infinite value that is practically impossible. Hence, a capacitor is impossible to have an abrupt change in its voltage except an infinite current is applied

G Practical MRI 1 – 26 th March 2015 Capacitance The ability of a capacitor to store energy in an electric field Capacitance is measured in Farad (F) – A farad is the charge in coulombs which a capacitor will accept for the potential across it to change 1 volt. A coulomb is 1 ampere second: 1 F = 1 A  s/V British scientist, Chemist, physicist and philosopher 22 nd September th August 1867 Michael Faraday

G Practical MRI 1 – 26 th March 2015 How Capacitors Are Made A capacitor consists of two conducting plates separated by an insulator (or dielectric) ε = ε r  ε 0 ε 0 = × (F/m)

G Practical MRI 1 – 26 th March 2015 Energy Stored in a Capacitor power Energy stored in an inductor

G Practical MRI 1 – 26 th March 2015 Resonance in Electric Circuits Any passive electric circuit will resonate if it has an inductor and capacitor Resonance is characterized by the input voltage and current being in phase – The driving point impedance (or admittance) is completely real when this condition exists “RLC Circuit”

G Practical MRI 1 – 26 th March 2015 Series Resonance The input impedance is given by: The magnitude of the circuit current is:

G Practical MRI 1 – 26 th March 2015 Resonant Frequency Resonance occurs when the impedance is real: We define the Q (quality factor) of the circuit as: Q is the peak energy stored in the circuit divided by the average energy dissipated per cycle at resonance – Low Q circuits are damped and lossy – High Q circuits are underdamped “Resonant Frequency”

G Practical MRI 1 – 26 th March 2015 Parallel Resonance The relation between the current and the voltage is: Same equations as series resonance with the substitutions: – R  1/R, L  C, C  L:

G Practical MRI 1 – 26 th March 2015 Problem Determine the resonant frequency of the RLC circuit above

G Practical MRI 1 – 26 th March 2015 Transmission Lines Fundamental component of any RF system – Allow signal propagation and power transfer between scanner RF components All lines have a characteristic impedance (V/I) – RF design for MRI almost always use Z 0 = 50 Ω Input and output of transmission lines have a phase difference corresponding to the time it takes wave to go from one end to the other Length is usually given with respect to λ

G Practical MRI 1 – 26 th March 2015 Geometry

G Practical MRI 1 – 26 th March 2015 Transmission Line Reflections A wave generated by an RF source is traveling down a transmission line The termination impedance (Z load ) may be a resistor, RF coil, preamplifier or another transmission line In general there will be reflected and transmitted waves at the load Z0Z0

G Practical MRI 1 – 26 th March 2015 Circuit Model Small sections of the line can be approximated with a series inductor and a shunt capacitors The transmission line is approximated as a series of these basic elements

G Practical MRI 1 – 26 th March 2015 Reflection and Transmission Forward wave (forward power): Reflected wave (reflected power): Transmitted wave (transmitted power): reflection coefficient transmission coefficient S 11 S 21

G Practical MRI 1 – 26 th March 2015 High Γ in High Power Applications Decreases the power transfer to load consequently causing loss of expensive RF power Increases line loss: 3 dB power loss can increase to > 9 dB with a severely mismatched load Causes standing waves and increased voltage or current at specific locations along the transmission line

G Practical MRI 1 – 26 th March 2015 Useful Facts to Remember If Z load = Z 0 then there is no reflected wave If the length of the line is λ/4 (or odd multiples) – Short one end, open other end – Can be considered a resonant structure with high current at shorted end and high voltage at open end If the length of the line is λ/2 (or multiples) – Same impedance at both ends – With open at both ends this can also considered a resonant structure with high current at center and high voltage at ends

G Practical MRI 1 – 26 th March 2015 Impedance Transformation Given the importance of reflections, it is generally desirable to match a given device to the characteristic impedance of the cable Can use broadband or narrowband matching circuits – Most MRI systems operate over a limited bandwidth, so narrowband matching works fine for passive devices such as RF coils There are many circuits that can be used for impedance transformation

G Practical MRI 1 – 26 th March 2015 The Smith Chart A useful tool for analyzing transmission lines reflection coefficient at the load Input impedance of a line of length d, with Z 0 and Z load Smith Chart is the polar plot of Γ with circles of constant r and x overlaid reflection coefficient at distance d from the load On the Smith Chart we can convert Γ to Z (or the reverse) by graphic inspection

G Practical MRI 1 – 26 th March 2015 Smith Chart Interpretation Circles correspond to constant r the centers are always on the horizontal axis (i.e. real part of the reflection coefficient) Partial circles correspond to constant x The intersection of an r circle and an x circle specifies the normalized impedance The distance between such point and the center of chart is the reflection coefficient (real + imaginary). Any point on the line can be on the circle with such radius

G Practical MRI 1 – 26 th March 2015 Smith Chart Interpretation The termination is perfectly matched (i.e. reflection coefficient is zero) for a point at the center of the Smith Chart (i.e. r = 1, x = j0 and radius of the circle = 0) Question: which point correspond to the termination being an open circuit? Answer: The right most point on the x-axis, which corresponds to infinite z and Γ = 1. The left most point corresponds to z = 0 and reflection coefficient Γ = –1.

G Practical MRI 1 – 26 th March 2015 Smith Chart Example 1 Locate these normalized impedances on the simplified Smith Chart: z = 1 + j0 z = j0.5 z = 0 + j0 z = 0 - j1 z = 1 + j2 z = ∞

G Practical MRI 1 – 26 th March 2015 Smith Chart Example 2 Graphically find the admittance corresponding to the impedance: z = j0.5 In fact: y = 1/(0.5 + j0.5) = 1 – j1 1.Locate the impedance 2.Draw a circle centered at the center of the Smith Chart and passing through the impedance 3.Plot a straight line through the impedance and the center of the Smith Chart 4.The intersection of the line with the circle yields the value for the admittance

G Practical MRI 1 – 26 th March 2015 Any questions?

G Practical MRI 1 – 26 th March 2015 See you next lecture!