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G16.4427 Practical MRI 1 Review of Circuits and Electronics G16.4427 Practical MRI 1 – 26th 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 André-Marie Ampère 20th January 1775 - 10th June 1836 French physicist and mathematician G16.4427 Practical MRI 1 – 26th 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 Alessandro Volta 28th February 1745 - 5th March 1827 Italian physicist, inventor of the battery G16.4427 Practical MRI 1 – 26th 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 Georg Simon Ohm 26th March 1789 - 6th July 1854 German physicist G16.4427 Practical MRI 1 – 26th 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 I V = R× I V I R V I= R R V R= I G16.4427 Practical MRI 1 – 26th 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 Gustav Kirchhoff 12th March 1824 - 17th October 1887 German physicist G16.4427 Practical MRI 1 – 26th March 2015 Kirchhoff’s Voltage Law (KVL) _ v2 + + v1 + _ _ • _ v3 + v3 + v 4 – v2 – v1 = 0 G16.4427 Practical MRI 1 – 26th March 2015 v4 Kirchhoff’s Current Law (KCL) i2 i3 i1 i2 + i3 – i1 = 0 G16.4427 Practical MRI 1 – 26th March 2015 Problem Use Kirchhoff's Voltage Law to calculate the magnitude and polarity of the voltage across resistor R4 in this resistor network G16.4427 Practical MRI 1 – 26th March 2015 Problem Use Kirchhoff's Voltage Law to calculate the magnitudes and directions of currents through all resistors in this circuit G16.4427 Practical MRI 1 – 26th 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 I L V dI V=L dt t 1 I(t) = ò V (s) ds + I(t0 ) Lt 0 G16.4427 Practical MRI 1 – 26th 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 G16.4427 Practical MRI 1 – 26th 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 Vs/A Joseph Henry 17th December 1797 - 13th May 1878 American scientist, first secretary of the Smithsonian Institution G16.4427 Practical MRI 1 – 26th March 2015 How Inductors Are Made • An inductor is made of a coil of conducting wire N 2m A L= l μ = μrμ0 μ0 = 4π × 10-7 (H/m) G16.4427 Practical MRI 1 – 26th March 2015 Energy Stored in an Inductor æ dI ö P =V ×I = ç L ÷ ×I è dt ø t power t æ dI(s) ö W (t) = ò P(s) ds = ò ç L × × I(s) ds ÷ ds ø è -¥ -¥ I (t ) 1 1 2 = L ò I dI = L × I(t) - L × I(-¥)2 2 2 I (-¥) I(-¥) = 0 1 W (t) = L × I(t)2 2 G16.4427 Practical MRI 1 – 26th March 2015 Energy stored in an inductor 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 I C V dV I =C dt t 1 V (t) = ò I(s) ds + V (t0 ) Ct 0 G16.4427 Practical MRI 1 – 26th 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 G16.4427 Practical MRI 1 – 26th 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 As/V Michael Faraday 22nd September 1791 - 25th August 1867 British scientist, Chemist, physicist and philosopher G16.4427 Practical MRI 1 – 26th March 2015 How Capacitors Are Made • A capacitor consists of two conducting plates separated by an insulator (or dielectric) C= eA d ε = εr ε0 ε0 = 8.854 × 10-12 (F/m) G16.4427 Practical MRI 1 – 26th March 2015 Energy Stored in a Capacitor æ dV ö P = V × I = V ×ç C è dt ÷ø t power t æ dV (s) ö W (t) = ò P(s) ds = ò V (s) × ç C × ds ÷ ds ø è -¥ -¥ V (t ) 1 1 2 = C ò V dV = C ×V (t) - C ×V (-¥)2 2 2 V (-¥) V (-¥) = 0 Energy stored in an inductor 1 W (t) = C ×V (t)2 2 q(t)2 W (t) = 2C (q = C ×V ) G16.4427 Practical MRI 1 – 26th 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 R V I “RLC Circuit” L C G16.4427 Practical MRI 1 – 26th March 2015 Series Resonance I R V L C • The input impedance is given by: æ 1 ö Z = R + jçw L w C ÷ø è • The magnitude of the circuit current is: I= I = V æ 1 ö R2 + ç w L w C ÷ø è 2 G16.4427 Practical MRI 1 – 26th March 2015 Resonant Frequency • Resonance occurs when the impedance is real: wL = 1 wC w0 = 1 LC “Resonant Frequency” • We define the Q (quality factor) of the circuit as: Q= w0 L R = 1 1 L = w 0 RC R C • 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 G16.4427 Practical MRI 1 – 26th March 2015 Parallel Resonance V I R L C • The relation between the current and the voltage is: æ1 1 ö I = V ç + jw C + w L ÷ø èR • Same equations as series resonance with the substitutions: – R 1/R, L C, C L: w0 = 1 LC Q = w 0 RC = R C L G16.4427 Practical MRI 1 – 26th March 2015 Problem R I V L C Determine the resonant frequency of the RLC circuit above G16.4427 Practical MRI 1 – 26th 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 Z0 = 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 G16.4427 Practical MRI 1 – 26th March 2015 Geometry er er Z0 = 276 × ( log D a er ) Z0 = 138 × ( ) log b a er G16.4427 Practical MRI 1 – 26th March 2015 Transmission Line Reflections Z0 • A wave generated by an RF source is traveling down a transmission line • The termination impedance (Zload) may be a resistor, RF coil, preamplifier or another transmission line • In general there will be reflected and transmitted waves at the load G16.4427 Practical MRI 1 – 26th March 2015 Circuit Model I L V C Z load • 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 G16.4427 Practical MRI 1 – 26th March 2015 Reflection and Transmission Transmitted wave Reflected wave (reflected power): (transmitted power): Forward wave (forward power): Z ( P× (Z Pi - Z0 load 0 i Z ( G= (Z T= (Z ) +Z ) load - Z0 load 0 2Z load load + Z0 ) ) +Z ) load Pi × (Z 2Z load load + Z0 ) reflection coefficient S11 transmission coefficient S21 G16.4427 Practical MRI 1 – 26th 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 G16.4427 Practical MRI 1 – 26th March 2015 Useful Facts to Remember • If Zload = Z0 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 G16.4427 Practical MRI 1 – 26th 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 G16.4427 Practical MRI 1 – 26th March 2015 The Smith Chart • A useful tool for analyzing transmission lines Z ( G= (Z ) +Z ) load - Z0 load 0 G(d) = Ge -2 jb d æ Z load ö ç Z - 1÷ zload - 1 è 0 ø = = æ Z load ö zload + 1 + 1 ç Z ÷ è 0 ø ( ( z(d) - 1) ( == ( z(d) + 1) ) ) reflection coefficient at the load z = reflection coefficient at distance d from the load R X + j = r + jx Z0 Z0 b = w LC Smith Chart is the polar plot of Γ with circles of constant r and x overlaid 1+ G(d) z(d) = 1- G(d) Zload + jZ0 tan( b d) Zin = Z0 Z0 + jZload tan( b d) Input impedance of a line of length d, with Z0 and Zload On the Smith Chart we can convert Γ to Z (or the reverse) by graphic inspection G16.4427 Practical MRI 1 – 26th 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 G16.4427 Practical MRI 1 – 26th 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. G16.4427 Practical MRI 1 – 26th March 2015 Smith Chart Example 1 Locate these normalized impedances on the simplified Smith Chart: • z = 1 + j0 • z = 0.5 - j0.5 • z = 0 + j0 • z = 0 - j1 • z = 1 + j2 •z=∞ G16.4427 Practical MRI 1 – 26th March 2015 Smith Chart Example 2 Graphically find the admittance corresponding to the impedance: • z = 0.5 + j0.5 1. 2. 3. 4. Locate the impedance Draw a circle centered at the center of the Smith Chart and passing through the impedance Plot a straight line through the impedance and the center of the Smith Chart The intersection of the line with the circle yields the value for the admittance In fact: y = 1/(0.5 + j0.5) = 1 – j1 G16.4427 Practical MRI 1 – 26th March 2015 Any questions? G16.4427 Practical MRI 1 – 26th March 2015 See you next lecture! G16.4427 Practical MRI 1 – 26th March 2015