Transcript Lecture 10

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
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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
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Kirchhoff’s Voltage Law (KVL)
_ v2
+
+
v1
+
_
_
•
_
v3
+
v3 + v 4 – v2 – v1 = 0
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v4
Kirchhoff’s Current Law (KCL)
i2
i3
i1
i2 + i3 – i1 = 0
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Problem
Use Kirchhoff's Voltage Law to calculate the magnitude
and polarity of the voltage across resistor R4 in this
resistor network
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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
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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
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 = μrμ0
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
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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
Michael Faraday
22nd September 1791 - 25th August 1867
British scientist,
Chemist, physicist and
philosopher
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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)

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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 )
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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
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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
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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
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Problem
R
I
V
L
C
Determine the resonant frequency of the RLC circuit above
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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
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Geometry
er
er
Z0 = 276 ×
(
log D a
er
)
Z0 = 138 ×
( )
log b a
er
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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
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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
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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
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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
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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.
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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=∞
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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
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Any questions?
G16.4427 Practical MRI 1 – 26th March 2015
See you next lecture!
G16.4427 Practical MRI 1 – 26th March 2015