Electronics A
Joel Voldman
Massachusetts Institute of Technology
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 1
Outline
> Elements of circuit analysis
TODAY
> Elements of semiconductor physics
• Semiconductor diodes and resistors
• The MOSFET and the MOSFET inverter/amplifier
> Op-amps
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 2
Elements of circuit analysis
> There are many ways to analyze circuits
> Here we’ll go over a few of them
•
•
•
•
Elements laws, connection laws and KVL/KCL
Nodal analysis
Intuitive approaches
Superposition
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 3
Lumped elements in circuits
v∫
> Circuit elements (R, L,
closed
surface
C) are lumped
approximations of
complex devices
ε EA = Q
> The electrical capacitor
between Q and V?
⇒ E(r, t ) = −∇V (r, t )
b
a
V (b) − V (a) = V = Eg
A
+
V
-
∇×E = 0
V (b) − V (a ) = − ∫ E ⋅ dl
• What is the relation
I
ε E ⋅ da = Q
g
Q = ε AV
g
=
εA
g
εA
C=
g
⇒ E =V
g
V = CV
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 4
Lumped elements in circuits
> The electrical capacitor
• We can replace all of field theory with terminal relations
• And introduce an element with an element law
• As long as capacitor size << wavelength of electrical signal
» In general, MEMS are small
e.g., λ=50 μm Î 600 GHz
I
C
+
V
-
Q = CV
dQ d
I=
= (CV )
dt dt
dV
I =C
dt
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 5
Elements and element laws
> Do this with all three basic
I
elements
> Resistor
+
> Capacitor
> Inductor
I
+
I
R
V
V = RI
-
+
C
V
dV
I =C
dt
-
L
V
-
dI
V =L
dt
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 6
Source elements
> We need elements to provide energy into the circuit
> Two common ones are voltage source and current
source
I
+
+
V V0
V = V0
I
+
V I0
I = I0
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 7
KVL and KCL
> These are continuity laws that allow us to solve
circuits
> Kirchhoff’s voltage law
• The oriented sum of voltages around a loop is zero
> Kirchhoff’s voltage law
• The sum of currents entering a node is zero
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 8
Complex impedances
> For LTI systems, use
complex impedances
instead
• Implicitly working in
I
C
+
V
I
1
V = IZ C = I
Cs
-
L
frequency domain
> Much easier circuit
+
I
analysis
the same, like
“resistors”
R
+
+
> All elements treated
-
V
-
V
V
Z
V = IZ L = ILs
V = IZ R = IR
V (s) = I ( s) Z ( s)
i
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 9
Let’s analyze a circuit
1. Figure out what are
you trying to determine
+
V0
-
2. Replace elements with
3. Assign across and
4. Use KVL
5. Substitute in element
laws
6. Solve
L
R
complex impedances
through variables
C
iV
i C + VC -
+
VV V0+
-
ZC
ZL
ZR
- VR +
VL
+
iL
iR
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 10
Let’s analyze a circuit
1. Figure out what are
you trying to determine
2. Replace elements with
complex impedances
iV
i C + VC -
+
VV V0+
-
ZL
ZR
- VR +
3. Assign across and
through variables
ZC
VL
+
iL
iR
VV − VC + VL − VR = 0
4. Use KVL
V0 − iC Z C + iL Z L − iR Z R = 0
5. Substitute in element
iC = −iL = iR
laws
6. Solve
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 11
Let’s analyze a circuit
1. Figure out what are
iV
you trying to determine
+
VV V0+
-
2. Replace elements with
complex impedances
3. Assign across and
through variables
4. Use KVL
5. Substitute in element
laws
i C + VC ZC
ZL
ZR
- VR +
VL
+
iL
iR
V0 − iR Z C − iR Z L − iR Z R = 0
iR =
V0
=
ZC + Z L + Z R 1
V0
Cs
+ Ls + R
Cs
iR =
V0
2
LCs + RCs + 1
6. Solve
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 12
Example #1
> Solve for VL
+
V0
-
+
C
L
R
VL
-
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 13
Nodal analysis
> Element law approach becomes tedious for circuits
with multiple loops
> Nodal analysis is a KCL-based approach
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 14
Nodal analysis
1. Figure out what are you
+
VC
-
trying to determine
2. Replace elements with
complex impedances
V0
C
+
-
L
R
3. Assign node voltages &
ground node
4. Write KCL at each node
5. Solve for node voltages
6. Use node voltages to
v2
v1
V0
+
-
C
R
L
find what you care about
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 15
Nodal analysis
1. Figure out what are you
trying to determine
2. Replace elements with
complex impedances
i1
v1
V0
i2
i3
C
+
-
v2
L
R
3. Assign node voltages &
ground node
4. Write KCL at each node
5. Solve for node voltages
6. Use node voltages to
Node 1:
Node 2:
v1 = V0
i1 + i2 + i3 = 0
v1 − v2 0 − v2 0 − v2
+
+
=0
ZC
ZL
ZR
find what you care about
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 16
Nodal analysis
1. Figure out what are you
trying to determine
2. Replace elements with
complex impedances
i1
v1
V0
+
-
i2
v2
i3
C
R
L
3. Assign node voltages &
⎛ 1
1
1 ⎞ V0
v2 ⎜
+
+
⎟=
⎝ ZC Z L Z R ⎠ ZC
4. Write KCL at each node
v2 ( Z L Z R + Z C Z R + Z L Z C ) = V0 Z L Z R
ground node
5. Solve for node voltages
6. Use node voltages to
ZLZR
v2 = V0
Z L Z R + ZC Z R + Z L ZC
find what you care about
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 17
Nodal analysis
1. Figure out what are you
trying to determine
2. Replace elements with
complex impedances
i1
v1
V0
+
-
i2
v2
i3
C
L
R
3. Assign node voltages &
ground node
6.
LRs + 1
R + Ls 1
Cs
LRCs 2
Solve for node voltages v2 = V0 LRCs 2 + Ls + R
4. Write KCL at each node
5.
v2 = V0
LRs
Cs
2
LRCs
Use node voltages to
VC = v1 − v2 = V0 − V0
2
LRCs
+ Ls + R
find what you care about
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 18
Example #2
R1
+
-
+
V0
C
R2
L
VL
-
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 19
Intuitive methods
> Instead of “solving” the circuit using
equations, use series/parallel tricks to
analyze the circuit by inspection
> Current divider & impedances in
parallel
• Both elements have SAME voltage
• Terminals connected together
+
V
-
i
i1
i2
Z1
Z2
+
i
Z
V
-
Z2
i1 = i
Z1 + Z 2
Z1
i2 = i
Z1 + Z 2
Z1Z 2
V = i1Z1 = i
Z1 + Z 2
Z1Z 2
= Z1 // Z 2
Z=
Z1 + Z 2
1 1
1
= +
Z Z1 Z 2
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 20
Intuitive methods
> Voltage divider & impedances in
series
• Both elements have SAME current
i
+
V1
+
V2
-
Z2
V2 = V
Z1 + Z 2
+
Z1
Z2
-
V
V1
V
=
=i
i1 =
Z1 Z1 + Z 2
-
Z = Z1 + Z 2
+
i
V
Z1
V1 = V
Z1 + Z 2
Z
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 21
Intuitive methods
> Examples of elements NOT in series OR parallel
Z1
Z2
Z1
Z3
Z4
Z3
Z1 and Z3 in series
Z2 and Z4 in series
Z1 and Z2 NOT in parallel
Z3 and Z4 NOT in parallel
Z4
Z3 and Z4 in parallel
Z1 and Z3 NOT in series
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 22
Intuitive methods
iL
> Let’s use this approach to solve
a circuit
V0
+
-
C
L
R
1. Figure out what are you
trying to determine
+
2. Replace elements with
complex impedances
V0
+
-
C
Za
4. Re-expand to find signal of
interest
Z a = Z R // Z L
-
3. Collapse circuit in terms of
series/parallel relations till
circuit is trivial
Va
Za
Va = V0
Z a + ZC
Va
iL =
ZL
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 23
Intuitive methods
> Let’s use this approach
to solve a circuit
Za
Z R // Z L
1
1
iL = V0
= V0
Z a + ZC Z L
Z R // Z L + Z C Z L
1. Figure out what are you
trying to determine
2. Replace elements with
complex impedances
3. Collapse circuit in terms of
series/parallel relations till
circuit is trivial
4. Re-expand to find signal of
interest
ZRZL
ZR + ZL
1
= V0
ZRZL
+ ZC Z L
ZR + ZL
= V0
= V0
ZRZL
1
Z R Z L + ( Z R + Z L ) ZC Z L
RLs
RLs + ( R + Ls ) 1
Cs
1
Ls
RCs
iL = V0
RLCs 2 + Ls + R
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 24
Example #3
R1
+
-
+
V0
C
R2
VR
-
L
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 25
Superposition
> These equivalent circuits are linear
and obey the principles of
superposition
• This can be useful
> For circuits with multiple sources,
• Turn off all independent sources
•
•
except one
Solve circuit
Repeat for all sources, then add
responses
V0
+
-
short
> Turning off a voltage source gives
a short circuit
> Turning off a current source gives
I0
open
an open circuit
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 26
Superposition
v2
v1
> For circuits with multiple
sources,
• Turn off all independent
C
I0
R
sources except one
• Solve circuit
• Repeat for all sources, then add
v2 = I 0 Z R // Z C
responses
v2
v1
V0
+
-
V0
C
R
Find v2
I0
v2
v1
+
-
C
R
v2 = V0
ZR
Z R + ZC
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
JV: 2.372J/6.777J Spring 2007, Lecture 6E - 27
Superposition
> For circuits with multiple
sources,
• Turn off all independent
sources except one
ZR
v2 = I 0 Z R // Z C + V0
Z R + ZC
Z Z
ZR
• Solve circuit
v2 = I 0 R C + V0
• Repeat for all sources, then add
Z R + ZC
Z R + ZC
responses
v2
v1
V0
+
-
C
R
I0
v2 =
I0 R 1
Cs
R+ 1
+ V0 R
Cs
I 0 R + V0 RCs
v2 =
RCs + 1
Find v2
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
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JV: 2.372J/6.777J Spring 2007, Lecture 6E - 28
Conclusions
> There are many ways to analyze equivalent circuits
> Use the simplest method at hand
> Element laws & connection laws are OK for simple
ckts
> Nodal analysis works for most any circuit, but will be
tedious for complicated circuits
> Try to use intuitive approaches whenever possible
Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
JV: 2.372J/6.777J Spring 2007, Lecture 6E - 29