6.002 Circuits and Systems
Final Review
Jason Kim
Page -1
6.002 Circuits and Systems: Outline of Topics
•
•
•
Resistor Networks
concepts:
1st Order Circuits
concepts:
2nd Order Circuits
concepts:
Node Method, KCL, KVL, Superposition, Thevenin and Norton equivalent circuit
models, Multiport Networks, Transformers, Power.
Consituitive Laws, RC netwroks, RL networks, Homogenous and Particular solution,
Time constant τ , response to an impulse, power, Low/High/Band/Notch pass filters,
sinusodial steady-state, transients, Impedance Model, Bode Plots(Magnitude and Phase).
LC networks, LRC networks, damping coefficient α , natural frequency ω o ,
Underdamped/Critical/Overdamped Systems, Resonance, Q factor, Half-power point,
Sinusodial steady-state, transients, Power(real/reactive), Impedance Model, ELI ICE,
Bode Plots(Magnitude and Phase).
•
Digital Abstraction
concepts:
Boolean Logic(truth table, formula, gates, and transistor level), primitive laws,
DeMorgan’s Law(formula and gate equivalent), MSP(minumum sum of products),
MPS(minimum product of sums).
•
MOSFET Transistors
concepts:
Large Signal Model(S, SR, SCS, SVR), Small Signal Model, Saturation/Linear Region,
Ron, Loadline, Operating Point, Input/Output Resistence, Current Gain, Power Gain,
Noise Margin.
•
Op-Amp Circuits
concepts:
•
Single/Multiple input op-amp circuits, differentiator, integrator, adder, subtractor, opamp with R,L,C, negative feedback, positive feedback, Bode Plots(Magnitude and
Phase), Input/Output resistence, Schmitt Trigger, Hysteresis, Cascaded stages.
Diodes
concepts:
i-v characteristics, model, diode w/ R, L, C, and op-amp, Peak detection, Clipper circuits,
Incremental Analysis.
Page 0
1. Primitive Elements
Resistor:
time domain: V = IR
freq. domain: Z = R
Series: R 1 + R 2
1 2
Parallel: R 1 || R 2 = ------------------
R R
R1 + R2
R1
Vi
+
-
Ii
Vo
R2
R2
V o = -------------------- V i
R1 + R2
Voltage Divider
I2
Vi
+
-
R1
R2
R1
I 2 = -------------------- I i
R1 + R2
Current Divider
• Resistor Network
N-unknowns & N-equations (most primitive approach)
Node Analysis (KCL, KVL)
*KCL: Sum of all currents entering and leaving a node is zero.
*KVL: Sum of all voltages around a closed loop path is zero. (be careful with polarities)
1) LABEL all current directions and voltage polarities. (Remember, currents flow from + to -)
2) write KCL & KVL Equation.
3) substitute and solve.
Simplification (by inspection)
• Thevenin Equivalent Circuit Model
Norton Equivalent Circuit Model
Rth
Vth
+
Voc=Vth
-
+
-
same i-v characteristics at the ports
IN
RN
+
Isc=IN
-
* Three variables: Vth=Voc, Rth=RN, and IN=Isc . They are related by Voc=Isc*Rth .
* Vth=Voc: Leave the port open(thus no current flow at the port) and solve for Voc.
For a resistive network, this gives you a point on the V-axis of the i-v plot.
* IN=Isc: Short the port(thus no voltage across the port) and solve for Isc flowing out of + and
into -. For a resistive network, this gives you a point on the I-axis of the i-v plot.
* Rth: Set all sources to zero, except dependent sources. Solve the resistive network.
When setting sources to zero, V source becomes short and I source becomes open.
Rth may also be found by attaching Itest and Vtest and setting Rth = (Vtest)/(Itest).
If dependent sources are present, set only the independent sources to zero and attach Itest
and Vtest. Use KCL and KVL to find the expression (Vtest)/(Itest)=Rth.
Itest
Original
Circuit
N
+
-
Vtest
note the direction of Itest
and polarity of Vtest.
V test
R th = ----------I test
* For both Thevenin & Norton E.C.M.’s, they have the same power consumption at the port as
the original circuit, but not for its individual components. Power dissipated at Rth does not
equal power dissipated at the resistors of the original circuit. Same holds for the power
delivered by the sources in the Thevenin model and the original circuit. Thevenin and Norton
E.C.M.’s are for terminal i-v characteristics only.
• Power = Real Power = IV = I2R = V2/R (energy dissipated as heat)
T
1
• ⟨ Power⟩ = --- ∫ I ( t )V ( t ) dt
T
0
Page 1
• Superposition
"In a linear network with a number of independent sources, the response can be found by
summing the response to each independent sources acting alone, with all other independent
sources set to zero."
VR when only V1 is on.
+
Linear Network
with n sources
VR = VR
VR
R
_
1
V2 ∼ n = 0
+ VR
2
V 1 ,V 3 ∼ n = 0
… + VR
n
V1 ∼ n – 1 = 0
1) Leave one source on and turn off all other sources.
(Voltage source "off" = short & Current source "off" = open)
2) Find the effect from the "on" source.
3) Repeat for each sources.
4) Sum the effect from each sources to obtain the total effect.
For cases where a linear dependent source is present along with multiple independent sources,
DO NOT turn off the dependent source. Leave the dependent source on and carry it in your
expressions. Tackle the dependent source term last by solving linear equations. Remember that
the variable which the dependent source is depended on is affected by the individual independent
sources that you are turning on and off.
• Multiport Network
i1
V1
+
-
+
+
V
- 2
M
M
R11
i1
i2
R12i2
V1
R22
+
-
+
-
i2
+
R21i1
V2
-
-
z-parameter model
* By attaching V1, I1 and V2,I2 at the ports,
V 1 = R 11 I 1 + R 12 I 2
V 2 = R 21 I 1 + R 22 I 2
The first subscript denote the place(port) of voltage measurement.
The second subscript denote the place(port) of current source.
Thus, R12 means the resistance V1/I2 when voltage V1 is measured at open port 1 and I2
current source is placed at port 2.
* If all the Z-parameters(Rxx’s) are identical, then the networks can’t be differentiated using only
the i-v characteristics at the ports.
* Reciprocity: R12=R21.
* Using the definitions from above,
R11-R12
i1
M
R22-R21
+
V1
i2
+
+
V2
R12=R21
-
-
M
i1
G11+G12
V1
-G12=-G21
-
Page 2
+
G22+G21
V
-
π
T model
i2
model
• Transformer
i1
N1:N2
i2
+
+
V2
V1
V
V
------1- = ------2N1
N2
N
N2
& N 1 i 1 = – N 2 i 2 . Turns Ratio = ------1-
_
_
Reflected Resistance: the equiv. resistance seen at port 1, of the resistance on the other side, port2.
i1
N1:N2
i2
+
+
V1
+
R
V2
2
N1
R
N2
V1
_
_
Reflected Resistance
i1
_
Transformer vs. Voltage Divider : Voltage divider uses a series of resistors to divide and obtain a
scaled-down voltage. In doing so, power is dissipated not only at the load but also in the series
resistors as well, resulting in unwanted power dissipation. On the other hand, a transformer
converts its voltage through magnetic coupling and power is dissipated only in the load. Thus, no
unwanted power dissipation. (assuming ideal transformers)
Capacitor:
time domain: I = C dV
dt
1
Cs
1
jwC
freq. domain: Z = ------ = -----------
High Frequency = Short & Low Frequency = Open
C C
C1 + C2
1 2
Series: ------------------
Parallel: C 1 + C 2
Vc(0-)=Vc(0+) except when the input source is an impulse.
Vc(t) is continuous while Ic(t) may be discontinuous.
"I C E" : Current(I) LEADS Voltage(EMF) by 90o.
Energy Conservation: ∫ I dt = Q = CV .
1
2
Energy Stored: E = --- CV . (no energy dissipation)
2
R
Note: Two identical capacitors, C1 and C2, in series may act
like an open circuit after a long time where the total
voltage across the capacitors are split between the two
cap’s. For this case, Vc does not discharge to zero but
VC1=VC2.
Inductor:
time domain: V = L
dI
dt
freq. domain: Z = Ls = jwL
High Frequency = Open & Low Frequency = Short
Series: L 1 + L 2
L L
L1 + L2
1 2
Parallel: -----------------
IL(0-)=IL(0+) except when the input source is an impulse.
IL(t) is continuous while VL(t) may be discontinuous.
Page 3
+
VC1
C1
C2 +
_
_
VC1(0-) = VC2(0-)
VC2
"E L I" : Voltage(EMF) LEADS Current(I) by 90o.
Energy Conservation: ∫ V dt = λ = LI .
1 2
Energy Stored: E = --- LI . (no energy dissipation)
2
Note: Two identical inductors, L1 and L2, in parallel may act
like a short circuit after a long time and have current
flow through this loop indefinitely. For this case, iL is
not zero but iL1=-iL2.
L1
i1
L2
R
i2
iL1(0-) = iL2(0-)
LC
Series@ ω o = Short
Parallel@ ω o = Open
&
Energy transfers between inductors and capacitors sinusodially (180o out of phase).
Also, Vc peaks when IL=0 and IL peaks when Vc=0.
2. First Order Circuits
RC Network
RL Network
R
iL
R
+
+
-
Vi
C
+
Vi
Vc=Vo
-
+
-
L
Vo
-
V
di L i L
+ --------- = -----i
d t  L
L
-- R
dV c V c
V
+ -------- = -------idt
RC
RC
Time Constant: τ = --L-
Time Constant: τ = RC
R
*one time constant τ charges/discharges 63% of the maximum value. Small time constant shows the
effect of the exponential curve sooner, while big time constant shows a linear line for a long time
before the exponential effect becomes visible.
Homogeneous:
V_i
R
Vi
Vi
iC
VL
t
t
t
t
iL
VC
V
- _i
R
Particular:
Vi
VC
V_i
R
iC
t
iL
Vi
V_i
R
VL
t
Page 4
t
t
Total:
V_i
R
Vi
VC
V_i
R
iL
Vi
iC
VL
t
t
t
t
* Notice the Duality between Vc and IL & Ic and VL.
Approach to First Order Circuits
1) Set up differential equations, using KCL and KVL. Use the constitutive laws for C and L.
dV c V c
V
+ -------- = -------idt
RC
RC
2) Find Homogeneous solution by setting input to zero and substituting Vh=Aest as a solution.
1
st
st
Ase + -------- Ae = 0
RC
1
∴s = – -------RC
3) Find Particular solution. Remember the output follows the form of the input.
V p = Vi ;t > 0
4) Combine Homogeneous and Particular solutions, and use initial conditions to find missing
variables. (Be careful when input is an impulse!)
V o = V h + V p = Ae
–t
-------RC
+ Vi
V o = A + V i = 0 at t=0+. ∴ A = – V i
–t
--------

RC
V o = V i1 – e 


at t=0+, Vo=0 and t= ∞ , Vo=Vi.
*After a long time, transients(homogeneous solution) die away, leaving only the particular
solution. Thus, the output may look different from the input in the beginning but slowly
catches up to follow the input.
Impulse as Input
ex. RC circuit: V i = Qδt
and V o ( 0
0
1
Q
dV o + -------- V o dt = -------- δt dt ➔
RC
RC
V o(0
plus
Q
) – 0 = -------- ➔
RC
plus
0
minus
∴V o ( 0
) = 0
1
dV o + -------RC
∫
0
minus
plus
plus
∫
0
dV o V o
V
+ -------- = -------i- .
RC
RC
dt
and
0
Q
V o dt = -------RC
minus
plus
∫
0
δt dt ➔ V o ( 0
plus
) – V o(0
minus
Q
) = -------RC
3. Second Order Circuits
LRC Network
#1
ii
R
iL
L
2
+
C Vc=Vo
-
∂ Vo
∂t
Page 5
2
1 ∂i i
1 ∂V o V o
+ -------+ ------- = ---C ∂t
RC ∂ t
LC
minus
Q
) = -------RC
Vo
jωLR
------ = ----------------------------------------------------2
ii
( jω ) LCR + jωL + R
1
LC
Time Constant: τ = ------------
H ( jω )
∠H ( jω )
R
"Band Pass Filter"
+90o
w
1
wC
wL
+1
-1
-90o
w
1
ω o = -----------LC
1
ω o = -----------LC
L
#2
Vi
iL
+
-
+
C Vc=Vo
-
R
Vo
R
------ = ----------------------------------------------------2
Vi
( jω ) LRC + jωL + R
2
∂ Vo
∂t
2
V
1 ∂V o V o
+ -------+ ------- = -------i
RC ∂ t
LC
LC
1
Time Constant: τ = ----------LC
H ( jω )
∠H ( jω )
"Low Pass Filter"
0o
w
-2
-180o
w
1
ω o = -----------LC
1
ω o = -----------LC
L
#3
Vi
R
+
C Vc=Vo
-
iL
+
-
i
jωC
----L- = ----------------------------------------------------2
Vi
( jω ) LC + jωRC + 1
2
∂ iL
∂t
2
1 dV i
R ∂i L i L
+ --+ ------- = --L dt
L ∂t
LC
1
Time Constant: τ = ----------LC
H ( jω )
∠H ( jω )
1
R
"Band Pass Filter"
+90o
w
wC
+1
-1
1
wL
-90o
w
1
ω o = -----------LC
1
ω o = -----------LC
Page 6
ω
2α
• "peakiness" Q = ------o-
Half-Power Point = ω o ± α
&
α
α
Vpeak
Vpeak
2
2
w
ωo
ωo – α
V peak 2 1
V peak
Power peak
- = -------------------------Power =  -------------- --- = ------------ 2  R
2R
2
ωo + α
Hence, Half-power Point.
Characteristic Equation (Homogeneous Solution)
Obtain a characteristic equation by setting input to zero and replacing d with s to get into the form:
dt
2
2
ωo
s + 2αs +
= 0
where ω o is the "resonant" or "natural" frequency and α is the damping coefficient.
solving s with quadratic formula,
2
2
2
2
s 1, 2 = – α ± α – ω o = – α ± j ω o – α = – α ± jω d ; ω d is the damped natural frequency
1) "Overdamped"
α > ωo
s1 ≠ s2
s 1, 2 are real
2) "Critically Damped"
α = ωo
s1 = s2
s 1, 2 are real
3) "Underdamped"
α < ωo
s1 ≠ s2
s 1, 2 are complex
Approach (Transients)
1) Set up 2nd order differential equations.
2
∂ iL
∂t
2
1 dV i
R ∂i L i L
+ --+ ------- = --(ex. circuit #3)
L dt
L ∂t
LC
2) Obtain characteristic equation and identify ω o and α .
1 2
R
2
s + 2  ------ s +  ------------ = 0 ;
 2L
 LC
1
R
∴ω o = ------------ and ∴α = -----2L
LC
3) Find Homogeneous solution.(for the appropriate system type)
2
Overdamped:
V h = Ae
Critically Damped:
V h = Ae
Underdamped:
V h = Ae
2
– ( α – α – ω o )t
– αt
– αt
+ Bte
2
+ Be
2
– ( α + α – ω o )t
– αt
cos ω d t + Be
– αt
sin ω d t
4) Find Particular solution. Remember the output follows the form of the input.
V p = V i (usually a step)
5) Combine Homogeneous and Particular solutions, and use initial conditions( v ( 0 ), i ( 0 ) ) to find
missing variables. (Be careful when input is an impulse!)
V o = V p + V h = V i + Ae
V o(0) = V i + A = 0
– αt
cos ω d t + Be
– αt
sin ω d t
∴A = –V i
Thus, V o = V i ( 1 – e
– αt
cos ω d t )
i L = Ce
– αt
( – Aω d sin ω d t + Bω d cos ω d t )
i L ( 0 ) = CBω d = 0
& i L = V i Cω d e
Page 7
– αt
sin ω d t
∴B = 0
Approach (Sinusodial Steady-State)
1) Replace the primitive elements with their impedance models and solve the network like a resistor
network.
ii
R
iL
L
+
C Vc=Vo
-
jωL
1
V˜o = ----------------------------------- i˜i = ------------------------------------------ i˜i
2
L
1
1
 ----1
–
ω
LC + jω --- + sC + --R
 sL
R
V o = V cos ( ωt + φ )
ii = I cos(wt)
2) Put the equation into the form of a real amplitude and a complex phase.
ωL
V˜o = ------------------------------------------------------------ I e
2
2
L 2
( 1 – ω LC ) +  ω ---
 R
  ω --L- 
  R 
j 0 – atan  -----------------------------
 ( 1 – ω 2 LC )


imaginary part
real part
3) Match the coefficient V and phase shift φ
ωL
V = ------------------------------------------------------------ I
2
2
L 2
( 1 – ω LC ) +  ω ---
 R
and
  ω --L- 
  R 
φ = – atan  -----------------------------
 ( 1 – ω 2 LC )


1
*notice that at the resonant frequency ω o = ------------ , the magnitude of the output is just R, and the LC
LC
circuit just behaves as if nothing is there (=open circuit). See circuit #1.
If the input was a sine wave instead of a cosine wave, then we would add an additional phase shift of
π
90 degrees, since sin( ω t)=cos( ω t- --- ). Thus, our new phase shift would be φ' = π--- – φ
2
2
4. Bode Plots
Poles & Zeros
obtain a transfer function of the system by replacing d/dt with s, which is also equal to j ω ,
( 10 + 100sτ )
10 + j ( 100ωRC )
H ( jω ) = ------------------------------- = -----------------------------------------sτ ( 10 + sτ )
jωτ ( 10 + jωRC )
roots of the numerator are "zeros"
roots of the denominator are "poles"
Break Point Frequency: ω BP
Poles and Zeros are break point frequencies. Equate real and imaginary parts for each term in the
numerator and denominator (i.e., factor and find the roots) to get the break point frequencies:
numerator:
denominator:
1
ω z1 = -------------10RC
10
ω p1 = 0 and ω p2 = -------RC
Page 8
Magnitude Plot
short cut: "zeros" give you a +1 slope. (constant before ω z , +1 slope after ω z ; in log scale)
"poles" give you a -1 slope. (constant before ω p , -1 slope after ω p ; in log scale)
Thus,
Z1:
P1:
P2:
0
0
-1
+1
0
-1
ω p1
∑:
ω z1
+1
-1
-1
ω p2
-1
0
-1
Since ω p1 = 0 , which is the left most end of the frequency axis (in log scale), the magnitude plot
starts off at -1 slope.
H ( jω )
actual
asymptote
-1 slope
*both axes are in
log scale.
difference = 3dB
100
constant
10
-1 slope
1
ω z1
endless
ω p2
1
-------------10RC
1
-------RC
10
-------RC
ω
100
--------RC
Approach to Plotting Magnitude
1) Find all ZEROs and POLEs. (i.e.,
ω BP ’s)
2) Draw and Label Axes. Denote break point frequencies.
3) Beginning at the far end of the ω -axis, plot for ω very small
if ZERO at ω =0, start with +1 slope, (+2 slope if double Zero@ ω =0, and +3 for triple, ...)
POLE at ω =0, start with -1 slope,
(-2 slope if double Pole@ ω =0, and -3 for triple, ...)
otherwise, start with a constant. (label!)
4) As you come up on the frequency axis, for each POLE you hit, go down by -1 slope, for each
ZERO you hit, go up by +1 slope.
5) At
ω BP , your actual curve will be off from the sharp asymptote corner by 3dB.
Phase Plot
π
2
short cut: "zeros" give you a + --- phase over ± decade.
π
2
"poles" give you a - --- phase over ± decade.
∠H ( jω )
example,
*note that the asymptote is back to a constant
unlike the magnitude case, where the slope continued
π
---
2
π
--4
0
starts a decade before
ωz
------
ends a decade after
ωz
10 ω z
10
Page 9
ω
Thus,
Z1:
0
0
+ π---
+ π---
+ π---
+ π---
P1:
- π--2
- π--2
- π--2
- π--2
- π--2
- π--2
P2:
0
0
0
0
- π---
- π---
ω p1
- π---
∑:
2
4
1
----------------100RC
2
ω z1
- π--2
1
-------RC
- π---
0
ω z1
1
-------RC
4
2
2
4
2
ω p2
100
--------RC
- π---
- π---
4
2
∠H ( jω )
1
----------------100RC
100
--------RC
ω p2
actual
ω
0
– π--4-
asymptote
difference = 6o
– --π2
Approach to Plotting Phase
1) Find all ZEROs and POLEs. (i.e.,
ω BP ’s)
2) Draw and label all axes and break point frequencies, including their +/- decades.
3) Beginning at the far left of the ω -axis, plot for ω very small
------ for triple, ...)
ZERO at ω =0, start with + π--- phase, (+ π slope if double Zero@ ω =0, and + 3π
if
POLE at ω =0, start with
2
2
- π--2
3π
- ----2
phase, (- π slope if double Zero@ ω =0, and
for triple, ...)
otherwise, start at 0 phase.
4) As you come up the ω -axis, for each POLE you hit, make a change of - π--- phase over a ± decade.
2
ωp
( ------10
π
2
∼ ω p ∼ 10 ω p ). For each ZERO, make a change of + --- phase over a ± decade.
5) Be careful when the regions overlap, the effect from each pole and zero may cancel each other out.
6) At
ω BP , the actual curve is off by 6o.
Building Blocks
H ( jω )
1
--S
S
+1
+1
+1
1
-----------------S + 100
S + 100
+1
H ( jω )
100
100
∠H ( jω )
π
--2
π
--4
π
--2
– π--2-
10 100 1000
0
– π--4-
0
10 100 1000
Page 10
– π--2-
5. Digital Abstraction
Boolean Logic
*become comfortable expressing in different forms: truth table ↔ formula ↔ gates ↔ transistor level.
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
OUT
1
0
1
0
1
0
0
0
pull-up resistor
OUT = ( A ⋅ B ) + C
↔
↔
A
B
C
OUT
↔
A
OUT
A
C
C
C
B
OUT
B
A
B
*Note that in the transistor-level implementation, transistors in series form an AND expression and those
in parallel form an OR expression. The final output is always inverted in the pull-up implementation,
equivalent to adding a NOT(inverter) to the output.
Primitive Rules
A + (B + C) = (A + B) + C
A+B=B+A
A + (B ⋅ C) = (A + B) ⋅ (A + C)
A+1=1
A+0=A
A+A=1
A+A=A
A + (A ⋅ B) = A
A + (A ⋅ B) = A + B
A
A
A
A
A
A
A
A
A
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(B ⋅ C) = (A ⋅ B) ⋅ C
(associative)
B=B ⋅ A
(communative)
(B + C) = (A ⋅ B) + (A ⋅ C) (distributive)
0 =0
1=A
A=0
(complement)
A=A
(idempotence)
(A + B) = A
(absorption)
(A + B) = A ⋅ B
(absorption)
DeMorgan’s Laws
1) A ⋅ B = A + B
Gate Equivalent:
2) A + B = A ⋅ B
Gate Equivalent:
A
B
A
B
C
=
A
C
=
A
C
B
C
B
ex) (A ⋅ B) + (B ⋅ C)
ex) (A+B) ⋅ (B+C)
Minimum Sum of Products(MSP): (1)first level AND’s, (2) combine with OR’s.
Minimum Product of Sums(MPS): (1)first level OR’s, (2) combine with AND’s.
Common Gates
A
B
C
AND
A
0
0
1
1
B
0
1
0
1
A
B
C
NAND
C
0
0
0
1
A
0
0
1
1
B
0
1
0
1
C
1
1
1
0
A
B
C
A
B
OR
A
0
0
1
1
B
0
1
0
1
C
A
0
0
1
1
Page 11
B
0
1
0
1
C
XOR
NOR
C
0
1
1
1
A
B
C
1
0
0
0
A
0
0
1
1
B
0
1
0
1
A
B
C
XNOR
C
0
1
1
0
A
0
0
1
1
B
0
1
0
1
C
1
0
0
1
6. MOSFET Transistor (n-channel)
Large Signal Model
S Model:
SR Model:
D
D
Off
G
D
On
G
Ron
S
VGS ≥ VT
VGS < VT
When (VDS ≥ VGS - VT),
When (VDS < VGS - VT),
SCS Model (Saturation):
SVR Model (Linear):
D
D
iDS=
Off
On
G
S
VGS ≥ VT
VGS < VT
G
Off
G
S
S
D
K(VGS-VT)2
2
On
G
D
D
Off
G
On
G
Ron=
S
S
S
VGS ≥ VT
iDS
S
VGS ≥ VT
VGS < VT
VDS = VGS - VT
VGS < VT
Load Line
iDS
VS
RL
VGS5
Operating Point
-1
RL
VGS5
Lin
ear
Reg
ion
Saturation Region
1
K(VGS-VT)
VDS < VGS - VT
VGS4
VGS4
VGS3
VGS2
VGS1
VGS3
VGS2
VGS1
VO
VDS
VDS > VGS - VT
MOSFET characteristics
VDS
VS
Load Line superimposed on MOS characteristics
for operating point analysis of MOSFET Amplifier
7. MOSFET Amplifier
VS
VS
vO
RL
VI +
_
VI+vi +
_
vO
vO
vO
iD=
Linear for small vi
id=K(vI-vT)vi
K(VI-vT)2
2
vO
(VI,VO)
RL
vi +
_
vi
MOSFET Amplifier
Small Signal gain:
Large Signal Model(SCS)
Small Signal Model
vo
----- = – K ( V I – v T )R L = – g m R L
vi
vI
Operating Point@VI
*Note that the output is a function of both VI(DC) and vi(AC).
Page 12
Parameter
Value (for small signal model)
Parameter
Value (for small signal model)
Input Resistance
∞
Current Gain
Ri
K ( V I – v T ) ( R L || R O ) -------- ; = ∞ if Ri= ∞
RO
Output Resistance
RL
Power Gain
2 Ri
[ K ( V I – v T ) ( R L || R O ) ] -------- ; = ∞ if Ri= ∞
RO
8. Noise Margins
(1)
for large noise margins
VO
need high gain
Noise Margin "Low"
VOH
Valid
Low
(2)
VOL
Noise Margin "High"
Forbidden Zone
VIH VOH
VOL VIL
VIL
VIH
Valid
High
Noise Margins
VI
Inverter characteristic curve
To improve noise margins: (1) increase VIL and lower VIH, and/or (2) increase VOH and lower VOL.
(1) : increase vT of the MOSFET. This would make the transistor to turn on at a higher threshold
voltage, increasing VIL and lowering VIH.
(2):
9.
R
L
W
on
- and R on ∝ ----- .
increase load resistance RL or (W/L) ratio of the MOSFET since V O ∝ ---------------------
R on + R L
Op-Amp Circuits
Model
V+
+
V-
-
V+
Vout
V-
+
-
+
-
A(V+-V-)
Vout
• Operations
* The Difference (V+ - V-) is amplified by A(~106) as Vout. The + and - terminal inputs have infinite
impedance, so no current flows into the terminals. Most Op-Amp circuits makes use of negative
feedback where output tries to follow the input. A typical op-amp has output limits imposed by the
power supply of +/- 15V.
• Approach
1) Set V+=V-. (ideal op-amp w/ negative feedback)
2) Do KCL at + and - terminal. (no current flows into the + and - terminals)
3) Solve for the expression Vo/Vi.
Page 13
Negative Feedback
• Inverting Op Amp:
Rf
Rf
Vin
Rf
Ri
Ri
-
Vout
+
+
-
+
op-amp abstraction
rt
Ri
Vout
-
Vin
Vin
A(V+-V-)
op-amp model
ri
+
+
-
Vout
A(V+-V-)
more accurate op-amp model
* No current flows into the + and - terminals.
* Vout=A(V+-V-) => V+-V- = Vout/A ~ 0. Thus, V+=V-.
R
Ri
Z
Zi
f
* KCL @ V-, V out = – ------V
in .
V
V in
Z
Zi
f
out
f
In General, V out = – ------V
in thus, ----------- = H ( jω ) = – ------
R +r
A
R +r
A
f
f
-t ≈ ----------------t
* Input Resistance: R i = r i || ( R f + r t ) ||  ---------------
r
r
Rs
A -----------------Rs + R f
t
t
- ≈ ----------------------* Output Resistance: R o = -------------------------------
Rs
1 + A -----------------Rs + R f
• Non-inverting Op Amp:
Vin
+
Vout
-
Vin
R1
rt
Vout
+
-
+
-
A(V+-V-)
Vin
R1
+
-
ri
+
-
A(V+-V-)
R2
R2
R2
op-amp abstraction
op-amp model
more accurate op-amp model
R +R
R2
1
2
- V in
* V out = ------------------
AR
R1 + r t + R2
2
* Input Resistance: R i ≈ r i ----------------------------
r
R2
A ------------------R1 + R2
t
* Output Resistance: R o ≈ ----------------------
Important Circuits
Adder
V1
V2
Subtractor
R3
R1
R2
+
V1
V2
Vout
R1
-
R2
+
R1
R3
R3
V o = –  ------ V 1 + ------ V 2
 R1
R2 
R2
R2
V o = ------ ( V 2 – V 1 )
R1
Page 14
Vout
R1
Vout
Integrator
Differentiator
Cf
-
Vin
Vout
+
Rf
Ci
Ri
Vin
+
Integrator
dV i
dt
Vout
+
Double Differentiator
Lf
Ci
Ri
Vin
-
Vout
+
R
= --- ∫ V i dt
L
Vo
Vout
1
V o = ------- ∫ ∫ V i dt dt
LC
Lf
Rf
-
+
Differentiator
Li
Vo
Vin
Vout
V o = RC
Cf
Li
-
1
V o = -------- ∫ V i dt
RC
Vin
Double Integrator
Vin
-
Vout
+
2
L dV i
= --Rdt
V o = LC
d Vi
dt
2
Several of these op-amp circuits may be cascaded in stages. For such cases, you may find the Vo/Vi
relationship for each stage using impedances and multiply them together to obtain the total Vo/Vi.
example:
R
L
+
∫
R
V a = – --- V o dt
L
Va
R
dV o
V b = – V i – RC
dt
+
Vo
Va + Vb
V o = --------------------2
C
R
+
Thus,
R
Vb
Vi
dV i
dt
2
d Vo
dV o R
= – RC
–2
– ---V o
2
dt
L
dt
10. Non-linear Circuits
Diode
• Model
id
"on"
+ VD slope=gd=1/Rd
0.6V
+-
Rd
+-
id
"off"
VD
0.6V
0.6V
OR
DC Analysis: For the large signal model, the diode acts like a short with a voltage drop of 0.6V when
the diode is on(Vd>0.6V), and like an open when the diode is off(Vd<0.6V).
qV D
iD
 ---------
kT
= I se
– 1


where I s ∼ 10
Page 15
– 12
Amps for Silicon diodes.
AC Analysis: For the incremental model, the diode acts like a current dependent resistor. The resistor
26mV
KT
value is set by the nominal(DC) operating current Id. r d = -------- = --------------Id
qI d
• Circuit Analysis
One Diode: subst. the model for "on" and "off" state and solve using KCL and KVL.
Multiple Diodes:
1) Setup cases of "on" and "off" states for each diode. Total of 2n cases for n diodes.
2) For each case, subst. the model and solve using KCL and KVL.
3) Obtain asymptote for each case.
4) determine which case (i.e., asymptote) is appropriate for each region of operation.
* ex) For asymptote crossing the V=0 axis, see which case is valid for V=0 and choose
the asymptote for that case. repeat for other regions of operation.
example,
V3
0.1
-1.4
4K
Vs +
_
2K
+
_ V3
1K
(0.7, 0.1)
Vs
0.7
(-1.4, -0.2)
-0.2
another example,
L
V o ---C
iD
Vo
L
iL
Vs +
_
L
V o ---C
+
iL
iD
_ Vc
C
Vo
Vc
π
---------2ω o
• Half-wave Rectifier: Clips the negative portion of the input wave and passes only the positive portion
of the input wave with a diode drop of 0.6V.
V
+ D Vo
Vi
0.6V
+
Vi
+
_
R
_
Vo
t
"off"
"on"
"on"
– V i + iD R + V D = 0
• Peak Detector
+ VD -
Vo
*using ideal diode
Vi
Vi
+
_
R
C
+
Vo
_
t
"off"
"on"
Page 16
"on"
11. Electrical/Mechanical Analogs
Electrical
Through Variable
Capacitor State
d/dt
q
d/dt
i
dv
capactior: i = C
dt
capactior: q = Cv
v=iR
inductor: λ = Li
λ
d/dt
di
dt
Power=iv
KVL:
∑v
= 0
∑i
= 0
loop
inductor: v = L
v
di
dt
d/dt
dv
dt
KCL:
df
dt
Power=uf
node
Across Variable
Inductor State
Mechanical
Through Variable
Mass State
d/dt
p
d/dt
f
∑u
mass: f = ma
mass: p = ma
f=Bu
spring:
spring: f = kx
x
d/dt
Spring State
= 0
loop
u
df
= ku
dt
d/dt
Across Variable
du
dt
∑
f = 0
node
Electrical/Mechanical Pairs
Capacitor/Mass
Inductor/Spring
Resistor/Damper
q = Cv vs. p = mu
1
λ = Li vs. x = --- f
k
1
v = Ri vs. u = --- f
B
References
Agarwal, Anant and Jeff Lang. 6.002 Notes (1998)
Leeb, Steve. 6.002 Recitation Notes (fall ’97)
Schlecht, Martin. 6.002 Lecture Notes (fall ’97)
Searle, Campbell. 6.002 Notes (1991)
Good Luck!
12. 8. 1998. JK.
Page 17