MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.002 Circuits and Systems
Midterm Review
by Jason Kim
1. Primitive Elements
Resistor:
time domain: V = IR
freq. domain: Z = R
• 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 Eq.
3) subst. 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.
Page 1
• Power = Real Power = IV = I2R = V2/R (energy dissipated as heat)
T
1
• ⟨ Power⟩ = --- ∫ I ( t )V ( t ) dt
T
0
• 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 dependent source is present along with multiple 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
-
-
* 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
-
Pi model
T model
i2
time domain: I = C dV
Capacitor:
dt
1
freq. domain: Z = ---------jwC
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
+
VC1
C2 +
C1
_
VC2
_
VC1(0-) = VC2(0-)
freq. domain: Z = 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.
"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.
2.
L1
i1
L2
i2
iL1(0-) = iL2(0-)
First Order Circuits
RC Network
RL Network
R
Vi
+
-
iL
R
C
+
Vc=Vo
-
Vi
dV c V c
V
+ -------- = -------idt
RC
RC
+
-
L
di L i L
Vi
+ --------- = ----d t  L
L
-- R
Page 3
+
Vo
-
R
L
Time Constant: τ = ---
Time Constant: τ = RC
R
Homogeneous:
V_i
R
Vi
Vi
iC
VL
t
t
t
t
iL
VC
V
- _i
R
Particular:
Vi
VC
V_i
R
iL
Vi
iC
V_i
R
VL
t
t
t
t
Total:
V_i
R
Vi
VC
iL
Vi
iC
VL
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
Vo
V_i
R
--------

RC
= 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.
Page 4
t
Impulse as Input
ex. RC circuit: V i = Qδt
0
1
Q
dV o + -------- V o dt = -------- δt dt ➔
RC
RC
3.
plus
Q
) – 0 = -------- ➔
RC
minus
plus
0
1
dV o + -------RC
∫
0
V o(0
V o(0
and
minus
∴V o ( 0
plus
) = 0
plus
∫
0
V
dV o V o
+ -------- = -------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
minus
Q
) = -------RC
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
↔
OUT = ( A ⋅ B ) + C
A
B
↔
↔
OUT
C
OUT
A
C
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, 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
B
C
B
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.
Page 5
C
ex) (A ⋅ B) + (B ⋅ C)
ex) (A+B) ⋅ (B+C)
4.
MOSFET Transistor
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
5.
VDS
VS
Load Line superimposed on MOS characteristics
for operating point analysis of MOSFET Amplifier
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 6
Parameter
6.
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
Noise Margins
(1)
for large noise margins
VO
need high gain
Noise Margin "Low"
VOH
(2)
VOL
Valid
Low
VIH
Valid
High
Forbidden Zone
VOL VIL
VIL
Noise Margin "High"
VIH VOH
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):
R
R on + R L
L
W
on
- and R on ∝ ----- .
increase load resistance RL or (W/L) ratio of the MOSFET since V O ∝ ---------------------
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!
10. 27. 1998. JK.
Page 7