6.

002 Circuits and Systems Final Review

Jason Kim

Page -1

6.002 Circuits and Systems: Outline of Topics

Resistor Networks concepts:

Node Method, KCL, KVL, Superposition, Thevenin and Norton equivalent circuit models, Multiport Networks, Transformers, Power.

1st Order Circuits concepts:

Consituitive Laws, RC netwroks, RL networks, Homogenous and Particular solution, Time constant , response to an impulse, power, Low/High/Band/Notch pass filters, sinusodial steady-state, transients, Impedance Model, Bode Plots(Magnitude and Phase).

2nd Order Circuits concepts:

LC networks, LRC networks, damping coefficient , natural frequency 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, DeMorgans 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 Series: R 1 + R 2
R1 Vi

freq. domain: Z = R
1 2 Parallel: R 1 || R 2 = ------------------

R R R1 + R2

Ii R2 Vo R2 -V V o = ------------------R1 + R2 i I2 Vi

+ -

+ -

R1

R2

R1 -I I 2 = ------------------R1 + R2 i

Voltage Divider

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
Rth Vth + Voc=Vth same i-v characteristics at the ports IN

Norton Equivalent Circuit Model

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 note the direction of Itest and polarity of Vtest.

+ -

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)
1 Power = --- I ( t ) V ( t ) dt T
0 T

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

+
R VR

VR = VR

V2 n = 0

+ VR

V 1 ,V 3 n = 0

+ VR

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 + i2 i1 + V1 R12i2 R11 + -

M
+ -

R22 R21i1

i2 + V2 -

+ V - 2

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(Rxxs) are identical, then the networks cant be differentiated using only the i-v characteristics at the ports. * Reciprocity: R12=R21. * Using the definitions from above,
i1 + V1 R12=R21 R11-R12

R22-R21

i2 + V2

i1 + V1 G11+G12

M
-G12=-G21 G22+G21

i2 + V -

T model

model

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 Transformer i1 + V1 _ _ N1:N2 i2 + V2
V1 V2 ------ = -----N1 N2
1 & N 1 i 1 = N 2 i 2 . Turns Ratio = ------

N N2

Reflected Resistance: the equiv. resistance seen at port 1, of the resistance on the other side, port2. i1 + V1 _ N1:N2 i2 + V2 _ R V1 _ + i1
Reflected Resistance

N1 R N2

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 C C C1 + C2

freq. domain: Z = ------ = ----------

1 Cs

1 jwC

High Frequency = Short & Low Frequency = Open

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 - CV . (no energy dissipation) Energy Stored: E = -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 caps. For this case, Vc does not discharge to zero but VC1=VC2. Inductor: time domain: V = L
dI dt L L L1 + L2

+
VC1

C1

C2 +

VC2

_
VC1(0-) = VC2(0-)

freq. domain: Z = Ls = jwL

High Frequency = Open & Low Frequency = Short Series: L 1 + L 2

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.

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"E L I" : Voltage(EMF) LEADS Current(I) by 90o. Energy Conservation: V dt = = LI .

1 2 - LI . (no energy dissipation) Energy Stored: E = -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. LC Series@ o = Short & Parallel@ o = Open

L1 i1 i2

L2 R

iL1(0-) = iL2(0-)

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
R + Vi

RL Network
R iL + Vi L Vo -

+ -

Vc=Vo -

+ -

dV c V c Vi + ------- = ------dt RC RC

Vi di L i L + -------- = ---d t L L - R L Time Constant: = -R

Time Constant: = RC

*one time constant 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:
Vi V _i R Vi iC VL

t
VC

t
V - _i R

iL

Particular:
Vi VC V _i R iC Vi VL iL V _i R

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Total:
Vi VC V _i R Vi iC VL iL V _i R

* 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 Vi + ------- = ------dt RC RC

2) Find Homogeneous solution by setting input to zero and substituting Vh=Aest as a solution.
1 st st - Ae = 0 Ase + ------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
------ RC V o = V i1 e t

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 = Q t and V o ( 0
0
plus

minus

) = 0
0
plus

and

dV o V o Vi + ------- = -------. RC RC dt
0
plus

1 Q - V dt = ------- t dt dV o + ------RC o RC
0

minus

1 dV o + ------RC
0 plus

minus

Q V o dt = ------RC
0

minus

t dt V o ( 0

plus

) V o(0

minus

Q ) = ------RC

V o(0

plus

Q ) 0 = ------- RC

V o ( 0

Q ) = ------RC

3. Second Order Circuits

LRC Network #1
ii R iL L + C Vc=Vo -

Vo t
2

1 i i 1 V o V o + ------+ ------- = --C t RC t LC

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Vo j LR ----- = ---------------------------------------------------2 ii ( j ) LCR + j L + R

H ( j) R "Band Pass Filter" 1 wC +1 -1 1 o = ----------LC

Time Constant: = ----------H ( j ) +90o

1 LC

w
-90o

wL

w
1 o = ----------LC

L iL + C Vc=Vo -

#2

Vi

+ -

Vo t
2

Vi 1 V o V o + ------+ ------- = ------RC t LC LC

Vo R ----- = ---------------------------------------------------2 Vi ( j ) LRC + j L + R

H ( j) "Low Pass Filter"

1 Time Constant: = ----------LC

H ( j )

0o -2

w
-180o 1 o = ----------LC

w
1 o = ----------LC

L iL

R + C Vc=Vo -

#3

Vi

+ -

iL t
2

1 dV i R i L i L + -+ ------- = -L dt L t LC

iL jC ---- = ---------------------------------------------------2 Vi ( j ) LC + j RC + 1
H ( j) 1 R 1 wL "Band Pass Filter"

1 Time Constant: = ----------LC

H ( j ) +90o

w
wC +1 -1 -90o

w
1 o = ----------LC

1 o = ----------LC

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o "peakiness" Q = ------

&

Half-Power Point = o

Vpeak Vpeak 2
o

w
o +

V peak 2 1 V peak Power peak - = ------------- -- = -------------------------Power = ------------ 2 R 2R 2

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 s + 2s +
2 2 o

= 0

where o is the "resonant" or "natural" frequency and is the damping coefficient. solving s with quadratic formula,
s 1, 2 = o = j o = j d ; d is the damped natural frequency
2 2 2 2

1) "Overdamped" 2) "Critically Damped" 3) "Underdamped" Approach (Transients)

> o = o < o

s1 s2 s1 = s2 s1 s2

s 1, 2 are real s 1, 2 are real s 1, 2 are complex

1) Set up 2nd order differential equations.

iL t
2 2

1 dV i R i L i L + -+ ------- = -(ex. circuit #3) L dt L t LC

2) Obtain characteristic equation and identify o and .

1 2 R 2 - s + ----------- = 0 ; s + 2 ---- 2 L LC
( o ) t t t
2 2

1 R o = ----------- and = ----2L LC

2 2

3) Find Homogeneous solution.(for the appropriate system type) Overdamped: Critically Damped: Underdamped:
V h = Ae V h = Ae V h = Ae + Be
( + o ) t

+ Bte

t t

cos d t + Be

sin 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
t

cos d t + Be

sin d t

i L = Ce

( A d sin d t + B d cos d t ) B = 0

A = V i
t

i L ( 0 ) = CB d = 0

Thus, V o = V i ( 1 e

cos d t )

& iL = V i C d e

sin d t

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Approach (Sinusodial Steady-State) 1) Replace the primitive elements with their impedance models and solve the network like a resistor network.
ii R ii = I cos(wt) iL L + C Vc=Vo V o = V cos ( t + )
jL 1 - i - i = ----------------------------------------Vo = ---------------------------------2 L i 1 1 i ----1 LC + j -- + sC + -R sL R

2) Put the equation into the form of a real amplitude and a complex phase.
L - R - j 0 atan --------------------------- ( 1 2 LC )

imaginary part real part

L -Ie Vo = ----------------------------------------------------------2 2 2 L - ( 1 LC ) + - R

3) Match the coefficient V and phase shift

L -I V = ----------------------------------------------------------2 2 L 2 ( 1 LC ) + - R L - R - = atan --------------------------- ( 1 2 LC )

and

1 *notice that at the resonant frequency 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
- ). Thus, our new phase shift would be ' = -- 90 degrees, since sin( t)=cos( t- -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 ,
( 10 + 100 s ) 10 + j ( 100 RC ) H ( j ) = ------------------------------- = -----------------------------------------s ( 10 + s ) j ( 10 + j RC )

roots of the numerator are "zeros" roots of the denominator are "poles"

Break Point Frequency: 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 z 1 = ------------10 RC 10 p 1 = 0 and p 2 = ------RC

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Magnitude Plot short cut: "zeros" give you a +1 slope. (constant before z , +1 slope after z ; in log scale) "poles" give you a -1 slope. (constant before p , -1 slope after p ; in log scale) Thus, Z1: P1: P2:
p1

0 0 -1
z 1

+1 0 -1
p2

+1 -1 -1

:
starts off at -1 slope.
H ( j)

-1

-1

Since p 1 = 0 , which is the left most end of the frequency axis (in log scale), the magnitude plot

actual -1 slope asymptote

difference = 3dB

*both axes are in log scale.

100 10 1 constant

-1 slope

endless

z 1
1 -------------10 RC 1 ------RC

p2
10 ------RC 100 -------RC

Approach to Plotting Magnitude 1) Find all ZEROs and POLEs. (i.e.,

BP s)

2) Draw and Label Axes. Denote break point frequencies. 3) Beginning at the far end of the -axis, plot for very small if ZERO at =0, start with +1 slope, (+2 slope if double Zero@ =0, and +3 for triple, ...) POLE at =0, start with -1 slope, (-2 slope if double Pole@ =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

BP , your actual curve will be off from the sharp asymptote corner by 3dB.
2

Phase Plot
- phase over decade. short cut: "zeros" give you a + -- phase over decade. "poles" give you a - -H ( j )
-*note that the asymptote is back to a constant unlike the magnitude case, where the slope continued

example,

2
-4 0

starts a decade before

ends a decade after

z ----10

10 z

Page 9

Thus,

Z1: P1: P2:

0
-- 2

0
-- 2

-+ 4 -- 2

-+ 2 -- 2

-+ 2 -- 2 -- 4

-+ 2 -- 2 -- 2

0
p1

0
z 1
-- 4

1 ----------------100 RC

1 ------RC

p2
-- 4

100 -------RC

:
H ( j )

-- 2

-- 2

-- 2

1 ----------------100 RC 0

z 1

1 ------RC

p2

100 -------RC

actual

asymptote

- 4 --

difference = 6o

Approach to Plotting Phase 1) Find all ZEROs and POLEs. (i.e.,

BP s)

2) Draw and label all axes and break point frequencies, including their +/- decades. 3) Beginning at the far left of the -axis, plot for very small if
3 -- phase, (+ slope if double Zero@ =0, and + ----- for triple, ...) ZERO at =0, start with + 2 2

POLE at =0, start with

-- 2

phase, (- slope if double Zero@ =0, and

3 - ----2

for triple, ...)

otherwise, start at 0 phase.

-- phase over a decade. 4) As you come up the -axis, for each POLE you hit, make a change of - 2
p ( ------10

- phase over a decade. p 10 p ). For each ZERO, make a change of + --

5) Be careful when the regions overlap, the effect from each pole and zero may cancel each other out. 6) At

BP , the actual curve is off by 6o.

Building Blocks
H ( j) S 1 -S
+1

S + 100
+1

1 ----------------S + 100
+1 100

+1

H ( j)
100

H ( j )

-2 - 2

-2 -4 0 10 100 1000

10 100 1000 0 - 4

- 2

Page 10

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

OUT = ( A B ) + C
OUT

pull-up resistor

A C

B OUT

OUT A C B

A B C

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 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 DeMorgans Laws 1) A B = A + B
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)

Gate Equivalent:

2) A + B = A B

Gate Equivalent:

A B

A B

Minimum Sum of Products(MSP): (1)first level ANDs, (2) combine with ORs. Minimum Product of Sums(MPS): (1)first level ORs, (2) combine with ANDs. Common Gates
A B AND A 0 0 1 1 B 0 1 0 1 C 0 0 0 1 C A B NAND A 0 0 1 1 B 0 1 0 1 C 1 1 1 0 C A B OR A 0 0 1 1 B 0 1 0 1 C 0 1 1 1 C A B NOR A 0 0 1 1 B 0 1 0 1 C 1 0 0 0 A B XOR A 0 0 1 1 B 0 1 0 1 C 0 1 1 0

ex) (A B) + (B C) ex) (A+B) (B+C)

A B XNOR A 0 0 1 1 B 0 1 0 1 C 1 0 0 1

Page 11

6. MOSFET Transistor (n-channel)

Large Signal Model S Model:
D D

SR Model:
D D

G S VGS < VT

Off

G S

On

G S VGS < VT

Off

On
Ron S VGS VT

VGS VT

When (VDS VGS - VT), SCS Model (Saturation):

D D iDS= G S K(VGS-VT)2 2

When (VDS < VGS - VT), SVR Model (Linear):

D D

Off

G S

On

G S

Off

On
Ron= S 1 K(VGS-VT)

VGS < VT
VDS = VGS - VT

VGS VT

VGS < VT

VGS VT

iDS

iDS Saturation Region VS RL -1 RL

Load Line Operating Point VGS5 VGS4 VGS3 VGS2 VGS1

VGS5 VGS4 VGS3 VGS2 VGS1 VDS

Lin

ear

Reg

ion

VDS < VGS - VT

VDS > VGS - VT

VO

VS

VDS

MOSFET characteristics

Load Line superimposed on MOS characteristics for operating point analysis of MOSFET Amplifier

7. MOSFET Amplifier
VS VS RL vO iD= vO K(VI-vT)2 2
vi + _ vi vI vO Linear for small vi

vO id=K(vI-vT)vi RL
vO (VI,VO)

VI+vi + _

VI + _

MOSFET Amplifier

Large Signal Model(SCS) vo ---- = K ( V I vT ) RL = gm RL vi

Small Signal Model

Operating Point@VI

Small Signal gain:

*Note that the output is a function of both VI(DC) and vi(AC).

Page 12

Parameter Input Resistance

Value (for small signal model)

Parameter Current Gain

Value (for small signal model)

Ri - ; = if Ri= K ( V I v T ) ( R L || R O ) ------RO 2 Ri - ; = if Ri= [ K ( V I v T ) ( R L || R O ) ] ------RO

Output Resistance

RL

Power Gain

8. Noise Margins
(1)

VO

for large noise margins

need high gain VOH

Noise Margin "Low"

Noise Margin "High"

(2)

Valid Low

Forbidden Zone VOL VIL

Valid High

VOL VI

VIH VOH

VIL

VIH

Noise Margins

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):
on - and R on ----. increase load resistance RL or (W/L) ratio of the MOSFET since V O ---------------------

R on + R L

L W

9.

Op-Amp Circuits
Model V+ V+ Vout V+ 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 Vin Ri + Vout Ri Vin +

+ -

Rf Vout A(V+-V-) Ri Vin +

Rf
rt ri + -

Vout

A(V+-V-)

op-amp abstraction

op-amp model

more accurate op-amp model

* No current flows into the + and - terminals. * Vout=A(V+-V-) => V+-V- = Vout/A ~ 0. Thus, V+=V-.
f V in . * KCL @ V-, V out = -----

R Ri

f out f In General, V out = -----V in thus, ---------- = H ( j ) = ------

Z Zi

V V in

Z Zi

f t f t --------------- ---------------* Input Resistance: R i = r i || ( R f + r t ) || t t - ---------------------* Output Resistance: R o = -------------------------------

R +r A

R +r A

Rs 1 + A -----------------Rs + R f

r Rs A -----------------Rs + R f

Non-inverting Op Amp: Vin + Vout R1 R2 op-amp abstraction

1 2 - V in * V out = -----------------2 * Input Resistance: R i r i ---------------------------t * Output Resistance: R o ----------------------

Vin

Vout + + -

rt

A(V+-V-)

R1 R2

Vin

+ -

ri

+ -

A(V+-V-)

Vout R1 R2

op-amp model
R +R R2

more accurate op-amp model

AR R1 + r t + R2

r R2 A -----------------R1 + R2

Important Circuits Adder V1 V2 R1 R2 +

R3 R3 - V + -----V V o = ---- R 1 1 R 2 2

Subtractor R3 V1 V2 R1 + R1 R2 R2 Vout

Vout

R2 -(V V 1) V o = ----R1 2

Page 14

Integrator Cf Vin Ri + Vout Vin Ci

Differentiator Rf +
dV i dt

Double Integrator Li Vout Vin +

1 V o = ------- V i dt dt LC

Cf

Vout

1 - V dt V o = ------RC i

V o = RC

Integrator Li Vin + Rf Vout Vin

Differentiator Lf Ri + Vout Vin

Double Differentiator Lf Ci +
2

Vout

Vo

R - V dt = -L i

Vo

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:
L R Va R - V dt V a = -L o dV o V b = V i RC dt Vo Va + Vb V o = -------------------2 dV i dt 2 d Vo dV o R -V = RC 2 -2 dt L o dt

+
C

+
R Vi

R Thus, Vb

10. Non-linear Circuits

Diode Model id
slope=gd=1/Rd 0.6V

+ VD id

"on"
0.6V 0.6V Rd

+"off"

OR

+-

VD

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).
iD --------- kT = I se 1
qV D

where I s 10

12

Amps for Silicon diodes.

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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 = ------qI d Id

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

(0.7, 0.1)
0.7 -0.2

Vs + _

4K

2K 1K

+ _ V3

Vs

(-1.4, -0.2)

another example,
L V o --C

iL

iD Vo Vs + _ iL

L + C _ Vc

L V o --C

iD

Vo ---------2 o

Vc

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

+ _

Vo "off" "on" "on"

t

V i + iD R + V D = 0

Peak Detector + VD R C + Vo _ "off" "on" "on"

t Vo Vi Vi

*using ideal diode

+ _

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11. Electrical/Mechanical Analogs

Electrical Capacitor State q
d/dt
capactior: q = Cv

Through Variable i v=iR

inductor: = Li inductor: v = L

d/dt dv capactior: i = C dt di dt

di dt

Power=iv KVL:

loop

v i

= 0

d/dt

v Across Variable

d/dt

dv dt

KCL:

= 0

node

Inductor State Mechanical Mass State p

d/dt
mass: p = ma

Through Variable f f=Bu

spring: f = kx spring: df = ku dt

d/dt
mass: f = ma

df dt

Power=uf

loop

= 0

x Spring State

d/dt

u Across Variable

d/dt

du dt

f = 0

node

Electrical/Mechanical Pairs Capacitor/Mass

q = Cv vs. p = mu

Inductor/Spring
1 -f = Li vs. x = -k

Resistor/Damper
1 -f v = Ri vs. u = -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.

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