CHAPTERS 7 & 8CHAPTERS 7 & 8

NETWORKS 1: NETWORKS 1: 0909201-010909201-01 4 December 2002 – Lecture 7b

ROWAN UNIVERSITYROWAN UNIVERSITY

College of EngineeringCollege of Engineering

Professor Peter Mark Jansson, PP PEProfessor Peter Mark Jansson, PP PEDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERINGDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING

Autumn Semester 2002Autumn Semester 2002

networks I

Today’s learning objectives – review op-amps introduce capacitance and

inductance introduce first order circuits introduce concept of complete

response

THE OP-AMPFUNDAMENTAL CHARACTERISTICS

_

+

INVERTING INPUT NODE

NON-INVERTING INPUT NODE

OUTPUTNODEi1

i2

io

vo

v2

v1

Ro

Ri

Op-Amp Fundamentals

for KCL to apply to Op-Amps we must include all currents: i1 + i2 + io + i+ + i- = 0 When power supply leads are omitted

from diagrams (which they most often are) KCL will not apply to the remaining 3 nodes

are Op-Amps linear elements?

yes .. and no…

three conditions must be satisfied for an op-amp to be a linear element: |Vo | <= Vsat

| io | <= isat

Slew rate >= | dVo/dt |

Example from Text

the A741 when biased +/- 15 V has the following characteristics: vsat = 14 V isat = 2 mA SR = 500,000 V/S

So is it linear? When RL = 20 kOhm or 2 kOhm?

Using Op-Amps

Resistors in Op-Amp circuits > 5kohm Op-Amps display both linear and non-linear behavior

Remember: for Ideal Op-Amp

node voltages of inputs are equal currents of input leads are zero output current is not zero

One more important Amp

difference amplifier See Figure 6.5-1, page 213

Practical Op-Amps

characteristic ideal practical sample

Bias current 0 > 0 0.1-80 nA

Input resistance

infinite finite 2-106 Mohm

Output resistance 0 > 0 60-1Kohm

Differential gain infinite finite 100-5000V/mV

Voltage saturation infinite finite 6V-15V

What you need to know

Parameters of an Ideal Op Amp Types of Amplification Gain (K) vs. Which nodes and Amps circuits are needed to achieve same How to identify which type of circuit is in use (effect) How to solve Op Amp problems

new concepts from ch. 7

energy storage in a circuit capacitors series and parallel

inductors series and parallel

using op amps in RC circuits

DEFINITION OF CAPACITANCE

Measure of the ability of a device to store energy in the form of an electric field.

CAPACITOR: IMPORTANT RELATIONSHIPS:

+ –

i

+

+

_

_ d

AC

Cvq

dt

dvCi

CALCULATING ic FOR A GIVEN v(t)

Let v(t) across a capacitor be a ramp function.

t

v

v

t

tt0t

vCic

ci0tAs

Moral: You can’t change the voltage across a capacitor instantaneously.

VOLTAGE ACROSS A CAPACITOR

dt

dvCi c

c dtC

idv c

c

dC

idv

tc

c

diC

1dv

t

cc

oc

t

tcc tvdi

C

1v

o

ENERGY STORED IN A CAPACITOR

divtwt

ccc

dd

dvCvtw

tc

cc

dvvCtw)t(v

)(vcc

2cc Cv

21

tw)t(v

0)(v

2cc Cv

2

1tw

CAPACITORS IN SERIES

+–

C1 C2 C3

+ v1 - + v2 - + v3 -

i

0tvdiC

1tvdi

C

1tvdi

C

1v o3

t

t3o2

t

t2o1

t

t1 ooo

o3o2o1

t

t321tvtvtvdi

C

1

C

1

C

1v

o

vKVL

oc

t

teqtvdi

C

1v

o

321eq C

1

C

1

C

1

C

1 o3o2o1oc tvtvtvtv

Capacitors in series combine like resistors in parallel.

CAPACITORS IN SERIES

CAPACITORS IN PARALLEL

C1 C2 C3

i

i1i2 i3KCL

0dt

dvC

dt

dvC

dt

dvCi 321

dt

dvC

dt

dvCCCi eq321

Capacitors in parallel combine like resistors in series.

HANDY CHARTELEMENT CURRENT VOLTAGE

R

C

L

R

VI RIV

dt

dvCi c

c dtiC

1v

t

cc

DEFINITION OF INDUCTANCE Measure of the ability of a device to store

energy in the form of a magnetic field.

INDUCTOR: IMPORTANT RELATIONSHIPS:

i

+ _

LiN dt

diLv

v

CALCULATING vL FOR A GIVEN i(t)

Let i(t) through an inductor be a ramp function.

t

i

i

t

ttt

iLvL

0

Lvt 0As

Moral: You can’t change the current through an inductor instantaneously.

CURRENT THROUGH AN INDUCTOR

dt

diLv L

L dtL

vdi L

L

dL

vdi

tL

L

dvL

1di

t

LL

oL

t

tLL tidv

L

1i

o

ENERGY STORED IN AN INDUCTOR

divtwt

LLL

dd

diLitw

tL

LL

dviLtwti

iLL

)(

)( 2

2

1)(

0)(

LL Litwti

i

2

2

1LL Litw

INDUCTORS IN SERIES

L1 L2 L3

+ v1 - + v2 - + v3 -

iKVL

0321 dt

diL

dt

diL

dt

diLvi

dt

diL

dt

diLLLv eqi 321

Inductors in series combine like resistors in series.

INDUCTORS IN PARALLEL

L1 L2 L3v

i1 i2i3

KCL+–

0tidvL

1tidv

L

1tidv

L

1i o3

t

t3o2

2o1

t

t1v

oo

o3o2o1

t

t321v tititidv

L

1

L

1

L

1i

o

oL

t

teqv tidv

L

1i

o

321eq L

1

L

1

L

1

L

1 o3o2o1oL titititi

INDUCTORS IN PARALLEL

Inductors in parallel combine like resistors in parallel.

HANDY CHARTELEMENT CURRENT VOLTAGE

R

C

L

R

VI RIV

dt

dvCi c

c dtiC

1v

t

cc

dt

diLv L

L dtvL

1i

t

LL

OP-AMP CIRCUITS WITH C & L

_

+

i1

i2

io

vo

v2

v1

Cf

Ri

+–

vs

Node a

0

dt

vvdCi

R

vv o1f1

i

1s

12

2

1

vv

0i

0i

dt

vdC

R

v of

i

s

fi

so

CR

v

dt

vd

dtCR

vvd

fi

so dt

CR

vvd

fi

so dtv

CR

1v s

fio

QUIZ: Find vo= f(vs)

_

+

i1

i2

io

vo

v2

v1

Rf

+–

vs

Node a

Li

ANSWER TO QUIZ

12

2

1

vv

0i

0i

0R

vvidtvv

L

1

f

o111s

i

0R

vdtv

L

1

f

os

i

dtvL

1

R

vs

if

o dtvL

Rv s

i

fo

IMPORTANT CONCEPTS FROM CH. 7

I/V Characteristics of C & L.

Energy storage in C & L. Writing KCL & KVL for circuits with C & L. Solving op-amp circuits with C or L in feedback loop. Solving op-amp circuits with C or L at the input.

new concepts from ch. 8

response of first-order circuits the complete response stability of first order circuits

1st ORDER CIRCUITS WITH CONSTANT INPUT

+–

t = 0

R1 R2

R3 Cvs

+v(t)-

s321

3 vRRR

R0v

Thevenin Equivalent at t=0+

Rt

C+–

Voc

+v(t)-

32

32t RR

RRR

s

32

3oc v

RR

RV

KVL 0)t(vR)t(iV toc

i(t)

+ -

0)t(vdt

)t(dvCRV toc CR

V

CR

)t(v

dt

)t(dv

t

oc

t

SOLUTION OF 1st ORDER EQUATION

CR

V

CR

)t(v

dt

)t(dv

t

oc

t

CR

)t(v

CR

V

dt

)t(dv

tt

oc dtCR

)t(vV)t(dv

t

oc

dtCR

1

)t(vV

)t(dv

toc

dt

CR

1

V)t(v

)t(dv

toc

DdtCR

1

V)t(v

)t(dv

toc

SOLUTION CONTINUED

DCR

tV)t(vln

toc

DdtCR

1

V)t(v

)t(dv

toc

D

CR

texpV)t(v

toc

CR

texpDexpV)t(v

toc oc

tV

CR

texpDexp)t(v

oct

VCR

0expDexp)0(v

ocV)0(vDexp

SOLUTION CONTINUED

oct

oc VCR

texpV)0(v)t(v

CR

texpV)0(vV)t(v

tococ

WITH AN INDUCTOR

+–

t = 0

R1 R2

R3 Lvs

21

s

RR

v0i

i(t)

Norton equivalent at t=0+

RtIsc

+v(t)-

L i(t)

32

32t RR

RRR

2

ssc R

vI

KCL 0)t(iR

)t(vI

tsc

0)t(idt

)t(diL

R

1I

tsc sc

tt IL

R)t(i

L

R

dt

)t(di

SOLUTIONsc

tt IL

R)t(i

L

R

dt

)t(di

CR

V

CR

)t(v

dt

)t(dv

t

oc

t

CR

1

L

R

t

t

CR

texpV)0(vV)t(v

tococ

tL

RexpI)0(iI)t(i t

scsc

HANDY CHARTELEMENT CURRENT VOLTAGE

R

C

L

R

VI RIV

dt

dvCi c

c dtiC

1v

t

cc

dt

diLv L

L dtvL

1i

t

LL

IMPORTANT CONCEPTS FROM CHAPTER 8

determining Initial Conditions setting up differential equations solving for v(t) or i(t)