Topic 3: Op-Amp: Golden Rules of OP Amp

1. iin=0, no current flow into op amp.

2. V+=V-

• Typically one end of op amp is connected to ground, therefore, V+=V-= 0V, virtual ground. Often V+ is connected to ground to avoid stability problem.

3. Various circuits: amplifier, sum and abstract, differentiator, integrator, etc.

Example

• What does the following circuit do? That is, find an expression for the gain and give the circuit a suitable name.

Example II

(a) (7pts)The circuit is an integrate circuit. Derive vout(t). RS=10 kΩ, CF=0.008 µF.

(b) (5pts)If the input signal vin=sin(2000πt) V, calculate vout(t) and the peak amplitude.

(c) (3pts)If the input signal vin=10 mV, calculate vout(t) and vout(t=100ms)

(d) (3pts)Op-amp is saturated when the output voltage reach the power supply voltage,

±15V in this case. At what time does the integration of the DC input cause the op-amp to

saturated fully?(e) (7pts) If the input signal is now

vin=0.01+ sin(2000πt) V, describe what happens to the output waveform till the op-amp is fully saturated.

2000cos2 125)( e.

ms 120)(12.0 d.

(V) 5.12)100(

(V) 125)( c.

(V) 2000cos2 .

125001

a.

tttv

Sect

msv

ttv

tvb

dtvdtvCR

v

out

out

out

out

ininFS

out

Example 4Design an op-amp circuit to convert the triangular waveform v1 in the following Figure into the

square wave vo shown. Use 0.1 F capacitor. (Hints: First quantitatively determine the

mathematical expression of vo in terms of v1 ).

kR

R

dt

dvCRv

F

F

SF

25101

2.0101.05.0

36

10

v1(t)

Topic 4: Filters

• Low-pass filters (LP)

RCe

RC

eRC

e

e

RCRCj

RCj

CjjH

jVZZ

ZjV

jV

jVjH

j

RCj

RCj

j

V

iRC

C

iV

1 ;

1

1

1

1

1

1

1

1

1

1

0/tan

2

tan

2

1tan

0

2

0

0

01

1

1

Active Filter

S

F

S

out

Z

Z

V

V

S

F

S

out

Z

Z

V

V1

It is phasor time again

Active Low Pass Filter

FF

SF

FF

F

FFF

S

F

S

out

CRj

RRjA

CRj

R

CjRZ

Z

Z

V

VjA

1

/)(

1

1||

)( • Amplification: RF/RS

• low pass factor 1/(1+jRFCF)• Cut off frequency: RFCF=1

RF/RS=10, 1/RFCF=1

Band-pass Filters and Resonant Circuits

0

0

arctan

2

0

0

2

0

0

1

1

1

1

Q

V

e

Q

jQ

jH

LC

10

12

000

BR

LQ

Resonant frequency

Quality factor

RLC Resonant

CL ZZ

CjLj

CL

1

1

when 0

R

V

CLR

VI

CLjR

V

ZZZ

jVjI

CLR

22 1

1

)(

At resonance:• current is max.• Zeq =R• current and voltage are in phase.• the higher Q, the narrower the resonant peak.

Applications: tuning circuit

Example

RC

jCjR

R

v

vVH

in

out

1

11

980102802

1

11

6

R

RCkHzf

Req

67.11026.472

1

6.4798050

98050

6

a. Write the transfer function for this filter. (Both Vin and Vout are referenced to ground).

b). Design a high-pass RC filter with a characteristic frequency of 80 Hz using a capacitor of 2.0 F.c). What is the new characteristic frequency when the filter is loaded with 50 ohm?

How do you solve the problem:

Use small R or use active filter

SS

S

F

SS

F

S

F

S

out

CRjR

R

CjR

R

Z

Z

V

VjA

11

11

)(

Example 2

The circuit shown in an active filter. Determine: The voltage transfer function. The pass-band gain.

RF

Ri

v+

6.145

6811

11

11

1

0

0

0

1

i

F

i

i

F

i

F

Fci

R

RG

CRjR

R

v

v

CjR

R

v

vv

R

vv

ZR

v

vv

Exam 3

kRkR

CR

R

R

FS

FF

S

F

10 ;1 :say slet'

,1

1010

100

1

11.0101010 641

FFCR

The input signal can be written as Vi=10mV(sin10t+sin10,000t). Design a circuit so that the output signal is V0=-100mVsin10t. In another words, the high frequency signal has to be much smaller (<1%) than the low frequency part.

FF

SF

S

out

CRj

RR

V

VjA

1

/)(

FC

CR

CR

CRj

RR

F

FF

FF

FF

SF

11010

100

100

100

1

1

1

10

1

1

/

44

2

22

2

Low frequency

high frequency

Check low freq.

Topic 5: Electromachine

• Motion Electricity (generator): Magnetic Induction, Faraday’s law

• Electricity Motion (motor): force on a current, torque• Three phase power: structure, advantages.