Slide009 Op-Amps Applications

7/27/2019 Slide009 Op-Amps Applications

1/45

Op-Amps Applications

1. Voltage to current converters,

2. Current to voltage converters,

3. Instrument Amplifiers,

4. Sample-Hold circuits,

5. Active Low-pass Filter6. Schmitt Trigger Comparator


2/45

x Iv Ri

xv

(Trans-resistance amplifier)


3/45

(Trans-conductance amplifier)


4/45

Practical Op Amp Limitations Finite gain

Finite input impedance

Out put impedance is not zero

Input bias current is not zero

Band width is limited

Common mode rejection ratio is limited

1

1

o d

I o o d

i a R r A

v R a r R r r

|| 1o d oR R r a r

0I D L o O Dv v v r i av

0I D D d Ov v R v r i Eliminating v

D

At output loop

At node x.

ro

Load

rd

R

vI avD

vL

io

vD

x


5/45


6/45

R3 R4

R1

R2

RO

vt

vt

vt

it

4 3

0t x tv v vR R

2 1

t x tt

v v vi

R R

x

v

2

2 1 4 3

tO

t

v RR

i R R R R

Output Resistance


7/45


8/45


9/45


10/45

Difference Amplifier and Common-Mode

Rejection Ratio (CMRR)

A(orAdm) = differential-mode gain

Acm = common-mode gain

vid= differential-mode input voltage

vic = common-mode input voltage

A real amplifier responds to signal

common to both inputs, called the

common-mode input voltage (vic).

In general, v

o

Adm

v

id

Acm

vic

Adm

Adm

v

id

vic

CMRR

CMRRAdm

Acm

and CMRR(dB) 20log10

(CMRR)

An ideal amplifier hasAcm = 0, but for a

real amplifier,

21 id

v

icvv 22 id

v

icvv

voAdm(v

1v

2)Acm

v1

v2

2

voAdm(vid

)Acm(vic)


11/45

4 1 2 22 1

1 3 4 1o

R R R Rv v v

R R R R

xv

4 1 2 4 1 22 2

1 3 4 1 1 3 4 1

1

2o dm cm

R R R R R RR Rv v v

R R R R R R R R


12/45


13/45

Difference Amplifier: Example

Problem: Determine vo

Given Data: R1= 10kW, R2 =100kW, v1=5 V, v2=3 V

Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.

Analysis: Using dc values,

Adm

R

2R

1

100kW10kW

10

VoAdmV1

V2

10(5 3)

Vo20.0 V

HereAdmis called the differential mode voltage gain of the difference amplifier.


14/45

Finite Common-Mode Rejection Ratio:

Example

Problem: Find output voltage error introduced by finite CMRR.

Given Data: Adm= 2500, CMRR = 80 dB, v1 = 5.001 V, v2 = 4.999 V

Assumptions: Op amp is ideal, except for CMRR. Here, a CMRR in dB of80 dB corresponds to a CMRR of 104.

Analysis:

The output error introduced by finite CMRR is 25% of the expected idealoutput.

5.001V 4.999V=0.002V

5.001V 4.999V 5.000V2

5.0002500 0.002 V 6.25VCMRR 410

In the "ideal" case, 5.00 V

vid

vic

vicv A vo dm id

v A v

o dm id

% output error 6.255.005.00

100%25%


15/45

uA741 CMRR Test: Differential Gain


16/45

Differential Gain Adm = 5 V/5 mV = 1000


17/45

uA741 CMRR Test: Common Mode Gain


18/45

Common Mode Gain Acm = 160 mV/5 V = .032


19/45

CMRR Calculation for uA741

4

10

1000CMRR 3.125 10

.032

CMRR(dB) 20log CMRR 89.9 dB

dm

cm

A

A


20/45

1v

2v

xv

xv


21/45

2 ( )2 1

1

Rv v vo o oR

11 23 2 3

1 2 1 0

G

Rv v voR R R R

232 1 ( )

2 1

1

G

RRv v vo RR

1v

2v

xv

xv2 12

3 3

1 10o

G G

v vv

R R R R

1 21

3 3

1 10o

G G

v vv

R R R R

1( )

2 1 2

Rv v vo

o o R

1 2 1 2 1 23 3

1 1 1 10

o oG G

v v v v v vR R R R


22/45

Instrumentation Amplifier

Combines 2 non-inverting amplifiers

with the difference amplifier to

provide higher gain and higher input

resistance.

)b

va(v

3

4v R

Ro

bv

2i)

1i(2

2iav RRR

12

2

v

1

v

iR

)2

v1

(v

1

21

3

4v

R

R

R

R

o

Ideal input resistance is infinite

because input current to both op

amps is zero. The CMRR is

determined only by Op Amp 3.

NOTE


23/45

Instrumentation Amplifier: Example

Problem: Determine Vo

Given Data: R1 = 15 kW, R2 = 150 kW, R3 = 15 kW,R4 = 30 kWV1 = 2.5 V, V2 =

2.25 V

Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.

Analysis: Using dc values,

150k30k24 1 1 22( ) 15k15k1 2 13

( ) 22(2.5 2.25) 5.50V1 2

RRvoAdm v v RR

V A V V o dm

WW

WW


24/45


25/45

exercise

Exercise 1

Find VO?


26/45

exercise

Exercise 2

Find V2 and V3?


27/45

exercise

Exercise 3

Find VO?


28/45

exercise

Exercise 4

Find VO?


29/45

Cascaded Amplifiers

Connecting several amplifiers in cascade (output of one stage connected tothe input of the next) can meet design specifications not met by a singleamplifier.

Each amplifer stage is built using an op amp with parametersA, Rid

, Ro

,called open loop parameters, that describe the op amp with no externalelements.

Av, Rin, Routare closed loop parameters that can be used to describe eachclosed-loop op amp stage with its feedback network, as well as the overallcomposite (cascaded) amplifier.


30/45


31/45

Fig. 2.35


32/45

1

1

1

1ref

Rv

R RV

1 2

1 1

1

1ref ref

Rv v R

R R R RV V

1 2 1

1 12

ref ref

v v RR

V V R R R R R

Analysis

1 2 1

1

||

ref

v v R R

R R RV

1 2 1

1 12 2 21

ref

v v RR

V R R RR R R R

1 2 1 12 11

Vrefv v

R R R R R R

1 2 1 1 12 1 21 1

o ref ref v A v v A V A V R R R R R R R R R R

2

1ref

v R

R RV


33/45

Example 2.12

Let the transducer of fig. 2.35 be the Pt RTD (platinum resistance temperature detector)

and equal to 100 ohm with temperatur coefficient of 0.00392 per oC and let Vref=15 V.

(a) specify values of R1 and A suitable for achieving an output sensitivity of 0.1 V/oC near0 oC. to avoid self-heating in the RTD, limit its power dissipation to less than 0.2 mW. (b)

Compute vo (100 oC) and estimate the equivalent error, in degrees Celsius, in making the

approximation of Equation.


34/45


35/45

The Active Low-pass Filter

Use a phasor approach to gain analysis of

this inverting amplifier. Let s = jw.

Avvo

(jw)

v(jw)

Z2(jw)

Z1(jw)

Z1jw R1

Z2(jw)

R2

1

jwC

R2

1jwC

R2jwCR

21

AvR2

R1

1

(1 jwCR2

)R2

R1

ej

(1

jw

wc

)

wc2fc1

R2C

fc1

2R2C

fcis called the high frequency cutoff of

the low-pass filter.


36/45

Active Low-pass Filter (continued)

At frequencies belowfc(fH in thefigure),the amplifier is aninverting amplifier with gain setby the ratio of resistors R2 andR

1

.

At frequencies abovefc, theamplifier response rolls off at-20dB/decade.

Notice that cutoff frequencyand gain can be independently

set.

AvR2

R1

ej

(1 jwwc

)

R2

R1

12 w

wc

2

ej

ejtan1(w/w

c)

R2

R1

1 wwc

2ej[ tan1(w/w

c)]

magnitude phase


37/45

Active Low-pass Filter: Example

Problem: Design an active low-pass filter

Given Data: Av= 40 dB, Rin= 5 kW,fH = 2 kHz

Assumptions: Ideal op amp, specified gain represents the desiredlow-frequency gain.

Analysis:Input resistance is controlled by R1 and voltage gain is set by R2 / R1.

The cutoff frequency is then set by C.

The closest standard capacitor value of 160 pF lowers cutofffrequency to 1.99 kHz.

100dB20/dB4010 vA

W k51 in

RR

Av

R

2R

1

R2

100R1

500kW

C 12fH

R2

12(2kHz)(500kW)159pF

and


38/45

Low-pass Filter Example PSpice Simulation


39/45

Output Voltage Amplitude in dB


40/45

Output Voltage Amplitude in Volts (V) and Phase in Degrees (d)


41/45

non-linear application:schmitt trigger

VV

VtVVandVVstateinitialwith

VVVandRRassume

VRR

RV

S

o

EECCf

O

f

5.7)15(2

1

sin1015

151

1

1

w

Non-linear application

Schmitt Trigger (Pemicu Schmitt)

(a) Transfer Characteristic of Schmitt Trigger

(c) Output Voltage of Schmitt Trigger

(b) Input Voltage of Schmitt Trigger

1R

2R

ov

iv

ov

iv


42/45


VV


VVVandRRassume

VRR

RV

S

o

EECCf

O

f

5.7)15(2

1

sin1015

151

1

1

w



Vo(V)

15

-15

t

VS(V)

t

7.5

-7.5

Vo(V)

VS(V)-7.5 7.5-10 10

15

-15



(b) Input Voltage of Schmitt Trigger


43/45


VV


VVVandRRassume

VRR

RV

S

o

EECCf

O

f

5.7)15(2

1

sin1015

151

1

1

w





1R

2R ov

iv

ov

iv


44/45

Examples and Problems

Examples: 2.1, 2.2, 2.3, 2.4, 2.8, 2.9, 2.10, 2.11, 2.12, 2.13

Problems: 2.5, 2.14, 2.28, 2,39, 2,49, 2.50.


45/45

Assignment # 03

Problems: 1.8, 1.16, 1.31, 1.43, 1.53, 1.60, 2.14, 2.28, 2,39, 2.50.

Documents

Slide009 Op-Amps Applications