Operational Amplifiersjoelsd/electronicsnew/Classnotes... · 2020. 9. 29. · Op-amp • Because we assumed that the Op-amp was ideal, we found that with negative feedback we can

OperationalAmplifiers

Lecture4

BME372NewSchesser 94

95

Ideal Amplifiers •  Parameters:

vs vo vi

Ro

Ri

Rs

RL Avovi

io ii + +

- -

+

-

vovs

=Ri

Ri + RsAvo

RLRL + RO

This states the gain of the amplifier depends on the external components.This is BAD!!!!!

However, if Ri →∞ and RO → 0; then Ri

Ri + Rs→1 and

RLRL + RO

→1.

Therefore, vovs

→ Avo and the gain of the amplifier is independent of the external components.

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Operational Amplifiers

•  An operational Amplifier is an ideal differential with the following characteristics: –  Infinite input impedance –  Infinite gain for the differential signal – Zero gain for the common-mode signal – Zero output impedance –  Infinite Bandwidth vo-

+v1v2

Inverting input

Non-Inverting input

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Operational Amplifier Feedback •  Operational Amplifiers are used with negative feedback •  Feedback is a way to return a portion of the output of an

amplifier to the input –  Negative Feedback: returned output opposes the source signal –  Positive Feedback: returned output aids the source signal

•  For Negative Feedback –  In an Op-amp, the negative feedback returns a fraction of the

output to the inverting input terminal forcing the differential input to zero.

–  Since the Op-amp is ideal and has infinite gain, the differential input will exactly be zero. This is called a virtual short circuit

–  Since the input impedance is infinite the current flowing into the input is also zero.

–  These latter two points are called the summing-point constraint.

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Operational Amplifier Analysis Using the Summing Point Constraint

•  In order to analyze Op-amps, the following steps should be followed:

1.  Verify that negative feedback is present 2.  Assume that the voltage and current at the input

of the Op-amp are both zero (Summing-point Constraint

3.  Apply standard circuit analyses techniques such as Kirchhoff’s Laws, Nodal or Mesh Analysis to solve for the quantities of interest.

BME372NewSchesser

99

Example: Inverting Amplifier 1.  Verify Negative Feedback: Note that a

portion of vo is fed back via R2 to the inverting input. So if vi increases and, therefore, increases vo, the portion of vo fed back will then have the affect of reducing vi (i.e., negative feedback).

2.  Use the summing point constraint.

3.  Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output

RLvin

R2

R1

vivo

-+

0 i1=vin/R1 i2

inverted) :input the toopposite isoutput that the(note oft independen is which -

zero is since 0constraintpoint -summing the todue

constraintpoint -summing the todue zero is since 0

1

2

220

21

11

Lin

i

iin

RvRR

vRivii

vRiv

=

+−==

+=1

1

11

1

RiRi

ivZ inin ===

-+ ++

--

BME372NewSchesser

100

Op-amp

•  Because we assumed that the Op-amp was ideal, we found that with negative feedback we can achieve a gain which is:

1.  Independent of the load 2.  Dependent only on values of the circuit

parameter 3.  We can choose the gain of our amplifier by

proper selection of resistors.

BME372NewSchesser

101

Non-inverting Amp 1.  First check: negative feedback? 2.  Next apply, summing point constraint 3.  Use circuit analysis

+

-- vo

+

--

RL R2

R1

+

--

vi

+

--

vf

io

Iin= 0

+

-

vin

1

1 2

2 1 2

1 1

0

;

1

Since 0;

in i f f f

f o in

ov

in

inin in

in

v v v v vRv v v

R Rv R R RAv R R

vi Zi

= + = + =

= =+

+= = = +

= = = ∞Note:1.  Thegainisalwaysgreaterthanone2.  Theoutputhasthesamesignasthe

input

BME372NewSchesser

102

Non-inverting Amp Special Case What happens if R2 = 0?

+

-- vo

+

--

RL

+

--

vi

io

Iin= 0

+

-

vin

∞===

=+==

==+

=

=+=+=

in

ininin

in

o

inoof

fffiin

ivZi

RR

vvAv

vvvRRv

vvvvv

;0 Since

10

;0

0

1

1

1

1

This is a unity gain amplifier and is also called a voltage follower.

BME372NewSchesser

103

Special Amplifiers

•  Summer (Homework Problem) •  Instrumentation Amplifier – Uses 3 Op-amps – One as a differential amplifier – Two Non-inverting Amps using for providing gain

BME372NewSchesser

104

Medical Instrumentation Amplifier

Differential Amplifier

+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1

R3

R4 +

-

vo

R5

R6

Non-inverting Amplifier


BME372NewSchesser

105


+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1

R3

R4

vo

R5

R6




v2D

v1D

+

-

BME372NewSchesser

106

Medical Instrumentation Amplifier 2

6 5

2

6 6 5 5

5 62

6 6 5 5

41

3 4

5 62 41

6 3 4 6 5 5

5 5 6 41 2

6 5 3 4

5 6 4

5 3 4

51 2

6

1 1( )

( )

( )

( )

Chose 1

( )

D x x o

oDx

oDx

y D x i x

oDD

o D D

o D D

v v v vR R

vv vR R R R

R R vv vR R R R

Rv v v v vR R

R R vv R vR R R R R R

R R R Rv v vR R R RR R RR R R

Rv v vR

− −=

−− + =

+ −− =

= = − =+

+ −− =+

+= −+

+ =+

= −

R3

R4

vo

R5

R6


v2D

v1D

vx

+

-

4

3

5

6

3564

45356454

43

4

5

65

1 Chose

RR

RR

RRRRRRRRRRRR

RRR

RRR

=

=+=+

=+

+

vi=0

vy

BME372NewSchesser

107


+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1

R3

R4 +

-

vo

R5

R6




v2D

v1D

BME372NewSchesser

108


+

--

R2

R1

+

-

v2

+

--

R2

R1

+

-

v1 Non-inverting Amplifier


v2D

v1D vA

AD

AD

AD

AD

vRRv

RRRv

vRRv

RRRv

vRRv

RRRv

Rvv

Rvv

1

21

1

211

1

22

1

212

1

22

12

22

1

2

2

22

Likewise

)11(

−+=

−+=

−+=

−=−

BME372NewSchesser

109


))(1())((

)]([

)(

21

1

2

6

521

1

21

6

5

1

22

1

21

1

21

1

21

6

5

1

21

1

211

1

22

1

212

21

6

5

vvRR

RRvv

RRR

RRv

vRRv

RRRv

RRv

RRR

RRv

vRRv

RRRv

vRRv

RRRv

vvRRv

o

AAo

AD

AD

DDo

−+=−+=

−+−−+=

−+=

−+=

−=

+

-- R2

R1

+

-

v2

+

--

R2

R1 + -

v1

R3

R4

vo

R5

R6




+

-

-

+

BME372NewSchesser

110

Uses of the Differential Amplifier

BME372NewSchesser

111

Integrators and Differentiators

∫∫ −=−=

==

t

in

t

o

in

dxxvRC

dxxiC

v

tiRtvti

002

21

)(1)(1

)()()(

vin

C

Rvi

vo

-+

0 i2 i1

vin

R

Cvi

vo

-+

0 i2 i1

dttdvRCRtiv

tidttCdvti

ino

in

)()(

)()()(

2

21

−=−=

==

BME372NewSchesser

112

Frequency Analysis

vinvi

vo

-+

0 i1=vin/Z1

i2

1in

o

Lin1

o

1

11in

ZZ

VV

ZVZZ

ZIVII

ZIV

2

2

22

2

- )()(

oft independen is which )(-

zero )(virtually is since 0)()(constraintpoint -summing the todue )()(

zero )(virtually is since 0)()()(

=

=

+−==

+=

ωω

ω

ωωωω

ωωω

jj

j

vjjjj

vjjj

i

i

ZLvin

C

Rvi vo

-+

0i1=vin/Z1 i2

ZLvin

R

Cvi vo

-+

0i1=vin/Z1 i2

integratoran 1- )()( 2

RCjjj

ωωω −==

1in

o

ZZ

VV tordifferenia a - )(

)( 2 RCjjj ωωω −==

1in

o

ZZ

VV

Z2

Z1

BME372NewSchesser

113

Frequency Response

vin

CRvi

vo

-+

0 i2 i1

vin

RCvi

vo

-+

0 i2 i1

1- )()( 2

RCjjj

ωωω −==

1in

o

ZZ

VV

RCjjj ωωω −==

1in

o

ZZ

VV 2-

)()(

ω

ω

BME372NewSchesser

114

Frequency Response

vin

C2R1vi

vo

-+

0 i2 i1

vin

R2

C1vi

vo

-+

0 i2

i1

R2 VoutVin

= −Z2Z1

= −R2R1

1(1+ jωC2R2 )

=R2R1

1

1+ (ωC2R2 )2∠π − tan−1(ωC2R2 )

0

500

1,000

1,500

2,000

2,500

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ

R151C110mfR2100k

R1

12 2 1 1 2 1 11 12

1 1 1 1 1 1 1

tan ( )(1 ) 21 ( )

out

in

V Z R j C R R C R C RV Z R j C R R C R

ω ω π ωω ω

−= − = − = ∠− −+ +

0

500

1,000

1,500

2,000

2,500

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ

R151C110mfR2100k

BME372NewSchesser

115

Homework •  Probs 2.2, 2.5, 2.6, 2.10, 2.22, 2.24, 2.25,

2.28 •  Calculate and plot the output vs frequency

for these circuits. R1=1k, R2=3k, C=1µf Use Matlab to calculate the Bode plot

vin

C

R1vi

vo-+

R2

vinCR1

vivo

-+

R2

vin

C

R1vi

vo-+

R2R2

vinC

R1

vivo

-+

BME372NewSchesser

Documents

Operational Amplifiersjoelsd/electronicsnew/Classnotes... · 2020. 9. 29. · Op-amp • Because we assumed that the Op-amp was ideal, we found that with negative feedback we can