Op-Amp Applications: Voltage to Current Converters, Instrument Amplifiers, and Sample-Hold, Slides of Analog Electronics

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Op

‐Amps

Applications

Voltage

to

current

converters,

2. Instrument

Amplifiers,

3. Sample

‐Hold

circuits.

x^

I

v^

Ri

v^ x

(Trans

‐resistance

amplifier)

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

2

d

L I^

o^

o^

d

a^

R^

r

v

A^

v^

R^

a^

r^

R^

r^

r

^

^

^

^

^

^

^

ro Load

rd

R

I^ v

av

D io vL

^





||^

1

o^

d^

o

R^

R^

r^

a^

r

^

^



0

I^

D^

L^

o O

D

v^

v^

v^

r i

av

^

^

^

^



Difference

Amplifier

and

Common

‐Mode

Rejection

Ratio

(CMRR)

A (or

A

dm

)^ =

differential

‐mode

gain

A^ cm

=^

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

( v

) .ic

In

general,

vo

^

Adm

vid

A

cm

vic Adm

 

Adm

vid

vic CMRR





CMRR

^

AdmAcm

and CMRR(dB)

^ 20log

(CMRR)

An

ideal

amplifier

has

A

cm

=^

0,^

but

for

a

real

amplifier,

vid vic v^

^

vid vic v^

vo

^

Adm

( v^1

 v

^ A

cm

v^1

 v

 



vo

^

Adm

( vid

Acm

( vic

docsity.com

^

^

4

1

2

2 2

1

1

3

4

1

o

R^

R^

R^

R

v^

v^

v

R^

R^

R^

R



^





v^ x

^

^

^

^

4

1

2

4

1

2

2

2

1

3

4

1

1

3

4

1

1 2 o^

dm

cm

R^

R^

R^

R^

R^

R

R^

R

v^

v^

v

R^

R^

R^

R^

R^

R^

R^

R

^

^

^



^



^

^

^



^

^

^



^



^

^

^



Difference

Amplifier:

Example

•^

Problem

:^ Determine

vo

•^

Given

Data

:^ R

10k

,^ R

=100k 2

v^1

V,

v^2

V

•^

Assumptions:

Ideal

op

amp.

Hence,

v‐

=^ v

and +

i=‐

i+

=^ 0.

•^

Analysis:

Using

dc

values,^ A

dm

R^2^ R^1

100k

10k

Vo

^

Adm

V^1

 V



^ 3)

Vo

20.0 V

Here

A

dm

is called

the

“differential

mode

voltage

gain”

of

the

difference

amplifier.

Finite

Common

‐Mode

Rejection

Ratio:

Example

•^

Problem

:^ Find

output

voltage

error

introduced

by

finite

CMRR.

•^

Given

Data

:^ A

dm

=^ 2500,

CMRR

=^

dB,

v^1

=^

V,

v^2

=^

V

•^

Assumptions:

Op

amp

is

ideal,

except

for

CMRR.

Here,

a^

CMRR

in

dB

of

dB

corresponds

to

a^

CMRR

of

•^

Analysis: The

output

error

introduced

by

finite

CMRR

is

of

the

expected

ideal

output.

5.001V

4.999V = 0.002V

5.001V

4.999V

5.000V

V^

6.25V

CMRR

In the "ideal" case,

5.00 V

vid vic

vic

v^

A^

v

o^

dm

id

v^

A^

v

o^

dm id

^

^

^



^

^

^



^

^

^



^



^



^

^

^

^

^

^

^

% output error

^

^ 5.005.

^ 25%

Differential

Gain

A^ dm

=^ 5 V/

mV

=^

1000

uA

CMRR

Test:

Common

Mode

Gain

CMRR

Calculation

for

uA

^



4

10 1000

CMRR

3.125 10

.

CMRR(dB)

20log

CMRR

89.9 dB

A^ dm Acm ^

^

^



^



v^1 v^2

vx vx