Tài liệu Circuits & Electronics P19 pdf

6.002 CIRCUITS ANDELECTRONICS The Operational Amplifier Abstraction... „ Can use as an abstract building block for more complex circuits of course, need to be careful about input and ou

6.002 CIRCUITS AND

ELECTRONICS

The Operational Amplifier

Abstraction

„ MOSFET amplifier — 3 ports

power port

input

port

output port +

–I

v

+ –

O

v

+

–

S V

„ Amplifier abstraction

+

–

I

v

+

–

S V

+ –v O I

Function of v I

Review

„ Can use as an abstract building block for more complex circuits (of course, need

to be careful about input and output)

„ Today

Introduce a more powerful amplifier

abstraction and use it to build more

complex circuits

Reading: Chapter 15 from A & L.

I

Function of v I

Review

Operational Amplifier

Op Amp

OUT

v

+ –

+ –IN

v

More abstract representation:

input

port

S

V

output port

power port

S

V

−

+ –

+ –

+ –

Circuit model (ideal):

i.e  ∞ input resistance

 0 output resistance

 “A” virtually ∞

 No saturation

O

v

Av

∞

→

A

+ –

+ –

v

v+

v –

0

=

i+

0

=

i–

(Note: possible confusion with MOSFET saturation!)

Using it…

+ –

V

V S = − 12

O

v

+

–

12V

+–

12V V S = 12V

Demo

IN

v

μV 10 μV

10

−

O

v

V 12

V 12

−

6

10

~

A

but unreliable, temp dependent

saturation active region

IN

v

Let us build a circuit…

Circuit: noninverting amplifier

Equivalent circuit model

1

R

O

v

+ –

2

R

IN

v

+

v

−

v

(v+ − v−)

A

+ –

0

=

i+

0

=

i–

op amp

1

R

O

v

+ –

2

R

IN

v

+ –

+

v

−

v

Let us analyze the circuit:

Find v O in terms of v IN, etc.

What happens when “A” is very large?

( + − −)

= A v v

v O

⎟

⎠

⎞

⎜

⎝

⎛

+

−

=

2 1

2

R R

R v

v

A IN O

IN 2

1

2

R R

AR 1

⎠

⎞

⎜

⎝

⎛

+ +

2 1

2

IN O

R R

AR 1

Av v

+ +

=

Let’s see… When A is large

Gain:

„ determined by resistor ratio

2 1

2

IN O

R R

AR 1

Av v

+ +

=

2

2

1 IN

R

R

R

≈

gain

Demo

Suppose A = 10 6

R 9

R 1 =

R

R2 =

R R

9

R

10 1

v

10

6 O

+ +

⋅

=

10 v

v O ≈ IN ⋅

10

1 10

1

v 10

6 IN 6

⋅ +

⋅

=

2 1

2 IN

R R

AR

Av

+

≈

e.g v IN = 5V

Suppose I perturb the circuit…

(e.g., force v O momentarily to 12V somehow).

Stable point is when v+ ≈ v-

Key: negative feedback Æ portion of

output fed to –ve input.

e.g Car antilock brakes

Æ small corrections

Why did this happen?

Insight:

+ –

R

IN

v =

+

–

R

IN

v

+

v

−

v

negative feedback

2

v O

5V

5V

10V

0

i–=

A

12V

6V 6V

Question: How to control a

high-strung device?

Antilock brakes

Michelin

k

yes/no

is it turning?

it’s all about control

di sc

v v powerful brakes

apply

release

More op amp insights:

Observe, under negative feedback,

0 A

v R

R R

A

v v

v

IN 1

2 1

⎟

⎠

⎞

⎜

⎝

=

=

− − +

− + ≈ v

v

We also know

i+ ≈ 0

i - ≈ 0

Æ yields an easier analysis method

(under negative feedback).

Insightful analysis method

under negative feedback

+

–

1

R

O

v

+

–

2

R

IN

v

IN

v

c

2

2

1 IN O

R

R

R v

g

IN

v

b

0

=

i

e

2

IN

R

v

d

2

IN

R

v

f

0 i

0 i

v v

≈

≈

≈

− +

− +

IN

v

a

+

v

+ –

IN

v

+

v

−

0

1 =

R

∞

=

2

R

2

2 1

R

R

R v

v O = IN + or

with

IN

v ≈

IN

v

c

IN

v

b

IN

v

a

voltage gain = 1

input impedance = ∞

output impedance = 0

current gain = ∞

power gain = ∞

+

v

+ –

IN

v

IN

v ≈

Why is this circuit useful?