Amplifiers --small Signal Model

6.002

CIRCUITS AND

ELECTRONICS

Amplifiers -Small Signal Model

6.002 Fall 2000

Lecture 10

1

Review „

MOSFET amp VS RL

vO vI

„

iDS

Saturation discipline — operate MOSFET only in saturation region

„Large signal analysis 1. Find vO vs vI under saturation discipline.

2. Valid vI , vO ranges under saturation discipline.

Reading: Small signal model -- Chapter 8

6.002 Fall 2000

Lecture 10

2

Large Signal Review 1

vO vs vI K (vI −1)2 RL

2

valid for vI ≥ VT and vO ≥ vI – VT K 2 (same as iDS ≤ vO ) 2

vO = VS −

6.002 Fall 2000

Lecture 10

3

Large Signal Review

2

Valid operating ranges

V S

v O

5V

corresponding interesting region for vO

vO > vI −VT

vO = vI −VT vO < vI −VT

1V

vI

VT

1V

2V

“interesting” region for vI . Saturation discipline satisfied.

6.002 Fall 2000

Lecture 10

4

But… VS 5V

vO

vO = vI −VT

vO 1V

vI VT

1V

Demo

vI

2V

Amplifies alright, but distorts

vI

vO

t

Amp is nonlinear … / 6.002 Fall 2000

Lecture 10

5

Small Signal Model vO

~ 5V VS

Focus on this line segment

(VI , VO )

~ 1V vI

VT 1V

~ 2V 2 K (vI − VT ) vO = VS − RL 2 Amp all right, but nonlinear! Hmmm … So what about our linear amplifier ???

Insight: But, observe vI vs vO about some point (VI , VO) … looks quite linear ! 6.002 Fall 2000

Lecture 10

6

Trick ∆vO

vo VO

vi

(VI ,VO )

looks linear

VI ∆vI ™

Operate amp at VI , VO Æ DC “bias” (good choice: midpoint of input operating range)

™ Superimpose small signal on top of VI

™ Response to small signal seems to be approximately linear

6.002 Fall 2000

Lecture 10

7

Trick ∆vO

vo VO

vi

(VI ,VO )

looks linear

VI ∆vI

™

Operate amp at VI , VO Æ DC “bias” (good choice: midpoint of input operating range)

™ Superimpose small signal on top of VI ™ Response to small signal seems to be approximately linear Let’s look at this in more detail — I graphically II mathematically III from a circuit viewpoint 6.002 Fall 2000

Lecture 10

next week 8

I Graphically

We use a DC bias VI to “boost” interesting input signal above VT, and in fact, well above VT .

VS RL

interesting input signal

∆vI + – VI + –

vO

Offset voltage or bias

6.002 Fall 2000

Lecture 10

9

Graphically

VS RL

interesting input signal

vO

∆vI + – VI + –

VS

vO

operating point

VO

0

VI , VO

VT

Good choice for operating point:

midpoint of input operating range

6.002 Fall 2000

vO = vI −VT

vI

Lecture 10

VI

10

Small Signal Model

aka incremental model aka linearized model

Notation — Input:

vI = VI + vi

total DC small variable bias signal (like ∆vI) bias voltage aka operating point voltage Output: vO = VO + vo Graphically, vI

vO

vi

vo

VI

VO

0

6.002 Fall 2000

t

0

Lecture 10

t

11

II Mathematically

(… watch my fingers)

RL K 2 vO = VS − (vI −VT ) VO = VS − RL K (VI −VT )2

2

2 “

substituting vI = VI + vi vi << VI

RL K vO = VS − 2

( [VI + vi ] − vT )2

RL K 2

( [VI −VT ] + vi )2

= VS −

(

RL K [VI −VT ]2 + 2 [VI − vT ]vi + vi 2 = VS − 2

RL K VO + vo = VS − (VI − VT )2 − RL K (VI −VT ) vi

2 From “,

)

vo = −RL K (VI −VT ) vi

gm

6.002 Fall 2000

related to

Lecture 10

VI

12

Mathematically vo = −RL K (VI −VT ) vi

gm

related to

VI

vo = −g m RL vi

For a given DC operating point voltage VI,

VI – VT is constant. So,

vo = − A vi

constant w.r.t. vi

In other words, our circuit behaves like a linear amplifier for small signals

6.002 Fall 2000

Lecture 10

13

Another way RL K vO = VS − (vI −VT )2

2

(

)

R K 

 2 L v −V VS −  I T 2 d   vo = dv I

⋅ vi

v = V

I I

slope at VI

vo = −RL K (VI −VT ) ⋅ vi

g m = K (VI −VT ) A = −g m RL

amp gain

Also, see Figure 8.9 in the course notes for a graphical interpretation of this result 6.002 Fall 2000

Lecture 10

14

More next lecture … Demo

iDS load line

input signal response operating point VI

VO

vO

How to choose the bias point: 1. Gain component g m ∝ VI 2. vi gets big Æ distortion. So bias carefully 3. Input valid operating range. Bias at midpoint of input operating range for maximum swing. 6.002 Fall 2000

Lecture 10

15