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1 ECE 255 Cascode Amplifiers (Sedra and Smith, 7 th Ed., Sec. 8.5) Mark Lundstrom School of ECE Purdue University West Lafayette, IN USA ECE 255: Fall 2019 Purdue University Lundstrom: Fall 2019

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### Transcript of L32 Cascode Amplifiers - nanoHUB.org

1

ECE 255

Cascode Amplifiers (Sedra and Smith, 7th Ed., Sec. 8.5)

Mark Lundstrom School of ECE

Purdue University West Lafayette, IN USA

ECE 255: Fall 2019 Purdue University

Lundstrom: Fall 2019

2 Lundstrom: Fall 2019

+

υi

+

υ0

D S

VDD

G

IO

Ideal current source

RO = ∞ Aυo= υo

υi

= −gmro

Rin = ∞

Ro = ro

But this gain is not large enough

Aυo= −A0

3 Lundstrom: Fall 2019

Examples: CE:CE CC:CE CE:CC

CE:CB or CS:CG = “cascode”

Stage 2 Stage 1

Basic cascode

4 Lundstrom: Fall 2019

+

υi

υ0

D S

G

VG2

grid

plate

cathode

Q1

Q2

Cascode amplifiers

5

CG CS

1) Discrete transistor version first 2) Then MOS IC cascodes 3) Finally, BJT cascodes

Outline

6

1)  Introduction

2)  Discrete cascode!

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: Fall 2019

Discrete CS:CG cascode

7

+

υi

υ0

VDD

RD

RSυsig

Rsig

CS

R1

R2

R3

CG

CC1

CC2

RL

Q1

Q2

Ignore ro

CS:CG discrete cascode (without bias resistors)

8

+υi

Aυo= Aυ1

× Aυ2

RD

+υo

Aυ1= ?

Aυ2= ?

Rin = ∞

Ro = RD

Q1

Q2

Rin2 =1gm2

*

CS:CG discrete cascode (without bias resistors)

9

+υi

Aυo= Aυ1

× Aυ2

RD

+υo

Aυo= −gm1

1gm2

⎛⎝⎜

⎞⎠⎟× gm2RD

Rin = ∞

Ro = RD

Q1

Q2

Aυo= −gm1RD

Aυ1= − gm1 gm2 Aυ2

= +gm2RD

CS:CG cascode summary

10 Lundstrom: Fall 2019

CS CG

•  Input resistance of CS •  Output resistance of CS •  Voltage gain of CS (not gain squared!) •  No Miller effect

(excellent high frequency response)

Aυo≈ −1 Aυo

= +gmRD

Outline

11

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: Fall 2019

CS:CG cascode (IC)

12 Lundstrom: Fall 2019

+υi

υo

Rin = ∞

Q1

Q2

Aυo= Aυ1

× Aυ2

Aυ1= −gm1 ro1 || Rin2( )

Rin2 Rout2

Aυ2= +gm2Rout2

Recall L31

13 Lundstrom: Fall 2019

RL

RS

Rout = ro + 1+ gmro( )Rseries

Rout ≈ ro + gmro( )RS ≈ gmro( )RS

Rin =ro + RL( )1+ gmro( )

Rin ≈1gm

+ RL

gmro

The easy way

14

+υi

Q1

Q2

Rin2Rout2

gm1υiυo = − gm1υi( )Rout2

Rout 2 ≈ gm2ro2( )ro1

Aυi= − gmro( )2

Aυo= −A0

2

gm1υi

Lundstrom: Fall 2019

Rout ≈ gmro( )RS

*

Aυo= − A0( )2Cascode:

Recall: CS amplifier with active load

15 Lundstrom: Fall 2019

+

υi

+

υ0

D S

VDD

G

IO

Ideal current source

RO = ∞

Aυo= −gmro

Aυo= −A0

Outline

16

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: Fall 2019

Implementation

17

VDD

R1

Q2

IREF

Q3

cascode amplifier

υo

Implementation

18 Lundstrom: Fall 2019

+υi

υ0

VG2

Q1

Q2

VDD

IO

Aυo= −gm gmr0( )ro⎡⎣ ⎤⎦ || rop

Aυo≈ −gmro = −A0

PNP current mirror

The solution is to use a cascode current course.

(S&S p. 549)

19

+υi

υo

Q1

Q2

Rin2

Rout2 RL

Aυo= −gm1Rout2 || RL

Rout2 = gm2ro2( )ro1

Can you find the gain of stage 1 (it depends on RL!). See S&S, Sec. 8.5.3

Aυo≈ −gm1RL

Double cascode

20

+υi

Q2

Q1

+υo

Q3

⇐ R02 ⇐ R03

Aυo= −gmRo3

*

Double cascode

21

+υi

Q2

Q1

+υo

Q3

⇐ R02 = gm2ro2( )r01 ⇐ R03 = gm3ro3( )R02R03 = gm3ro3( ) gm2ro2( )r01

Aυo= −gmRo3 Aυo

= − gmr0( )3

Outline

22

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: Fall 2019

BJT cascode

23

+υi

Q1

Q2

Rout2

gm1υi

υo = − gm1υi( )Rout2gm1υi

+υo

Recall L31

24

RL

RS

Rout = ro2 + 1+ gm2ro2( ) RS || rπ 2( )

Q2 Rout ≈ gm2ro2( ) RS || rπ 2( )

cannot be as larges as for the MOS case

Lundstrom: Fall 2019

BJT cascode

25

+υi

Q1

Q2

Rout2

gm1υi

υo = − gm1υi( )Rout2

Rout ≈ gm2ro2( ) ro2 || rπ 2( )

Aυo≈ −gm1 gm2ro2( )rπ 2

Aυo= −βgmro = −βA0

Lundstrom: Fall 2019

gm1υi

Rout ≈ gm2ro2( ) RS || rπ 2( )

Rout ≈ gm2ro2( )rπ 2

BJT cascode

26

+υi

Q1

Q2

Rout2

gm1υi

Aυo= −βA0

See the discussion in Sec. 8.5.6 of S&S

Lundstrom: Fall 2019

Outline

27

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: Fall 2019

Cascode:

Aυo= − gmro( )2

MOS summary

28 Lundstrom: Fall 2019

+

υi

+

υ0

D S

VDD

G

Aυo= −gmro

IO

Ideal current source

RO = ∞Common Source:

Double Cascode:

Aυo= − gmro( )3

Cascode:

Aυo= −β gmro( )

BJT summary

29 Lundstrom: Fall 2019

+

υi

+

υ0

C E

VDD

B

Aυo= −gmro

IO

Ideal current source

RO = ∞ Common Emitter:

Questions

Lundstrom: Fall 2019 30

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascodes

6)  Summary