1

ECE 255: L25

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

Mark Lundstrom School of ECE

Purdue University West Lafayette, IN USA

Spring 2019 Purdue University

Announcements

2

1)  HW8 due at 5:00 PM Friday, March 29

2)  Exam 3 is at 6:30 PM, Tuesday, April 2

3)  Spice Project 3 will be due on April 17

Lundstrom: 2019

CS amplifier

3 Lundstrom: 2019

+

υi

−

+

υ0

−

D S

VDD

RD

G

Aυo= υo

υi

= −gmRD

Rin = ∞

Ro = RD

CS amplifier with active load

4 Lundstrom: 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

Solution: cascade amplifiers

5 Lundstrom: 2019

Examples: CE:CE CC:CE CE:CC

CE:CB or CS:CG = “cascode”

Stage 2 Stage 1

Basic cascode

6 Lundstrom: 2019

+

υi

−

υ0

D S

G

VG2

“cascaded cathode”

grid

plate

cathode

Q1

Q2

Cascode amplifiers

7 Lundstrom: 2019

CG CS

Discrete transistor version first Then MOS IC cascodes

Outline

8

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: 2019

Discrete CS:CG cascode

9

+

υi

−

υ0

VDD

RD

RSυsig

Rsig

CS

R1

R2

R3

CG

CC1

CC2

RL

Q1

Q2

CS:CG discrete cascode (without bias resistors)

10

+υi

−

Aυo= Aυ1

× Aυ2

RD

+υo

−

Aυo= −gm1

1gm2

⎛⎝⎜

⎞⎠⎟× gm2RD

Aυo= −gm1RD

Rin = ∞

Ro = RD

Q1

Q2

CS:CG cascode summary

11 Lundstrom: 2019

CS CG

•  Input resistance of CS •  Output resistance of CS •  Voltage gain of CS

(excellent high frequency response)

Aυo≈ −1 Aυo

= +gmRD

Outline

12

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: 2019

CS:CG cascode (IC)

13 Lundstrom: 2019

+υi

−

υo

Rin = ∞

Q1

Q2

Aυo= Aυ1

× Aυ2

Aυo= −gm1 ro1 || Rin2( )× gm2Rout2

Rin2

Rout2

Recall L24

14 Lundstrom: 2019

RL

RS

Rout = ro + 1+ gmro( )Rseries

Rout ≈ ro + gmro( )RS

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

Rin ≈1gm

+ RL

gmro

Rout ≈ gmro( )RS

First stage gain

15 Lundstrom: 2019

+υi

−

υo

Rin = ∞

Q1

Q2 Aυ1

= −gm1 ro1 || Rin2( )

Rin2

Rout2

Rin2 ≈1gm

+ RL

gmro

Aυ1= −gm1 1 gm2( ) ≈ −1

Second stage gain

16

+υi

−

υo

Rin = ∞

Q1

Q2

Aυ2= +gm2Rout2

Rin2

Rout2

Aυ2= +gm2 gm2r02r01( )

Rout 2 ≈ ro2 + gm2ro2( )RSRout 2 ≈ gm2ro2( )ro1

Total gain

17

+υi

−

υo

Rin = ∞

Q1

Q2

Rin2

Rout2

Aυo= − gm1r01( ) gm2r02( )

Aυo= Aυ1

× Aυ2

Aυ1= −gm1 1 gm2( )

Aυ2= +gm2 gm2r02r01( )

Total gain of cascode

18

+υi

−

υo

Rin = ∞

Q1

Q2

Rout2

Aυo= − gmr0( )2

Aυo= −A0

2

An easier way

19

+υi

−

Q1

Q2

Rin2

Rout2

gm1υi

υo = − gm1υi( )Rout2

Rout 2 ≈ gm2ro2( )ro1

Aυi= − gmro( )2

Aυo= −A0

2

gm1υi

Recall: CS amplifier with active load

20 Lundstrom: 2019

+

υi

−

+

υ0

−

D S

VDD

G

IO

Ideal current source

RO = ∞

Aυo= −gmro

Aυo= −A0

Aυo= − A0( )2Cascode:

Outline

21

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: 2019

Implementation

22

VDD

R1

Q2

IREF

Q3

cascode amplifier

υo

Implementation

23 Lundstrom: 2019

+υi

−

υ0

VG2

Q1

Q2

VDD

IO Aυo= − gmr0( )2

Aυo= −gm gmr0( )ro⎡⎣ ⎤⎦ || rop

Aυo≈ −gmro = −A0

PNP current mirror

The solution is to use a cascode current course.

(S&S p. 549)

Include a load resistor

24

+υi

−

υo

Q1

Q2

Rin2

Rout2 RL

Aυo= −gm1Rout2 || RL

Rout2 = gm2ro2( )ro1

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

Double cascode

25

+υi

−

Q2

Q1

+υo

−

Q3

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

Aυo= −gmRo3 Aυo

= − gmr0( )3

Outline

26

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: 2019

BJT cascode

27

+υi

−

Q1

Q2

Rout2

gm1υi

υo = − gm1υi( )Rout2

Recall L24

28

RL

RS

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

Q2 Rout = ro2 + 1+ gm2ro2( ) ro2 || rπ 2( )

Rout ≈ gm2ro2( ) ro2 || rπ 2( )

cannot be as larges as for the MOS case

BJT cascode

29

+υi

−

Q1

Q2

Rout2

gm1υi υo = − gm1υi( )Rout2

Rout ≈ gm2ro2( ) ro2 || rπ 2( )

Aυo= −gm1 gm2ro2( ) ro2 || rπ 2( )

Aυo max= −gm1 gm2ro2( )rπ 2

Aυo max= −βgmro = −βA0

BJT cascode

30

+υi

−

Q1

Q2

Rout2

gm1υi

Aυo max= −βA0

See the discussion in Sec. 8.5.6 of S&S

Outline

31

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascode

6)  Summary

Lundstrom: 2019

Cascode:

Aυo= − gmro( )2

Summary

32 Lundstrom: 2019

+

υi

−

+

υ0

−

D S

VDD

G

Aυo= −gmro

IO

Ideal current source

RO = ∞Common Source:

Double Cascode:

Aυo= − gmro( )3

Questions

Lundstrom: 2019 33

1)  Introduction

2)  Discrete cascode

3)  IC MOS cascode

4)  Discussion

5)  BJT cascodes

6)  Summary