EE2603-03 1

3. Cascade and Cascode configurations (12.2 - 12.5)

•  Cascaded amplifiers •  Cascode amplifiers •  Darlington pair •  Feedback pair

EE2603-03 2

Cascaded amplifiers Cascaded amplifiers consists of two or more stages of the basic BJT and/or FET circuits each of which utilize individual power supply VCC

Cascaded amplifiers are used to obtain higher voltage amplification, or matching of the input impedance with the transducer and matching the output impedances with the loading of the following stage.

Cascade amplifiers

+Vcc

-Vcc or "0"

Load RL

Transducer ROI1 I2 I3

Rin

Ro

Vin

Vo

EE2603-03 3

0.2K15k

VCC=12V

8.2k

b=100

47k

Vin

Rin

Iin

0.2K15k

VCC=12V

8.2k

b=100

47k VO

RO2Rin2

10k

IL

Example:1

.amplifierCascadeBJTBJTfolowingtheofii

A,vvA,R,RFind

1in

Li

1in

2ovT1in2o −==

Ω=====

+⎟⎠

⎞⎜⎝

⎛ +=+⎟⎟⎠

⎞⎜⎜⎝

⎛+

β==

===+

×=+

=

7.3mA03.7mV26

ImV26rmA03.7

k313.0V2.2)mA(I

7.0k2.0100

k37.11)mA(I7.0RR

I9.2V

k37.1115//47R,V9.24715

1512RR

RVV

CeC

CEB

CBB

B21

1CCBB

(a) Find re for both BJT

EE2603-03 4

(b) Find Rin, Ro, Av, and Ai of the cascade amplifier

32.9037.0345.0

037.036.0//2.8// 2

1

11 −==−=−==

kkk

rRR

vvA

einC

ino

vVoltage gain 1

kkrRivR eBinin

in 36.07.3100//37.11//1 =×=== βInput resistance 1

Output resistance 2

Voltage gain 2

Voltage gain overall ( ) ( ) 24.11358.12132.9AAvv

vv

vvA 2v1v

1o

2o

1in

1o

1in

2ovT =−×−=×=×==

0.2K15k

VCC=12V

8.2k

b=100

47k

Vin

Rin

Iin

0.2K15k

VCC=12V

8.2k

b=100

47k VO

RO2Rin2

10k

IL

k2.8RivR Co

oo ===

8.121037.0

10//2.8//

2−=−=−==

kkk

rRR

vvA

eLC

ino

v

87.4010

36.024.1135=

×−==

×⎟⎠⎞⎜

⎝⎛

===k

kRRA

R

Rvv

Rv

Rv

ii

ALinv

L

inino

inin

Lo

inL

iCurrent gain

EE2603-03 5

1k1M

1k

1k

12V

VO2

RO2Rin2

1k1M

1k

12V

Rin

VinIDSS=8mAVP=-8V

Example:2

.amplifierCascadeFETFETfolowingtheofvvA,R,RFindin

2ovTin2o −=

(a) Find gm of the FET ( )

( )

( ) ( )( )

mS235.1806.31

882

VV

1VI2

g

neglectedisitV8VbeyondisV94.20VButV94.20k1mA94.20orV06.3k1mA06.3V

94.20or06.32

88.17242

25657624I

064I24II16I64I8

I8648

8I

18VV

1II

IRIVbutk1R,mA8I,V8V

P

GS

P

DSSm

PGS

GS

D

D2DD

2DD

2D

2D

2

P

GSDSSD

DSDGSSDSSp

=⎟⎠

⎞⎜⎝

⎛−

−−

−

×−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=∴

∴−=−=

−=×−−=×−=∴

=±

=−±−−

=

=+−⇒−+=

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=∴

−=−===−=

EE2603-03 6

1k1M

1k

1k

12V

VO2

RO2Rin2

1k1M

1k

12V

Rin

Vin IDSS=8mAVP=-8V

( ) 235.1k1mS235.1M1//k1gvv

A m1in

1o1v −=×−=−==Voltage gain 1

Ω== MRR Gin 1Input resistance 1

Ω== kRR Do 12Output resistance 2

Voltage gain 2

Voltage gain overall ( ) ( ) 76.06175.0235.1AAvv

vv

vv

A 2v1v1o

2o

1in

1o

1in

2ovT =−×−=×=×==

( ) 6175.0k5.0mS235.1k1//k1gvv

A m1in

1o2v −=×−=−==

(b) Find AVT of the FET-FET cascade amplifier

EE2603-03 7

0.2K15k

VCC=12V

8.2k

b=100

47k

Vin

Rin

Iin

VDD=12V

RS=2.4k

IDSS=8mAVP=8V

RL=2.4kVo

Ro

Rin

RG=1M

Example:3

.amplifierCascadeFETBJTfolowingtheofvvA,R,RFind

1in

2ovT1in2o −=

Ω=====

+⎟⎠

⎞⎜⎝

⎛ +=+⎟⎟⎠

⎞⎜⎜⎝

⎛+

β==

===+

×=+

=

7.3mA03.7mV26

ImV26rmA03.7

k313.0V2.2)mA(I

7.0k2.0100

k37.11)mA(I7.0RR

I9.2V

k37.1115//47R,V9.24715

1512RR

RVV

CeC

CEB

CBB

B21

1CCBB(a) Find re of the BJT

EE2603-03 8

( )

( )

( ) ( )

mSVV

VI

g

neglectedVVbeyondVVButVkmAorVkmAV

orI

IIIII

II

VVII

IRIVbutkRmAIVV

PGS

P

DSSm

PGS

GS

D

DDDDD

DD

PGS

DSSD

DSDGSSDSSp

94.0825.41

8821

281.15

1.154.229.625.44.277.1

29.677.152.11

04.264.4652.11

6.147421534.46

0644.4676.54.3876.5648

4.28648

84.2

181

4.24.2,8,8

22

222

=⎟⎠

⎞⎜⎝

⎛−

−−

−

×−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−

−=∴

∴−=−=

−=×−−=×−=∴

=±

=−±−−

=

=+−⇒−+=

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=∴

−=−===−=

VDD

RS=2.4k

IDSS=8mAVP=8V

RL=2.4k

Vin

Vo

Ro

Rin

RG=1M

(b) Find gm of the FET

EE2603-03 9

0.2K15k

VCC=12V

8.2k

b=100

47k

Vin

Rin

Iin

VDD=12V

RS=2.4k

IDSS=8mAVP=8V

RL=2.4kVo

Ro

Rin

RG=1M

6.221037.0

1//2.8// 2

1

11 −=−=−==

kMk

rRR

vvA

einC

ino

vVoltage gain 1

kkrRivR eBinin

in 36.07.3100//37.11//1 =×=== βInput resistance 1

( ) Ω==== kkkkkgRR mSo 74.006.1//4.294.01//4.2/1//2

( ) 53.013.213.1

)4.2//4.2(94.01)4.2//4.2(94.0

//1)//(

2

22 ==

+=

+==

kkkk

RRgRRg

vvA

LSmLSm

ino

v

Output resistance 2

Voltage gain 2

Voltage gain overall 45.11753.06.221AAvv

vv

vvA 2v1v

1o

2o

1in

1o

1in

2ovT −=×−=×=×==

Ω== 1M R R Gin2

(c) Find AVT of the BJT-FET cascade amplifier

EE2603-03 10

Cascode amplifiers

Cascode amplifiers

+Vcc

-Vcc or "0"

Load RL

Transducer RO

Rin

Ro

Vin

VoI1

I1

Cascode amplifiers consists of two or more stages of the basic BJT and/or FET circuits each of which utilize single power supply VCC for both stages

Cascaded amplifiers are used to obtain higher voltage amplification, or matching of the input impedance with the transducer and matching the output impedances with the loading of the following stage.

EE2603-03 11

1.1k4.7k

b=200Vin

Rin

IinVB1

VE1

VC1

5.6k

VCC=18V

1.8k

b=200

6.8k VO

RO2

10k

IL

VC2

VB2

VE2

IC1=IC2

Example:4 – (page 549 text)

amplifier.CascodeBJTBJTfolowingtheofii

A,vvA,R,RFind

in

Li

in

o2vTino2 −==

Ω===⇒==

=−

=⇒=++

×=

8.69.3

268.3

8.31.1

7.09.49.48.66.57.4

7.418

2121

11

mAmVrrmAII

mAk

VVIVkkk

kV

eeCC

CB

(a) Find re

32.224k0068.0k10//k8.1

rR//R

A)Cwith(CBisstageSeconde

LC2VB ===∴

1rrAloadasrwithCEisstageFirste

e1V2e −=−=∴

32.22432.2241212 −=×−=×== VVin

oVT AA

VVA

Ω=×== kkkrRR eBin 887.00068.0200//6.5//7.4// 1β Ω== k8.1RR 2C2o

(b) Find Ro2, Rin, AVT, Ai

89.1910

887.032.224−=

×−==

kk

RRAAL

inVTi

EE2603-03 12

Example:4 – (page 301 text) amplifier.CascodeBJTFETfolowingtheof

ii

A,vvA,R,RFind

in

Li

in

o2vTino2 −==

VCC=20V

1.1k

b=160

91k VO

RO2

10k

IL

VC2

VB2

VE2

ID=IC

RS=1.2k

IDSS=6mAVP=6V

Vin

Rin

RG=18k

Iin

(a) Find VGS and gm of FET

( )

( )

( ) ( )( )

mS586.1624.11

662

VV

1VI2

g

V24.1VisanswerCorrectneglectedisitV6VbeyondisV768.15VBut

V768.15k2.1mA89.153.3orV24.1k2.1mA78.33.3V

mA89.15ormA78.388.2

43.1732.2888.2

2.49880232.28I

049.86I32.28I44.1I32.22I44.149.86I6

I2.13.36366

6I2.13.3

16VV

1II

V3.3k91k18k1820whereV

I2.13.3RIVVbutk2.1R,mA6I,V6V

P

GS

P

DSSm

GS

PGS

GS

D

D2DD

2DD

2D

2D

2

P

GSDSSD

G

DSDGGSSDSSp

=⎟⎠

⎞⎜⎝

⎛−

−−

−

×−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=∴

−=

∴−=−=

−=×−−=×−=∴

=±

=−±−−

=

=+−⇒−+=

+−−=⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=∴

=+

×=

−=−===−=

EE2603-03 13

(c) Find Ro2, Rin, AVT, Ai

(b) Find re

Ω==⇒== 88.6mA78.3mV26rmA1.2II eCD

Ω== kkkRin 1591//18Ω== kRR Co 1.12

144k0068.0k10//k1.1

rR//R

A)Cwith(CBisstageSeconde

LC2VB ===∴

0109.088.610586.1rgAloadasrwithCSisstageFirst 32em1V2e −=××−=−=∴ −

56.11440109.0AAVV

A 2V1Vin

2oVT −=×−=×==

34.2k10

k1556.1RRA

AL

inVTi −=

×−==

VCC=20V

1.1k

b=160

91k VO

RO2

10k

IL

VC2

VB2

VE2

ID=IC

RS=1.2k

IDSS=6mAVP=6V

Vin

Rin

RG=18k

Iin

EE2603-03 14

Darlington pair Type 2N999 – NPN Darlington-connected Silicon

Transistor Package

Parameter Test conditions Min Max

VBE IC=100mA 1.8V

hFE (βD) IC=10mA IC=100mA

4000 7000

70,000

BEEEB

BDBDE

EDB

BECCB

VRIV,II)1(I

,RR

VVI

+=

β=+β=

β+

−=

Q2

Q1

E

C

BQD

E

CB

Q2

Q1

E

C

B

VCC

IB

IE

RE

RB

IC

DC analysis of the Darlington pair

EE2603-03 15

Q2

Q1

E

C

B

18V

IB

IE

3.3MW

IC

RB

bD = 8000VBE = 1.6V

390W

Vin

VoRin

Ro

mA48.20A56.28001I8001I)1(I BBDE =µ×==+β=

amplifier.DarlingtonfolowingtheofV,V,I,IFind BEEB

V6.96.18VVV BEEB =+=+=

A56.23908000M3.3

6.118RR

VVI

EDB

BECCB µ=

Ω×+Ω

−=

β+

−=

V8390mA48.20RIV EEE =Ω×==

EE2603-03 16

RE

riRB

vo

Rin Ro

vin

iin ib CB

EbDib

ioIQ2

Q1

E

C

B

18V

iin

IC

RB

bD = 8000VBE = 1.6VVin

Vo

Rin

Ro

RE

ioI

EDiBb

inBin

EDib

inEbDibin

i

oinb

Rr//RiV//RRbut

RriVRiriVand

rVVi

β+==

β+=⇒β+=−

=

AC analysis of the Darlington pair

( )( )

( ) [ ]E

EDB

E

EDiB

EDi

ED

E

inv

E

inino

inin

Eo

in

oi R

R//RR

Rr//RRr

RRRA

RRV/V

R/VR/V

iiA

β=

β+×⎥⎥⎦

⎤

⎢⎢⎣

⎡

β+

β==

×===

[ ] )Rrif(1Rr

RRriRi

vvA EDi

EDi

ED

EDib

EbD

in

ov β<<=

β+

β=

β+

×β==

D

iE

bD

ibE

bD

oE

o

oE

oo

r//Riri

//Ri

v//Ri

v//RIvR

β=

β−

−=

β−=

−==

EE2603-03 17

Q2

Q1

E

C

B

18V

IB

IE

3.3MW

IC

RB

bD = 8000VBE = 1.6V

390W

Vin

VoRin

Ro

ri = 5kW

amplifier.DarlingtonfolowingtheofiiA,

vvA,R,RFind

in

oi

in

ovino ==

( ) Ω==×+=β+== M605.1M125.3//M3.3k39.08000k5//M3.3Rr//RiV//RR EDiBb

inBin

4107k39.0

k1605998.0RRA

iiA

E

inv

in

oi =

×===

1998.0k3125k5

k3125k39.08000k5

k39.08000Rr

RA

EDi

EDv ≈=

+=

×+

×=

β+

β=

Ω=Ω==β

= 625.0625.0//k39.08000

k5//k39.0r//RRD

iEo

EE2603-03 18

Feedback pair Q2

Q1

C

E

BbD

C

E

B

VCC

IB

IE

RC

RB

IC

Q2

Q1

C

E

B

IB2

[ ]

( )7.0RIVVRIVRIV

III1I,II

,RR

VVI

ECCC1BECCCCBBB

B212B22B2EB12B

21DCDB

1EBCCB

−−=−−==

ββ=β≈+β=β=

β×β=ββ+

−=

DC analysis of the Feedback pair

EE2603-03 19

amplifier.pairFeedbackfolowingtheofV,V,I,IFind BCEB

18V

IB

IE

RC

RB

IC

Q2

Q1

C

E

B

IB2

2MW

75Wb1=140b2=180

( ) mA1.112A45.425200IIII1I BDB212B22B2E =µ×=β=ββ=β≈+β=

V6.975mA1.11218RIVV CCCCC =Ω×−=−=

[ ]2520180140A45.47525200M2

7.018RR

VVI D

CDB

1EBCCB =×=βµ=

Ω×+

−=

β+

−=

V9.8M2A45.4RIV BBB =Ω×µ==

EE2603-03 20

RC

ri1RB vo

Rin Ro

vin

iin ib C1B1

E1 b1ib

IRC I

C2B2

E2 b2ib2

ib2ri2 -b2ib2

Q2

Q1

E

C

B

VCC

iin

RB

b1 = 140

Vin

Vo

Rin

Ro

RE

iob2 = 180RC

( ) ( ) ( )

( ) C21BC211iBb

inBin

C211ibC2b11ibb12b2C1ibin

R//RRr//RiV//RRbut

RriR1iriiiRriV

ββ=ββ+==

ββ+=+ββ+=β+β−+=

AC analysis of the Feedback pair

( )( )

( ) [ ]

( ) ( )C

C21B

C21

C21B21

C

C21B

C21i

C21

C

inv

C

inino

inin

Co

in

oi

RR//R

RR//R

RR//R

RrR

RRA

RRV/V

R/VR/V

iiA

ββ=

ββ

ββββ=

ββ×⎥⎦

⎤⎢⎣

⎡

ββ+

ββ=

−=

×−=

−==

[ ] )Rrif(1Rr

RRriRi

vvA C21i

C21i

C21

C21ib

Cb21

in

ov ββ<<=

ββ+

ββ=

ββ+

×ββ==

Ai of the Feedback pair

Rin of the Feedback pair

Av of the Feedback pair

EE2603-03 21

RC

ri1RB vo

Ro

vin=0

ib C1B1

E1 b1ib

I

C2B2

E2 b2ib2

ib2ri2 -b2ib2

IRC

( )( )

21

1iC

b21

1ibC

b21

oC

RC

o

RC

oinoo

b21RC2b1RC

b1b12RCb12b2RC

r//Ri

ri//R

iv//R

IIv//

Iv

I)0Vat(vR

iI1iIiiIiiII

ββ=

ββ−

−=

ββ−=

−=

==

ββ−=−ββ−=

β+β−×β+=β+β+=

Ro of the Feedback pair