7/27/2019 Lecture23(20-3-11)

http://slidepdf.com/reader/full/lecture2320-3-11 1/4

Understanding of concepts related to:

o Frequency Response CE Amplifier

Lecture GoalsEE 332

DEVICES AND CIRCUITS II

Lecture 23

FREQUENCY RESPONSE (3)

FREQUENCY RESPONSE OF CE AMPLIFIER

VCC

v i

R2

R1

C1

C2

vo

Rs

RC

RE CE

Q1

, ,0.7 ; 0.1 ;

1(2 3904) 100; 150 ; 100 ;

1 ; 1

B E on CE sat

F A

je jc

V V V V

Q N ps V V

C pF C pF

β τ

= = ⎫

⎪ ⎪= = = ⎬

⎪ ⎪= =⎩ ⎭

1 2100 ; 10 ; 100 ;

10 ; 1 ; 10 ;s

C E L

R R k R k

R k R k R k

12 V1 2 10 ;E CC C C C F V

= Ω = Ω = Ω ⎫

⎪ ⎪= Ω = Ω = Ω ⎬

⎪ ⎪= = = = ⎭

FREQUENCY RESPONSE OF CE AMPLIFIER

Three frequency regions: high frequency (consider internal capacitors)Low frequency (consider external capacitors)Medium frequency: don’t have to consider capacitors as before

Consider medium frequencies:

Step 1: DC analysis: Find IC, IB, and VCE

391 ; 3.91 ;C B I A I Aμ=

7.70CE V V

⎧ ⎫⎨ ⎬= ⎭

98

7/27/2019 Lecture23(20-3-11)

http://slidepdf.com/reader/full/lecture2320-3-11 2/4

FREQUENCY RESPONSE OF CE AMPLIFIER

15.6 ; 6.4 ; 256

2.34 ; 3.34 ; 1

m o

b

g mS r k r k

C pF C pF C pF

π

π μ

= = Ω = Ω⎫

⎨ ⎬= = =⎩ ⎭

Step 2: SS parameters

5722 ( )

m T

g f MHz

C C μ π

= =+

FREQUENCY RESPONSE OF CE AMPLIFIER

rπ

r o

gm vπ

v i

R1 //R 2

Rs

RLRCv o

Low ‐Mid frequencies: include the effects of C1, C2, CE

Use time ‐constant short circuit method tofind the pole frequencies for each capacitor

Consider C1 and let C2, CE short

, 1 1 2

1 , 1

/ / / / 3.86

1 38.6

1

E Q C s

E Q C

1

1

4.122

R R R R r k

R C m s

π

f H z

τ

π τ

= + = Ω

= × =

C1 C2

RE CE

= =

FREQUENCY RESPONSE OF CE AMPLIFIER

rπ

ro

gm vπ

v i

R1//R 2

Rs

RLRCvo

Low ‐Mid frequencies: include the effects of C1, C2, CE

Use time ‐constant short circuit method tofind the pole frequencies for each

Consider C2 and let C1, CE short

C1 C2

RE CE

capacitor

, 2

2 , 2

/ / 19.6

2 196

1

E Q C L C o

EQ C

22

0.812

R R R r k

R C m s

f H z

τ

πτ

= + = Ω

= × =

= =

FREQUENCY RESPONSE OF CE AMPLIFIER

[ ( / / / / )]

3.76( 76.5) 74.5

3.86

i vs m o C L

i s

R A g r R R

R R

k

k

= −+

= − = −

Step 3: Small signal analysis: Find source gain

rπ

ro

gm vπ

vi

R1//R 2

Rs

RLRCvo

Ri

99

7/27/2019 Lecture23(20-3-11)

http://slidepdf.com/reader/full/lecture2320-3-11 3/4

FREQUENCY RESPONSE OF CE AMPLIFIER

Consider CE and let C1, C2 short

1 2,

,

, 3 1 2

/ / / // /[ ]

1

1000/ /64.4 60.5

605

1263

2

268

s EQ CE E

E EQ CE E

E E

L dB E

r R R R R R

R C s

f Hz

f f f f Hz

π

β

τ μ

πτ

−

+=+

= = Ω= × =

⇒ = =

= + + =

f 1

f 2

f E37.5

,vs A dB

f, log

FREQUENCY RESPONSE OF CE AMPLIFIER

(1 )

3.34 1.0 (1 ( 76.5))

in v C C C A

3.34 77.5 80.8 pF

π μ+ −

= + × − −= + =

rπ

ro

gmvπvi R1//R 2

RLRC

Cπ

Cμ 2C

μ 1

1(1 )

11.0 (1 )

76.5

out v

C C A

−

= × +

1.01 pF

FREQUENCY RESPONSE OF CE AMPLIFIER

, 1 2

,

/ / / / / / 97.4

7.87

120.22

EQ in s

i n E Q in in

in in

R R R R r

R C n s

f M H z

π

τ

πτ

= = Ω

= × =

⇒ = =

,

,

1

/ / / / 4.9

4.95

132.2

2

1 1

E Q ou t C L o

ou t E Q ou t ou t

ou t ou t

R R R r k

R C ns

f M H z

, 3 ( ) 12.4H db in ou t

f M H z f f

τ

πτ

−−

= = Ω

= × =

⇒ = =

= + =

f in f out

37.5

,vsdB

f, log

FREQUENCY RESPONSE OF CE AMPLIFIER

Mid ‐high frequencies: include the effects of Cπ

and Cμ

Use Miller’s Theorem to simplify the Cμ effects

rπ

r o

gm vπ

v iR1//R 2

Rs

RLRCC

π

Cμ

rπ

ro

gm vπv i R1//R 2

Rs

RLRC

Cπ

Cμ 2C

μ 1

100

7/27/2019 Lecture23(20-3-11)

http://slidepdf.com/reader/full/lecture2320-3-11 4/4

End of Lecture 23

101