Problems for Amplifier Section - University of California ...aries.ucsd.edu/NAJMABADI/CLASS/ECE65/12-W/Slides/... · Problems for Amplifier Section Lecture notes: Sec. 6 ... (emitter

Problems for Amplifier Section

Lecture notes: Sec. 6

F. Najmabadi, ECE65, Winter 2012


Exercise 1: Find the bias point and the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a common collector amplifier (emitter follower) .

o Input at the base, output at the emitter.

It has a emitter-degeneration bias with a voltage divider.


Bias circuit Thevenin form of the Voltage divider

V 95.49k 18k 22

k 22

k 90.9k 18||k 22

=×+

=

==

BB

B

V

R

Real circuit



Bias Calculations

A 3.20/ mA 05.4

107.0)1/(109.9 4.95

4.95 :KVLBE33

µβ

β

==≈=

+++×=

++=−

CB

CE

EE

EEBEBB

IIII

IIRIVRI

V 7.0 and 0 V, 7.0: Activein is BJT Assume

≥>= CECBE VIV

V 0.7 V 5 10104 9

9 :KVLCE

0

33

=>=××+=

+=−−

DCE

CE

EECE

VVV

RIV

V 95.4k 9.9 == BBB VR

k 28.1 r

k 0.3710x 4

150r

mA/V 156 1026104.05

3o

3

3

===

==≈

=××

==

mB

T

-C

A

-

-

T

Cm

gIVIVVIg

βπ


Thevenin form of the Voltage divider


Signal circuit Real circuit



11511

151

151k)100||k1||k8.38(10156)||||(

)||||(1)||||(

3

≈+

=

=×=

+==

−

v

LEom

LEom

LEom

i

ov

A

RRrgRRrg

RRrgvvA

k 28.1rk 8.38rmA/V 156

o

===

π

mg[ ]

[ ] k 9.9k194k3.1 ||k9.9 )||||)(1( ||

≈+=++=

i

LEoBi

RRRrrRR βπ

Ω==+

= 4.6||||

|| ββππ rr

RRrrR o

sigBoo

Amplifier Parameters This is a common collector amplifier (emitter follower)



Ω==

13k9.9

o

i

RR

k 28.1rk 8.38rmA/V 156

o

===

π

mg

Amplifier Parameters (Cut-off frequency)

Hz 39.31047.0)10100(6.4 2

1

)( 21

Hz 2.341047.0)0109.9( 2

1

)( 21

632

22

631

11

=×××+

=

+=

=××+×

=

+=

−

−

π

π

π

π

p

cLop

p

csigip

f

CRRf

f

CRRf

Hz 6.374.32.34 21 =+=+≈ ppp fff



This is a common emitter amplifier with RE . o Input at the base, output at the collector.

It has a emitter-degeneration bias with a voltage divider.




Bias circuit Thevenin form of the Voltage divider

V 22.215k 9.5k 34

k 9.5

k 0.5k 9.5||k 34

=×+

=

==

BB

B

V

R

Real circuit


Bias calculations Thevenin form of the Voltage divider

A 2.14/ mA 84.2

5107.0)1/(100.5 2.22

2.22 :KVLBE3

µβ

β

==≈=

+++×=

++=−

CB

CE

EE

EEBEBB

IIII

IIRIVRI


≥>= CECBE VIV

V 0.7 V 5.10 )51010(1084.215

15 :KVLCE

0

33

=>=+××+=

++=−−

DCE

CE

EECECC

VVV

RIVRI

V 22.2k 0.5 == BBB VR

k 83.1 r

k 8.5210 84.2

150r

mA/V 10.9 1026102.84

3o

3

3

===

=×

=≈

=××

==

mB

T

-C

A

-

-

T

Cm

gIVIVVIg

βπ







k 83.1r k 8.52rmA/V 10.9

o ===

π

mg

Amplifier Parameters This is a CE amplifier with RE

[ ][ ] k 8.4k 51.0201k 83.1 ||k 0.5

)1( || =×+=

++≈

i

EBi

RRrRR βπ

64.18.52/990.0510109.101

990109.10

0.990kk 100||k 1|| /)||(1

)||(

3

3

−=+××+××

−=

==++

−≈

−

−

i

o

LC

oLCEm

LCm

i

o

vv

RRrRRRg

RRgvv

k 0.1

||1 ||

=≈

+++≈

Co

sigBE

EoCo

RR

RRRrRrRR

π

β

59.1)64.1(100800,4

800,4

−=−×+

=

×+

==

v

i

o

sigi

i

sig

ov

A

vv

RRR

vvA



k 0.1 k 8.4 == oi RR

Amplifier Cut-off frequency (2 caps: 2 poles)

Hz 8.1510100)1010010( 2

1

)( 21

Hz 91.6107.4)100108.4( 2

1

)( 21

9332

22

631

11

=×××+

=

+=

=××+×

=

+=

−

−

π

π

π

π

p

cLop

p

csigip

f

CRRf

f

CRRf

Hz 7.228.159.6 21 =+=+≈ ppp fff

k 83.1r k 8.52rmA/V 10.9

o ===

π

mg



This is a PNP common emitter amplifier (no RE ). o Input at the base, output at the collector (RE is

shorted out by a cap)

It has a emitter-degeneration bias with two voltage sources.



Bias circuit Real circuit

A 0.5/ mA 0.1 103.2 3 :KVLBE 3

µβ ==≈=

+×=−

CB

CE

EBE

IIII

VI


≥>= ECCEB VIV

V 0.7 V 4.1 10106.46

3103.2103.23 :KVLCE

0

33

33

=>=××−=

−×++×=−−

DEC

EC

CECE

VVV

IVI

k 26.5 r

k 15010150r

mA/V 38.5 1026

10

3-o

3-

-3

===

==≈

=×

==

mB

T

C

A

T

Cm

gIVIVVIg

βπ




3.85

3.85)k 100||k 3.2||k 150(1038.5

)||||(

3

−==×+

=

−=×−=

−=

−

i

o

i

o

sigi

iv

i

o

LComi

o

vv

vv

RRRA

vv

RRrgvv

k 26.5r k 150rmA/V 38.5

o ===

π

mg

Hz 821667 21 =+=+≈ ppp fff* Show

Amplifier Parameters This is a CE amplifier with no RE

k 27.2k 3.2||k 150 ||

===

o

oCo

RrRR

k 26.5 || === ππ rrRR Bi


This is a MOS common source amplifier (no RS ). o Input at the gate, output at the drain (RS is

shorted out by a cap)

It has a source-degeneration bias with voltage divider.

Exercise 4: Find the bias point and the amplifier parameters of this circuit. (µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1. Assume capacitors are large and ignore channel width modulation in biasing. )


Bias (capacitors are open circuit):

V 353.18.110033

100V 0 G =+

=⇒=kk

kIG

Assume Saturation

232 10x3 5.0 OVOVoxnD VVL

WCI −== µ

V 803.0=+= tOVGS VVV

Saturation ⇒> OVDS VVmA 275.010x3 23 == −

OVD VI

V 550.0=−= GSGS VVV

V 249.18.1 =−= DDD IRV

V 699.0=−= SDDS VVV

V 303.0=OVV

DStOVDSGSG IRVVIRVV ++=+=

233 103 102 5.0353.1 OVOV VV −×××++=

0853.06 2 =−+ OVOV VV

GS-KVL:




k 3.3610 752.0

10r

mA/V 1.82 0.303

10 0.275 2 2

3o

3

=×

==

=××

==

-D

A

-

OV

Dm

IVV

Ig

45.3

45.3)||k 2||k 3.36(101.82

)||||(

3

−==×+

=

−=∞×−=

−=

−

i

o

i

o

sigi

iv

i

o

LDomi

o

vv

vv

RRRA

vv

RRrgvv

Amplifier Parameters This is a CS amplifier with no RS

k 90.1k 2||k 3.36 ||

===

o

Doo

RRrR

k 8.24 == Gi RRk 8.24k33||k100 ==GR



This is a common collector amplifier (emitter follower). o Input at the base, output at the emitter.

It is biased with a current source!




V 7.00

−=−=

E

EBE

VVV


≥>= CECBE VIV

k 21.1 r

k 9.34104.3

150r

mA/V 16.5 1026104.3

3o

3

3

===

=×

=≈

=××

==

mB

T

-C

A

-

-

T

Cm

gIVIVVIg

βπ

A 5.21/mA 3.4

µβ ==≈=

CB

CE

IIII

V 0.7 V 7.44 0 =>=−= DECE VVV

EV




k 21.1rk 9.34rmA/V 16.5

o

===

π

mg

14271

427

427)||||( k9.25k100||||k9.34||||

)||||(1)||||(

≈+

==

==∞=

+=

i

ov

LEom

LEo

LEom

LEom

i

o

vvA

RRrgRRr

RRrgRRrg

vv

Amplifier Parameters This is an emitter follower (RE = ∞ )

Ω=≈+

+≈ 6.60

1||

|| ββππ rRRr

RR sigBEo

[ ]M 2.5k9.25201k2.1

)||||)(1( || =×+=

++=

i

LEoBi

RRRrrRR βπ



Signal circuit

k 21.1rk 9.34rmA/V 16.5

o

===

π

mg

Cut-off frequency

Hz 34.31047.0)6110100( 2

1

][ 21

63

2

=××+×

=

+=

−π

π

p

coLp

f

CRRf

Exercise 6: Find the bias point and the amplifier parameters of the circuit below. (µpCox (W/L) = 400 µA/V2, Vtp = − 4 V, λ = 0.01 V-1. Ignore channel width modulation in biasing).

This is a PMOS common drain amplifier. o Input at the gate, output at the source.

It has a source-degeneration bias with two voltage sources.



Assume Saturation

092

09104005.010

||101013

2

264

44

=−+

=−+×××

++×=+×=−

OVOV

OVOV

tpOVDSGD

VVVV

VVIVI

GS-KVL:

5.0 2OVoxpD V

LWCI µ=

V 9.5=+= tOVSG VVV

Saturation ⇒> OVDS VV

mA 71.0104005.0 26 =××= −OVD VI

V 9.5 0 =→−= SSSG VVV

V 9.10)5( =−−= SSD VV

V 9.1=OVV

k 141100.7101.0

11r

mA/V 0.747 0.303

100.71 2 2

3o

3

=××

==

=××

==

−

−

D

OV

Dm

I

VIg

λ



k 2.1k 34.1||k 101 ||

88.0)||||(1

)||||(

===

∞==

=+

=

mSo

Gi

LSom

LSomv

gRR

RRRRrg

RRrgA

Amplifier Parameters This is a common collector Amp.

Cut-off frequency

Hz 39.31047.0)102.110100( 2

1

][ 21

633

2

=×××+×

=

+=

−π

π

p

coLp

f

CRRf

Exercise 7: Find the bias point and the amplifier parameters of the circuit below. (µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1. Ignore channel width modulation in biasing).

This is a NMOS common gate amplifier. o Input at the source, output at the drain.

It has a source-degeneration bias with voltage divider.

Bias circuit Assume Saturation

054

05108005.010

10106

2

264

44

=−+

=−+×××

×++=×+==−

OVOV

OVOV

DtOVDGSG

VVVV

IVVIVVGS-KVL:

5.0 2OVoxnD V

LWCI µ=

V 0.2=+= tOVGS VVV


mA 40.0108005.0 26 =××= −OVD VI

V 74415101015 44

=−−=×++×=

DS

DDSD

VIVI

V 0.1=OVV


GV

V 615M8.1M2.1

M2.1=×

+=GV

DS-KVL:

k 10)]||(1([ ||

k 1.1/)||(1 ||

7.7)||k 10||k 250(108.0

)||||( 3

≈+=

=

+=

=∞×=

=−

sigSmoDo

m

oLDSi

v

LSomv

RRgrRRg

rRRRR

ARRrgA

Amplifier Parameters This is a common gate Amp.

Cut-off frequency

kHz 47.1Hz 22kHz 45.1

][ 21

][ 21

21

22

11

=+=+=

+=

+=

ppp

coLp

csigip

fffCRR

fCRR

fππ


k 250100.401.0

11r

mA/V 0.80 1

100.4 2 2

3o

3

=××

==

=××

==

−

−

D

OV

Dm

I

VIg

λ


This is a NMOS common drain amplifier. o Input at the gate, output at the source.

It is biased with a current source.

Bias circuit

Assume Saturation

32 100.71 5.0 -OVoxnD V

LWCI ×== µ

V 0.4=+= tOVGS VVV


V 0.1=OVVSV

mA 71.0=DI


V 0.40 −=−=→−= GSSSGGS VVVVV

V 94) (5 =−−=−= SDDS VVV


704 704|| 1 ||

99.0)||||(1

)||||(

Ω=∞==

∞==

=+

=

mSo

Gi

LSom

LSomv

gRR

RRRRrg

RRrgA

Amplifier Parameters This is a common collector Amp.

Cut-off frequency

Hz 36.31047.0)70410100( 2

1

][ 21

63

2

=××+×

=

+=

−π

π

p

coLp

f

CRRf


k 141100.7101.0

11r

mA/V 1.42 1

100.71 2 2

3o

3

=××

==

=××

==

−

−

D

OV

Dm

I

VIg

λ


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Problems for Amplifier Section - University of California ...aries.ucsd.edu/NAJMABADI/CLASS/ECE65/12-W/Slides/... · Problems for Amplifier Section Lecture notes: Sec. 6 ... (emitter