ELEN 326 – ELECTRONIC CIRCUITS

111Electronic Circuits

ELEN 326 – ELECTRONIC CIRCUITS

FREQUENCY RESPONSE SECTION

OBJECTIVES:

1. Interpret and sketch the frequency response of a network using Bode plots;

2. Analyze and interpret the frequency response of single-stage amplifiers;

3. Interpret and determine small-signal parameters of BJTs and MOSFETs;

4. Determine the unity gain frequency of BJTs and MOSFETs;

5. Apply the dominant pole approximation;

6. Determine the frequency response, input and output impedances ofamplifiers;

7. Use SPICE for analysis and design


III – FREQUENCY RESPONSE

III.1. Bode plots

H(jωωωω) V2(jωωωω)V1(jωωωω)++

--

Frequency domain transfer function:

(V1/V2)(jω) = H(jω)

:)j(H)(j

e)(H)j(1

V)j(2

Vωωωωθθθθωωωω⋅⋅⋅⋅ωωωω====ωωωω

ωωωω

)(je)(H)j(H

ωωωωθθθθ⋅⋅⋅⋅ωωωω====ωωωω

transfer function

t

v1(t)

v2(t)A

(((( ))))OjHA ωωωω

( )1 O 2 O Ov (t)=Asinω t v (t)=A H(ω ) sin ω t+θ→

Bode magnitude and phase plots:

20log H(ω)

θθθθ(ωωωω)

vs ωωωω(log)


V1

+

-

H2H1VIN V2

++

--

21 Alog20Alog20Alog20 ++++====

jIN2 eAVV ⋅⋅⋅⋅⋅⋅⋅⋅==== θθθθ

Cascade amplifiers

e)(H1)j(H1 ωωωω====ωωωω jθθθθ1 (ωωωω)e)(H2)j(H2 ωωωω====ωωωω jθθθθ2 (ωωωω)

IN

1

1

2

IN

2

V

V

V

V

V

V⋅⋅⋅⋅====

1

IN

V

V= 2

1

V

V=

1 2θ=θ +θ

If one wants to calculate the overall

cascade amplifier gain, are H1 and H2 the

open-loop gains of networks 1 and 2 ?


Bode plots H(jω) = ± A . (a) Magnitude. (b) Phase.

H(jωωωω) = ±±±± A

A)j(H ====ωωωω

ππππ−−−−====ωωωωθθθθ

0)j(

180°°°°

0°°°° ωωωωlog

A)j(H −−−−====ωωωω

A)j(H ====ωωωω0je

ππππ−−−−je

)j( ωωωωθθθθ

(b)

A>0

)j(Hlog20 ωωωω in dB

1A ====

1A <<<<ωωωωlog

(a)

0dB

1A >>>>


Bode plots H(jωωωω) = .a

jωωωω

±±±± (a) Magnitude. (b) Phase.

(b)

-90°°°°

ωωωωlog0°°°°

90°°°° aj)j(Hωωωω

++++====ωωωω

aj)j(Hωωωω

−−−−====ωωωω

a

2j

e

ππππ++++

2j

e

ππππ−−−−


Bode plots H(jωωωω) = .a

jωωωω

±±±±


ωωωωloga0.1a

-20

2020dB/decade

(a)

10a



ωωωωloga

-20

20

0.1a

10a

-20dB/decade

(a)

ωωωω====ωωωω

a)j(H

2)(j

ππππ±±±±====ωωωωθθθθ

Bode plots H(jωωωω) = aj .ω

±

ωωωωlog

(((( ))))ωωωωθθθθ j

90°°°°

-90°°°°

0°

2j

e

ππππ++++ a

H(jω)=+jω

a2

j

e

ππππ−−−−

(b)

aH(jω)= -j

ω

Bode plots H(jωωωω) = aj .ω

±


( )2

1

C

V 1H(jω)= jω =

ωV1+jω

)j(V1 ωωωω+

-

R

C )j(V2 ωωωω+

-

(a)

( )2

1/ CH(jω) 1+ ω ω=-1

cθ(jω) -tg (ω/ω )=

3dB error

-20

0.1 1 10

exact

response -20dB/decade

log ω/ ωC0


(b)

Bode plots for first-order low-pass filter. (a) Filter (b) Magnitude (c) Phase

c

1ω =

RC

-45°

-90°

0.11 10

log ω/ ωC

0

(c)

5.7°error

-45°/decade

exact

response

-5.7° error



RC

1 where

j1

1

RC

j1

1)j(

V

V)j(H

C

C1

2

====ωωωω

ωωωω

ωωωω−−−−

====

ωωωω−−−−

====ωωωω====ωωωω

(b)

C

RV1(jωωωω) V2(jωωωω)

+ +

- -

(a)

(c)

Bode plots for first-order high-pass filter: (a) Filter. (b) Magnitude. (c) Phase.

in dB)j(Hlog20 ωωωω

-20

0.1 10

log ωωωω/ωωωωC

01

3dB error-5.7O error


0.1 1 10

5.7O error

0

45°

90°

log ωωωω/ωωωωC


III.2. The amplifier transfer function

Direct-coupled or

DC amplifiers

3 dB

AO

2/AO

ωωωωωωωωH

Capacitively coupled or

AC amplifiers

(Not used for IC)

2/AO Midband

OA

ωωωωL ωωωωH

Bandwidth BW = ωH - ωL

Gain-bandwidth product GBW = AOBW


20 dB/dec -20 dB/dec

10-2 10-1 100 101 102 103 104 105 106 ωωωω

40

60

80

dB)j(H ωωωω

)10s)(1s(

s10H(s) :Ex

5

9

++++++++====


III. 3. The low-frequency response of the CE amplifier

Use of RE produces stable bias BUT

results in lower amplifier gain.

RC

+VCC

+

-vbe

VBE

vBE

iC

+-

+

-

+

-

iB

vCE=vO

Stable IC is required for:

•stable voltage drop RCIC (keep the BJT in the active mode)

•stable voltage gain (=-gmRC=- RCIC/φt)

However is very sensitive to:

•VBE (4% variation /mV);

•temperature (IS varies ≈ 6 to 8%/K);

•BJT (IS is relatively unknown)

C S BE tI =I exp(V / )φ


RC

+VCC

+

-vbb

VBB

vBE

iC

+-

+

-

+

-

iB

vCE=vO

RE

Use of RE produces stable bias

BUT

results in lower amplifier gain.

C E

BB BEE

E

I I

V -VI =

R

≅

IC is stable if VBB>>variations in VBE

BUT

the AC input voltage vbb is divided


Frequency response of the CE amplifier with bypass capacitor

RC

+VCC

+

-vi

VBB

vBE

iC

+-

+

-

+

-

iB

vCE=vO

RE

RS

CE

CE acts as an open-circuit for DC and as a

short-circuit for the (AC) signal frequencies,

i. e., CE bypasses the AC signal.


vi

RS

rππππ

vππππ

RE CE

gmvππππRC

vo

+ +

- -

( )

O O C

i S π O E

E E

E

E E E E

v -β R=

v R +r +(β +1)Z

1Z =R //

sC

Z =R / R C s+1

++++ββββ

++++====ωωωω====ωωωω

ππππ

1

rR//RC

1;

CR

1

O

sEE

P

EE

Z

Frequency response of the CE amplifier with bypass capacitor

RC

+VCC

+

-vi

VBB

vBE

iC

+-

+-

+

-

iB

vCE=vO

RE

RS

CE

Small-signal equivalent circuit

(((( ))))(((( ))))

(((( ))))(((( ))));s1

s1

R1rR

Rs

v

v

p

z

EoS

Co

i

o

ωωωω++++

ωωωω++++⋅⋅⋅⋅

++++ββββ++++++++

ββββ−−−−====

ππππ


( )

+β

+=ω=ω

ω+

ω+⋅

+β++

β−=

ππ

1

rR//RC

1 ;

CR

1 ;

/s1

)/s1(

R)1(rR

R)s(

V

V

O

sEE

PEE

ZP

Z

EOs

CO

I

O

ωωωωZ ωωωωP

20 dB/dec

EOs

CO

R)1(rR

RA

++++ββββ++++++++

ββββ====

ππππ

ππππ++++

ββββ====

rR

RA

s

CO


Inclusion of coupling capacitor

RE

RS

RC

CEvi

vOR1

R2

VCC

CS

Transmission zero at ωωωω=0. (why?)

20 dB/dec

20 db/dec

capacitive

coupling

(CS)

DC coupling

midband

ωωωω


Simplified analysis of amplifiers in the frequency domain

Cinternal∞∞∞∞∞∞∞∞High-frequency

range

0∞∞∞∞∞∞∞∞Midband

0CbypassC couplingLow-frequency

range

CinternalCbypassC coupling


Base-charging (diffusion) capacitance Cb sb

be

qC

v=

S F CQ τ I=

BE

t

v

C Si I eφ=

C C

s S CF

be BE BEI I

q dq diτ

v dv dv= =

Cb F

t

IC τ

φ

=

Ex: τF = 4 x10-10s, IC = 1 mA

→ Cb = 16 pF

Minority carrier concentration in the base and the

corresponding change in the active region. QS is the

minority-carrier charge stored in the base.

Collector-base

depletion region

N NPbase

np(0,VBE)

np(0,VBE+vbe)

0 W x

QS

q s

Emitter-base

depletion region

III.4. BJT small-signal equivalent circuit

2

F

n

Wτ

2D=


BJT small-signal

equivalent circuit

E

rµµµµ

B’B C

rππππCππππ vi gmvi rO

Cµµµµ

Ccs

rcrb

-

+

bje

jc

CCC

CC

++++====

====

ππππ

µµµµ

Integrated-circuit npn bipolar transistor structure

showing parasitic elements. (Not to scale.)


Simplified small-signal model of the BJT

B

E

C

rO

gmvππππ

vππππrππππCππππ

Cµµµµ

+

-

I

Vr

gr

Ig

C

Ao

m

o

t

Cm ============

βπ

φjcjeb CC CCC ====++++==== µπ

mFb gττττC ====Cje and Cjc depend on bias

Small-signal BJT parameter set: jcjeFAo C,C,ττττ,V,β


iO

ii

AC schematic for measuring fT.

Small-signal circuit for

determining fT.

µµµµππππ ++++====ωωωω

CC

gmT

(((( ))))µµµµππππµµµµππππ

ππππββββ ββββ

ωωωω====

++++ββββ====

++++====ωωωω

CC

g

CC

g

O

T

O

m

ππππ

µµµµππππππππ

µµµµ

++++++++

−−−−

========

g

CCs1g

g

Cs1g

i

i

i

i m

m

b

c

i

O==== 0rc

The transition frequency (fT)

The magnitude of the short-circuit, common

-emitter current gain falls to unity at fT.(fT is

also called unity-gain frequency)

Cµµµµrb

Cππππrππππ

v1

+

- gmv1

rO

rc

ii iOCcs


Magnitude of small-signal current gain

versus frequency for a typical BJT.

)j( ωωωωββββ

ωωωωββββωωωωT

ωωωω (log scale)1

10

100

1000

)j( ωωωωββββ

ββββO

-6 dB/octave

high frequency poles/zeros

Tω

10


low currents: ( IC << ICM )

fT α IC

moderate currents ≈ ICM

fT ≈ independent of IC

high-level injection: IC > ICM

fT decreases with IC

fT

log IC

fTmax

ICM

Dependence of fT on collector current

high-level

injection

Dependence of fT on collector current


NMOS transistor with

bias voltages applied

III.5. MOSFET small-signal equivalent circuit


Small-signal equivalent

circuit of the MOSFET

Strong inversion and saturation

b

g

Cgs

Cgd

gmvgs

Cgb

Cbs

Cbd

d

gmbvbs

vbss+

-

vgs

+

-

rO

gs OX OV

gd OV

2C WLC C

3

C C

≅ +

≅

jsbsjdbd

OXgb

CC CC

WLCn

1n

3

1C

≈≅

⋅−

⋅≈( )

1.5 to2.1n

g1nV

ig

KI2V

ig

m

QBS

Dmb

D

QGS

Dm

≈

−=∂

∂=

=∂

∂=


Simplified small-signal

equivalent circuit of the

MOSFET for vbs=0

Strong inversion and saturation

gs OX OV

gd OV

2C WLC C

3

C C

≅ +

≅

s≡≡≡≡b

g

Cgd

gmvgs

Cgb+

Cgs

Cbd

d

vgs

+

-

rO

1.5 to2.1n

I

Vr

KI2V

ig

D

Ao

D

QGS

Dm

≈

=

=∂

∂=

CC

WLCn

1n

3

1C

jdbd

OXgb

≅

⋅−

⋅≈


III.6. Frequency response of differential amplifiers

(a) Differential-mode half-circuit.

RS

RS

-VEE

IEE

E

RL RL

vo

vid

VCC

+ -

RS

RL

vod/2

vid/2+-

(b) Small-signal equivalent circuit of (a).

Cµµµµ

vid/2vod/2

RS rb

rππππ Cππππ

i1

gmv1

RL+

+

--

A

A

+

v1

_


A capacitance CM=(1+gmRL)Cµ is seen looking across AA.

|Av|=gmRL is the low-frequency voltage gain from the internal base to the collector.

Since Av >>1→ CM >> Cµ

Equivalent circuit for

computing the gain. CM

is the Miller capacitance.

Cµµµµ

vi

vo

RS rb

rππππ Cππππ

i1

gmv1

RL+

+

--

A

A

+

v1

_

RS rb +

vππππ Ct=CM+Cππππ gmvππππ

RL vOvi

+

-rππππ

(((( )))) (((( )))) 0sCvvR

vvg ; sCvvi 1o

L

o1mo11 ====−−−−++++++++−−−−==== µµµµµµµµ

Assuming (((( )))) 1Lmo1oLo1m vR-gv CvvRvvg ====ωωωω−−−−>>>>>>>>++++ µµµµ

(((( )))) (((( )))) sCRg1v

i svCRg1i Lm

1

11Lm1 µµµµµµµµ ++++====++++====

2vv

2vv

idi

odo

====

====


( )1

b

pr ) // r ] r

π

π π

+ += = −

+ +S b

S t S b t

R r r1 1- [(R C R r C

(((( )))) (((( ))))

dB3-

LmbS

bS1dB3-

KGBW

CRg1C

1

rrR

rrRp

ωωωω⋅⋅⋅⋅≅≅≅≅

++++++++++++

++++++++========ωωωω

µµµµππππππππ

ππππ

K

0 dBωωωω-3dB GBW

dm

KA

1=

1

s-p

K π

π

=+ +

m L

S b

r-g R

R r r


Assume φt=26mV

Example: CE amplifier / differential amplifier

RS=1kΩIC=1mA

rb=200Ωdifficult to evaluate

β=100

range

fT=400MHz(@IC=1mA) Cµ=0.5pF

depends on VCB

RL=5KΩ

gm=1mA/26mV

( )

O m

m

T

M m L

M

r g 2.6k

gC C 15.3 0.5 14.8pF

2 f

5kC 1 g R C 1 0.5

26

C 96.7pF C

ππ µ

π µ

µ

π

β ω

π

= = Ω =+

= − = − =

Ω = + = + ⋅

Ω

= >>

mT

g

C C40

0104 108

20

dB

42.4 dB = 20log131.6

-6dB/octave

106

1.74MHz

228MHz

ω

m L

S b

rK g R

R r r

π

π

= −

+ +

Atenção: verificar consistência

do gráfico com os dados

|K|


(((( ))))

bS

ii

bS

rR

vi

r //rRR

++++====

++++==== π

Small-signal equivalent circuit using a Norton equivalent circuit at the input.

Summation of currents at B’ gives Summation of currents at C gives

(((( )))) (((( ))))B 0sCvvR

vvg 1O

L

O1m ====−−−−++++++++ µµµµ

From (A) and (B)

(((( ))))(((( )))) ππππµµµµµµµµππππµµµµµµµµ

µµµµ

++++++++++++++++++++

−−−−====

CRCRsRCRgRCRCRCs1

sCgRR

i

v

L2

LmL

mL

i

O

(((( )))) ππππµµµµµµµµππππµµµµµµµµ

µµµµ

++++++++++++++++++++

−−−−

⋅⋅⋅⋅++++

−−−−====CRCRsRCRgRCRCRCs1

sg

C1

rR

RRg

v

v

L2

LmL

m

bS

Lm

i

O

iiR Cππππ

Cµµµµv1

gmv1

RL vO

B’

E

C

+ +

- -

The low-frequency gain is

ππππ

ππππ

++++++++−−−−====

rrR

rRg

v

v

bSLm

i

O

(((( )))) )A( sCvvsCvR

vi o11

1i µµµµππππ −−−−++++++++====


The dominant pole approximation:

(((( )))) (((( )))) (((( ))))I pp

s

p

1

p

1s1sD

ps1

ps1sD

21

2

2121

++++

++++−−−−====

−−−−⋅⋅⋅⋅

−−−−====

p2 p1

s plane

( )2 1Assume p p . Then I becomes:>>

( ) ≈2

2

1 1 2

s sD s 1- + =1+as+bs

p p p

p1 and p2 are readily determined from:

→1 1 2 2

1 1 ap =- , p p = p =-

a b b


p2 p1

s plane

( ) ( ) 2L m L LD s 1 s C R C R C R g R RC s R RC C

p2 p1 ,

µ µ π µ µ π= + + + + +

>> then

Assuming that in

( )1

Lm L

1 1p

RRC C 1 g R

Rπ µ

≅ − ⋅

+ + +

m2

L L

1 1 1 gp

R C RC R C Cµ π π π

≅ − + + +

Using the data given in the previous example:

(((( ))))

MHz7.12

p MHz476

2

p

8202600//1200r//rRR

12

bS

−−−−====ππππ

−−−−====ππππ

ΩΩΩΩ========++++==== ππππ

( ≈ the value computed using the

Miller approximation)

Note: zero at

GHz13C

g

2

1f mz ≈≈≈≈⋅⋅⋅⋅

ππππ====

µµµµ


The common-mode gain

KHz16CR2

1

2pFC

M5R:Ex

EE

E

E

====ππππ

====

ΩΩΩΩ====

(((( )))) (((( )))) (((( ))))RsC1R2

Rs

v

vsA EE

E

L

ic

occm

++++−−−−≈≈≈≈====

VCC

RS

RL

2RE CE/2

-VEE

υυυυic

υυυυoc+

-

(a)

RS rbCµµµµ

voc

2RE CE/2

RLCππππ rππππ

gmv1

v1vic

+ +

- -

(a) Common-mode circuit (b)Small-signal equivalent circuit for

determining the common-mode gain

(b)


Variation of the CMRR with frequency

cmA

dmA

CMRR

dB

dB

dB

scale log

f

scale log

f

scale log

f

octave/dB6

octave/dB6−−−−



EECR2

1

ππππ

tRC2

1

ππππ

(a)

(b)

(c)

Variation with frequency of the gain parameters of the circuit.

(a) Common-mode gain. (b) Differential-mode gain. (c) Common-mode rejection ratio.


RS

I RE

-VCC

VCC

III. 7. The emitter - follower

RS

RE

+

-

vO

vi

Common-collector (or emitter-follower) circuit AC schematic of the common-collector circuit

vi

vO

+

-

RE

Rb

rππππ Cππππ

+

-vππππ

B’ C

E

gmvππππ

Cµµµµ

ii

Rb=RS+rb

Zππππ

(III) R

vvgi

(II) Ziv

(I) vvRiv

rsC1

r Z;RrR

E

omi

i

obii

Sbb

=+

⋅=

++=

+=+=

π

ππ

π

ππ

ππ


+

+=

=ω−=−≅

−

−

⋅+

++

+

=

π

ππ

π

π

Em

Eb1

11T

m1

1

1

EbEm

EEm

i

O

Rg1

RR//rR

RC

1-p

C

gz

p

s1

z

s1

r

RRRg1

r

RRg

v

v :Gain

Input impedance: ( )1i b m EZ r Z g Z Rπ π= + + +

Output impedance:1

S bO

m

Z R rZ

g Z

π

π

+ +=

+

vi

vO

+

-

RE

Rb

rππππ Cππππ

+

-vππππ

B’ C

E

gmvππππ

Cµµµµ

ii

Rb=RS+rb

Zππππ

1

1

1T

pC Rπ

ω= − ≈ − if Rb << RE and gmRE >> 1

Vs

Rs

Re (load)ZiZo


Emitter - follower

(((( ))))

(((( ))))

Eb

E

Em

Em

Embi

RsCR1

Rr

R

rRg1Rg1

Cs1

rRg1rZ

++++++++

++++====

++++

++++++++

++++

++++++++====

ππππππππ

ππππ

rbB B’

RE

Zi

EmRg1

C

++++ππππ

(((( )))) ππππ++++ rRg1 Em

(((( )))) ππππ++++==== rRg1R Em

EmRg1

CC

++++==== ππππ

Cµµµµ=1pF

i

O

υυυυ

υυυυ

dBf

f-3dB=712MHz

1MHz 10MHz 100MHz 1GHz

fT=612Hz

Cµµµµ=0bE

E

RR

R

++++-1

-2

-3

-4

-5

-6

-7

-8

Input impedance of

the emitter-follower

jωωωω

1

-1-4

2s plane

××××108 rad/sec

o x

o zero

x pole

-3

10

2

50

E

S

C pF

R k

R

π =

= Ω

= Ω

150

100

1

b

C

r

I mA

β

= Ω

=

=

0

1C

pFµ

=

O ERβ≈

if gmRE >>1


R1

R2

L

ZO

O

b

b2

O

b

m

1

RrCL

RR

R

g

1R

ββββ====

====

ββββ++++====

ππππππππ

Equivalent circuit for the output impedance

of an emitter-follower at moderate current levels

o

bb

bo

b

o

b

m

o

bb

m

b

o

RrsCR

RR

rsCR

g

1

rsC1

RrsCRr

rsC1

rg1

RrsC1

r

Z

ββββ++++

ββββ++++

ββββ++++

≈≈≈≈

++++++++ββββ

++++++++====

++++++++

++++++++

====

ππππππππ

ππππππππ

ππππππππ

ππππππππππππ

ππππππππ

ππππ

ππππππππ

ππππ


III.6. The unity gain current mirror

current

mirroriin vO

iO

AC analysis

gggrg1/m2m1m1o1m ≈≈≈≈====<<<<<<<<

CCCCCC2gb2gs1gb1gs1db

++++++++++++++++====

Ov1mg

1

1or C gsv

2gdC

gs2m vg2or

2dbC

Oi

+

-

1gdC

ini1 2

M1M2

iin

1:1


(((( ))))[[[[ ]]]] [[[[ ]]]]

(((( )))) [[[[ ]]]]B sCvvsCr

1vvgi:2 node

A vsCvCCsgi:1 node

2gdgsO2db2o

OgsmO

O2gdgsgd2min

−−−−++++

++++++++====

−−−−++++++++≅≅≅≅

Ov1mg

1

1or C gsv

2gdC

gs2m vg2or

2dbC

Oi

+

-

1gdC

ini1 2

current

mirroriin vO

iO

0ovin

o1

i

iA

====

====

0ii

vZ 0v

i

vZ in

o

ooo

in

inin ================


a) short-circuit current gain:

useful

range

MOSFET lumped model

is not valid

-20dB/dec

0

dBin

O

i

i

m

gd2

g

C C+( )log scaleωωωω

o

gd2

o m

gd2in υ =0

m

C1-s

i g=

C+Ci1+s

g


b) input impedance:

c) output impedance:

-20dB/dec

2dboCr

1(((( ))))logf2ππππ

ro

ZO

2gd

m

2dbo CC

g

Cr

1

++++<<<< useful frequency range

(((( ))))

++++++++

============

m

2gdm

0ovin

in

g

CCs1g

1

i

vZin

(((( ))))

o2db

oo2db

o

1o

0inio

o

m

2gd

2gd2db

o

1o

rsC1

rZsC

r

1Z

i

v

g

CCs1

sCsC

r

1Z

++++≈≈≈≈→→→→++++≈≈≈≈

====++++

++++

++++++++====

−−−−

====

−−−−

Documents

ELEN 326 – ELECTRONIC CIRCUITS