ECE 342 – Jose Schutt‐Aine ECE 442 Solid‐State Devices & Circuits 10. Frequency Response of Amplifiers

ECE 342 – Jose Schutt‐Aine 1

ECE 442Solid‐State Devices & Circuits

10. Frequency Response of Amplifiers

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]

1


Low-Pass Circuit

In frequency domain:1

1i

oVV

j CRj C

ωω

= ⋅+

11 1

i oo v

i

V VV Aj RC V j RCω ω

= ⇒ = =+ +

2

1 11 1 /vA

j RC jf fω= =

+ +

2


21 1

2 2f

RCπ πτ= =

2 time constantRCτ π= =

Low-Pass Circuit

3


High-Pass Circuit

1 11i i

oV R VV

Rj C j RCω ω

= =+ + 2

1 11 1 /1

2

ov

i

VAV jf fj

fRCπ

= = =−−

21

2f

RC=

π

4


If f2 = 2f1, then f2 is one octave above f1

If f2 = 10f1, then f2 is one decade above f1

2 22 10

1 1

# log 3.32logf fof octavesf f

= =

210

1

# log fof decadesf

=

2 GHz is one octave above 1 GHz

10 GHz is one decade above 1 GHz

Octave & Decade

5


3dB points are points at which the magnitude is1/ 2

20log(1.414) 3dBA dB= =

1 2o oV V

j=

+

that at midband frequencies.

3 dB Definition

From which

Power is halved. Voltage is scaled as:

6


Amplifier has intrinsic gain Ao

11 / hijf f+

Low-pass characteristics is:

High-pass characteristics is:

Overall gain A(f) is:

/1 /

lo

lo

jf fjf f+

/ 1( )1 / 1 /

loo

lo hi

jf fA f Ajf f jf f

= ⋅ ⋅+ +

Gain

7


At very high frequencies, STC goes as

220log 1 ( / )oG ω ω= − +

20log( / ) 20 log( / )o oG X where Xω ω ω ω− = − =

Slope of curve is –20; so if X=1 (ω=10ωo), decrease is –20 dB -20 dB/decade

Gain

8


Three Frequency Bands


Model for general Amplifying Element

Cc1 and Cc2 are coupling capacitors (large) μF

Cin and Cout are parasitic capacitors (small) pF

10


Midband Frequencies- Coupling capacitors are short circuits

- Parasitic capacitors are open circuits

out in LMB

in g in out L

v R RA Av R R R R

= =+ +

11


Low Frequency Model- Coupling capacitors are present- Parasitic capacitors are open circuits

1

1

1

1 1 ( )in in in c in

abc g in

g inc

v R v j C Rvj C R RR R

j C

ωω

ω

= =+ ++ +

1

1

( )1 ( )

c g ininab in

g in c g in

j C R RRv vR R j C R R

ωω

+= ⋅

+ ⎡ ⎤+ +⎣ ⎦

12


Low Frequency Model

( ) ( )1 221

1 122l l

L out cg in c

define f and fR R CR R C ππ

= =++

1

1

/1 /

in lab in

g in l

R jf fv vR R jf f

= ⋅+ +

2

2

/,1 /

lLout ab

L out l

jf fRSimilarly v AvR R jf f

= ⋅+ +

13


1 2

1 2

/ /1 / 1 /

out in l lL

in g in L out l l

v R jf f jf fROverall gain Av R R R R jf f jf f

= = ⋅ ⋅ ⋅ ⋅+ + + +

1 2

1 2

/ /1 / 1 /

out l lMB

in l l

v jf f jf fAv jf f jf f

= ⋅ ⋅+ +

Low Frequency Model

14


Rout = 3 kΩ, Rg=200 Ω, Rin=12 kΩ, RL=10 kΩCc1=5 μF and Cc2=1 μF

1 6

1 2.612 (12,200 5 10 )lf Hzπ −= =

× ×

Example

2 6

1 12.22 (13,000 10 )lf Hzπ −= =

×

15


High Frequency Model- Assume coupling capacitors are short- Account for parasitic capacitors

1th g inR R R=

1in in

thg in

v RVR R

=+

Potential Thevenin equivalent for input as seen by Cin

16


1

11

in inab

g in in th

v RvR R j C Rω

= ⋅+ +

11 1

1 11 / 2

in inab h

g in h th in

v Rv where fR R jf f R Cπ

= ⋅ =+ +

2

11

ab Lout

out L out th

Av RLikewise vR R j C Rω

= ⋅+ +

2th out Lwith R R R=

22 2

1 11 / 2

ab Lout h

L out h th out

Av Rv where fR R jf f R Cπ

= ⋅ =+ +

High Frequency Model

17


1 2

1 11 / 1 /

o in L

i in g L out h h

v R RAv R R R R jf f jf f

= ⋅ ⋅ ⋅ ⋅+ + + +

1 2

1 11 / 1 /

oMB

i h h

v Av jf f jf f

= ⋅ ⋅+ +

Overall gain is:

or

High Frequency

18


Example: Rout = 3 kΩ, Rg=200 Ω, Rin=12 kΩ, RL=10 kΩCin=200 pF and Cout=40 pF

1 10

1 4.052 2 10 (12,200 200)hf MHzπ −= =

× × ×

2 12

1 1.722 40 10 (10,000 3,000)hf MHzπ −= =

× × ×

64.05 10log 5.52 512.2

decades⎛ ⎞×

=⎜ ⎟⎝ ⎠

Summary: low-frequency <12.2 Hz, High frequency > 1.72 MHz

Example

19


Triode region:

Saturation region:

Cutoff:

12gs gd oxC C WLC= =

23gs oxC WLC= 0gdC =

0gd gsC C= =

gb oxC WLC=

MOSFET - Gate Capacitance Effect

20


ov ov oxC WL C=Overlap capacitance (gate-to-source):

1

sbosb

SB

o

CCVV

=+

1

dbodb

DB

o

CCVV

=+ Body

CSB

CGS CGD CGB

CDB

RG

Gate

DrainSourceRS

MOSFET – Junction Capacitances

21


22 Dm n ox eff n ox D

eff

W W Ig C V C IL L V

μ μ= = =

MOSFET High-Frequency Model

2 2mb m mF sb

g g gV

γχφ

= =+

1/ds A DD

r V IIλ

= =

23gs ox ov oxC WLC WL C= +

gd ov oxC WL C=

1

sbosb

SB

o

CCVV

=+

1

dbodb

DB

o

CCVV

=+

22


CS - Three Frequency Bands


o m gs gd gsI g V sC V= −

Unity-Gain Frequency fTfT is defined as the frequency at which the short-circuit current gain of the common source configuration becomes unity

(neglect sCgdVgs since Cgd is small)

o m gsI g V ( )i

gsgs gd

IVs C C

=+

( )o m

i gs gd

I gI s C C

=+

s jω=Define:

24


For s=jω, magnitude of current gain becomes unity at

( )2m m

T Tgs gd gs gd

g gfC C C C

ωπ

= ⇒ =+ +

fT ~ 100 MHz for 5-μm CMOS, fT ~ several GHz for 0.13μm CMOS

Calculating fT

25


CS - High-Frequency Response

26


1 2=GR R R


27


'sig sig GR R R=

( ) ( )'gd gd gs o gd gs m L gsI sC V V sC V g R V⎡ ⎤= − = − −⎣ ⎦

( )'1gd gd m L gsI sC g R V= +


' =L ds D LR r R R

28


( )'1= +eq gs gd m L gssC V sC g R V

( )'1 Miller Capacitance= + =eq gd m LC C g R

g sigin

g sig

R Vv

R R=

+

'= −o m L gsV g R V

Define Ceq such thatCS – Miller Effect

29


11 /

⎛ ⎞= ⎜ ⎟⎜ ⎟+ +⎝ ⎠

G siggs

G sig o

R VV

R R jf f

fo is the corner frequency of the STC circuit

'

12π

=oin sig

fC R

( )'1= + = + +

Miller

in gs eq gs gd m LC C C C C g R

CS – Miller Effect

30


' 11 /

⎛ ⎞= −⎜ ⎟⎜ ⎟+ +⎝ ⎠

o Gm L

sig G sig o

V R g RV R R jf f

1 /=

+o M

sig H

V AV jf f

'

12π

= =H oin sig

f fC R

CS – Miller Effect

31


Rsig = 100 kΩ, RG=4.7 MΩ, RD =15 kΩ, gm=1mA/V, rds=150 kΩ, RL=10 kΩ, Cgs=1 pF and Cgd=0.4 pF

' 4.7 1 7.14 74.7 0.1

= − = − × × = −+ +

GM m L

G sig

RA g RR R

( )': 1eq M m L gdMiller Cap C C g R C= = +

Example

' 150 15 15 7.14= = = ΩL ds D LR r R R k

( )0.4 1 7.14 3.26MC pF= × + =

32


( )1

2Hin sig G

fC R Rπ

=

Upper 3 dB frequency is at:

1.0 3.26 4.26= + =inC pF

( )12 6

1 3.822 4.26 10 0.1 4.7 10Hf kHzπ −= =

× × × ×

3.82Hf kHz=

Example (cont’)

33


≡ nde

BE

dQCdv

BJT Capacitances

Base: Diffusion Capacitance: Cde (small signal)

where Qn is minority carrier charge in base

ττ τ= = =C Fde F F m

BE T

diC gdv V

where τF is the forward transit time (time spent crossing base)

34


Base-emitter junction capacitance:

1

jeoje m

BE

oe

CC

VV

=⎛ ⎞

−⎜ ⎟⎝ ⎠

Cjeo is Cje at 0 V. Voe is EBJ built in voltage ~ 0.9 V

BJT Capacitances

35


BJT CapacitancesIn hybrid pi model, Cde+Cje=Cπ

Collector-base junction capacitance

1

om

CB

oe

CC

VV

μμ =

⎛ ⎞+⎜ ⎟

⎝ ⎠Cμo is Cμ at 0 V. Voc is CBJ built in voltage ~ 0.9 V

Cπ is around a few tens of pFCμ is around a few pF

36


High-Frequency Hybrid-π Model

, ,ACm o

T C m

VIg r rV I gπ

β= = =

, ,2

mde je de F m

T

gC C C C C C gfπ μ π τ

π+ = = + =

, 0.3 0.5

1

jcom

CB

oe

CC m

VV

μ = = −⎛ ⎞

+⎜ ⎟⎝ ⎠

( )2m

Tgf

C Cπ μπ=

+

37


CE - Three Frequency Bands


CE High-Frequency Model

39



1 2BR R R=

40


( )' B

sig sigB sig x sig B

rRV VR R r r R R

π

π

= ⋅ ⋅+ + +


41


'L o C LR r R R=

'o m LV g v Rπ−

( )'sig x sig BR r r R Rπ

⎡ ⎤= +⎣ ⎦


42


The left hand side of the circuit at XX’ knows the existence of Cμ only through the current Iμ replace Cμ with Ceq from base to ground

Bipolar Miller Effect

43


( ) ( )'o m LI sC v v sC v g R vμ μ π μ π π

⎡ ⎤= − = − −⎣ ⎦

( )'1 m LI sC g R vμ μ π= +


( )'1eq m LsC v I sC g R vπ μ μ π= = +

( )'1 , Miller capacitance for BJTeq m LC C g Rμ= +

44



' 11 /sig

o

v vjf fπ =

+

'

12o

in sig

fC Rπ

=

45


( )'1in eq m Lwhere C C C C C g Rπ π μ= + = + +

( )' 1

1 /o m LB

sig B sig ox sig B

V r g RRV R R jf fr r R R

π

π

⎡ ⎤ ⎡ ⎤= ⋅⎢ ⎥ ⎢ ⎥+ ++ +⎢ ⎥ ⎣ ⎦⎣ ⎦

Bipolar Miller Effect (cont’)

11 /

oM

sig o

V AV jf f

=+

'

12H o

in sig

f fC Rπ

= =

46


( )C mI g sC vμ π= −

Short-Circuit Current Gain

1BIv

sC sCr

π

μ ππ

=+ +

47


Define hfe as short-circuit current gain

( )1mC

feB

g sCIhI s C C

r

μ

π μπ

−= =

+ +

( )1C m

feB

I g rhI s C C r

π

π μ π

= =+ +

at freq. of interestmg sCμ

Short-Circuit Current Gain (Cont’)

48


Short-Circuit Current Gain (con’t)

( )1o

fehs C C rπ μ π

β=

+ +

Define hfe has a single pole (or STC) response. Unity gain bandwidth is for:

( ) ( )1 11 2

m mfe

T

g r gh ors C C r f C C

π

π μ π π μπ= = =

+ + +

In some cases, if Cμ is known, then

49


( )2m

Tgf

C Cπ μπ=

+

From which we get

2m m

T T

g gC Cfπ μ π ω

+ = =

, m m

T T

g gThus C C C Cπ μ π μω ω+ = ⇒ = −

Short-Circuit Current Gain (con’t)

50


CS – Miller Effect – Exact Analysis

51


ii

1GR

= =DD

1GR

=gg

1GR

=dsds

1gr

( )'i D m gdo

' ' 2 'i i g gs gd gd D i g gd m D gd gs D

G R g sCvv G G s C C sC R G G sC g R s C C R

−= −

⎡ ⎤ ⎡ ⎤+ + + + + + +⎣ ⎦ ⎣ ⎦


'D D ds

D ds

1R R rG g

= =+

52


2 ' 'gd gs D gd m D gs ms C C R sC g R or sC g

We neglect the terms in s2 since

( )( ) ( )'

i D m gdo' '

i i g gs gd m D gd D i g

G R g sCvv G G s C C 1 g R C R G G

−= −

⎡ ⎤+ + + + + +⎣ ⎦

ii

1RG

=If we multiply through by


53


( )( ) ( ){ }

'D m gdo

' 'i i g i gs gd m D gd D s g

R g sCvv 1 R G s R C C 1 g R C R 1 R G

−= −

⎡ ⎤+ + + + + +⎣ ⎦

( ) ( ){ }i g

H ' 'i gs gd m D gd D i g

1 R Gf

2 R C C 1 g R C R 1 R G

+=

⎡ ⎤+ + + +⎣ ⎦π

From which we extract the 3-dB frequency point


54


( ){ }H ' 'i gs gd m D gd D

1f2 R C C 1 g R C R⎡ ⎤+ + +⎣ ⎦π

H 'gd D

1f2 C Rπ

If Gg is negligible

If Ri =0


55


ii

1GR

= CC

1GR

=

1gr

=ππ

oo

1gr

=

BJT-CE – Miller Effect – Exact Analysis

'C C o

C o

1R R rG g

= =+

56


( )[ ]

's C mo

' ' 2 'i i C i m C C

G R g sCvv G g s C C sC R G g sC g R s C C R

−= −

⎡ ⎤+ + + + + + +⎣ ⎦

μ

π π μ μ π μ μ π


57


2 ' 'C m C ms C C R sC g R or sC gμ π μ π

( )( ) ( )

'i C mo

' 'i i m C C i

G R g sCvv G g s C C 1 g R C R G g

−= −

⎡ ⎤+ + + + + +⎣ ⎦

μ

π π μ μ π

( )( ) ( ){ }

'C mo

' 'i i i m D C i

R g sCvv 1 R g s R C C 1 g R C R 1 R g

−= −

⎡ ⎤+ + + + + +⎣ ⎦

μ

π π μ μ π


We neglect the terms in s2 since

If we multiply through by ii

1RG

=

58


( ) ( ){ }i

H ' 'i m C C i

1 R gf2 R C C 1 g R C R 1 R g

+=

⎡ ⎤+ + + +⎣ ⎦

π

π μ μ ππ

H 'C

1f2 C Rμπ

If Ri = 0


59


Example For the discrete common-source MOSFET amplifier shown, the transistor has VT= 1V, mCox(W/L) = 0.25 mA/V2, λ = 0, Cgs = 3 pF, Cgd = 2.7 pF and VA = 20 V. Assume that the coupling capacitors are short circuits at midband and high frequencies.

(a)Find the 3dB bandwidth if Ri=0

(b) Find the 3dB bandwidth if Ri= 100 Ω

60


If Ri =0,3 '

12dB

gd D

fC R

=π

20 131.516

Ads

D

Vr k

I= = = Ω

' 13 2 1.736D D dsR R r k= = = Ω

3 12 3

1 33.952 2.7 10 1.736 10dBf MHz−= =

× × ×π

Example – Part (a)

61


If Ri =100 Ω,

Example Part (b)

'm Dg R 0.870 1.736 1.51= × =

( ){ }H ' 'i gs gd m D gd D

1f2 R C C 1 g R C R⎡ ⎤+ + +⎣ ⎦π

( ){ }H1f 28 MHz

2 0.1 3 2.7 1 1.51 2.7 1.736=

+ + + ×⎡ ⎤⎣ ⎦π

62


Common Base (CB) Amplifier


High-Frequency Analysis of CB Amplifier

in 3dBx

S E

1r rC R R

1 1π

π

ω

β β

− =⎡ ⎤ ⎡ ⎤

+⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

out 3dBL

1C Rμ

ω − =

The amplifier’s upper cutoff frequency will be the lower of these two poles.

Exact analysis is too tedious approximate

From current gain analysis


Emitter Follower


Emitter Follower High-Frequency Exact analysis is too tedious approximate

( )

' m L mv

Em L

m E

sC1g R gA ( s ) sC R1 g R 1

1 g R

π

π

+=

+ ++

m E m3dB T

E

1 g R gR C C Cπ π μ

ω ω+= =

+


CE Cascade Amplifier

Exact analysis too tedious use computerCE cascade has low upper‐cutoff frequency


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

1 2

1 2

...( )

...m

v mm

s Z s Z s ZA s a

s P s P s P− − −

=− − −

Transfer Function Representation

Z1, Z2,…Zm are the zeros of the transfer function

P1, P2,…Pm are the poles of the transfer function

In general, the gain of an amplifier can be expressed as

68

s is a complex number s = σ + jω


• The poles of a multistage amplifier are difficult to obtain analytically

• An approximate value for the 3dB upper frequency point ω3dB can be obtained by assigning an open circuit time constant τio to each capacitor Ci

Bandwidth of Multistage Amplifier


• The time constant τio is the product of the capacitance and the resistance seen across its terminals with:All other internal capacitors open circuitedAll independent voltage sources short circuitedAll independent current sources opened

• The upper 3dB frequency point ω3dB is then found by using :

31

dBio

ωτ

=∑

Bandwidth of Multistage Amplifier


Cascode Amplifier

First stage is CE and second stage is CB

1 2v m LA g Rα= −

If Rs << rπ1, the voltage gain can be approximated by


Cascode Amplifier – High Frequency

High‐frequency incremental model


( )' 's i s 1 1 1 a 1 bG V G g s C C V sC Vπ π μ μ⎡ ⎤= + + + −⎣ ⎦

( ) ( ) ( )m1 1 a 2 m2 2 1 b 2 m2 2 d0 g sC V g g s C C V g g sC Vμ π π μ π π⎡ ⎤= − + + + + − + −⎣ ⎦

( ) ( )2 2 b x2 2 2 2 d 2 o0 g sC V g g s C C V sC Vπ π π π μ μ⎡ ⎤= − − + + + + −⎣ ⎦

( ) ( )m2 b m2 2 d L 2 o0 g V g sC V G sC Vμ μ= − + − + +

Applying Kirchoff’s current law to each node:

Find solution using a computer



gm=0.4 mhos β=100rπ=250 ohms rx=20 ohmsCπ=100 pF Cμ=5 pfGL=5 mmhos GS’=4.5 mmhos

sa=8.0 sd=‐0.0806sb=‐2.02 + j5.99 se=‐0.644sc=‐2.02 – j5.99 sf=‐4.05

sg=‐16.45

POLES (nsec‐1)ZEROS (nsec‐1)

As an example use:Cascode Amplifier – High Frequency


If one pole is at a much lower frequency than the zeros and the other poles, (dominant pole) we can approximate ω3dB

93dB 0.0806 10 rad / secω ×

3dBf 12.9 MHz

For the same gain, a single stage amplifier would yield:

93dB 0.0169 10 rad / secω ×

3dBf 2.7 MHz

Second stage in cascode increases bandwidth


ECE 342 – Jose Schutt‐Aine ECE 442 Solid‐State Devices & Circuits 10. Frequency Response of Amplifiers

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