Freq Response

ALP_Rotondaro EE5321/EE7321 1

EE5321/EE7321Semiconductor Devices and Circuits

Frequency Response Part1


Impedance network transfer function • Impedance network transfer function:

where H(ω), Vout(ω) and Vin(ω) are phasors

( ) ( )( )ωωω

in

out

VV H =

( ) ( )( ) C R j 1

1 C j

1 R

C j1

VV H

in

out

ωω

ωωωω

+=

+

==


H(ωωωω) in polar coordinates• H(ω) is represented by its amplitude and phase• Amplitude |H(ω)|

• Phase ∠θ

• If H(ω) = N(ω) / D(ω) then:

Re[H(ω)] = Re[N(ω)·D*(ω)]

Im[H(ω)] = Im[N(ω)·D*(ω)]

( ) ( ) )(HH H * ωωω •=

( ) ( )[ ]( )[ ]

=

ωωωθ

HReHImarctan


H(ωωωω) for the RC circuit• Amplitude

• Amplitude in Decibels

• For a -3dB reduction on the magnitude

( ) ( )

−

•

+

=•C R j1

1C R j1

1 HH *

ωωωω

( ) ( ) ( )

+=• 2

*

C R 11 HH

ωωω

( ) ( )[ ]ωω Hlog20 H dB •=

( )[ ]dB3Hlog20 3 ω•=− ( ) 0.7079H 3 =dBω


Bode Plot RC circuit – Amplitude

( ) ( ) 0.7079 C R 1

1 H 23

3dB =

+=

dBωω CR

1 p3dB == ωω

( )

+

= 2

3

1

1 H

dBωω

ω For ω >> ω3dB ( )

≈ 2

3

1 H

dBωω

ω

( )ω

ωω dB3 H ≈

Amp drops by 2 when f doubles

Amp drops by 10 every decade


Bode Plot RC circuit - Phase

• H(ω) phase ( ) ( )[ ]( )[ ]

=

ωωωθ

HReHImarctan

( ) 2

3dB

3dB

3dB

3dB

3dB3dB 1

j 1

j 1

j 1

j 1

1 j 1

1 H

+

−=

−

−•

+=

+=

ωω

ωω

ωω

ωω

ωω

ωωω

( ){ } 2

3dB

1

1 H Re

+

=

ωω

ω ( ){ } 2

3dB

3dB

1

H I

+

−

=

ωω

ωω

ωm


Bode Plot RC circuit - Phase• And the Phase is given by:

( ) ( )[ ]( )[ ]

=

−=

=

3dB3dB

-arctanarctan HReHImarctan

ωω

ωω

ωωωθ


RC circuit – sine wave• The output wave has amplitude and phase altered by

the circuit

In Out

( )C R j 1

1 Hω

ω+

=


Bode Plots – 1 pole• RC circuit

( )ω

ωω dB3 H ≈

( )

=3dB

-arctan ωωωθ

CR1 p3dB == ωω


SPICE SIM – RC circuit• Run AC Sweep with 1V amplitude and freq: 10Hz to

100MHz• Output DB[V2(C1)/V1(V1)] and P[V2(C1)]


SPICE SIM – RC circuit• ωp = 1/RC = 10k → fp = 1.6kHz


RC circuits in series – 2 poles• The combination of two RC circuits in series is going

to result in 2 poles

( )

p2p1

j 1

1

j 1

1 H

ωω

ωωω

+•

+= where: ωp1 = 1/R1C1 and

ωp2 = 1/R2C2


RC circuits in series – 2 poles• Overall transfer function ( ) ( ) ( )ωωωωω p2p1 HHH •=

( ) ( )[ ] ( )[ ]p2p2p1p1 jexpH jexpH H ωωωω θθω •••=

( ) [ ]( )p2p1p2p1 jexpH H H ωωωω θθω +••=

( ) ( ) ( )( )ωθωω jexpH H •=

( ) ( ) p2p1p2p1 and H H H ωωωω θθωθω +=•=∴


Amplitude Bode Plot – 2 poles• Second pole “accelerates” the amplitude reduction

20dB/Dec

40dB/Dec

( ){ } ( ) ( ){ } HH log20 H log20 21 pp ωωω ••=•

( ){ } ( ){ } ( ){ } H log20 H log20 H log20 21 pp ωωω •+•=•


Phase Bode Plot – 2 poles• Second pole adds to the phase shift

( ) p2p1 ωω θθωθ +=


2 poles circuit – 180° phase shift• A phase shift of 180° can be a problem

• If in a feedback loop, a 180° phase shift will turn a negative feedback into a positive feedback

• This results in an unstable system if the loop gain is > 1


Bode Plots – 3 Superimposed Poles• The phase shift

is quite “fast” and “strong”

• When used in a feedback loop will probably result in an unstable circuit


C R circuit – H(ωωωω)• Circuit has:

1 Zero at ω = 01 Pole at ω = 1/R C

( )

1/RC j 1

CRj C R j 1

C Rj C j

1 R

R H ωω

ωω

ω

ω+

=+

=

+

=

( ) ( )( )ωωω

in

out

VV H =

( ) ( )2C R 11C R H

ωωω

+•=


C R circuit – Bode plot amplitude• At ω = 0 → |H(ω)| = 0 and since

-3dB

( ) ( )[ ]ωω Hlog20H dB •=

( ) ( ) ∞→

+••= -

C R 11C R log20 H 2dB ω

ωω

( ) ( ) 3dB- 1 1

11 log20 H 2dB=

+••=pω

( )( )

01log20

HdB

=•=

>> pωω


SPICE SIM – C R circuit• ωp = 1/RC = 10k → fp = 1.6kHz


Zero’s phase response• The phase response of a Zero depends on which

half plane the Zero is located

( )zss-1 sH = ( )

zss1 sH +=zz j-s ω=


Zero’s gain response• For Zero in either half plane the amplitude

response is the same

0

( )2

z

1 H

+=

ωωω ( )

+•=

2

zdB 1 log10 H

ωωω

( )

dB/dec20

log20

H

z

dB

•≈

>>

ωω

ωω z


Transfer function – Other circuits• 1 Pole

• 1 Pole, 1 Zero

( ) ( )CR||Rj11

RRRH

2121

2

ωω

+•

+=

( ) ( )CR||Rj1CRj1

RRRH

21

1

21

2

ωωω

++

•+

=


1 Pole, 1 Zero response• The response depends

on the relative location of the Pole and the Zero ( )

p

z

j1

j1 H

ωωωω

ω+

+=


MOSFET capacitances - circuit• Specs: tox (Cox), CGSO, CGDO, CGBO, CJ, PB (φB)

• Typical Values

Cox = 10-4 F/m2

CGSO = 5x10-10 F/m

CGDO = 5x10-10 F/m

CGBO = 4x10-10 F/m

CJ = 10-4 F/m2

PB = 0.8 V


MOSFET capacitances - equationsSaturation Linear

with: PS = Perimeter of Source, AS = Area of Source

MJ = ½ (default), MJSW = 3 (default)

CGB = CGBO · L

WCGSOWLC32 C oxGS •+•••= WCGSO

2WLC C ox

GS •+••=

WCGDO CGD •= WCGDO2

WLC C oxGD •+••=

MJSWBS

MJBS

SB

PBV1

PCJSW

PBV1

ACJ C

+

•+

+

•= SS a similar equation is used to calculate CDB


MOSFET – classic layout• Area of Source = AS = 4λ·W

• Area of Drain = AD = AS = 4λ·W

• Perimeter of Source = PS = 8λ+W

• Perimeter of Drain = PD = 8λ+W


MOSFET – SPICE attributesM1 1 2 3 4 NMOS L=2U W=2U+ AS=4p AD=4p PS=6U PD=6U

• Overlap capacitances are calculated using W

• Capacitance to body have area and perimeter terms


Miller approximation• Capacitance between

input and output appears multiplied by the gain at the input

inout vA-v •=

( )

( )

( )dtdvA1C i

vAvdtdC i

v-vdtdCi

inc

ininc

outinc

•+•=

•+•=

•=


Miller approximation – Common source

( )[ ]CRRg1CRj1Rg

vv

outoutmin

outm

in

out

+•++•

=ω

( )outminp Rg1 CR

1•+••

=ω

Miller Capacitor


Common Source• CD can be ignored

sometimes

• Rout = RL || ro

• CG = CGB + CGS

Rout


Common source – small signal• Using impedances

Rout

0Z

v-vZv

Rv-v

GD

out1

G

1

in

in1 =++

( )[ ]{ } GDGinout2

GDLinGoutmGD

m

GD

outmin

out

CCRRCRRCRg1C j1gC-1

R-gvv

ωω −+•+•++••=

p2p1

2

p2p1

m

GD

p2p1

m

GD

in

out

1111j1

gCj-1

j1j1

gCj-1

vv

ωωω

ωωω

ω

ωω

ωω

ω

••−

++

=

+•

+

=


Common source – Poles and Zeros• From the transfer function:

( )[ ] GDLGoutmGDinp1 CRCRg1CR

1-++•+•

=ω

( ) Ggm1

inoutGDoutp2 C ||R ||R

1CR1- −=ω

GD

mz C

g=ω

( )

++

+

+

=

p2p1

z

j1j1

j1H

ωω

ωω

ωω

ω


Common source – Poles and Zeros• Converting to s space:

sz = -jωz sp1 = -jωp1 sp2 = -jωp2

( )

++

+

+

=

p2p1

z

j1j1

j1H

ωω

ωω

ωω

ω ( )

+

=

p2p1

z

-1-1

-1H

ss

ss

ss

s


Diode connected and Pole Splitting


Common source – Capacitance Cases• Relative magnitude of

the capacitors result in different scenarios

• Case1: Miller Cap small

RinCπ, RoutCD >> RinCMiller

π

ωCR1

inp1 =

DCR1

outp2 =ω oLout r||R R =


Common source – Small Miller capacitance• Output Impedance, Zout

• Stage gain, Av

• Output pole

Dout

DoLout Cj

1 ||R Cj1 ||r ||R Z

ωω==

Dout

out

Dout

D

out

moutmv CRj1R

Cj1 R

CjR

g- Zg- Aω

ω

ω+

=+

•=•=

Doutp CR

1 =ω


Common source – Small Miller capacitance• Input transfer function

• Input pole

π

π

π

ωω

ωCRj1

1

Cj1 R

Cj1

vv

inin

in

'in

+=

+=

π

ωCR1

inp =


Common source – Other cases• Case 2: Large CD

RoutCD >> RinCMiller, RinCπ

• Case 3: Large Cµ

RinCMiller >> RoutCD, RinCπ

Doutp1 CR

1 =ω ( )µπ

ωCCR

1 in

p2 +=

( ) µ

ωCRg1R

1 outmin

p1 +=

CMiller

( ) ( )D

m

Dm

p2 CCg

CCg1

1 +

=+

=π

π

ω


Poles and Zeros• Usually the multiplying factor on the Miller

capacitor results in poles far apart from each other than in other cases.

• The pole splitting is used to compensate the circuit.

MILLERinp1 CR

1 ≈ω


Common drain (source follower)• Small circuit analysis

vOUT

( ) outoutinin

g vv-vCRj1

1v +•

+

=πω


Common drain – Small signal analysis

( ) outmgmoutg

s

out vg1-Vg

Cj1v-v

Rv ••+•+= χ

ω π

( ) ( ) gmms

out vgCjCjg1R1v •+=

+•++• ππ ωωχ

( )( ) sm

smin

m

sm

sm

s

out

Rg11Rg1CRj1

gCj1

Rg11Rg

Rv

•++•+•+

+

••++

=

χχω

ω

χπ

π

π

ωCg m

z = ( )A1CR1

inp1 −

=π

ω ( ) sm

sm

Rg11Rg Aχ++

=


Common drain (source follower)• Effect of CSB

• The Body is Grounded


Common drain – small signal

( ) GSoutgGgin

gin Cjv-vCjvR

v-vωω •+•=

( ) ( ) SBouts

outoutgmGSoutg Cjv

Rvv-vg-Cjv-v ωω •+=••


Common drain – Small signal analysis

( ) ( )

+

++−

+

++

+++

+•

+=

sm

GSGSBGGSins

2

sm

SBGSs

sm

GSinGin

m

GS

sm

sm

in

out

Rg1CCCCCRR

Rg1CCR

Rg1CRCRj1

gCj1

Rg1Rg

vv

ωω

ω

• Having the denominator to be in the format:

• The poles are:

p2p1

2

p2p1p2p1

11 j1 j1j1ωω

ωωω

ωωω

ωω −

++=

+•

+

( ) ( )SBGSOsm

GSinGin

sm

SBGSs

sm

GSinGin

p1CCR

Rg1CRCR

1

Rg1CCR

Rg1CRCR

1

+++

+=

+++

++

=ω

( )[ ]GSSBGSBGGSinO

SBGSOsm

GSinGin

p2 CCCCCCRR

CCRRg1

CRCR

++

+++

+=ω s

mO R ||

g1R =


Common drain - Cases

• Case 1:

• Case 2:

( )SBGSOsm

GSGin CCR

Rg1CCR +>>

+

+

+

+=

sm

GSGin

p1

Rg1CCR

1ωsm

sm

Rg1Rg A

+= ( )A-1CC GSMiller •=

( )

+

+>>+sm

GSGinSBGSO Rg1

CCR CCR

( )SBGSOp1 CCR

1+

=ω


Common Gate

• Assuming ro → ∞


Common gate – small signal• Using KCL @ vs and @ vout

• No Zeros

smsss

sin vgCjvR

v-v+•= ω

L

outDoutsm R

vCjv vg +•= ω

( )

+

+•+

+=

sm

ssDL

sm

Lm

in

out

Rg1CRj1CRj1

Rg1Rg

vv

ωω

DLp1 CR

1=ω

sm

sssm

sp2

Cg1 ||R

1

CRg1

R1

=

+

=ω

Freq Response

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