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(), Vout() and Vin() are phasors

( ) ( )( )

in

out

VV H =

( ) ( )( ) C R j 11

C j1 R

C j1

VV H

in

out

+=

+

==


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

Phase

If H() = N() / D() then:

Re[H()] = Re[N()D*()]

Im[H()] = Im[N()D*()]

( ) ( ) )(HH H * =

( ) ( )[ ]( )[ ]

=

HReHImarctan


H() for the RC circuit Amplitude

Amplitude in Decibels

For a -3dB reduction on the magnitude

( ) ( )

+

=C R j1

1C R j1

1 HH *

( ) ( ) ( )

+= 2

*

C R 11 HH

( ) ( )[ ] Hlog20 H dB =

( )[ ]dB3Hlog20 3 = ( ) 0.7079H 3 =dB


Bode Plot RC circuit Amplitude

( ) ( ) 0.7079 C R 11 H 2

33dB =

+=

dB CR

1 p3dB ==

( )

+

= 2

3

1

1 H

dB

For >> 3dB ( )

2

3

1 H

dB

( )

dB3 H

Amp drops by 2 when f doubles

Amp drops by 10 every decade


Bode Plot RC circuit - Phase

H() phase ( ) ( )[ ]( )[ ]

=

HReHImarctan

( ) 23dB

3dB

3dB

3dB

3dB3dB 1

j 1

j 1

j 1

j 1

1 j 1

1 H

+

=

+=

+=

( ){ } 23dB

1

1 H Re

+

=

( ){ } 2

3dB

3dB

1

H I

+

=

m


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

( ) ( )[ ]( )[ ]

=

=

=

3dB3dB

-arctanarctan HReHImarctan


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

the circuit

In Out

( )C R j 1

1 H

+

=


Bode Plots 1 pole RC circuit

( )

dB3 H

( )

=

3dB

-arctan

CR1 p3dB ==


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 p = 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

+

+= where: p1 = 1/R1C1 and

p2 = 1/R2C2


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

( ) ( )[ ] ( )[ ]p2p2p1p1 jexpH jexpH H =( ) [ ]( )p2p1p2p1 jexpH H H +=( ) ( ) ( )( ) jexpH H =

( ) ( ) p2p1p2p1 and H H H +==


Amplitude Bode Plot 2 poles Second pole accelerates the amplitude reduction

20dB/Dec

40dB/Dec

( ){ } ( ) ( ){ } HH log20 H log20 21 pp =( ){ } ( ){ } ( ){ } H log20 H log20 H log20 21 pp +=


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

( ) p2p1 +=


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() Circuit has:

1 Zero at = 01 Pole at = 1/R C

( )1/RC j 1

CRj C R j 1

C Rj C j

1 R

R H

+

=

+=

+

=

( ) ( )( )

in

out

VV H =

( ) ( )2C R 11C R H

+

=


C R circuit Bode plot amplitude At = 0 |H()| = 0 and since

-3dB

( ) ( )[ ] Hlog20H dB =( ) ( )

+= -

C R 11C R log20 H 2dB

( ) ( ) 3dB- 1 111 log20 H 2dB =

+=p

( )( )

01log20

HdB

=

=

>> p


SPICE SIM C R circuit p = 1/RC = 10k fp = 1.6kHz


Zeros 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 =


Zeros gain response For Zero in either half plane the amplitude

response is the same

0

( )2

z

1 H

+=

( )

+=

2

zdB 1 log10 H

( )

dB/dec20

log20

H

z

dB

>>

z


Transfer function Other circuits 1 Pole

1 Pole, 1 Zero

( ) ( )CR||Rj11

RRRH

2121

2

+

+=

( ) ( )CR||Rj1CRj1

RRRH

21

1

21

2

+

+

+=


1 Pole, 1 Zero response The response depends

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

p

z

j1

j1 H

+

+

=


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

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 = (default), MJSW = 3 (default)

CGB = CGBO L

WCGSOWLC32 C oxGS += WCGSO2

WLC C oxGS +

=

WCGDO CGD = WCGDO2WLC 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 = 4W Area of Drain = AD = AS = 4W Perimeter of Source = PS = 8+W Perimeter of Drain = PD = 8+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

+++

=

( )outminp Rg1 CR1

+=

Miller Capacitor


Common Source CD can be ignored

sometimes

Rout = RL || ro

CG = CGB + CGS

Rout


Common source small signal Using impedances

Rout0

Zv-v

Zv

Rv-v

GD

out1

G

1

in

in1=++

( )[ ]{ } GDGinout2GDLinGoutmGDm

GD

outmin

out

CCRRCRRCRg1C j1gC-1

R-gvv

++++=

p2p1

2

p2p1

m

GD

p2p1

m

GD

in

out

1111j1

gCj-1

j1j1

gCj-1

vv

++

=

+

+

=


Common source Poles and Zeros From the transfer function:

( )[ ] GDLGoutmGDinp1 CRCRg1CR1-

+++=

( ) Ggm1inoutGDoutp2 C ||R ||R1

CR1- =

GD

mz C

g=

( )

++

+

+

=

p2p1

z

j1j1

j1H


Common source Poles and Zeros Converting to s space:

sz = -jz sp1 = -jp1 sp2 = -jp2

( )

++

+

+

=

p2p1

z

j1j1

j1H

( )

+

=

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

RinC, RoutCD >> RinCMiller

CR1

inp1 =

DCR1

outp2 = 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

==

Dout

out

Dout

D

out

moutmv CRj1R

Cj1 R

CjR

g- Zg- A

+=

+==

Doutp CR

1 =


Common source Small Miller capacitance Input transfer function

Input pole

CRj11

Cj1 R

Cj1

vv

inin

in

'in

+=

+=

CR1

inp =


Common source Other cases Case 2: Large CD

RoutCD >> RinCMiller, RinC

Case 3: Large CRinCMiller >> RoutCD, RinC

Doutp1 CR

1 = ( ) CCR1

inp2 +=

( ) CRg1R1

outminp1 +=

CMiller

( ) ( )Dm

Dm

p2 CCg

CCg1

1 +

=

+=


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


Common drain (source follower) Small circuit analysis

vOUT

( ) outoutinin

g vv-vCRj11v +

+

=


Common drain Small signal analysis

( ) outmgmoutgs

out vg1-Vg

Cj1v-v

Rv

++=

( ) ( ) gmms

out vgCjCjg1R1v +=

+++

( )( ) sm

smin

m

sm

sm

s

out

Rg11Rg1CRj1

gCj1

Rg11Rg

Rv

++

++

+

++

=

Cg mz = ( )A1CR

1 in

p1

=

( ) smsm

Rg11Rg A++

=


Common drain (source follower) Effect of CSB

The Body is Grounded


Common drain small signal

( ) GSoutgGgin

gin Cjv-vCjvR

v-v +=

( ) ( ) SBouts

outoutgmGSoutg CjvR

vv-vg-Cjv-v +=


Common drain Small signal analysis

( ) ( )

+

++

+

++

+++

+

+=

sm

GSGSBGGSins

2

sm

SBGSs

sm

GSinGin

m

GS

sm

sm

in

out

Rg1CCCCCRR

Rg1CCR

Rg1CRCRj1

gCj1

Rg1Rg

vv

Having the denominator to be in the format:

The poles are:

p2p1

2

p2p1p2p1

11 j1 j1j1

++=

+

+

( ) ( )SBGSOsm

GSinGin

sm

SBGSs

sm

GSinGin

p1CCR

Rg1CRCR

1

Rg1CCR

Rg1CRCR

1

+++

+=

+

++

++

=

( )[ ]GSSBGSBGGSinO

SBGSOsm

GSinGin

p2 CCCCCCRR

CCRRg1

CRCR

++

+++

+

= sm

O R ||g1R =


Common drain - Cases

Case 1:

Case 2:

( )SBGSOsm

GSGin CCR Rg1

CCR +>>

+

+

+

+

=

sm

GSGin

p1

Rg1CCR

1

sm

sm

Rg1Rg A

+= ( )A-1CC GSMiller =

( )

+

+>>+sm

GSGinSBGSO Rg1

CCR CCR

( )SBGSOp1 CCR1+

=


Common Gate

Assuming ro


Common gate small signal Using KCL @ vs and @ vout

No Zeros

smsss

sin vgCjvR

v-v+=

L

outDoutsm R

vCjv vg +=

( )

+

++

+=

sm

ssDL

sm

Lm

in

out

Rg1CRj1CRj1

Rg1Rg

vv

DLp1 CR

1=

sm

sssm

sp2

Cg1 ||R

1

CRg1

R1

=

+

=

Freq Response

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