Freq Rsp Ssa

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Microelectronic

BITS PilaniCam us Anu Gu ta


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anPilani Campus


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•

amplifier

achieve desired response


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Key questions

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956


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Why high frequency


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Motivation


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Intrinsic frequency response


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Why f t is chosen as figure of

Why current gain of common source---Because it is related

to transit time from source to drain which determines

max speed of mosfet

Why short circuit- because current gain is maximum for

short circuited output port

BITS Pilani, Pilani Campus


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Intrinsic frequency response


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MOS unity gain frequency wT

Limits for MOSFETs:

Metric –C.S short-circuit current gain unit gain freq.:

wT = (gm-SCgd)/[s(Cgs+Cgd)]

wT is approximately = gm /Cgs

= - 2

Where gm = (W/L) unCox(VGS -VT) and

Cgs = (2/3)WLCox

so wT≈ 3 μn(VGS -VT)/2L2

Design lessons – to increase wT bias at large, overdrive voltage ID –swing reduces


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minimize L (w in as L2) , λ (= 1/L)increases, ROUT dec.

use n-channel over p-channel , NOISE increases


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Freq. Response of Common source

C = C +C

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Two closely spaced pole frequencies (s) and one zero transmission freq

Rd CLRs Rd

Rd CLCLRs


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Roots—1+as+bs2

=

2

b

ss

2

2,1

Rd CLRs Rd


s


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Exact Expressions



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Pole estimation Intuitively

Zero has to be found out through observing circuit


f f


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Physical significance of pole

Gain de radation---C s im edance reduces v s lits

between Rs and 1/sCgs vgs ↓ gm vgs ↓ vout ↓

Phase shift---due to time constant of intermediate node----charging/discharging takes time- output waveform

shifts on time axis


D i hi l i f


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Dc gain--- graphical view of

Low freq

High freq


G i d ti /ti hift


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Gain reduction/time shift —

Low fre

High freq---gain

degradation, phase shift



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Example

g = 0.1mA/v

Rd= 100kR = 1k

GBW=wT = 4 x 109 rad/sec

Cgd= 5fF

=Cdb= 0.1fF

= - 9 9. .

s= -55 x 109, -2 x 109 Hence, Critical wp1


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F ( ith t


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Frequency response (without

Two poles

Multiple poles


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If poles are close to each other


f


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Easy way to find poles—dominant pole

approx.

= w-3dB

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---------------- approx. estimate


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= 56 x 109

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—

• --- = =n , p - n

• Case2----wout is small, then w 1= w-3db=wout


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C dominates– sim lified ex ression

wp

1

2

Ld

Then

Cgd AC Rwp

gss )1(

1

Cgs dominates DOMINANT PO E


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Cgs dominates— DOMINANT POLE

wp 11 gss , ------ 0 -3dB

W-3dB =wH = 1/ Rs Cgs

Why


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Why

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UGB=UGB ≈ Ao / [R C ]


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UGB=UGB csa≈ Ao / [Rs Cgs]

≈ wtmos d s

=w = 1/ R C-

When Cgs (win) dominates---UGB csa may not

.

As UGB = Ao / (Cgs +Ao Cgd) Rs

a er may ec. ue o nc. n

hence Cgs, Cgd.


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Impact of increasing AO on freq.

various pole frequency conditions

Using Bode plots


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• increasing gm,• ncreas ng out,

• increasing both

We are changing the design of amplifier

6 cases


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6 cases

Single pole response before UGB

• ----A is increased by increasing gm,---- out,

• ----- A is increased by increasing both

Two pole response before UGB

• ----A is increased by increasing g ,

• ----A is increased by increasing Rout

Cgs dominates


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Cgs dominates,ase --- om nan po e response --- po e

as A is increased by increasing gm, keeping Rout same

Dominant pole response

p2

WLog scale

UGB

Wp1

gs


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• ---

may not increase with increase in Ao.

a er may ec. ue o nc. n ence

Cgs, Cgd.

Cgs dominates,


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Cgs dominates,

Case 2--- win=wp1 ; wp2 pole > UGB

as A is increased by increasing Rout only

Ld C Rwp2

UGB wp2

WLog scaleWp1

= 1/ Rs C + AC d

Cgs dominates,


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Cgs dominates,

Case 3--- win=wp1 ; wp2 pole > UGB

as A is increased by increasing gm, and Rout both

UGB wp2

WLog scaleWp1

= 1/ Rs C + AC d

Cgs dominates,


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--- —as A is increased, keeping Rout same: (wp2 remains same),

UGB decreases

|A|2 pole response


w Log scale

wp

UGB

Wp1

= 1/ [Rs Cgs+ ACgd]

Cgs dominates,


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--- —as A is increased, increasing Rout only: (wp2 reduces ), UGB

decreases1

|A|2 pole response

Ld C R

wp2


w Log scale

wp2

UGB

Wp1

= 1/ [Rs Cgs+ ACgd]

Cgs dominates,


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--- —as A is increased, increasing both: (wp2 reduces ), UGB

decreases1

Ld C R

wp2

|A|2 pole response


w Log scale

wp2

UGB

Wp1

= 1/ [Rs Cgs+ ACgd]


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=

1

Ld C Rwp

Wp1= 2x 109

gssC R

wp2 Wp2 = 50 x 109


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=m .

Rd= 100ks

Cgd= 5fF Small, but Rd

Cdb

+CL is large

gs=

Cdb+ CL= 0.1fF

= 50 pf

gs ,

Unity gain bandwidth UGB (f t)


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U y g UG ( t)

• A(s) = Ao wH here wH= w-3db

• For dominant pole, we can neglect second

pole

wz

-

• s= a s = = ------ wt = 0 w-3dB

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t m d d L m LHere output pole is dominant


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1

= wH /2π

Ld

H

C R

wpw

wH = w-3dB

UGB=Wt csa≈ gm / CL

= =nmos

High f T means high UGB and high w-3dB bandwidth

- u

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Hence scaling is beneficial as small mos, so Cgs↓ gm

If CL dominates---UGB csa increases with increase


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As UGB = A / CL Rout

------

1--But as A increases (redoing the design keeping.,

also constt. )

gm UGB increases

But gm ↑ Cgs ↑ Rs Cgs ↑ second pole freq. reduces

ma become dominant UGB ma reduce

2--A increases due to Rout ↑

first pole freq.

reduces UGB reduces

If CL dominates, Dominant pole response


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L ,case1-as A is increased, keeping Rout same, UGB increases

|A|


Wp2

UGB’

Log scaleWp1

= 1/ Rout CL

CL dominates, Dominant pole response


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case --as s ncrease y ou , wp re uces soUGB remains same, wp2 nearly same

|A|


’ Log scale

Wp1

= 1/ Rout CL

CL dominates, Dominant pole response


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case --as s ncrease y nc. o , wp re uces soUGB reduces

|A|


w Log

scale

’Wp1

= 1/ Rout CL

If CL dominates, 2 pole response


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-- , ,

UGB increases

ANon Dominant pole response


w Log scale

wp2Wp1

= 1/ Rout CL

To keep power, output swing constt. , gm is to increase to increase A, or

w/L ↑ Cgs ↑ wp2 ↓

OTE ----- , , L

↑ , so wp1 ↓ slightly

If CL dominates, 2 pole response


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---- , , . ,

increases

ANon Dominant pole response


w Log scale

wp2Wp1

= 1/ Rout CL


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If Cgs dominates---UGBcsa

may not increase

As UGB = A / (Cgs +A Cgd) Rs

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•

always remains constant

• When we redo the design and increase

ga n, ncrease n may cause ncrease n

bandwidth. Don’t confuse it with above

concep


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y oes ga n a

Zin, Zout Signal gets reduced

Time constant of these nodes shd. be

evaluated without finding transfer function

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Approx.


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Application of OCTC to evaluate bandwidth of


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= 1.7 x 109

= 2 x 109

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Millers theorem


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Vy - Vx Vy

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2

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is implemented by a non-inverting amplifier with Av > 1.e curren c anges s rec on as e ou pu vo age

is higher than the input voltage.

• If the input voltage source has some internal

impedance Zint or if it is connected through another

impedance element, a positive feedback appears

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,

process, the dual version is a powerful tool for designing and understanding circuits based on

modifying impedance by additional current.

• Typical applications are some exotic circuits with

,

capacitance neutralizers

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-


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R1 is the only path from x to y. Another dominant path should beresent

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we canot apply miller th. in backward manner i.e o/p to i/p as

vo=A vin not valid.


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Z in parallel to main signal path

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Miller theorem applied to CSA


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= 13.3 x 109

This is Wp1 if RsCgs >> RoutCL

= 1.96 x 109

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wp1= 2 x 109

w 2= 55 x 109

= 1.96 x 109

win = 13.3 x 10

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Zin


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• Zin=∞ at s= small

• When s large

• Zin1/ s[Cgs + (1 +A ) Cgd] ; A = gm

(Rd

||r o

|| 1/sCdb

)

• When s very large Rd and Cdb comes into effect

• Zin (1 /gm) ||r o || Rd || 1/sCdb


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Low freq. value= ∞

Drops to low value quickly because Cgs is large value

Zout

Z t


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Zout

=out d 0

When s large

in d r o s db s gd ; = gm d r o s db

Zin (1 /gm) ||r o || Rd || 1/sCdbSame as Zin

---


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Low freq. value

Low value at high freq

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Zero transmission frequency

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-wz

• 20 lo A = 20 lo R

Zero increases gain magnitude

Why?

» +√(1+ w2/wz2)

- p1

» -√(1+ w2/wp22)]

Physical explanation

iouti


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• When Cgd path becomes effective , i starts flowing

throu h C d, thus addin to v current

• So iout increases, so Gm increases beyond gm, so gain

increases

• Gm =iout/vgs = (gm - sCgd );

• Gm =√ gm2 + (wCgd )2 ; > gm

Phase shift


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-zero

input

+zero

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Feed-forward path----origin of zero


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Estimation of zero---easy way

iouti


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wz-

vgs

Estimation of zero---easy way

From transfer function at s w ; A(s)0


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From transfer function, at s wz; A(s)0

vout(s)0

So to estimate wz magnitude------ Short output to ground

Write KCL at out ut node

S Cgd (vgs-0) = gm vgs wz = gm/ Cgd

,

+ zero when currents meet in anti phase at a node -

For CSA wz = +2 x 1010 rad/sec


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Bode magnitude plot (with transmission

zero


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Bode plots— corner plots

Θ= - tan-1

(w/wp1) - tan-1

(w/wp2) - tan-1

(w/wz)|A|


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- 20dB er decade

- 40dB per decade

WP1 WP2WZ

0.1 WP1 10 WP1θ

-20dB per dec

-90

-180

~ -195

-270

Observations

• Gain starts falling at wp, Phase starts


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.

• At UGB total extra phase > 180o

• So total phase ≈ 360o output in phase

• If feedback is used, output will add to

• output amplitude will keep on growing till

Bode plots— corner plots of our examplep < wz< wp --- ncrease ug ut severe p ase egra at on ea s to nsta ty

|A|


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- 20dB er decade

UGB

- 20dB per decade

WP1 WP2WZ

0.1 WP1 10 WP1θ

-20dB per dec

-90

-180

-270

>-270

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Freq Rsp Ssa

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