8. Frequency Response - University of California, San...

8. Frequency Response

Reading: Sedra & Smith: Sec. 1.6, Sec. 3.6 and Sec. 9 (MOS portions),

(S&S 5th Ed: Sec. 1.6, Sec. 3.7 (capacitive effects), Sec. 4.8, Sec. 4.9, ,Sec. 6. (Frequency response sections,

i.e., 6.4, 6.6, …), Sec. 7.6

Typical Frequency response of an Amplifier

F. Najmabadi, ECE102, Fall 2012 (2/59)

Up to now we have “ignored” capacitors in circuits & computed mid-band properties. We have to solve the circuit in the frequency domain in order to see the impact of capacitors (a typical response is shown below): o Lower cut-off frequency: fL o Upper cut-off frequency: fH o Band-width: B = fH − fL

Observation on the frequency response of an Amplifier

Observations:

Analytical solution of amplifiers in frequency domain is complicated!

Response (e.g., gain) of an ideal linear amplifier should be independent of frequency (otherwise signal “shape” would be distorted by the amplifier). Thus: o A practical amplifier acts as an ideal linear amplifier only for a range of

frequencies, called the “mid-band”.

o The lower and the upper cut-off frequencies (fL and fH) identify the frequency range over which the amplifier acts linearly.

o Amplifier response at high frequencies (near the upper cut-off frequency , fH) is important for stability considerations (gain and phase margins).

Thus, we are mainly interested in mid-band properties (where capacitors can be ignored) and in poles and zeros of the amplifier response (due to capacitors).


What do we mean by “capacitors can be ignored?”

Capacitor impedance depends on the frequency: |Z| = 1/(jωC). o At high frequency |Z| → 0: capacitor acts as a short circuit.

o At low frequency |Z| → ∞: capacitor acts as an open circuit.

For the above two limits, circuit becomes a “resistive” circuit and we do NOT need to solve the circuit in the frequency domain.

Thus, ignoring capacitors means that we operate at either a high enough or at a low enough frequency such that capacitors become either open or short circuits, leading to a “resistive” circuit. o Note that the circuit is modified by the presence of the capacitors

(e.g., elements may be shorted out).

But “high” and/or “low” frequency compare to what?


Capacitor behavior depends on the frequency of interest.


) /1(|| || CRZ ω=

Capacitor approximates an open circuit at low frequencies:

)/1( ) /1() /1(|| || ) /1(

RCCRRCRZCR

<<→<<≈=→<<

ωωωω

Capacitor approximates a short circuit at high frequencies:

)/1( ) /1(0) /1() /1(|| || ) /1(

RCCRCCRZCR

>>→>>→≈=→>>

ωωωωω

We cannot ignore the capacitor when This defines the reference frequency for high-f and low-f

)/1(~ ) /1(~ RCCR ωω →

Note: The above circuit is like a low-pass filter with a cut-off frequency of 1/RC

Example:

Finding Frequency response of amplifiers

Capacitors typically divide into two groups: low-f capacitors (setting fL) and high-f capacitors (setting fH). o We need to identify low-f and high-f caps. We will use absolute

limits of f = 0 (ALL capacitors open) and f = ∞ (ALL capacitors short) for this purpose.

For bias (f = 0) all caps are open circuit!

For “mid-band” properties (fL << f << fH) o Low-f capacitors will be short circuit (because fL << f ). o High-f capacitors will be open circuit (because f << fH).

o The resulting “resistive” circuit gives mid-band properties.

We will use time-constant method to find fL and fH (separately) o To find fL all high-f capacitors will be open circuit (because fL << fH)

o To find fH all low-f capacitors will be short circuit (because fH >> fL)


Impact of various capacitors depend on the frequency of interest


f → ∞ All Caps are short. This limit is used to find high-frequency Caps.

f → 0 All Caps are open. This limit is used to find low-frequency Caps

Mid-band: High-f caps are open Low-f caps are short.

Computing fH : High-f caps are included. Low-f caps are short

Computing fL: High-f caps are open. Low-f caps included.

Impendence of capacitors (1/ωC)

How to find which capacitors contribute to the lower cut-off frequency


Cc1 open: vi = 0 → vo = 0 Contributes to fL

Example:

Consider each capacitor individually. Let f = 0 (capacitor is open circuit): o If vo (or AM) does not change, capacitor does NOT contribute to fL (i.e.,

it is a high-f cap) o If vo (or AM) → 0 or is reduced substantially, capacitor contributes to fL

(i.e., it is a low-f cap)

CL open: No change in vo Does NOT contribute to fL

How to find which capacitors contribute to the upper cut-off frequency


Cc1 short: No change in vo Does NOT contribute to fH

CL short: vo = 0 Contributes to fH

Consider each capacitor individually. Let f → ∞ (capacitor is short circuit): o If vo (or AM) does not change, capacitor does NOT contribute to fH

(i.e., it is a low-f cap) o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fH

(i.e., it is a high-f cap)

Example:

Constructing appropriate circuits


Example:

Cc1 : Low-f capacitor CL : High-f capacitor

Mid-band: High-f caps are open Low-f caps are short.

Computing fH : High-f caps are included. Low-f caps are short

Computing fL: High-f caps are open. Low-f caps included.


Low-Frequency Response

Low-frequency response of an amplifier


Each capacitors gives a pole.

All poles contribute to fL (exact value of fL from computation or simulation)

A good approximation for design & hand calculations:

fL ≈ fp1 + fp2 + fp3 + … If one pole is at least a factor of 4 higher than others (e.g., fp2 in the above

figure), fL is approximately equal to that pole (e.g., fL ≈ fp2 in above within 20%)

321

ppp

Msig

o

ss

ss

ssA

VV

ωωω +×

+×

+×=

Example: an amplifier with three poles

(Set s = jω to find Bode Plots)

Low-frequency response of a CS amplifier (from detailed frequency response analysis)


Cc1 open: vi = 0 → vo = 0

Cc2 open: vo = 0

Cs open: Gain is reduced substantially (from CS amp. to CS amp. With RS)

)||||(

321

LDomsigG

GM

pppM

sig

o

RRrgRR

RA

ss

ss

ssA

VV

+−=

+×

+×

+×=

ωωω

Lengthy calculations: See S&S pp689-692 for detailed calculations (S&S assumes ro → ∞ and RS → ∞ )

,)]1/()||[(||[

1

)||(

1 ,)(

1

2

23

11

omLDoSsp

LoDcp

sigGcp

rgRRrRC

RrRCRRC

++≈

+=

+=

ω

ωω

All capacitors contribute to fL (as vo is reduced when f → 0 or caps open circuit)

Finding poles by inspection

1. Set vsig = 0* 2. Consider each capacitor separately, e.g., Cn (assume all others are

short circuit!)

3. Find the total resistance seen between the terminals of the capacitor, e.g., Rn (treat ground as a regular “node”).

4. The pole associated with that capacitor is

5. Lower-cut-off frequency can be found from

fL ≈ fp1 + fp2 + fp3 + …


* Although we are calculating frequency response in frequency domain, we will use time-domain notation instead of phasor form (i.e., vsig instead of Vsig) to avoid confusion with the bias values.

nnpn CR

fπ2

1=

Example: Low-frequency response of a CS amplifier (from pole inspection)


Examination of circuit shows that ALL capacitors are low-f capacitors.

In the following slides with compute poles introduced by each capacitor. (Compare with the detailed calculations of slide 13!)

fL ≈ fp1 + fp2 + fp3

Low-frequency response of a CS amplifier (fp1)

1. Consider Cc1 :

∞

)( 21

11

sigGcp RRC

f+

=π

2. Find resistance between Capacitor terminals

Terminals of Cc1



1. Consider CS :

Terminals of CS

)]1/()||[(||[ 21

2omLDoSS

p rgRRrRCf

++=

π


om

LDo

rgRRr

++1

||

om

LDo

rgRRr

++1

||

om

LDo

rgRRr

++1

||



1. Consider Cc2 : Terminals of Cc2

)||( 21

23

oDLcp rRRC

f+

=π




High-Frequency Response

o Amplifier gain falls off due to the internal capacitive effects of transistors as well as possible capacitors in the circuit.

Capacitive Effects in pn Junction


Forward Bias Reverse Bias

02 jj CC ⋅≈

T

DTd V

IC ⋅=τ

0=dC

mR

jj VV

CC

)/1( 0

0

+=

rD

Cj + Cd

Charge stored in the pn junction, leading to a capacitance

For majority carriers, stored charge is a function of applied voltage leading to a “small-signal” junction capacitance, Cj

For minority carriers, stored charge depends on the time for these carriers to diffuse across the junction and recombine, leading to a diffusion capacitance, Cd

Both Cj and Cd depend on bias current and/or voltage. Junction capacitances are small and are given in femto-Farad (fF)

1 fF = 10−15 F

*See S&S pp154-156 for detailed derivations

High-f small signal model of diode

Capacitive Effects in MOS*


1. Capacitance between Gate and channel (Parallel-plate capacitor) appears as 2 capacitors: between gate/source & between gate/drain

3. Junction capacitance between Source and Body (Reverse-bias junction)

4. Junction capacitance between Drain and Body (Reverse-bias junction)

2. Capacitance between Gate & Source and Gate & Drain due to the overlap of gate electrode (Parallel-plate capacitor)

MOS High-frequency small signal model

*See S&S pp154-156 for detailed derivations

MOS high-frequency small signal model


For source connected to body (used by S&S)

Accurate Model (we use this model here)

Generally, transistor internal capacitances are shown outside the transistor so that we can use results from the mid-band calculations.

High-frequency response of a CG amplifier


gddbLL CCCC ++=′

sbgsin CCC +=

Cgs between source & ground

“Input Pole” “Output Pole”

Low-pass filter

Low-pass filter

Mid-band Amp

Cgd between drain & ground

High-f response of a CG amplifier – Exact Solution (1)


gddbLL CCCC ++=′

sbgsin CCC +=

0)(/1

: Node

0)(/1

: Node

=−

+−+′

+′

=−

+−−+−

o

ioim

L

o

L

oo

o

oiim

in

i

sig

sigii

rvvvg

Csv

Rvv

rvvvg

sCv

Rvv

vCan be solved to find vo/vsig




)/1||(11

/1/1

)1/(11

11

11

0)( : Node

msiginsigm

m

sig

i

sigmsiginsigmsiginsigmsig

i

isigminsigii

gRsCRgg

vv

RgRsCRgRsCRgvv

vRgsCvvv

+×

+=

++×

+=

++=

=++−

LL

Lm

i

oimoL

L

oo RCs

RgvvvgvCs

Rvv

′′+′

=⇒=−′+′ 1

0)( : Node

Voltage divider (Ri= 1/gm and Rsig) “Input Pole”

“Output Pole”

Mid-band Gain

LLmsiginLm

sigi

i

sig

o

RCsgRsCRg

RRR

vv

′′+×

+×′×

+=

11

)/1||(11)(

Compact solution can be found by ignoring ro (i.e., ro → ∞)

Open-Circuit Time-Constants Method


b1 can be found by the open-circuit time-constants method.

1. Set vsig = 0 2. Consider each capacitor separately, e.g., Cj (assume others are open circuit!)

3. Find the total resistance seen between the terminals of the capacitor, e.g., Rj (treat ground as a regular “node”).

4.

A good approximation to fH is:

1 21

bfH π

=

...11

).../1)(/1(...1

...1...1)(

211

21

221

221

221

++=

+++++

=++++++

=

pp

pp

b

sssasa

sbsbsasasH

ωω

ωω

jjnj CRb 11 =Σ=

High-f response of a CG amplifier – time-constant method (input pole)


1. Consider Cin :

)]1/()(||[1 omLosigin rgRrRC +′+=τ


Terminals of Cin

om

Lo

rgRr

+′+

1

gddbLL CCCC ++=′

sbgsin CCC +=

om

Lo

rgRr

+′+

=1

om

Lo

rgRr

+′+

=1

High-f response of a CG amplifier – time-constant method (output pole)


1. Consider C’L :

)]1(||[2 sigmoLL RgrRC +′′=τ


)1( sigmo Rgr +≈

)1( sigmo Rgr +

gddbLL CCCC ++=′

sbgsin CCC +=

High-f response of a CG amplifier – time-constant method


omLoi

Lomsigi

iM

rgRrR

RrgRR

RA

/)(

)||(

′+=

′+

+=

)( 21

21

)]1(||[

)]1/()(||[

211

2

1

ττππ

τ

τ

+==

+′′=

+′+=

bf

RgrRCrgRrRC

H

sigmoLL

omLosigin

gddbLL CCCC ++=′

sbgsin CCC +=

LLmsiginLm

sigm

m

sig

o

RCsgRsCRg

Rgg

vv

′′+×

+×′×

+=

11

)/1||(11)(

/1/1

Comparison of time-constant method with the exact solution (ro → ∞)

AM 11 /1:poleInput

τω =p 22 /1:poleOutput

τω =p

High-frequency response of a CS amplifier


dbL CC +

Csb is shorted out. Cgd is between output and input!

“Input Pole?” “Output Pole?”

Two methods to find fH 1) Miller’s theorem 2) Direct calculation of

resistance between terminals of Cgd (see Problem Set 8, Exercise 4)

Miller’s Theorem


12 VAV ⋅=

ZVA

ZVVI 121

1)1( ⋅−

=−

=

)1( ,

)1/( 11

111 A

ZZZV

AZVI

−==

−=

AZZ

ZV

AZAVI

/11 ,

)1/( 22

222 −

==−

=

AZVA

ZVA

ZVVI

)1( )1( 2112

2−

=−

=−

=

Consider an amplifier with a gain A with an impedance Z attached between input and output

V1 and V2 “feel” the presence of Z only through I1 and I2 We can replace Z with any circuit as long as a current I1 flows out of

V1 and a current I2 flows out of V2.

Miller’s Theorem – Statement

If an impedance Z is attached between input and output an amplifier with a gain A , Z can be replaced with two impedances between input & ground and output & ground.


12 VAV ⋅=

Other parts of the circuit

A

ZZ 112

−=

AZZ−

=11

12 VAV ⋅=

Example of Miller’s Theorem: Inverting amplifier


1

0101

00

11

100

)/()/()/(

RR

vv

ARRR

ARRARA

RRRA

vvA

vv

f

i

o

f

f

f

f

f

f

i

n

i

o

−≈

+

−=

+

−=

+

−=−=

nnpo vAvvAv ⋅−=−⋅= 00 )( :OpAmp

1RR

vv f

i

o −=

Recall from ECE 100, if A0 is large

Solution using Miller’s theorem:

ff

f RA

RR ≈

+=

02 /1100

1 1 AR

AR

R fff ≈

+=

11

1

f

f

i

n

RRR

vv

+=

AZZ

/112 −=

AZZ−

=11

Applying Miller’s Theorem to Capacitors


A

ZZ 112

−=

AZZ−

=11

12 VAV ⋅=

)/11( /11

)1( 1

1

22

11

CACA

ZZ

CACA

ZZ

CjZ

−=⇒−

=

−=⇒−

=

=ω

Large capacitor at the input for A >> 1

High-frequency response of a CS amplifier – Using Miller’s Theorem


Use Miller’s Theorem to replace capacitor between input & output (Cgd ) with two capacitors at the input and output.

)]||(1[)1(, Lomgdgdigd RrgCACC ′+=−=

*

)]||(/11[)/11(,

gd

Lomgdgdogd

CRrgCACC

≈

′+=−=

)||( Lomg

d RrgvvA ′−==

* Assuming gmR’L >> 1

igdgsin CCC ,+= LogddbL CCCC ++=′ ,

Note: Cgd appears in the input (Cgd,i) as a “much larger” capacitor.

High-f response of a CS amplifier – Miller’s Theorem and time-constant method


Output Pole (C’L ):

)||( 2 LoL RrC ′′=τ 1 siginRC=τ

Input Pole (Cin ):

)||( 21

1 LoLsiginH

RrCRCbf

′′+==π

or

∞

LgddbL

Lomgdgsin

CCCCRrgCCC

++=′

′++= )]||(1[

High-f response of a CS amplifier – Exact solution


Miller & time-constant method:

1. Same b1 and same fH as the exact solution

2. Although, we get the same fH, there is a substantial error in individual input and output poles.

3. Miller approximation did not find the zero!

Solving the circuit (node voltage method):

)||(

]))([(

)||(

1)/1()||(

2

1

221

Losig

gdgsgdgsdbL

LoLsigin

mgdLom

sig

o

RrRCCCCCCb

RrCRCbsbsb

gsCRrgvv

′

×+++=

′′+=

++

−×′−=

LgddbL

Lomgdgsin

CCCCRrgCCC

++=′

′++= )]||(1[

)||( 21

1 LoLsiginH

RrCRCbf

′′+==π

Miller’s Theorem vs Miller’s Approximation


For Miller Theorem to work, ratio of V2/V1 (amplifier gain) should be calculated in the presence of impedance Z.

In our analysis, we used mid-band gain of the amplifier and ignored changes in the gain due to the feedback capacitor, Cgd. This is called “Miller’s Approximation.” o In the OpAmp example the gain of the chip, A0 , remains constant when Rf is

attached (as the output resistance of the chip is small).

Because the amplifier gain in the presence of Cgd is smaller than the mid-band gain (we are on the high-f portion of the Bode gain plot), Miller’s approximation overestimates Cgd,i and underestimates Cgd,o o There is a substantial error in individual input and output poles. However, b1

and fH are estimated well.

More importantly, Miller’s Approximation “misses” the zero introduced by the feedback capacitor (This is important for stability of feedback amplifiers as it affects gain and phase margins). o Fortunately, we can calculate the zero of the transfer function easily (next slide).

Finding the “zero” of the CS amplifier


1) Definition of Zero: vo(s = sz) = 0 2) Because vo = 0, zero current will flow in ro, CL+Cdb and R’L

3) By KCL, a current of gmvgs will flow in Cgd.

4) Ohm’s law for Cgd gives:

gdz

gsmgs Cs

vgi Z v ==− 0

gsmvgi =0=i

gd

mz

gd

mz C

gfCgs

2 ,

π==

Zero of CS amplifier can play an important role in the stability of feedback amplifiers


zf 2pf 1pf zf 2pf 1pf

12 of Case ppz fff >>> 12 of Case pzp fff >>

Substantial change in gain and phase margins!

Note: Since the input pole is at

Small Rsig can push fp2 to very large values!

) 2/(1)1/(2 1 siginRCππτ =

Comparison of CS and CG amplifiers


Both CS and CG amplifiers have a high gain of

CS amplifier has an infinite input resistance while CG amplifier suffers from a low input resistance.

CG amplifier has a much better high-f response:

o CS amplifier has a large capacitor at the input due to the Miller’s effect: compared to that of a CG amplifier

o In addition a CS amplifier has a zero.

The Cascode amplifier combines the desirable properties of a high input resistance with a reasonable high-f response. (It has a much better high-f response than a two-stage CS amplifier)

)||( Lom Rrg ′

)]||(1[ Lomgdgsin RrgCCC ′++=

sbgsin CCC +=

Caution: Time-constant Method & Miller’s Approximation


The time constant method approximation to fH (see S&S page 724).

Since,

This is the correct formula to find fH

However, S&S gives a different formula in page 722 (contradicting formulas of pp724). Ignore this formula (S&S Eq. 9.68)

Discussions (and some conclusions re Miller’s theorem) in Examples 9.8

to 9.10 are incorrect. The discrepancy between fH from Miller’s approximation and exact solution is due to the use of Eq. 9.68 (Not Miller’s fault!)

...11112

32

221

+++=pppH ωωωω

...111 ...11

21211 ++≈⇒++=

ppHpp

bωωωωω

Hjj

nj CRb

ω1

11 ≈Σ= =

Caution: Miller’s Approximation


The main value of Miller’s Theorem is to demonstrate that a large capacitance will appear at the input of a CS amplifier (Miller’s capacitor).

While Miller’s Approximation gives a reasonable approximation to fH, it fails to provide accurate values for each pole and misses the zero.

o Miller’s approximation should be used only as a first guess for analysis. Simulation should be used to accurately find the amplifier response.

o Stability analysis (gain and phase margins) should utilize simulations unless a dominant pole far away from fH is introduced.

Miller’s approximation breaks down when gain is close to 1 (See source follower, following slides).

High-f response of a Source Follower


Cdb is shorted out.

Cgs between output and input

Cgd between gate and ground

Because mid-band gain is close to 1 and positive, Miller approximation will not work well. o Need to apply Miller’s theorem exactly o Lengthy calculations

High-f response of a source follower – Exact Solution


0)()()(||

: Node

0)( : Node

=−+−+++′

=−++−

iogsoimosbLoL

oo

oigsigdsig

sigii

vvsCvvgvCCsrR

vv

vvsCvsCR

vvv

2211

)/1()||(1

)||( sbsb

gsCRrg

Rrgvv mgs

Lom

Lom

sig

o

++

+×

′+′

= gs

mz

gs

mz C

gfCgs

2 , : Zero

π=−=

Mid-band gain Lengthy analysis is needed to find b1, b2, and two poles

High-f response of a source follower – time-constant method (1)


CL + Csb:

)||/1)(( 2 LmsbL RgCC ′+=τ 1 siggd RC=τ

Cgd :

mg/1∞

High-frequency response of a follower – Time-constant method (2)


3. Cgs (cannot use elementary R forms)

High-frequency response of a follower – Time-constant method (3)


3. Cgs (cannot use elementary R forms), continued

)||(1)||(

)]||([)]||(1[

)||()||(

))(||( KVL

oLm

oLsig

x

xgs

oLsigxoLmx

xoLmxoLsigxx

gsmxoLsigxx

gsx

rRgrRR

IVR

rRRIrRgVVrRgIrRRIV

vgIrRRIVvV

′+

′+==

′+=′+

′−′+=

−′+=

=

3 gsgs RC=τ

)||(1)||(

)||/1)(( 21

3211oLm

oLsiggsLmsbLsiggd

H rRgrRR

CRgCCRCbf ′+

′+×+′++=++== τττ

π

Finding the “zero” of a source follower


gsmvgi =

0=i

0=i

1) Definition of Zero: vo(s = sz) = 0 2) Because vo = 0, zero current will flow in ro, CL, Csb and R’L

3) By KCL, a current of gmvgs will flow in Cgs.

4) Ohm’s law for Cgs gives:

gsz

gsmgs Cs

vgi Zv ==− 0

gs

mz

gs

mz C

gfCgs

2 ,

π=−=


Examples of Computing high-f response of various amplifiers

Procedure:

1. Include internal-capacitances of NMOS and simplify the circuit.

2. Use Miller’s approximation for “Miller” capacitors in configurations with large (and negative) A.

3. Use time-constant method to find fH

4. Do not forget about zeros in CS and CD configurations.

High-f response of a CS amplifier with current-source/active load


Between output & ground

Miller’s

Between input & ground

212,1 dbdbgdogdLL CCCCCC ++++=′

igdgsin CCC ,11 +=

Csb1 , Csb2 , & Cgs2 are shorted out.

High-f response of a CS amplifier with current-source/active load


212,1

,11

1,1

11,1

)/11()/11(

)1()1(

dbdbgdogdLL

igdgsin

Lmgdgdogd

Lmgdgdigd

CCCCCCCCC

RgCACCRgCACC

++++=′

+=

′+=−=

′+=−=

LmLoomg

d RgRrrgvvA ′−≡−== )||||( 21

1

1

∞

ro1

sigin

sigsigin

RCRRRC

=⇒

=∞=

1

|| :

τ

)||||( |||| :

212

21

LooL

LooL

RrrCRrrRC

′=⇒=′

τ

gd

mz

gd

mz C

gfCgs

2 ,

π==

)||||( 21

211 LooLsiginH

RrrCRCbf

′+==π

High-f response of Cascode amplifiers


Between output & ground

Between D1 & ground

Miller’s

Between input & ground

22 dbgdLL CCCC ++=′

221,11 sbgsdbogd CCCCC +++=

igdgsin CCC ,11 +=

Csb1 is shorted out.

High-f response of cascode amplifiers


∞ ro1

Ri2

Ro2

)||(

1

&

)||(

2111

122

22212212

222

iomv

Lom

Loioomooo

Lomv

RrgA

Rrg

RrRrrgrrR

RrgA

−=

=+

+=++=

= Mid-band Gains & resistances

igdgsin

sbdbgsogd

dbgdLL

vgdogd

vgdigd

CCCCCCCC

CCCCACC

ACC

,11

212,11

22

1,1

11,1

)/11(

)1(

+=

+++=

++=′

−=

−= Miller’s Capacitors

3211

232

2112211

1

21

)||( || : )||( || :

|| :

τττπ

ττ

τ

++==

′=⇒=′=⇒=

=⇒=∞=

bf

RRCRRRCRrCRrRC

RCRRRC

H

LoLLoL

ioio

siginsigsigin

1

1

1

1

2 ,

gd

mz

gd

mz C

gfCgs

π==

High-f response of differential amplifiers

For a symmetric differential amplifier (i.e. when we can use the half-circuit concept):

o Therefore, the poles and zero of the differential amplifier

(differential gain) are the same as those of differential half circuit.

Detailed analysis is necessary for asymmetric differential amplifier (e.g., IC differential amplifier with a single-ended output of slides 21-29 of Lecture set 7B),


d

do

d

do

d

dod

dodododooo

vv

vv

vv

A

vvvvvv

5.02

2

1,1,,

,1,1,2,12

−=×−==

−=−=→−=

Dominant Pole Compensation


1. Very often we need to purposely introduce an additional “pole” in the circuit (in order to provide gain or phase margin in feedback amplifiers). This is called the dominant pole compensation.

2. This pole has to be the “dominant pole” (several octave below any zero or pole).

3. As such, we can ignore transistor internal capacitances in the analysis as poles introduced by these capacitances would be at higher frequencies and does not impact the dominant pole.

4. Dominant pole is introduce by the addition of a capacitor either a) Between output & ground or b) Between input & output of one stage (using Miller Effect)

Dominant Pole by adding CL


Typically the required CL is large and CL is located outside the

chip (i.e., between output terminal and ground).

)||( 2/1 oLLp RRCf π=

Ro

Generic Form: Example:

Dominant Pole via Miller’s effect


Ri

Generic Form: Example:

Dominant pole is produced by |A|CM at the input (not CM in the output). Otherwise we would have connected CM in the output and not as a Miller capacitor! )||( |A| 2/1 isigMp RRCf π=

Dominant Pole via Miller’s effect


Generic Form:

Note: fp1 is proportional to Rsig

1. Usually not used in the first-stage as we do NOT know what Rsig is.

2. For second or latter stages, Rsig is Ro of previous stage and can be large. Because CM appears as a large capacitor due to Miller’s effect, this is the preferred method for including a capacitor inside the chip.

3. Note Miller’s approximation does NOT give the correct value for poles. Simulation should be used to confirm the value of CM . Also note that CM introduces a zero.

)||( |A| 2/1 isigMp RRCf π=

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