Chapter 11 Frequency Response - ocw.snu.ac.kr

1 / 78

Chapter 11 Frequency Response

11.1 Fundamental Concepts

11.2 High-Frequency Models of Transistors

11.3 Analysis Procedure

11.4 Frequency Response of CE and CS Stages

11.5 Frequency Response of CB and CG Stages

11.6 Frequency Response of Followers

11.7 Frequency Response of Cascode Stage

11.8 Frequency Response of Differential Pairs

11.9 Additional Examples

2 / 78

Chapter Outline

CH 11 Frequency ResponseCH 11 Frequency Response

3 / 78CH 11 Frequency ResponseCH 11 Frequency Response

High Frequency Roll-off of Amplifier

As frequency of operation increases, the gain of amplifier

decreases. This chapter analyzes this problem.

4 / 78CH 11 Frequency Response

Example: Human Voice I

Audible sound spans a frequency range from 20Hz to 20kHz.

However, conventional telephone system passes

frequencies from 400Hz to 3.5kHz over which human voice

can produce.

Natural Voice Telephone System


Example: Human Voice II

CH 11 Frequency Response

Mouth RecorderAir

Mouth EarAir

Skull

Path traveled by the human voice to the voice recorder

Path traveled by the human voice to the human ear

Since the paths are different, the results will also be

different.


Example: Video Signal

Video signals without sufficient bandwidth become fuzzy as

they fail to abruptly change the contrast of pictures from

complete white into complete black. (The case with analog

raster scan)CH 11 Frequency Response

High Bandwidth Low Bandwidth


Gain Roll-off: Simple Low-pass Filter

In this simple example, as frequency increases the

impedance of C1 decreases and the voltage divider consists

of C1 and R1 attenuates Vin to a greater extent at the output.



Gain Roll-off: Common Source

The capacitive load, CL, is the culprit for gain roll-off since at high frequency, it will “steal” away some signal current and shunt it to ground.

1||out m in D

L

V g V RC s


Frequency Response of the CS Stage

At low frequency, the capacitor is effectively open and the gain is flat. As frequency increases, the capacitor tends to a short and the gain starts to decrease. A special frequency is ω=1/(RDCL), where the gain drops by 3dB.

2 2 2

1

1

1

out

in

m D

L

m D

D L

out m D

in D L

VH s s

V

g RC s

g R

R C s

V g R

V R C


Example: Figure of Merit

This metric quantifies a circuit’s gain, bandwidth, and power

dissipation. In the bipolar case, low temperature, supply, and

load capacitance mark a superior figure of merit.

. . .

1

1

1

m C

C L

C CC

CC

T C L

C CC

T CC L

Gain BandwidthF O M

Power Consumption

g RR C

I V

IR

V R C

I V

V V C


Example: Relationship between Frequency

Response and Step Response


2 2 21 1

1

1H s j

R C

0

1 1

1 expout

tV t V u t

R C

The relationship is such that as R1C1 increases, the

bandwidth drops and the step response becomes slower.


Bode Plot

When we hit a zero, ωzj, the Bode magnitude rises with a

slope of +20dB/dec.

When we hit a pole, ωpj, the Bode magnitude falls with a

slope of -20dB/dec

21

210

11

11

)(

pp

zz

ss

ss

AsH


Example: Bode Plot

The circuit only has one pole (no zero) at 1/(RDCL), so the slope drops from 0 to -20dB/dec as we pass ωp1.

1

1p

D LR C


Pole Identification Example I

1

1p

S inR C 2

1p

D LR C

2

2

22

1

2 11 pp

Dm

in

out Rg

V

V


Pole Identification Example II

1

1

1||

p

S in

m

R Cg

2

1p

D LR C


Circuit with Floating Capacitor

The pole of a circuit is computed by finding the effective

resistance and capacitance from a node to GROUND.

The circuit above creates a problem since neither terminal

of CF is grounded.


Miller’s Theorem

If Av is the gain from node 1 to 2, then a floating impedance

ZF can be converted to two grounded impedances Z1 and Z2.

1 2 1 1 2 2

1 2

11

1 2

22

1 2

,

1

1

1

F F

FF

v

FF

v

V V V V V V

Z Z Z Z

V ZZ Z

V V A

V ZZ Z

V V

A


Miller Multiplication

With Miller’s theorem, we can separate the floating capacitor. However, the input capacitor is larger than the original floating capacitor. We call this Miller multiplication.

1

0

1

1 1

F

v F

ZZ

A A C s

2

0

1

1 11 1

F

Fv

ZZ

C sA A


Example: Miller Theorem

FDmS

inCRgR

1

1

F

Dm

D

out

CRg

R

1

1

1


High-Pass Filter Response

12

1

2

1

2

1

11

CR

CR

V

V

in

out

The voltage division between a resistor and a capacitor can be configured such that the gain at low frequency is reduced.



Example: Audio Amplifier

,

12 20Hz

179.6nF

100 2 20

2 20kHz

139.8nF

200 2 20k

i i

i

mp out

L

L

R C

Ck

g

C

C

In order to successfully pass audio band frequencies (20 Hz-20 kHz), large input and small output capacitances are needed.

100 k

1/ 200

i

m

R

g



Capacitive Coupling vs. Direct Coupling

Capacitive coupling, also known as AC coupling, passes

AC signals from Y to X while blocking DC contents.

This technique allows independent bias conditions between

stages. Direct coupling does not.

Capacitive Coupling Direct Coupling



Typical Frequency Response

Lower Corner Upper Corner



High-Frequency Bipolar Model

At high frequency, capacitive effects come into play. Cb

represents the base charge, whereas C and Cje are the

junction capacitances.

b jeC C C


High-Frequency Model of Integrated Bipolar

Transistor

Since an integrated bipolar circuit is fabricated on top of a

substrate, another junction capacitance exists between the

collector and substrate, namely CCS.


Example: Capacitance Identification


MOS Intrinsic Capacitances

For a MOS, there exist oxide capacitance from gate to channel, junction capacitances from source/drain to substrate, and overlap capacitance from gate to source/drain.


Gate Oxide Capacitance Partition and Full Model

The gate oxide capacitance is often partitioned between source

and drain. In saturation, C2 ~ Cgate, and C1 ~ 0. They are in

parallel with the overlap capacitance to form CGS and CGD.


Example: Capacitance Identification

DB1 DB2

GS2

C +C

+C


Transit Frequency

Transit frequency, fT, is defined as the frequency where the

current gain from input to output drops to 1.

2 2 2 2 2

1,

1 1

1 1

2

The transit frequency of MOSFETs

is obtained in a similar fashion.

2

in out m in in

out m

in

outT

in

mT T

mT T

GS

Z r I g I ZC s

I g r

I r C s r C s

Ir C

I

gf

C

gf

C


Example: Transit Frequency Calculation

2

,

2 29 6

,1980

2

,

From Problem 11.28,

32

2

100 40059

65 10 1 10

If 400 / ,

226

nT GS TH

T today

T s

n

T today

f V VL

f

f

cm V s

f GHz


The minimum channel length of MOSFETs has been scaled

from 1μm in the late 1980s to 65nm today. Also, the

inevitable reduction of the supply voltage has reduced the

gate-source overdrive voltage from about 400mV to 100mV.

By what factor has the fT of MOSFETs increased?


Analysis Summary

The frequency response refers to the magnitude of the

transfer function.

Bode’s approximation simplifies the plotting of the

frequency response if poles and zeros are known.

In general, it is possible to associate a pole with each node

in the signal path.

Miller’s theorem helps to decompose floating capacitors

into grounded elements.

Bipolar and MOS devices exhibit various capacitances that

limit the speed of circuits.



High Frequency Circuit Analysis Procedure

Determine which capacitor impact the low-frequency region

of the response and calculate the low-frequency pole

(neglect transistor capacitance).

Calculate the midband gain by replacing the capacitors with

short circuits (neglect transistor capacitance).

Include transistor capacitances.

Merge capacitors connected to AC grounds and omit those

that play no role in the circuit.

Determine the high-frequency poles and zeros.

Plot the frequency response using Bode’s rules or exact

analysis.



Frequency Response of CS Stage

Ci acts as a high pass filter.

Lower cut-off frequency must be lower than the lowest

signal frequency fsig,min (20 Hz in audio applications).


1 21 2

1 21 2

1 1

iX

in i

i

R R C sR RVs

V R R C sR R

C s

,min

1 2

1

2sig

i

fR R C

Thus,


Frequency Response of CS Stage

with Bypassed Degeneration

1

1 1 1

m D S bout D

X S b m SS

b m

g R R C sV Rs

V R C s g RR

C s g

In order to increase the midband gain, a capacitor Cb is

placed in parallel with Rs.

The pole frequency must be well below the lowest signal

frequency to avoid the effect of degeneration.


1

out D

XS

m

V Rs

VR

g


Unified Model for CE and CS Stages


Unified Model Using Miller’s Theorem

,

1

1p in

Thev in m L XYR C g R C

,

1

11

p out

L out XY

m L

R C Cg R

r


Example: CE Stage

p,in

p,out

2 516 MHz

2 1.59 GHz

(a) Calculate the input and output poles if RL=2 kΩ. Which

node appears as the speed bottleneck?


,

1

1p in

S m LR r C g R C

,

1

11

p out

L CS

m L

R C Cg R

200 , 1 mA

100, 100 fF

20 fF, 30 fF

S C

CS

R I

C

C C


Example: CE Stage – cont’d

(b) Is it possible to choose RL such that the output pole

limits the bandwidth?

, ,

m L

1 1

1 11

If g R 1,

p in p out

S m L

L CS

m L

CS m S L S

R r C g R CR C C

g R

C C g R r C R R r C

With the values assumed in this example, the left-hand side is negative,

implying that no solution exists. Thus, the input pole remains the speed

bottleneck.


Example: Half Width CS Stage

W 2X

bias current 2X

2

21

2

1

221

2

1

,

,

XY

Lm

outL

outp

XYLminS

inp

C

Rg

CR

CRgCR


m n ox D

Wg 2 C I 2X

L

capacitances 2X

bandwidth 2X

gain 2X

gain bandwidth constant


Direct Analysis of CE and CS Stages

1

1At Node Y:

XY out

LX out XY m X out out X out

L XY m

C s C sR

V V C s g V V C s V VR C s g

2 1

XY m Lout

Thev

C s g RVs

V as bs

At Node X: X Thevout X XY X in

Thev

V VV V C s V C s

R

1

1XY out

ThevLout XY XY in out

Thev XY m Thev

C s C sVR

V C s C s C s VR C s g R

where ,

1

Thev L in XY out XY in out

m L XY Thev Thev in L XY out

a R R C C C C C C

b g R C R R C R C C


Direct Analysis of CE and CS Stages – cont’d

Direct analysis yields different pole locations and an extra zero.

22

1 2 1 2 1 2

1 1 1

2 1 p1 p2 p1

1

1

2

| |

1 11 1 1 1

1

1

1

1| |

mz

XY

p p p p p p

p p

p

p

m L XY Thev Thev in L XY out

m L XY Thev Thev in L XY o

p

g

C

s s sas bs s

if

b

g R C R R C R C C

g R C R R C R C Cb

a

ut

Thev L in XY out XY in outR R C C C C C C

Dominant-pole approximation


Example: Dominant-pole approximation

outinXYoutXYinOOS

outXYOOinSSXYOOmp

outXYOOinSSXYOOm

p

CCCCCCrrR

CCrrCRRCrrg

CCrrCRRCrrg

21

212112

21211

1

||

)(||||1

)(||||1

1

1

1

1 2 2

in GS

XY GD

out DB GD DB

C C

C C

C C C C


Example: Comparison Between Different Methods

,

,

2 571 MHz

2 428 MHz

p in

p out

2 264 MHz

2 4 53 GHz

p,in

p ,out .

2 249 MHz

2 4 79 GHz

p,in

p ,out .

1

200

250 fF

80 fF

100fF

150

0

2 k

S

GS

GD

DB

m

L

R

C

C

C

g

R

Miller’s Exact Dominant Pole

This error arises because we have multiplied by the midband

gain (1+ ) rather than the gain at high frequencies.

GD

m L

C

g R


Input Impedance of CE and CS Stages

rsCRgC

ZCm

in ||1

1

sCRgCZ

GDDmGS

in

1

1


Low Frequency Response of CB and CG Stages

out C m C i

1

in m S i mS i m

V R g R C ss

V 1 g R C s gR C s 1/ g

As with CE and CS stages, the use of capacitive coupling

leads to low-frequency roll-off in CB and CG stages

(although a CB stage is shown above, a CG stage is similar).



Frequency Response of CB Stage

X

m

S

Xp

Cg

R

1||

1,

CCX

,

1p Y

C YR C

CSY CCC

Or


Frequency Response of CG Stage

Or

X

m

S

Xp

Cg

R

1||

1,

SBGSX CCC

,

1p Y

D YR C

DBGDY CCC

Similar to a CB stage, the input pole is on the order of fT, so

rarely a speed bottleneck.

Or


Example: CG Stage Pole Identification

p,X

S SB1 GS1

m1

1

1R || C C

g

2211

2

, 1

1

DBGSGDDB

m

Yp

CCCCg


Example: Frequency Response of CG Stage

1

200

250 fF

80 fF

100 fF

150

0

2 kΩ

S

GS

GD

DB

m

D

R

C

C

C

g

R

,

,

11/ || 2 5.31 GHz

/ 2 442 MHz

p X S X

m

p Y L Y

R Cg

R C


Emitter and Source Followers

The following will discuss the frequency response of

emitter and source followers using direct analysis.

Emitter follower is treated first and source follower is

derived easily by allowing r to go to infinity.


Direct Analysis of Emitter Follower

1

2

1

with 1

out mm

in

Cs

V gr g

V as bs

At node X: 0out inout

S

V V V VV V C s V C s

R r

At output node: 1

out Lm out L

m

V V C sV C s g V V C s V

rC s g

r

mz T

gf

C

where

1

SL L

m

S LS

m m

Ra C C C C C C

g

C R Cb R C

g r g


Direct Analysis of Source Follower Stage

1

1

2

bsas

sg

C

V

V m

GS

in

out

SGD GS GD SB L GS SB L

m

GD SB LS GD

m

Ra C C C C C C C C

g

C C Cb R C

g


Example: Frequency Response of Source Follower

1

200

100 fF

250 fF

80 fF

100 fF

150

0

S

L

GS

GD

DB

m

R

C

C

C

C

g

21 2

11

1

2

2.58 10

5.8 10

/ 2 4.24 GHz

2 1.79 GHz 2.57 GHz

2 1.79 GHz 2.57 GHz

z m GS

p

p

a s

b s

g C

j

j


Example: Source Follower

1

1

2

bsas

sg

C

V

V m

GS

in

out

1

22111

2211111

1

))((

m

DBGDSBGDGDS

DBGDSBGSGDGSGD

m

S

g

CCCCCRb

CCCCCCCg

Ra


Input Capacitance of Emitter/Source Follower

1 1

1 1

1

Lv X v XY XY

m LL

m

XYin GD

m L

RA C A C C

g RR

g

CC C or C

g R


Example: Source Follower Input Capacitance

1 2

1 2

1

1 1

1 1

1 1 2

1

1

1

1 ||

O O

v

O O

m

in GD v GS

GD GS

m O O

r rA

r rg

C C A C

C Cg r r


Output Impedance of Emitter Follower

1

X m S X

S SX

X

I g V R V V

R r C s r RV

I r C s

1

1

X m

X

I g V r VC s

rV I

r C s


Output Impedance of Source Follower

1

with

1

S SX

X

m

S GSX

X GS m

R r C s r RV

I r C s

g r

R C sV

I C s g


Active Inductor

The plot above shows the output impedance of emitter and source followers. Since a follower’s primary duty is to lower the driving impedance (RS>1/gm), the “active inductor” characteristic on the right is usually observed.

1

S SX

X

R r C s r RV

I r C s

1S GSX

X GS m

R C sV

I C s g


Example: Output Impedance

33

321 1||

mGS

GSOO

X

X

gsC

sCrr

I

V

O3r


Frequency Response of Cascode Stage

For cascode stages, there are three poles and Miller

multiplication is smaller than in the CE/CS stage.

Assuming for all transistors,or

,1

2

x v XY XY

XY

C A C

C

1,

2

1mv XY

m

gA

g


Poles of Bipolar Cascode

111

,2||

1

CCrRS

Xp

121

2

,

21

1

CCCg

CS

m

Yp

22

,

1

CCR CSL

outp


Poles of MOS Cascode

1

2

11

,

1

1

GD

m

mGSS

Xp

Cg

gCR

1

1

221

2

,

11

1

GD

m

mGSDB

m

Yp

Cg

gCC

g

22

,

1

GDDBL

outpCCR


Example: Frequency Response of Cascode

1

200

250 fF

80 fF

100 fF

150

0

2 kΩ

S

GS

GD

DB

m

L

R

C

C

C

g

R

,

,

,

2 1.95 GHz

2 1.73 GHz

2 442 MHz

p X

p Y

p out


MOS Cascode Example

1

2

11

,

1

1

GD

m

mGSS

Xp

Cg

gCR

,

21 2 1 2 3 3

2 1

1

11

p Y

mDB GS GD SB GD DB

m m

gC C C C C C

g g

22

,

1

GDDBL

outpCCR


I/O Impedance of Bipolar Cascode

sCCrZ in

11

12

1||

sCC

RZCS

Lout

22

1||


I/O Impedance of MOS Cascode

sCg

gC

Z

GD

m

mGS

in

1

2

11 1

1

sCCRZ

DBGD

Lout

22

1||


Bipolar Differential Pair Frequency Response

Since bipolar differential pair can be analyzed using half-circuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of the half circuit’s.

Half Circuit


MOS Differential Pair Frequency Response

Since MOS differential pair can be analyzed using half-circuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of the half circuit’s.

Half Circuit


Example: MOS Differential Pair

,

1 1 3 1

,

31 3 3 1

3 1

,

3 3

1

[ (1 / ) ]

1

11

1

p X

S GS m m GD

p Y

mDB GS SB GD

m m

p out

L DB GD

R C g g C

gC C C C

g g

R C C


Common Mode Frequency Response

1

2 11 12

m D SS SSout D

CM SS SS m SS

SS

m SS

g R R C sV R

V R C s g RR

g C s

Css will lower the total impedance between point P to ground at high frequency, leading to higher CM gain which degrades the CM rejection ratio.


Tail Node Capacitance Contribution

Source-Body Capacitance of M1, M2

Drain-Body Capacitance of M3

Gate-Drain Capacitance of M3



Example: Capacitive Coupling

1 1

11

1

1 1 1

1

1 1

1

For , assuming 800 mV,

1.7 mA

ln / 748 mV

1.75 mA 14.9

1.49 k

BE

CC BEC

B

BE T C S

C m

Q V

V VI

R

V V I I

I g

r

2 2

2 2 2 2

22

2

1

2 2

2

For Q , assuming 800mV,

1.13 mA/

Iteration yields

1.17 mA, 22.2

2.22 k

BE

CC B B BE E C

CC BEC

B E

C m

V

V I R V R I

V VI

R R

I g

r

165 10 A

100

S

A

I

V


Example: Capacitive Coupling – cont’d

EBin RrRR 1|| 222

2

2 2

1

2 22.9 Hz

L

C inR R C

1

1 1 1

1

||

2 542 Hz

L

Br R C


2

1 2 2

1

2 6.92 MHz

L

D inR R C

1

1 1

1

1 1

1 1

1

1

1

2 42.4 MHz

L

S

m

m S

S

R Cg

g R

R C

2

2

2 2 2

2

1

6.67

1.30 k

Fin

v

v m D

in

RR

A

A g R

R

Example: IC Amplifier – Low Frequency Behavior


From P 35

Miller Effect

1.3k 8.7k


1 1 2

2 2

|| 3.77

6.67

25.1

out outX

in in X

Xm D in

in

outm D

X

out

in

v vv

v v v

vg R R

v

vg R

v

v

v

Example: IC Amplifier – Midband Behavior


Example: IC Amplifier – High Frequency Behavior

1

v2 GD2 GD2

With Miller effect,

1 A C 1.15 C

1

2

2 (308 MHz)

2 (2.15 GHz)

p

p

3

2 2 2

1

(1.15 )

2 (1.21GHz)

p

L GD DBR C C

28dB

42.4 MHz

308 MHz6.92 MHz

1.21 GHz

2.15 GHz

Chapter 11 Frequency Response - ocw.snu.ac.kr

Documents