Chapter 9 Cascode Stages and Current Mirrossoc.hanyang.ac.kr/kor/classes/undergraduate... · Chapter 9 Cascode Stages and Current Mirros ... MOS Cascode Differential Pair A v

1

Chapter 9 Cascode Stages and Current Mirros

Microelectronics, 2nd edition by Behzad Razavi

CH 9 Cascode Stages and Current Mirrors 2

Short-Circuit Transconductance

The short-circuit transconductance of a circuit measures its strength in converting input voltage to output current.

0=

=outvin

outm v

iG


Transconductance Example

1mm gG =


Derivation of Voltage Gain

By representing a linear circuit with its Norton equivalent, the relationship between Vout and Vin can be expressed by the product of Gm and Rout.

outminout

outinmoutoutout

RGvvRvGRiv

−=−=−=


Example: Voltage Gain

11 Omv rgA −=

6

Chapter 10 Differential Amplifiers


CH 10 Differential Amplifiers 7

Differential Pair

With the addition of a tail current, the circuits above operate as an elegant, yet robust differential pair.


MOS Differential Pair’s Common-Mode Response

Similar to its bipolar counterpart, MOS differential pair produces zero differential output as VCM changes.

2SS

DDDYXIRVVV −==


Equilibrium Overdrive Voltage

The equilibrium overdrive voltage is defined as the overdrive voltage seen by M1 and M2 when both of them carry a current of ISS/2.

( )

LWC

IVVoxn

SSequilTHGS

µ=−


Minimum Common-mode Output Voltage

In order to maintain M1 and M2 in saturation, the common-mode output voltage cannot fall below the value above.

This value usually limits voltage gain.

THCMSS

DDD VVIRV −>−2


Small-Signal Response

Similar to its bipolar counterpart, the MOS differential pair exhibits the same virtual ground node and small signal gain.

Dmv

P

RgAV

−==∆ 0


MOS Differential Pair Half Circuit Example I

−=

≠

133

1 ||||1

0

OOm

mv rrg

gA

λ


MOS Cascode Differential Pair

1331 OmOmv rgrgA −≈


MOS Telescopic Cascode

( )[ ])(|| 7551331 OOmOOmmv rrgrrggA −≈


Differential to Single-ended Conversion

Many circuits require a differential to single-ended conversion, however, the above topology is not very good.


Supply Noise Corruption

The most critical drawback of this topology is supply noise corruption, since no common-mode cancellation mechanism exists. Also, we lose half of the signal.


Better Alternative

This circuit topology performs differential to single-ended conversion with no loss of gain.


Active Load

With current mirror used as the load, the signal current produced by the Q1 can be replicated onto Q4.

This type of load is different from the conventional “static load” and is known as an “active load”.


Differential Pair with Active Load

The input differential pair decreases the current drawn from RL by ∆I and the active load pushes an extra ∆I into RL by current mirror action; these effects enhance each other.


Active Load vs. Static Load

The load on the left responds to the input signal and enhances the single-ended output, whereas the load on the right does not.


MOS Differential Pair with Active Load

Similar to its bipolar counterpart, MOS differential pair can also use active load to enhance its single-ended output.


Asymmetric Differential Pair

Because of the vastly different resistance magnitude at the drains of M1 and M2, the voltage swings at these two nodes are different and therefore node P cannot be viewed as a virtual ground.

23

Chapter 11 Frequency Response


CH 11 Frequency Response 24

High Frequency Roll-off of Amplifier

As frequency of operation increases, the gain of amplifier decreases. This chapter analyzes this problem.

Example: Human Voice I

Natural human voice spans a frequency range from 20Hz to 20KHz, however conventional telephone system passes frequencies from 400Hz to 3.5KHz. Therefore phone conversation differs from face-to-face conversation.


Natural Voice Telephone System

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.


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 DL

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.

1222 +=

ωLD

Dm

in

out

CRRg

VV


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

ωω

ωω


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.

LDp CR

11 =ω


Pole Identification Example I

inSp CR

11 =ω

LDp CR

12 =ω

( )( )22

221

2 11 pp

Dm

in

out RgVV

ωωωω ++=


Pole Identification Example II

inm

S

p

Cg

R

=

1||

11ω

LDp CR

12 =ω


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.

v

F

AZZ−

=11

v

F

AZZ

/112 −=


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.


Example: Miller Theorem

( ) FDmSin CRgR +=

11ω

FDm

D

out

CRg

R

+

=11

1ω

High-Pass Filter Response

121

21

21

11

+=

ωω

CRCR

VV

in

out

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



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

Frequency Response of CS Stagewith Bypassed Degeneration

( ) ( )11

+++−

=SmbS

bSDm

X

out

RgsCRsCRRgs

VV

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.



Unified Model for CE and CS Stages


Unified Model Using Miller’s Theorem


Direct Analysis of CE and CS Stages

Direct analysis yields different pole locations and an extra zero.

( ) ( )( ) ( )

( )outinXYoutXYinLThev

outXYLinThevThevXYLmp

outXYLinThevThevXYLmp

XY

mz

CCCCCCRRCCRCRRCRgCCRCRRCRg

Cg

++++++

=

++++=

=

1||

11||

||

2

1

ω

ω

ω


Example: CE and CS Direct Analysis

( )[ ] ( )( )[ ] ( )

( )( )outinXYoutXYinOOS

outXYOOinSSXYOOmp

outXYOOinSSXYOOmp

CCCCCCrrRCCrrCRRCrrgCCrrCRRCrrg

++++++

≈

++++≈

21

212112

212111

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

1

ω

ω

Example: Comparison Between Different Methods

( )( )MHz

MHz

outp

inp

4282

5712

,

,

×=

×=

πω

πω ( )( )GHz

MHz

outp

inp

53.42

2642

,

,

×=

×=

πω

πω ( )( )GHz

MHz

outp

inp

79.42

2492

,

,

×=

×=

πω

πω

( )

Ω==

Ω=

===

Ω=

−

KR

g

fFCfFCfFC

R

L

m

DB

GD

GS

S

20

150

10080250

200

1

λ

Miller’s Exact Dominant Pole



Input Impedance of CE and CS Stages

( )[ ] πµπ

rsCRgC

ZCm

in ||1

1++

≈ ( )[ ]sCRgCZ

GDDmGSin ++≈

11


I/O Impedance of MOS Cascode

sCggC

Z

GDm

mGS

in

++

=

12

11 1

1

( )sCCRZ

DBGDLout

22

1||+

=


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

( )33,

11

331

3

,

1311,

1

111

])/1([1

GDDBLoutp

GDm

mGSDB

m

Yp

GDmmGSSXp

CCR

CggCC

g

CggCR

+=

+++

=

++=

ω

ω

ω

Common Mode Frequency Response

( )12

1++

+∆=

∆∆

SSmSSSS

SSSSDm

CM

out

RgsCRCRRg

VV

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 and M3

Gate-Drain Capacitance of M3


54

Chapter 12 Feedback


CH 12 Feedback 55

Negative Feedback System

A negative feedback system consists of four components: 1) feedforward system, 2) sense mechanism, 3) feedback network, and 4) comparison mechanism.

CH 12 Feedback 56

Close-loop Transfer Function

1

1

1 KAA

XY

+=

CH 12 Feedback 57

Feedback Example

A1 is the feedforward network, R1 and R2 provide the sensing and feedback capabilities, and comparison is provided by differential input of A1.

121

2

1

1 ARR

RA

XY

++

=

CH 12 Feedback 58

Comparison Error

2

11RR

XY

+≈

CH 12 Feedback 59

Example: Phase Response

As it can be seen, the phase of H(jω) starts to drop at 1/10 of the pole, hits -45o at the pole, and approaches -90o at 10 times the pole.

CH 12 Feedback 60

Example: Three-Pole System

For a three-pole system, a finite frequency produces a phase of -180o, which means an input signal that operates at this frequency will have its output inverted.

CH 12 Feedback 61

Instability of a Negative Feedback Loop

Substitute jω for s. If for a certain ω1, KH(jω1) reaches -1, the closed loop gain becomes infinite. This implies for a very small input signal at ω1, the output can be very large. Thus the system becomes unstable.

)(1)()(

sKHsHs

XY

+=

CH 12 Feedback 62

“Barkhausen’s Criteria” for Oscillation

180)(1|)(|

1

1

−=∠=

ωωjKH

jKH

CH 12 Feedback 63

Time Evolution of Instability

CH 12 Feedback 64

Oscillation Example

This system oscillates, since there’s a finite frequency at which the phase is -180o and the gain is greater than unity. In fact, this system exceeds the minimum oscillation requirement.

CH 12 Feedback 65

Condition for Oscillation

Although for both systems above, the frequencies at which |KH|=1 and ∠KH=-180o are different, the system on the left is still unstable because at ∠KH=-180o, |KH|>1. Whereas the system on the right is stable because at ∠KH=-180o, |KH|<1.

CH 12 Feedback 66

Condition for Stability

ωPX, (“phase crossover”), is the frequency at which ∠KH=-180o.

ωGX, (“gain crossover”), is the frequency at which |KH|=1.

PXGX ωω <

CH 12 Feedback 67

Marginally Stable vs. Stable

Marginally Stable Stable

CH 12 Feedback 68

Phase Margin

Phase Margin = ∠H(ωGX)+180

The larger the phase margin, the more stable the negative feedback becomes

CH 12 Feedback 69

Phase Margin Example

45=PM

CH 12 Feedback 70

Frequency Compensation

Phase margin can be improved by moving ωGX closer to origin while maintaining ωPX unchanged.

CH 12 Feedback 71

Frequency Compensation Example

Ccomp is added to lower the dominant pole so that ωGX occurs at a lower frequency than before, which means phase margin increases.

CH 12 Feedback 72

Frequency Compensation Procedure

1) We identify a PM, then -180o+PM gives us the new ωGX, or ωPM.

2) On the magnitude plot at ωPM, we extrapolate up with a slope of +20dB/dec until we hit the low frequency gain then we look “down” and the frequency we see is our new dominant pole, ωP’.

CH 12 Feedback 73

Example: 45o Phase Margin Compensation

2pPM ωω =

CH 12 Feedback 74

Miller Compensation

To save chip area, Miller multiplication of a smaller capacitance creates an equivalent effect.

cOOmeq CrrgC )]||(1[ 655+=

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Chapter 9 Cascode Stages and Current Mirrossoc.hanyang.ac.kr/kor/classes/undergraduate... · Chapter 9 Cascode Stages and Current Mirros ... MOS Cascode Differential Pair A v