2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock...

2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock

Copyright © 2005 – The McGraw-Hill Companies srl

Chapter 13Frequency Response

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock



Chapter Goals• Review transfer function analysis and dominant-pole

approximations of amplifier transfer functions.• Learn partition of ac circuits into low and high-frequency

equivalents.• Learn short-circuit and open-circuit time constant methods to

estimate upper and lower cutoff frequencies.• Develop bipolar and MOS small-signal models with device

capacitances.• Study unity-gain bandwidth product limitations of BJTs and

MOSFETs.• Develop expressions for upper cutoff frequency of inverting, non-

inverting and follower configurations.• Explore gain-bandwidth product limitations of single and multiple

transistor circuits.



Chapter Goals (contd.)• Understand Miller effect and design of op amp frequency

compensation.

• Develop relationship between op amp unity-gain frequency and slew rate.

• Understand use of tuned circuits to design high-Q band-pass amplifiers.

• Understand concept of mixing and explore basic mixer circuits.

• Study application of Gilbert multiplier as balanced modulator and mixer.



Transfer Function Analysis

)()(

...2210

...2210

)()()(

sHFsLFmid

A

nsnbsbsbb

msmasasaa

sDsNsvA

Amid is midband gain between upper and lower cutoff frequencies.

HPl

sHP

sHP

s

HZl

sHZ

sHZ

s

sHF

LPk

sLPsL

Ps

LZk

sLZsL

ZssLF

1...

2

1

1

1

1...

2

1

1

1

)(

...21

...21)(

1)( jHF HPi

HZi,for ,i =1…l

)()( sLFmid

AsLA

1)( jLF for ,j =1…k)()( sHF

midAsHA

LPj

LZj,



Low-Frequency Response

2

2

)(

PLP

sssLF

Pole P2 is called the dominant low-frequency pole (> all other poles) and zeros are at frequencies low enough to not affect L.

If there is no dominant pole at low frequencies, poles and zeros interact to determine L.

21

21)()(

PsPsZsZs

midAsLF

midAsLA

For s=j, at L, 2)( mid

A

LjA

2

222

12

22

221

2

21

PLPL

ZLZL

4

22

21

2

22

211

4

22

21

2

22

211

21

L

PP

L

PP

L

ZZ

L

ZZ

Pole L > all other pole and zero frequencies

In general, for n poles and n zeros,

22

221

222

21 ZZPPL

n Znn PnL

222



Transfer Function Analysis and Dominant Pole Approximation Example• Problem: Find midband gain, FL(s) and fL for

• Analysis: Rearranging the given transfer function to get it in standard form,

Now,

and

Zeros are at s=0 and s =-100. Poles are at s= -10, s=-1000

All pole and zero frequencies are low and separated by at least a decade. Dominant pole is at =1000 and fL =1000/2= 159 Hz. For frequencies > a few rad/s:

100011.0

11002000)(

ss

sssLA

100010

100200)( ss

sssLA

200

)()(

midA

sLFmid

AsLA

)1000)(10()100()(

ss

sssLF

Hz158210020(2210002102

1 Lf

1000200)(

sssLA



High-Frequency Response

3

)3

/(1)(

PH

Ps

ssLF

Pole P3 is called the dominant high-frequency pole (< all other poles).

If there is no dominant pole at low frequencies, poles and zeros interact to determine H.

)2/(1)1/(1

)2/(1)1/(1

)()(

PsPsZsZs

midA

sHFmid

AsHA

For s=j, at H, 2)( mid

A

HjA

)2

2/2(1)2

1/2(1

)22

/2(1)21

/2(1

2

1

PHPH

ZHZH

22

21

4

22

2

21

21

22

21

4

22

2

21

21

21

PP

H

P

H

P

H

ZZ

H

Z

H

Z

H

Pole H < all other pole and zero frequencies

In general,

22

22

1

22

2

12

1

11

ZZPP

H

n Znn Pn

H

212

21

1



Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier

(s)gsV

)3(2

1)3(

33

)2

/1(3

(s)gsV3(s)oI(s)oV

RDRCs

sD

RRmg

RRsCR

DR

mgR

(s)gV

)/1(3

1

)3

/1(sV-gV(s)gsV

SRmgC

s

SRCs

IRGRGR

DRRmg

midA

(s)LFmid

A(s)vA

)3(

(s)i

V

(s)oV

(s)i

V1)(

1

1(s)gV

GR

IRCs

GRCs



Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier (contd.)

The three zero locations are: s = 0, 0, -1/(RS C3).The three pole locations are:

Each independent capacitor in the circuit contributes one pole and one zero. Series capacitors C1 and C2 contribute the two zeros at s=0 (dc), blocking propagation of dc signals through the amplifier. Third zero due to parallel combination of C3 and RS occurs at frequency where signal current propagation through MOSFET is blocked (output voltage is zero).

)3

(2

1

)/1(3

1)(

1

1

)3

/1(2

(s)

RD

RCs

SRmgC

s

GR

IRC

s

SRCss

LF

)3(2

1,)/1(3

1,)(1

1sRDRC

SRmgCGRIRC



Short-Circuit Time Constant Method to Determine L • Lower cutoff frequency for a

network with n coupling and bypass capacitors is given by:

where RiS is resistance at terminals of ith capacitor Ci with all other capacitors replaced by short circuits. Product RiS Ci is short-circuit time constant associated with Ci.

n

i iC

iSRL

1

1

Midband gain and upper and lower cutoff frequencies that define bandwidth of amplifier are of more interest than complete transfer function.



Estimate of L for C-E AmplifierUsing SCTC method, for C1,

For C2,

For C3,

)(2)(1 rBRRRCEinBRIRSR

CRRorCRRRCE

outCRRSR

3

)(3)(32

1

)(

13

o

BR

IRr

ER

o

thRr

ERRCCoutERSR

3

1

1

i iC

iSRL



Estimate of L for C-S AmplifierUsing SCTC method,

For C1,

For C2,

For C3,

GRSRRCSinGRIRSR )(1

DRRorDRRRCS

outDRRSR

3

)(3)(32

mgSRRCGoutSRSR 1

3



Estimate of L for C-B Amplifier

Using SCTC method,

For C1,

For C2,

)1()(1mgERIRRCB

inERIRSR

CRRRCBoutCRRSR 3)(32



Estimate of L for C-G Amplifier

Using SCTC method,

For C1,

For C2,

)1()(1mgSRIRRCG

inSRIRSR

DRRRCGoutDRRSR 3)(32



Estimate of L for C-C Amplifier

Using SCTC method,

For C1,

For C2,

31

)(1

RE

RorB

RIR

RCCinBRIRSR

13)(32o

thRr

ERRRCC

outERRSR



Estimate of L for C-D Amplifier

Using SCTC method,

For C1,

For C2,

GRIRRCDinGRIRSR )(1

mgSRRRCDoutSRRSR 1

332



Frequency-dependent Hybrid-Pi Model for BJT

Capacitance between base and collector terminals is:

Co is total collector-base junction capacitance at zero bias, jc is its built-in potential.

)/(1jcCB

VoC

C

Capacitance between base and emitter terminals is:

F is forward transit-time of the BJT. Cappears in parallel with r. As frequency increases, for a given input signal current, impedance of C reduces vbe and thus the current in the controlled source at transistor output.

FmgC



Unity-gain Frequency of BJT

1)(

1

(s)b

I

(s)cI)(

1)((s)

b)I(

(s)be

)V((s)cI

rCCsmg

sC

os

rCCs

rsCmg

sCmg

The right-half plane transmission zero Z = + gm/C occurring at high frequency can be neglected.

= 1/ r(C + C ) is the beta-cutoff frequency

where

and fT = T /2 is the unity gain bandwidth product. Above BJT has no current gain.

1)/(1)()(

so

rCCsos

s

Ts

os)(

CCmg

rCCo

oT

)(



Unity-gain Frequency of BJT (contd.)

Current gain is o = gmr at low frequencies and has single pole roll-off at frequencies > f, crossing through unity gain at T. Magnitude of current gain is 3 dB below its low-frequency value at f.

C

T

CI

C

T

mgC

40



High-frequency Model of MOSFET

)/(11

(s)gI

(s)d

I)(

)(

)((s)

bI

(s)gs)V((s)d

I

GDC

GSC

T

ssTs

GDC

GSCs

GDsCmg

GDsCmg

CCmg

rCCo

oT

)(

22

3

"(2/3)

"

L

TNV

GSVn

WLoxC

TNV

GSV

L

WoxCn

Tf



Limitations of High-frequency Models

• Above 0.3 fT, behavior of simple pi-models begins to deviate significantly from the actual device.

• Also, T depends on operating current as shown and is not constant as assumed.

• For given BJT, a collector current ICM exists that yield maximum fTmax.

• For FET in saturation, CGS and CGD are independent of Q-point current, so

DImgT



Effect of Base Resistance on Midband Amplifiers

Base current enters the BJT through external base contact and traverses a high resistance region before entering active area. rx models voltage drop between base contact and active area of the BJT.

To account for base resistance rx is absorbed into equivalent pi model and can be used to transform expressions for C-E, C-C and C-B amplifiers.

xrro

xrrr

mgmg

mgxrr

rmgmg

'

bev'

bevvi

xrrr ' oo '



Direct High-Frequency Analysis: C-E Amplifier

The small-signal model can be simplified by using Norton source transformation.

kΩ3.4kΩ1003 C

RRLR kΩ10kΩ3021 RRBR

BRIRBR

ivth

v BRIR

BRIRth

R

xrthR th

vsi )( xrth

Rror



Direct High-Frequency Analysis: C-E Amplifier (Pole Determination)

From nodal equations for the circuit in frequency domain,

High-frequency response is given by 2 poles, one finite zero and one zero at infinity. Finite right-half plane zero, Z = + gm/C > T can easily be neglected.

For a polynomial s2+sA1+A0 with roots a and b, a =A1 and b=A0/A1.

ogLgogL

CggL

gCL

gCs

LCC

LCCCs

mgsC

2

)-((s)sI(s)2V

LCC

mg

LCC

LCC

mgP

TCorA

A

P

orL

R

LCorL

R

LRmgCCTC

)/(12

1

1

01

1

Smallest root that gives first pole limits frequency response and determines H. Second pole is important in frequency compensation as it can degrade phase margin of feedback amplifiers.



Direct High-Frequency Analysis: C-E Amplifier (Overall Transfer Function)

)1/(1(s)th

V(s)oV

)(

)1

/(1

(s)th

V-(s)oV

)2

/(1)1

/(1

)/(1)(

(s)th

V(s)oV

)2

/(1)1

/(1

)-((s)th

V(s)oV

Psmid

As

vthA

Ps

orL

Rmg

xrthR

Ps

Psog

Lg

Zs

orLRmgxrth

R

Ps

Psog

Lg

mgsC

xrthR

rxrthR

LRo

midA

TCorP

1

1

Dominant pole model at high frequencies for C-E amplifier is as shown.



Direct High-Frequency Analysis: C-E Amplifier (Example)• Problem: Find midband gain, poles, zeros and fL.

• Given data: Q-point= ( 1.60 mA, 3.00V), fT =500 MHz, o =100, C

=0.5 pF, rx =250CL

• Analysis: gm =40IC =40(0.0016) =64 mS, r = o/gm =1.56 k.pF9.19

2 C

Tfmg

C

kΩ12.4kΩ3.4kΩ1003 C

RRLR

882ΩkΩ1kΩ5.7 I

RBRth

R

656)( xrthRror

pF1561

orL

R

LCorL

R

LRmgCCTC

MHz56.12

11

TCorPf

L

RmgormgCmg

CL

RP1111

2

153

rxrthR

LRo

vthA

GHz4.202

C

mgZfMHz603

22

2

P

Pf

Overall gain is reduced to -135 as vth =0.882vs.



Gain-Bandwidth Product Limitations of C-E Amplifier

• If Rth is reduced to zero in order to increase bandwidth, then ro would not be zero but would be limited to approximately rx.

• If Rth = 0, rx <<r so that rx = ro and

TCorrxrth

RL

RoHvA

1GBW

)( LRmgCTC

Cxr1GBW



High-Frequency Analysis: C-S Amplifier

GRIR

thR

3RDRLR

GRIRGR

ivth

v

TC

thRP

11

th

RL

R

LRmgGDCGSCTC 1

L

Rmgth

RmgGS

Cmg

GDC

LRP

11112

GDC

mgZ



Miller Multiplication

)1()(

1V

)(sI)(Y

)(1V)(oV

AsCs

ss

sAs

Total input capacitance = C(1+A) because total voltage across C is vc = vi(1+A) due to inverting voltage gain of amplifier.

)(oV)(1

V)(sI sssCs

For the C-E amplifier,

)1()1( LRmgCCACCTC



Miller Integrator

Assuming zero current in input terminal of amplifier,

inVoV

)oVin

V(inV

1V

A

sCR

osoA

ARCs

AA

RCsvA

)1(11

1

1V

oV)(

)1(1

ARCo where

For frequencies >> o, assuming A>>1,

which is the transfer function of an integrator.

sRCsoA

svA 1)(



Open-Circuit Time Constant Method to Determine H

At high frequencies, impedances of coupling and bypass capacitors are small enough to be considered short circuits. Open-circuit time constants associated with impedances of device capacitances are considered instead.

m

ii

Cio

RH

1

1

where Rio is resistance at terminals of ith capacitor Ci with all other capacitors open-circuited.

For a C-E amplifier, assuming CL =0oroR

)1(xixv

orL

R

LRmgoroR

TCorCoRCoRH

11



Gain-Bandwidth Trade-off Using Emitter Resistor

ER

LR

ERorxrth

RL

Romid

A

)1(

xrthRr for and

gain decreases as emitter resistance increases and bandwidth of stage will correspondingly increase.

To find bandwidth using OCTC method:

1ERmg

ERmg

ERxrth

ReqR

1xixv

ERmg

ERxrth

R

eqRroR

1



Gain-Bandwidth Trade-off Using Emitter Resistor (contd.)

Test source ix is first split into two equivalent sources and then superposition is used to find vx =(vb - vc).

Assuming that o >>1 and

xrthR

LR

ERmgLRmg

xrthRoR

11)(

xixv

ERorxrthR )1(

xrthR

LR

ERmgL

RmgC

xrthR

ER

ERmg

Cxrth

R

H

111

1)(

1



Dominant Pole for C-B Amplifier

IRER

thR

3RCRLR

thRmg

xrthR

thRmg

xrthR

rroR

11xixv

Using split-source transformation Assuming that o >>1 and rx << r

LRthRmg

LRmgxroR

11

xicvbv

LRC

thRmgL

RmgC

xrth

R

thRmg

Cxr

H

111

1

1

Neglecting first term of order of 1/ T and since last term is dominant.

LRCH

1



Dominant Pole for C-G Amplifier

IRR

thR 4

3RDRLR

mgth

GthRmg

thR

GSoR

11

LRGDoR

LR

GDC

LR

GDC

mgth

GGS

CH11



Dominant Pole for C-C Amplifier

IRBR

thR

3RERLR

LRmg

LRxrth

R

LRmg

LRxrth

RrroR

11xixv

)(

)1()()(

xrthR

LRorxrth

RRCCinxrth

RoR

Cxrth

R

LRmg

CL

RxrthR

H)(

1)(

1

A better estimate is obtained if we set RL =0 in expression for Ro.

C

LRmg

Cxrth

RH

1)(

1

xrCH

1)1(GBW



Dominant Pole for C-D Amplifier

Substituting ras infinite and rx as zero in expression for emitter follower,

IRGR

thR

3RSRLR

GDC

LRmg

GSC

thRth

RGD

CGS

C

LRmg

thRH

1

1

1

1



Frequency Response: Differential Amplifier

CEE is total capacitance at emitter node of the differential pair.

Differential mode half-circuit is similar to a C-E stage. Bandwidth is determined by the product. As emitter is a virtual ground, CEE has no effect on differential-mode signals.

For common-mode signals, at very low frequencies,

Transmission zero due to CEE is

TCor

12

)0( EE

RC

RccA

EER

EECZs 1



Frequency Response: Differential Amplifier (contd.)

Common-mode half-circuit is similar to a C-E stage with emitter resistor 2REE. OCTC for C and C is similar to the C-E stage. OCTC for CEE/2 is:

mgo

xrrEEREEOR 1

12

mgEE

C

xrC

R

EERmgC

RmgC

xrEE

R

EERmg

Cxr

P

2211

21

21

1

As REE is usually designed to be large,

)(1

)(2

1

xrCRC

xrCRC

mgEE

CCP



Frequency Response: Common-Collector/ Common-Base Cascade

REE is assumed to be large and neglected.

Sum of the OCTC of Q1 is:

1

111

111

mgo

xrrRCC

out

2

112

222

mgo

xrrRCB

in

12

111

2

11

1

11

CC

xrC

mgm

g

C

xr



Frequency Response: Common-Collector/ Common-Base Cascade (contd.)

Sum of the OCTC of Q2 is:

Combining the OCTC for Q1 and Q2, and assuming that transistors are matched,

2

21

222

22

1

12

11

2

1

12

1

22

xr

CR

CRmg

CC

xrCRC

mgm

g

CRmg

C

mgm

g

C

xr

xrC

RC

RmgCCxr

H

22

1



Frequency Response: Cascode Amplifier

OCTC of Q1 with load resistor 1/ gm2 :

As IC2 = IC1, gm2 = gm1, gain of first stage is unity. Assuming gm2 ro1 >>1,

OCTC of Q1, a C-B stage for ro1 >> RL and f>>1:

Assuming matched devices,

12

1

2

11111111111

or

mg

mg

mg

CCorTCorCoRCoR

12

1111 CCorTCor

2)2(

2

22222

CLRxr

mg

CCoRCoR

C

LRxrCC

orH

21

1



Frequency Response: MOS Current Mirror

1

1

mgor 2orLR

21 GSCGSCC 2GDCC

222

1

222

1

2

11

or

GDC

or

GDC

mg

GSCP

For matched transistors,



Frequency Response: Multistage Amplifier• Problem:Use open-circuit and short-circuit time constant methods to

estimate upper and lower cutoff frequencies and bandwidth.• Approach: Coupling and bypass capacitors determine low-frequency

response, device capacitances affect high-frequency response.

At high frequencies, ac model for multi-stageamplifier is as shown.



Frequency Response: Multistage Amplifier (Estimate of L)

SCTC for each of the six independent coupling and bypass capacitors has to be determined.

1.01MΩ

1MΩkΩ10)1

(1

in

RGRIRSR

7.66S01.0

1200

1

112

mgSRSR

kΩ69.22211

22113

rBR

or

DR

inR

BR

OR

DRSR

5711122 o

rD

RBRth

R

4.19

12

2224

o

rth

R

ERSR

kΩ4.18

)3

)(13

(3322

33225

LR

ER

or

BR

or

CR

inR

BR

OR

CRSR

k99.32233 o

rC

RBRth

R

311

13

3336

o

rth

R

ERLRSR

Hz5302

rad/s33001

1

LLf

n

i iC

iSRL



Frequency Response: Multistage Amplifier (Estimate of H)

OCTC for each of the two capacitors associated with each transistor has to be determined.For M1,

For Q2,

4782121 rIRLR

s71007.1

111

1111

thR

LR

LR

mg

GDC

GSC

thRTC

thR

5701122 o

rIRth

R

610)22

(22 xr

thRror

kΩ54.3

)3

)(13

(323

3232

LR

ER

or

IR

inR

IRLR

s71074.1

2

222

122222

or

LR

LR

mgCCorTCor

For Q3, k99.32233 o

rIRth

R

s81051.1

3)33(3

31

)33

(

3333

Cxrth

RC

EER

mg

xr

thR

CORCOR

kHz5382

rad/s61038.3

1

1

HHf

m

ii

Cio

RH



Single-pole Op Amp Compensation• Frequency compensation forces overall amplifier to have a

single-pole frequency response by connecting compensation capacitor around second gain stage of the basic op amp.

Bs

T

Bs

BoAsvA

)(



Three-stage MOS Op Amp AnalysisInput stage is modeled by its Norton equivalent- current source Gmvdm and output resistance Ro. Second stage has gain of gm5ro5= f5 and follower output stage is a unity-gain buffer.Vo(s) = Vb(s) = - Av2Va(s)

For large Av2

Bs

BoA

Bs

T

v2A

CCosR1

v2AoRmG

s

sv2

A

s

ssvA

)1()(dm

V

)(aV-

)(dm

V

)(oV)(

)1(v2

AC

CoR

1B

)1(v2

AC

Cv2

AmG

T

CC

mGT



Transmission Zeros in FET Op Amps

Incorporating the zero determined by gm5 in the analysis,

This zero can’t be neglected due to low ratio of transconductances of M2 and M5. Zero can be canceled by addition of RZ =1/ gm5.

)1/(1

)/(1)55()(

PsZs

ormgsvth

A

oR

or

fGDCCCGSCTC 55

1)5(5

TCoRP1

1

2

5

5

5

mg

mg

TGD

CC

Cm

g

Z

CC

ZR

mg

Z

)5

/1(



Bipolar Amplifier Compensation• Bipolar op amp can be

compensated in the same manner as a MOS amplifier

• Transmission zero occurs at too high a frequency to affect the response due to higher transconductance of BJT that FET for given operating current.

• Unity gain frequency is given by:

2

55

CIC

I

TC

Cm

g

Z

CC

CI

CC

CI

CCm

g

T1

202

402



Slew rate of Op Amp• Slew-rate limiting is caused by

limited current available to charge/discharge internal capacitors. For very large Av2, amplifier behaves like an integrator:

• For CMOS amplifier,

• For bipolar amplifier,

dt

todvCC

dt

tB

dv

CCCI)()(

1

1

/1

max

)(SR

ImGT

CCI

dt

todv

21

1/

SRnKI

TImGT

201

/SR T

ImGT



Tuned Amplifiers

• Amplifiers with narrow bandwidth are often required in RF applications to be able to select one signal from a large number of signals.

• Frequencies of interest > unity gain frequency of op amps, so active RC filters can’t be used.

• These amplifiers have high Q (fH and fL close together relative to center frequency)

• These applications use resonant RLC circuits to form frequency selective tuned amplifiers.



Single-Tuned Amplifiers• RLC network selects the

frequency, parallel combination of RD, R3 and ro set the Q and bandwidth.

• Neglecting right-half plane zero,

3

)/1()()(i

V

)(oV)(

GDGogPG

sLGD

CCsP

Gmg

GDsC

s

ssvA

22)(

oQoss

Qos

midAsvA

)(1

GDCCLo

Lo

PR

GDCCPRoQ

)(



Single-Tuned Amplifiers (contd.)

• At center frequency, s = jo, Av = Amid.

)3

( RD

RormgPRmgmid

A

PR

Lo

GDCC

PRQ

o2

)(1BW



Use of tapped Inductor- Auto Transformer

CGD and ro can often be small enough to degrade characteristics of the tuned amplifier. Inductor can be made to work as an auto transformer to solve this problem.

These results can be used to transform the resonant circuit and higher Q can be obtained and center frequency doesn’t shift significantly due to changes in CGD.

Similar solution can be used if tuned circuit is placed at amplifier input instead of output

)(sI)(1V2

/)(sI)(1nV

)(2I)(oV

ss

nns

sss )(2)( spZnssZ



Multiple Tuned Circuits• Tuned circuits can be placed at both input

and output to tailor frequency response.• Radio-frequency choke(an open circuit at

operating frequency) is used for biasing. • Synchronous tuning uses two circuits

tuned to same center frequency for high Q.

• Stagger tuning uses two circuits tuned to slightly different center frequencies to realize broader band amplifiers.

• Cascode stage is used to provide isolation between the two tuned circuits and eliminate feedback path between them due to Miller multiplication.

1/121BWnBW n



Mixers: Conversion Gain

• Amplifiers discussed so far have always been assumed to be linear and gain expressions involve input and output signals at same frequency.

• Mixers are nonlinear devices, output signal frequency is different from input signal frequency.

• A mixer’s conversion gain is the ratio of phasor representation of output signal to that of input signals, the fact that the two signals are at different frequencies is ignored.



Single-Balanced Mixer• Eliminates one of the two input

signals from the output.

• No signal energy appears at1 , but 2 appears in output spectrum, so circuit is single-balanced.

• Up-conversion uses component (2-1) and down-conversion uses (2+1) component.

tnoddn n

tv

tIEEIEEi

2sin4)(2

1sin1

tnC

RItnC

RI

tnC

REE

I

oddn ntoV

)12

cos(2

1)12

cos(2

1

2sin

4)(



Double-Balanced Mixer/ Modulator: Gilbert Multiplier

• Double-balanced mixers don’t contain spectral components at either of the two input frequencies.

• Modulator applications give double sideband suppressed carrier output signal. Amplitude-modulated signal can also be obtained if

tmRmV

BBICi sin

121 tmR

mVBBICi sin

122

tmcntmcn

oddn nRCR

mVtoV )cos()cos(4

1)(

tmcnMtmcnMtcn

oddn nRCR

mVtov )cos(2

)cos(2

sin4

1)(

)sin1(1 tmMmVv

2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock...

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Transcript of 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock...