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14-60 CHUNG-YU WU

1

vs. f

H(ejωT)

3. SC realization

(1) H0(z)= z +1

z − 0.9063

CD

Vin

 

+

-1/CE

C

-1

 

 

1-Z

+

S

(1+Z-1)

 

2

 

CS/2

φ1

φ2

Vin

 

CS

 

 

 

 

 

φ1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vout

φ2 CD

φ2 CE

A1

fp

0 f fp

H0(z)=-

Cs / 2

×

z +1

 

z −CE /(CD +CE )

 

CD +CE

Cs=2 arbitrarily chosen φ1=>CE=0.9063 , CD=0.0937

φ2

Vout

(2) H1(z)=

 

z2

+C z +1

Q1=

(a 2

+b

2 )

12

 

Low-Q

 

 

1

 

 

 

1

1

 

 

0.99

(1/ f

)z2

+ (e

/ f

)z +1

 

 

2

a1

 

 

 

 

 

 

 

 

 

 

1

 

1

1

 

 

 

 

 

 

 

 

 

 

The SCF is shown on P.14-21.

The component values are: C1"=a0=1, C1'=a2-a0=0,

C2=C3= b1 +b2 +1 = (e1 +1) / f1 +1 0.12436,

14-61 CHUNG-YU WU

C1=(a0+a1+a2)/C3 0.30358,

C4=b2-1=1/f1-1 0.12939, CA=CB=1.

(3) H2(z)=

 

 

z2 +C

2

z +1

Q2=

(a 2

+b

2 )

12

 

4.33 =>High-Q

 

 

 

 

 

2

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

( 1

f

)z2 + (e2 f2 )z +1

 

 

2

 

a2

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

The SCF is shown on P.14-23.

 

 

 

 

 

 

 

 

 

The component values are:

 

 

 

 

 

 

 

 

 

C1"= a2 / b2

= f2 0.96845

C1 '= (a1 −a0 ) / b2 c3

= 0,

C2 = C3 = (1+b1 +b2 ) / b2 = f2 +e2 +1 0.13795,

C1=(a0+a1+a2)/b2C3=(2+c2)f2/C3 0.58645, C4=(1-1/b2)/C3=(1-f2)/C3 0.22873.

(4)Overall SCF

* Ho (low-pass linear section) is placed first

=>High-frequency out-of-band signals and input noise can be attenuated. The antialiasing filter preceding the SCF has a lower requirement.

* H2 (high-Q section) is placed to the center=>good signal-to-noise ratio

 

 

 

 

CD

φ1

 

 

 

 

C2

 

φ1

 

CS/2

CE

φ

 

 

 

 

C4

 

φ2

 

 

 

 

2

 

 

 

 

 

 

V

φ

C

S

φ

 

C

 

 

 

 

CB

 

1

 

2

φ

φ

C

φ

C3

 

 

in

 

 

 

 

 

 

 

 

A1

2

1

2

A

φ2

 

 

 

φ2

 

 

 

A2

1

 

 

 

 

φ1

 

φ1

φ1

 

φ2

A3

A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

φ1

 

 

 

 

 

 

 

 

 

C1"

 

 

 

 

SECTION 1

 

SECTION 2 (HIGH Q)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

φ1

 

 

 

 

C

 

C4

 

φ

 

φ2

φ2

C1

A

 

 

CB

2

 

φ2

φ1

C3

 

 

A

 

 

φ2

 

 

 

A4

 

 

 

 

 

 

 

φ2

A5

Vout

 

 

 

 

 

 

 

φ1

φ1

φ2

 

 

 

φ1

 

 

 

C1"

SECTION 3 (LOW Q)

Thought the same procedures, we have CA 17.666,CB 7.7286 C1 1.9926,C2 =1
C3 2.1116,C4 = 2
C1" 6.3085
5 Final Design
Cmin is chosen as 0.5pF => C=1
op amp: gain 70dB bandwidth 3 MHz passband sensitivity to capacitance variation
§14-12.2 Bilinear Ladder SCF Design
1. The same filter specification.
Elliptic ladder filter is chosen(fifth-order). The result is

14-62 CHUNG-YU WU

4.Scaling

1)Vp1=occurs at dc where H0(1)=-CS/CD=-21.345

(1)We want an overall passband gain of 1. =>Ho(1)→-1 => CD=CS=2, CE 19.345, C1"20.672, C2 12.518

(Multiplying all capacitors connected or switched to the output node of op-amp A1 by 21.345)

(2)All capacitors at the input node of A1 should be scaled so that the

smallest (Cs

2

) equals 1. (O.K.)

 

 

2)Vp2 (peak output voltage of op-amp A2) occurs around fp2=1.10kHz

(1)Vp2 177.05 for Vin=1

Reducing Vp1/Vin to 1

=>CA and C3 are multiplied by 177.05=> CA 177.05, C3 24.424.

(2) Vp3 180.80 at 1.07kHz

=>CB,C2, and C4 are multiplied by 180.80=>CB 180.80, C2 24.941, C4 41.354.

(3)Minimize total capacitance=>C1, C2, C4, and CA at the input node of op-amp A2 are scaled to make C1=1

=>C1=1, C2 1.9926, C4 3.3036, CA 14.144

(4)Similarly, C1"=1, C3 1.1815, CB 8.7466. (The input of A3)

3)Vp4 503.57 and Vp5 230.14

0.2dB/1%

14-63 CHUNG-YU WU

 

RS

L2

L4

 

 

 

 

 

 

 

Vin

C1

C2 C3

C4

C5

RL

Normalized component values:

Rs=RL=1 C1=0.85535 C2=0.15367 L2=1.20763 C3=1.48438 C4=0.46265 L4=0.89794 C5=0.63702

ŵap=1 rad/s

2. Frequency prewarping and denormalization

ω

 

2

tan

ωpT

= 2 f

 

tan

πf p

6291.4667rad / s

 

2

 

 

ap

 

T

 

c

 

fc

Multiplying each resistor by z0, each inductor by L0=z0/ωap, and each capacitor by C0= 1z0 ωap. 50Ω

Usually choose z0=real source and termination resistance 100Ω 600Ω

Here, C0=1 is chosen => z0=

1

and L0=

1

 

ω

ap

ω

 

2

 

 

 

 

 

 

 

ap

We have the denormalized element values as:

C1=0.85535, C2=0.15367, L2=1.20763×Lo=3.05090×10-8,

C3=1.48438, C4=0.46265, L4=0.89794×Lo=2.26851×10-8,

C5=0.63702, Rs=RL=z0=1.58945×10-4

3. SC realization

Using the exact design technique of SC ladder filter (Section 14-10), the state equations are

-V = −

 

 

1

(

1

(V

−V ) − I

 

+ sC'

V ),

 

 

 

 

 

2

1

 

 

 

 

 

 

 

 

 

in

1

2

3

 

 

 

 

 

sC'1 RS

 

 

 

 

-I2=-(

 

 

1

 

− sCL2 ) (V1 V3),

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sL2

 

 

 

 

V3=

 

1

 

 

(−I2 − sC'2 V1 − sC4 'V5 + I4 ),

 

sC'3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

14-64

 

 

 

CHUNG-YU WU

I4=(

1

− sCL4 )(V3 −V5 ),

The signal flow diagram is:

 

 

sL4

 

-V5=

−1

(I4

+ sC'4 V3

−

V5

),

 

 

 

sC'5

 

 

RL

Vin 1/RS

where

CL2=

T 2

=0.003278,

sC'2 -I'2

4L

 

 

 

 

 

2

 

 

 

C'2=C2+CL2=0.15695,

 

C'1=C1+C'2=1.01230,

sC'4

CL4=

T 2

 

 

= 0.0044082,

 

4L4

 

 

 

 

C'4=C4+CL4=0.46706, C'3=C3+C'2+C'4=2.10839,

C'5=C5+C'4=1.10408.

C1

SCF:

Vin

φ

C

 

 

1

 

 

φ

arbitrarily chosen

 

2

 

 

 

 

 

1/RS

-1/sC'1

-V1

1-(sT/2)2

sC'2

sL2

 

-1/sC'3

V3

 

I4

 

1-(sT/2)2

 

sL4

 

 

sC'4

-1/sC'5

-V5

1/RL

 

φ2

 

C2

 

φ1

φ2

CA

φ2

φ1

 

 

A1

 

 

 

 

C=CL2=0.003278 C'=CL4=0.004408

 

Co1

 

 

 

 

 

 

 

 

 

Co2

Cs=

T

= 0.1258293,

 

φ1

φ2

CB

 

φ

φ

 

Rs

 

 

 

 

 

 

A2

1

2

 

 

 

 

 

 

 

 

 

 

 

CA=C1+C2+CL2- Cs =

C21

C22

 

 

 

 

 

 

 

 

0.94938,

 

Co3

 

 

 

C

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

o4

CB=

C

2

=

C

L 2

=0.0008195,

φ

φ

 

CC

φ2

φ1

 

 

 

 

1

2

A3

 

4CL 2

4

 

 

 

 

 

 

 

 

 

 

Co5

 

 

 

 

Cc=C'3=2.10839,

 

 

 

 

 

Co6

 

 

 

 

 

 

CD=

C'2

=

CL4

= 0.001102,

 

 

CD

 

φ1

φ2

4C

 

4

 

φ1

φ2

 

 

 

 

L4

 

 

 

 

 

 

 

A4

 

 

CE=C'5- CL

=1.041165,

C41

C42

Co7

 

 

 

 

 

 

 

 

2

 

 

 

CE

 

Co8

 

T

 

 

 

 

 

 

 

 

CL=

 

 

 

 

 

 

 

 

 

 

= 0.12583.

 

φ

φ

 

 

φ2

φ1

 

R

L

 

 

 

 

 

 

1

2

A5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

CL

Vout

 

 

 

 

 

14-65

4.Scaling

 

 

 

 

CHUNG-YU WU

 

 

 

 

 

Vin=1V, we have:

Vp1 0.92V,

 

C1=1.00000

C05=1.14172

A1: CA,C2,C21, and C02

 

Vp2

34V,

 

C2=1.83854

C06=1.52861

multiplied by Vp1

 

Vp3

0.764V,

C3=2.00000

C07=2.02212

A2: C3,C01, and C03

Vp4

28.86V,

CA=13.87171

C08=1.00000

multiplied by Vp2

Vp5

0.5V,

 

C01=1.77112

CE=8.27441

 

 

 

 

 

 

C02=1.20275

CD=14.43078

 

 

 

 

C03=1.00000

CL=1.00000

 

 

 

 

C04=1.00000

C41=2.09575

 

 

 

 

CB=11.11901

C42=5.67396

for dynamic range scaling

 

 

 

Cc=14.46156

C21=1.29480

 

 

 

 

C22=1.90667

minimum-capacitance scaling:

Cs

 

=13.87171

CA

2

 

 

 

 

 

 

5.Final design

Cmin , OP amp: 70dB 3 MHz

=>Passband ripple: 0.06dB minimum stopband loss 39.5dB

Maximum sensitivity: 0.05dB %

§14-12.3 LDI Ladder SCF Design

1. LCR prototype circuit

Fifth-order elliptic LC ladder filter with the same lowpass specifications.

2. Frequency prewarping and denormalization ωap ωp (for simplicity)

z0=1Ω => C0= 1

3

)

F, L0=

1

H

(2π103 )

(2π10

 

 

 

The denormalized element values:

Rs=1Ω,

C1=136.13318µF, C4=73.633034µF, C2=24.45734µF, L4=142.91159µH, L2=192.20028µH, C5=101.38488µH, C3=236.24641µF, RL=1 Ω.

State equations:

 

−V1 +Vin + sC V

 

 

 

-V =

 

1

 

(

− I

 

),

s(C

+C

)

 

1

 

Rs

2 3

 

2

 

 

1

2

 

 

 

 

 

 

 

-I2= V3 −V1 ,

 

 

 

 

sL2

 

 

 

 

 

 

 

1

 

 

 

 

V3= − s(C2 +C3 +C4 ) ( − I2 − sC2V1 − sC4V5 + I4 ),

I4= V3 −V5 ,

 

 

 

 

 

sL4

 

 

 

 

 

-V5= −

s(C4

1

(I4

+ sC4V3

−

V5 ).

 

+C5 )

 

 

 

RL

14-66 CHUNG-YU WU

3.SCF design

The flow diagram is shown on P.14-? whereas the active-RC circuit is given on P.14-?.

The SCF is shown on P.14-? where T=20µs is chosen and the component values are

C1+C2=160.59µF,

C2+C3+C4=334.34µF,

CS=

 

T

=20µF,

C4+C5=175.018µF,

 

 

 

 

 

Rs

 

 

 

C=

T

 

=20µF,

CL=

T

=20µf

 

 

 

1

 

 

 

 

RL

4.Scaling

Dynamic range scaling with Vpi listed: followed by minimum-capacitance scaling

Element values:

SCF:

C1=8.03214,

 

C3=12.97271,

C1A=1,

C3A=1.08390,

C1B=1.07930, C3B=1,

C1C=1.29263, C3C=1.02540,

C1D=1.13212, C3D=1.66885,

C2=13.42236, C4=15.76379,

C2A=1.08053, C4A=1.71203,

C2B=1,

 

C4B=1,

 

 

C5=8.75121,

 

 

C5A=2.20614,

 

 

C5B=1,

 

 

C5C=6.29664.

5.Final design

Cmin Passband ripple: 0.095dB>0.044dB Minimum stopband loss: 40.5dB OP amp: 70dB, 3MHz

Maximum passband sensitivity: 0.08dB%

for A1:Vp1=0.927 V at 1.182kHz. for A2:Vp2=1.198 V at 1.121 kHz. for A3:Vp3=0.857V at 1.061 kHz. for A4:Vp4=1.105 V at 1.061 kHz. for A5:Vp5=0.501 V at 967 kHz.

14-67 CHUNG-YU WU

§14-13 Nonideal Effects in Switched-Capacitor Filters

1. Switch Turn-On Resistance

The turn-On resistance of a MOSFET can be written as

Ron=

 

uco

 

1

 

2

w

(VG S

−VT )

φ

2

L

 

 

 

Vin

 

 

 

C

* Nonlinear behavior

 

 

 

 

signal voltage -Vss

 

 

 

 

 

 

 

 

 

The Ron effect on the simple SC integrator:

 

 

 

C2

 

φ 1

φ 2

 

 

 

 

φ1

Vin

 

-

 

C1

+

Vout

 

 

 

φ2

 

 

 

t=nT t=(n+1)T

C2

φ1

R1 R2

-φ2

+

V1

C1

+

Vout

 

t=nT

nT+T/2 (n+1)T

- Vin

 

 

 

At t=nT , V1(t)=V1(nT)=Vin(nT)(1-e −T2R1C1 )

Assume φ1

and φ2 are activated for T

2

.

 

 

 

 

 

 

 

 

 

 

∆Q (nT+

T

) =C1V1(nT )(1-e −T / 2R2C1 )=[Vout(nT+T)-Vout(nT)]C2

2

Let R1=R2=R

 

 

 

 

=> H(z)=

−(1 −e−T 2 RC1 )2

C C

 

 

 

 

 

z −1

1

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ideal:

 

 

H(z)=-

C1 C2

 

 

 

 

 

 

 

z −1

 

 

 

 

 

 

 

 

 

 

 

 

 

14-68 CHUNG-YU WU

Error: ε=1-(1-e −T / 2RC1 )2 2e −T / 2RC1

Usually ε<0.1%(cap. ratio error) is acceptable.

2e −T / 2RC1 ≤10−4

=>

RC1

=RC1fc ≤

1

0.05

T

2ln 20000

 

 

 

or RC1 ≤ 20T (= 201fc)

fc=500KHz, C1=5pF R ≤ 20KΩ ; fc=100MHz, C1=2pF, R ≤ 250Ω?

2.Clock Feedthrough Noise

*All switches directly connected to the integrating node generate clock feedthrough noises.

φ1

m

φ2 n

T

* All clock feedthrough noises are proportional to the sampling frequency. They may have a dc component.

* As soon as the clock feedthrough error voltage does not

14-69 CHUNG-YU WU

saturate the OP AMP, it can be eliminated at the output by reconstruction filters(LPF).

* The dc component cause offset voltage problems.

3.Junction Leakage

* Worst-case (100°C or 125°C) leakage at the integrating node: ~10 nA/mil2 5µm×5µm junction => 400 pA leakage

*fs, max is about 25KHz in this case to avoid significant errors.

*The leakage cause dc offset voltages.

4.DC offset Voltage of the OP AMP

-

-

+

pratical op amp

C2

φ2 φ2

-

C1 +

+

-

Voff

+

+ ideal op amp

-

Voff Voff = 5~20mv

Vout

* Vout=(1+ C1 ) Voff

C2

*Integrator-based design may have a dc offset problem if no other negative feedback paths exist.

*Too-low-frequency operation is not good.

C2

5. Finite Gain of the OP AMP.

 

φ2

 

 

 

φ2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vout(nT)=Vc2(nT)-

 

1

Vout(nT)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

 

 

 

C1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

O

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vout

C2[Vc2(nT)-Vc2(nT-T)]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

φ1

 

φ1

 

 

 

 

 

 

 

 

 

 

 

A0

+C1[Vin(nT)+

1

 

Vout(nT)]=0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

O

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

(z)

 

 

 

 

−(C

C

)[1

+ (1 +C

C

) / A

]−1 z

 

 

 

 

=>H(z)=

 

out

 

=

 

1

2

 

1

2

 

O

 

 

 

 

 

 

Vin (z)

 

z −(1 + 1

AO

) /[1 + (1 +C

C ) / A ]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

2

 

O

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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