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Chung-Yu Wu - Analog Circuit Design

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<<< < Предыдущая 25 26 27 28 29 30 31 32 33 34 35 36 37 38 3940 / 4940 41 42 43 44 45 46 47 48 49 > Следующая >>>

14-30 CHUNG-YU WU

Frequency response of high-Q Band-pass SCF biquad

CALCULATED

CENTER FREQUENCY: 1K Hz

SAMPLING FREQUENCY: 39.6 Hz

○ COMPUTED BY SWITCH CAP × EXPERIMENTAL

14-31 CHUNG-YU WU

§14-8 First-Order SCFs

Ha(s)=	K1S + K0	H(z)=	a1 z + ao
	S +ω0
			b1 z +1

1. Flow diagram

ω0

Vin	K0	+	-1/S

Vout

K1S

2. Active-RC design

		1/ω0
	1/K0	CA=1
Vin		-	Vout
	K1	+	Vout
		+
3. SCF
			φ1
	φ1		C2	φ1
	φ1			φ1
C1 K1		C'1		φ2
C1 K1		φ2	φ2	φ2
C1' TK0			φ2
C2 ω0T			CA
fs	Vin		CA
fc= 2πx	Vin	C	-	Vout
		C		Vout
		1	+
			+

14-32 CHUNG-YU WU

4. Z-domain block diagram
H(z)=Vout =	(C1	+C1 ')z −C1
V	(1	+C			)z −1
in				2
	C'1			+		C2
				+
Vin				+		-1 / CA	Vout
Vin		-1	)	+		1 - Z-1	Vout
C1(1-Z			)

§14-9 Switched-Capacitor Ladder Filters

§14-9.1 Approximate Design of SC Ladder Filters

(1) Third-order low-pass filter without finite transmission zeros

-V1= 1 (Vin −V1 − I2 )

SC1

Vout

-I2=

−1 (V1 −V3 )

SL2

V3= − 1

(−I2 +

V3 )

LCR prototype circuit

SC3

(no loss)

Loss

transmission zero

=> ∞

1/RS

- 1

-V1

Vin

SC1

-1

-I2

- 1

SL2

ωp

-1

Loss response

- 1

+V3

SC3

1/RL

Flow diagram

14-33 CHUNG-YU WU

		RS
RS		Active-RC realization
RS		C1
Vin		-
Vin		-V1
	1	+
	1

L2	-1
-I2
-		CS		CS
+		CS		CS
+	Vin	φ2
1		φ2
1
C3	-1	φ1		C1
			-	C1
-		φ1	-
		φ1	φ2
		Vout			φ		φ1
+			+		φ	2	φ1
			+			2

		C					C
RL				L2
				-	φ1		φ2
					φ1		φ2
		φ1	φ2	+
				+
	SCF	C					C
	SCF						C
			-	C3
			-
		φ1	φ2
			+		φ2		φ1
							Vout
SCF realization equations:
C= T		ωT<<1		CL
R

=> Cs RsT C T

CL T / RL

Due to the approximation made in finding C values, error still exists which may be refined by the z-domain analysis.

14-34 CHUNG-YU WU

(2) Third-order low-pass filter with transmission zeros

ωa1=-ωa2= L2C2

-V =

−1

Vin −V1

+ sC V

− I

s(C1 +C2 )

-I2=

−1

[V1 −V3 ],

sL2

+V3=

−1

− sC V

− I

s(C2 +C3 )

2 1

LCR Prototype circuit:
		C	IC2
		2
	RS	L2	I2	V3
		V1		V3
	~	1		3	+
					+
Vin		C1	C3	RL	Vout
Vin		C1	C3	RL	Vout
					-

			1/RS
Vin	1/RS	+	- 1			-V1
Vin		+	s(C	+C )		-V1
			1	2
Flow diagram:			SC2
			SC2			-1
						-1
		-I2	- 1			+
		-I2	- 1
			SL2
						-1
			SC2
		+	- 1		+V3	Vout
			S(C2+C3)			Vout
			S(C2+C3)

1/RL

Active-RC						RS
Active-RC					CA=C1+C2
Realization:					CA=C1+C2
Realization:				RS
		Vin		RS
		Vin			-			-V1
					+			-V1
					+
				R	C = L /R2			-R
					B	2
				-RI2		-
						-
		C	2	C2		+
			2

				R
					CC=C2+C3			-R
					-			Vout
					+		V3	Vout
					+		V3
						RL
				CS		CS
Vin	φ2
	φ2
SCF:		φ1				CA
SCF:						CA
				φ1	φ2	-
				φ1	φ2
						+		φ2	φ1
						+
									C
				C		CB
						CB	-
							-	φ1	φ2
								φ1	φ2
				φ1	φ2		+
							+
	C2	C2		C
				C					C
									C
						CC
				φ1	φ2	-
				φ1	φ2
						+		φ2	φ1
						CL

14-35 CHUNG-YU WU

Cs T / Rs,

CA=C1+C2,

C T ,

CB=L2,

Cc=C2+C3,

CL T / RL .

Vout

14-36 CHUNG-YU WU

	(3) Fourth-order Bandpass filter
					C2
	R	1		I2	L2	V3		3
	S	1		V1		V3		3
	~			I1			I3		+
Vin	~	C	L	1		L3	C3	RL	Vo
		1		1
									-

LCR Prototype Circuit

1/RS

Vin

1/RS

- 1

-V1

s(C

+C )

-V1= −1

Vin −V1

+ sC V

− I

2 3

-I1-I2

s(C1 +C2 )

-I

- 1

-1

-V1 -I1=

−V1 ,

SL1

sL1

-I2

-I2=

−V1 +V3 ,

SC2

sL2

I3= sL ,

-I2

- 1

-1

−1

SL2

V3=

+ I3 − I2

s(C2 +C3 )

− sC2V1

-I2

SC2

- 1

-1

SL3

I3-I2

- 1

Vout

S(C2+C3)

1/RL

14-37 CHUNG-YU WU

** The circuit has a stability problem at dc. Due to inductor loops!

HAB

− sL

Circuits below A and

B disconnected

S 2 L

(C +C

)

+ SL / Rs +1

HCD

− sL

Circuits below C and

disconnected

S 2 L (C

) + SL / R

When S→0, HAB→0, HCD→0

> There will be no dc feedback paths around the

center integrator which provides -I2.

> OP AMP will be in the open-circuit status with A→∞.

> Saturation occurs

How to solve this problem?

Don't model the inductor loop currents separately.

i.e. I1, I2, I3.
Only two inductive currents I 1 and I 3	entering nodes
1 and 3 are modeled.
> Vin =0 and S→0, I 1 =0 and I 2	=0

>No any instability at dc.

i.e.

Treat only two inductors as independent inductors. I 1 =I1+I2

I 3 =I2-I3

New flow diagram:

14-38 CHUNG-YU WU

-V1= s(C1−+1C2 ) VinR−s V1 + sC2V3 − I(11) ,

-I(1)

∆ −(I

+ I

) =

−1

V −

L1V3

sL12

L1 + L2

V3=

−1

− sC V

− I

(3)

2 1

s(C2 +C3 )

-I(3)

∆I3-I2=

−1

L1V1

−V3

sL12

L1 + L2

Where L12 ∆L1 L2 = L1 L2 /(L1 + L2 )

14-39 CHUNG-YU WU

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