active circuits lecture 9: active inductors, negative...

Active Circuits Lecture 9: Active Inductors, Negative

Resistance and LC-Ladder Filters ELC 302 – Spring 2013

Dr. Mohamed M. Aboudina [email protected]

Department of Electronics and Communications Engineering Faculty of Engineering – Cairo University

© Mohamed M. Aboudina, 2013

mailto:[email protected]


Other Alternatives

Vin Vout

R sL

1/sC

𝐻 𝑠 =

1𝐿𝐶

𝑠2 + 𝑠𝑅𝐿 +

1𝐿𝐶

Replace this RLC filter with an

Active Opamp RC filters

Use Active Circuits to Emulate “L” ≡ Replace L with a

circuit

Vin Vout

xR xsL

x/sC

𝑥 = 1/𝑠

1

𝑠2𝐶= −

1

𝜔2𝐶≡ FDNR

Frequency-Dependent Negative Resistance

Leapfrog Filters

Bruton’s transformation

Vin Vout

R/s L

1/s2C


Emulation of RLC Filters using Opamps

• Active Inductors – Inductor Emulations.

• FDNR: Frequency Dependent Negative Resistance

• (LC Ladder Filters) Leapfrog Filter Architectures.

GIC (General Impedance Converter)

GII (General Impedance Inverter)

Inductor Emulation Using Two-port Network

𝑇 = 𝐴 =

𝑘 0

0𝑘

𝑓 𝑠

𝑍𝑖1 =𝑉1

𝐼1= 𝐺𝑎𝑍𝐿 𝑠 while 𝐺𝑎 =

𝑎11

𝑎22= 𝑓(𝑠)

𝐴 =0 𝑎12

𝑎21 0


Gyrator

Floating inductor

General Impedance Inverter

Gyrator – Positive Impedance Inverter

𝑍𝑖𝑛 =1

𝑔1𝑔2.1

𝑍

If 𝑍 =1

𝑠𝐶 ⇒ 𝑍𝑖𝑛 = 𝑠

𝐶

𝑔1𝑔2= 𝑠𝐿𝑒𝑞

𝑍𝑖𝑛

𝐿𝑒𝑞 =𝐶

𝑔1𝑔2

Good For Grounded Inductors

𝑔 𝑔

𝐿𝑒𝑞 =𝐶

𝑔1𝑔2


Gyrator Example General Impedance Inverter

• 𝑉𝑥 = 2𝑉1

• 𝐼𝑎 = (𝑉1−𝑉𝑥)/𝑅

• 𝐼𝑏 =𝑉1−𝑉2

𝑅

• 𝐼𝑐 =𝑉𝑥−𝑉2

𝑅

• 𝐼1 = 𝐼𝑎 + 𝐼𝑏

• 𝑉𝑦 = 𝑉2 − 𝑅𝐼𝑐

• 𝐼𝑑 =𝑉2−𝑉𝑦

𝑅

• 𝐼2 = 𝐼𝑑 − 𝐼𝑏

• 𝐴 =0 −𝑅

−1

𝑅0

Gyrator Example

RR

R

R

R

R

R

I1

I2V1

V2

Vx

VyV2V2

V1

IaIb

Ic

Id

Gyrator Resistance = R


General Impedance Converters

Antoniou GIC

𝐿𝑒𝑞 = 𝐶𝑅2



• 𝑍𝑖𝑛 =𝑉1

𝐼1= 𝑠𝐶𝑅2

• 𝐿𝑒𝑞 = 𝐶𝑅2

Riordan Gyrator


Example

For Gyration resistance =1𝑘Ω


Filter Example


Antoniou GIC

• Note: Inductance emulation is optimum in case of no floating inductors i.e., LC high-pass filters

General Impedance Converters

Antoniou GIC

𝑓 𝑠 = 𝑠𝐶𝑅 = 𝑘𝑠

𝑘𝑠 𝑠𝑘𝑅𝐿 𝑅𝐿

𝒇 𝒔 = 𝒌𝒔 𝒇 𝒔 = 𝒌𝒔

𝑅𝐿 𝑠𝑘𝑅𝐿


Antoniou GIC General Impedance Converters

Antoniou GIC

𝒇 𝒔 = 𝒌𝒔 𝒇 𝒔 = 𝒌𝒔

𝑍𝐿 𝑠𝑍𝐿


3rd Order LPF

6th Order BPF

Example


Bruton’s transformation


FDNR

• Bruton’s inductor simulation based on FDNR

• Most suitable for LC LPF with minimum cap realization

FDNR : Frequency-Dependent Negative Resistance

𝑓 𝑠 =1

𝑠𝐶1𝑅4 ⇒ 𝑍𝑖𝑛 =

1

𝑠2𝐶1𝐶5𝑅4

𝐶 = 𝐶1𝐶5𝑅4


active circuits lecture 9: active inductors, negative...

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