Browse Prior Art Database

Active All Pass Network

IP.com Disclosure Number: IPCOM000092087D
Original Publication Date: 1968-Aug-01
Included in the Prior Art Database: 2005-Mar-05
Document File: 2 page(s) / 38K

Publishing Venue

IBM

Related People

West, LP: AUTHOR

Abstract

In integrated circuits, no convenient way of synthesizing all-pass networks without inductors is available. At low frequencies, the inductors are large. At high frequencies, the mutual coupling is a problem. The only technique known for performing the all-pass function without inductors involves the use of operational amplifiers, but none exist which operate satisfactorily at frequencies above 500 KHz.

This text was extracted from a PDF file.
At least one non-text object (such as an image or picture) has been suppressed.
This is the abbreviated version, containing approximately 70% of the total text.

Page 1 of 2

Active All Pass Network

In integrated circuits, no convenient way of synthesizing all-pass networks without inductors is available. At low frequencies, the inductors are large. At high frequencies, the mutual coupling is a problem. The only technique known for performing the all-pass function without inductors involves the use of operational amplifiers, but none exist which operate satisfactorily at frequencies above 500 KHz.

The normalized transfer function for this circuit is Eout/Ein = 1/(S/2/+K/(5)S+1). This is a known relationship which permits devising multipole active filters with excellent stability. Because of the unity gain requirements, emitter-followers are often used as amplifying elements, thus providing a filter prototype which operates at any frequency from a few hertz to many megahertz.

With regard to drawing 1, (E2/Ein)=(S+1)/(S/2/+K(5)S+1). defined as delta K(6)>(E2/Ein)-(Eout/Ein)|-(Ein/Ein), then Y(S)=>(K(6)S+K(6)-K(6))/ (S/2/+K(5)S+1)|-1 and letting K(6) = 2K(5) by resistor selection, Y(S)=>2K(5)S/ (S/2/+K(5)S+1)|-1=-(S/2/-K(5)S+1)/(S/2/+K(5)S+1) and Y(S) is the all-pass function. Drawing 2 shows an implementation of Y(S).

Referring to the second equation, and noting that, for the circuit of drawing 3, (E3/E2)=SCR2/>SC(R1+R2)+1|.

Then, letting C(R1+R2)=1, (E3/E2)=K(7)S/(S+1) and (E3/Ein)=(K(7)S)/ (S/2/+K(5)S+1) and, letting K(7) = 2K(5), (E3/Ein)- (Ein/Ein)=>2K(5)S/(S/2/+K(5)S+1)|-1 == (S/2/-K(5)S+1)/(S/2/+K(5)S+1) which is, again,...