Dismiss
InnovationQ will be updated on Sunday, Oct. 22, from 10am ET - noon. You may experience brief service interruptions during that time.
Browse Prior Art Database

Accurate Balanced Current Source

IP.com Disclosure Number: IPCOM000045644D
Original Publication Date: 1983-Apr-01
Included in the Prior Art Database: 2005-Feb-07
Document File: 2 page(s) / 42K

Publishing Venue

IBM

Related People

Carsalade, H: AUTHOR

Abstract

The purpose of this circuit is to provide a balanced and adjustable current or voltage threshold for multipurpose applications such as comparators with balanced or unbalanced hysteresis and accurate wave generators.

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 98% of the total text.

Page 1 of 2

Accurate Balanced Current Source

The purpose of this circuit is to provide a balanced and adjustable current or voltage threshold for multipurpose applications such as comparators with balanced or unbalanced hysteresis and accurate wave generators.

The circuit comprises a differential input stage 1, which controls the operation of two matched current sources T9 and T10 in output stage 2, such that a current I+ or I- flows in the load depending upon the values of input signals IN+ or IN-.

Collector currents of matched transistors T5 and T6 depend on the collector current of transistor T11. Since emitter resistors r of transistors T5 and T6 are equal, currents I(T5) and I(T6) are set equal to I. The collector current of transistor T5 is reflected through matched transistors T7 and T8 in the collector path of transistor T10.

When IN+ is higher than IN-, transistor T1 is OFF and transistor T2 is ON so that transistor T10 is made conductive and current I- flows in the load.

Conversely, when IN- is higher than IN+, transistor Tl is ON and transistor T2 is OFF so that transistor T9 is made conductive and current I+ flows in the load. Current I+ = I (1 + IBT9/I) and Current I- = I (1 - IBT10/I) where IBT9 and IBT10 are the base current of transistors T9 and T10, respectively.

Consequently: I+ = (1 + Epsilon)I- Epsilon = (IBT9 + IBT10) 1/I

Since I >>> IBT9, IBT10, Epsilon can be neglected and: I = I- with a discrepancy of 0.1%.

1

Page 2 of 2

2

[This page contains 3 pictur...