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Ternary Logic Switch

IP.com Disclosure Number: IPCOM000097964D
Original Publication Date: 1961-Oct-01
Included in the Prior Art Database: 2005-Mar-07
Document File: 2 page(s) / 31K

Publishing Venue

IBM

Related People

Baskin, HB: AUTHOR

Abstract

In the ternary logical circuit, particular values of biasing and supply potentials, and relative values of input resistors, are specified. The truth tables indicate that the point of intersection, or origin, is in the upper left corner. The input biasing network for transistor TA is designed so that it is switched from off to on as the signal at point X changes from a digital 1 to digital 0, i.e., from 0V to -6V. When the signal potential at input X is -6V, the base potential is then +4.8V. Thus, it is negative with respect to the potential at the emitter which is +6V. However, when the signal potential at input X is 0V, the base potential is then +7.2V. Thus, it is positive with respect to the emitter potential which is +6V.

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Ternary Logic Switch

In the ternary logical circuit, particular values of biasing and supply potentials, and relative values of input resistors, are specified. The truth tables indicate that the point of intersection, or origin, is in the upper left corner. The input biasing network for transistor TA is designed so that it is switched from off to on as the signal at point X changes from a digital 1 to digital 0, i.e., from 0V to -6V. When the signal potential at input X is -6V, the base potential is then +4.8V. Thus, it is negative with respect to the potential at the emitter which is +6V. However, when the signal potential at input X is 0V, the base potential is then +7.2V. Thus, it is positive with respect to the emitter potential which is +6V. A condition of reverse bias between base and emitter then obtains, and the transistor TA is cut off.

In a similar manner, the input biasing network for transistor TB is such that TB switches from off to on as the input value at Y changes from a digital 1 to a digital 0, or from 0V to -6V. When the input value at Y is 0V, the potential at the base of TB is +2 volts. It is, therefore, positive with respect to the emitter which is at ground potential. When the input value at Y is -6V, the potential at the base of TB is -2V, or negative with respect to the emitter.

Consider the output T and the truth table for T. Let it be assumed that the input at X has the value 0 and the input at Y also has the value 0. Both transistors a...