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Exclusive OR with PNP Amplifier

IP.com Disclosure Number: IPCOM000075377D
Original Publication Date: 1971-Sep-01
Included in the Prior Art Database: 2005-Feb-24
Document File: 2 page(s) / 25K

Publishing Venue

IBM

Related People

Hansen, AA: AUTHOR

Abstract

By using a PNP transistor as a pulse amplifier, the PNP can be integrated in the same epitaxial bed as the NPN's, hence decrease the area and reduce the cost. If both pulses A and B are at the same potential (either positive or negative) neither transistor T1 or T2 can be ON because of the base-emitter coupling. This coupling forces their levels to the same value, hence no signal on the output F. The junction F for an Exclusive-OR is determined by the formula: F = AB + AB.

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Exclusive OR with PNP Amplifier

By using a PNP transistor as a pulse amplifier, the PNP can be integrated in the same epitaxial bed as the NPN's, hence decrease the area and reduce the cost. If both pulses A and B are at the same potential (either positive or negative) neither transistor T1 or T2 can be ON because of the base-emitter coupling. This coupling forces their levels to the same value, hence no signal on the output F. The junction F for an Exclusive-OR is determined by the formula: F = AB + AB.

When pulse A is positive and pulse B is negative (AB) the base-emitter junction of T2 is forward biased and T2 is ON. The base-emitter of T1 is reversed biased and T1 is OFF. The collector of the NPN transistors are common to the base of the PNP (T3) which is integrated in the same epitaxial bed. Because T2 is ON, the NPN collectors are forced to a negative potential, forward biasing the PNP device. This forces the collector current of T3 through R3 forcing F positive.

To prevent each NPN from saturating, a Schottky-barrier diode (SB) is integrated into the structure. The same operation is repeated for AB, with the difference that T1 is ON and T2 is OFF. True and compliment values of F are available depending on the desired potential on the output.

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