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# FET Logic Bridge Circuit

IP.com Disclosure Number: IPCOM000089932D
Original Publication Date: 1968-Dec-01
Included in the Prior Art Database: 2005-Mar-05
Document File: 2 page(s) / 28K

IBM

## Related People

Gurski, CS: AUTHOR

## Abstract

This circuit for a complementary logic bridge using IGFET's provides a multiuse logic element whose logic statement can be altered by control of selected inputs. The circuit includes two complementary bridge circuits each using opposite type IGFET's. When the logic of the N channel portion of the network is conducting to the output 0, the complementing P channel portion is off. When the P channel is conducting, the N channel is off. Therefore, the only current drawn through the on transistors is that required to charge the gates of subsequent driven circuits. Since subsequent stages are also IGFET's, current is passed only until the gate, essentially a capacitor, is charged. Therefore, the output attains the source level of the conducting side regardless of the number of FET's in the conducting series string.

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FET Logic Bridge Circuit

This circuit for a complementary logic bridge using IGFET's provides a multiuse logic element whose logic statement can be altered by control of selected inputs. The circuit includes two complementary bridge circuits each using opposite type IGFET's. When the logic of the N channel portion of the network is conducting to the output 0, the complementing P channel portion is off. When the P channel is conducting, the N channel is off. Therefore, the only current drawn through the on transistors is that required to charge the gates of subsequent driven circuits. Since subsequent stages are also IGFET's, current is passed only until the gate, essentially a capacitor, is charged. Therefore, the output attains the source level of the conducting side regardless of the number of FET's in the conducting series string.

The logic equation for the bridge circuit is given by 0 = AxB + CxD + AxDxE + BxCxE. Connecting E to a positive bias changes the function of the circuit to 0 = (B + D)x(A + C). Upon connecting E to ground 0 = (AxB) + (CxD). Upon connecting E to A, 0 = AxB + CxD + AxD.

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