Electronics - Logic Gates - Discussion

Please give a clear explanation.

Use the Karnaugh Map to find the logical expression.

NAND gate IC pin configuration shows a clear explanation of this question.

Ic 7400 contains 4 logic gates. It is a two-input and one output gate. NAND gate ic has 14 total pin.

Pin1,2,4,5,9,10,12,13 are input pins and pin 3,6,8,11 are the output pins whereas pin 7 is ground and pin 14 is connected to VCC(supply voltage +5v) on the bread board or trainer kit.

Refer the pin diagram of the nand gate.....

The 7400, designates a device containing four two-input NAND gates. Each gate uses two pins for input and one pin for its output, and the remaining two contacts supply power (+5 V) and connect the ground. The 4000 series, the newer CMOS versions of the 7400 series are also usable as analog amplifiers using negative feedback (similar to operational amplifiers with only an inverting input).

7400ics pin has total 14 pin. Pin no 7 use for GND and pin no 14 used for +vcc, remaining pin used for connections. For a NAND gate two input is required and one output is obtained means for NAND gate 3 pin connection is required.

SO IN 7400ICS TOTAL 4 NAND GATE ARE CONNECTED.