Playing with Boolean Gates
If you used the table on the previous page to order your parts, you should have six different chips containing six different types of gates:
- 7400 - NAND (four gates per chip)
- 7402 - NOR (four gates per chip)
- 7404 - NOT (six gates per chip)
- 7408 - AND (four gates per chip)
- 7432 - OR (four gates per chip)
- 7486 - XOR (four gates per chip)
Let's start with a 7408 AND chip. If you look at the chip, there will normally be a dot at pin 1, or an indentation at the pin 1 end of the chip, or some other marking to indicate pin 1. Push the chip into the breadboard so it straddles the center channel. You can see from the diagrams that on all chips, pin 7 must connect to ground and pin 14 must connect to +5 volts. So connect those two pins appropriately. (If you connect them backward you will burn the chip out, so don't connect them backward. If you happen to burn a chip out accidentally, throw it away so you do not confuse it with your good parts.) Now connect an LED and resistor between pin 3 of the chip and ground. The LED should light. If not, reverse the LED so it lights. Your IC should look like this:
Here is what is happening. In TTL, +5 represents a binary "1" and ground represents a binary "0." If an input pin to a gate is not connected to anything, it "floats high," meaning the gate makes an assumption that there is a 1 on the pin. So the AND gate should be seeing 1s on both the A and B inputs, meaning that the output at pin 3 is delivering 5 volts. So the LED lights. If you ground either pin 1 or 2 or both on the chip, the LED will extinguish. This is the standard behavior for an AND gate, as described in How Boolean Logic Works.
Try out the other gates by connecting them on your breadboard and see that they all behave according to the logic tables in the Boolean logic article. Then try wiring up something more complicated. For example, wire up the XOR gate, or the Q bit of the full adder, and see that they behave as expected.