This is the second part of Digital Logic Gates, the first part is right here: Digital Logic Gates (Part 1).

In this instructable, we are going to cover the fundamentals of the following gates: NAND, NOR, XOR. Also, we are going to observe the operations of the NAND and NOR gates through implementing them on a breadboard.

Parts needed:

9V battery

Battery connector

5V regulator

2 IC Chips: 74LS00, 74LS02.

One LED (any color)

One 330 Ohm resistor

Wires as needed

## Step 1: NAND Gate

The NAND gate is equivalent to NOT AND; the output is the complement to that of the AND gate. Also, it is important to note that the output of a NAND gate will be Low (0) only if both inputs are High (1). Otherwise, the output will always be High (1). The behavior of a NAND gate is shown in the truth table in Table I.

Q = (A.B)’

The NAND gate is also referred to as a universal gate due to the fact that any boolean function can be implemented using NAND gates. Furthermore, the 74LS00 is an IC chip that comprises of 14 pins; four NAND gates, a supply voltage pin (Vcc), and a ground pin (GND). The supply voltage (Vcc) is connected to pin 14, and ground (GND) is connected to pin 7. Moreover, we are going to test the NAND gate by constructing a simple circuit and observing its output using an LED detector circuit; if the LED is on then the output = 1 and if the LED is off then the output = 0. For the 74LS00 chip used, the supply voltage (Vcc) should be between 4.75V and 5.25V in order for the chip to function properly. Finally, for the breadboard part of this step, the blue wire is Input 1 (A), the white wire is the Input 2 (B), and the LED is the output.

## Step 2: NOR Gate

The NOR gate is equivalent to NOT OR; the output is the complement to that of the OR gate. Also, it is important to note that the output of a NOR gate will be High (1) only if both inputs are Low (0). Otherwise, the output will always be Low (0). The behavior of a NOR gate is shown in the truth table in Table I.

Q = (A+B)’

The NOR gate is a universal gate as well because any boolean function can be implemented using NOR gates. Furthermore, the 74LS02 is an IC chip that comprises of 14 pins; four NOR gates, a supply voltage pin (Vcc), and a ground pin (GND). The supply voltage (Vcc) is connected to pin 14, and ground (GND) is connected to pin 7. Moreover, we are going to test the NOR gate by constructing a simple circuit and observing its output using an LED detector circuit; if the LED is on then the output = 1 and if the LED is off then the output = 0. Finally, for the breadboard part of this step, the blue wire is Input 1 (A), the white wire is the Input 2 (B), and the LED is the output.

## Step 3: Exclusive OR (XOR) Gate

The Exclusive OR gate (XOR) is used to compare two bits; the output of an XOR gate is High (1) only if the two inputs are different. The behavior of the XOR gate is shown in Table I.

Q = A ⊕ B

Also, the 74LS136 is an IC chip that comprises of 14 pins; four XOR gates, a supply voltage pin (Vcc), and a ground pin (GND). The supply voltage (Vcc) is connected to pin 14, and ground (GND) is connected to pin 7.

<p>if you have scope and signal generator, could you run speed test and tell results? how fast can switch from high to low? thanks!</p>
<p>TTL gates comprise of transistors, and they<br>take a finite amount of time to switch. Thus, the time it takes for the gate's<br>output to change after a change in the input is defined as propagation delay,<br>denoted as tpd. If you look at the manufacturer's datasheet for the IC chip you<br>are using, you will find the propagation delay range of the chip used.Yes, you can find the propagation delay using an oscilloscope just as<br>shown in the attached figure. You need to measure the tplh (Propagation Delay<br>Time Low-to-High Level Output) and tphl (Propagation Delay Time, High-to-Low<br>Level Output).</p><p>Finally, for instance,<br>the 74LS00 IC chip has a range of propagation delays ranging from 3ns - 10ns. </p><p>Note: ns is nanoseconds</p><p>Hopefully, this<br>answered your question :)</p>
<p>Again, you might want to explain the difference between LS and standard ICs. </p><p>The faster you push a signal through a gate, the more the output looks like a sine wave and the specification has not only rise and fall time but voltage threshold requirements too. That is how I. C.s are tested ( and a lot may fail at the manufacturing plant ).</p>
<p>Yes, that was usafeul. Thank you :)</p>