Click "Show more" to see the entire description. Hope this helps out, more videos coming up. /*--------------------Written Companion--------------------*/ How's it going everybody, in this video I will be teaching you about the 74LS86, which is the QUAD 2-INPUT Exclusive OR chipset, which is better known as the XOR Gate. This chipset contains a total of 14 pins, of which its 7th pin is reserved the Ground, and its 14th pin is reserved for the VCC. In it, you will find 4 XOR Gates, each of these taking two inputs and generating an output. In the bottom row, when holding the chipset properly, the first and second pins shall be for inputs 1A & 1B, followed by a third pin reserved for the output 1Y, which contains the resulting value from the Boolean algebra operation (A)*(B)' + (A)'*(B). This pattern continues throughout the chipset, meaning there is a 2A, 2B, and 2Y, so on and so forth, of course skipping pins 7 and 14. For the top row, right after the 14th pin, the 13th and 12th pin shall be the inputs 4B, 4A, and the 11th pin shall be its output 4Y, this pattern continues throughout that side of the chipset, in descending order. Now we shall proceed by generating the XOR gate's truth table. As we said we are going to establish the variable inputs A, B, and the output Y. We shall also add an extra column for comments, in which we shall explain the process that is occurring internally on the chipset, as the inputs enter the logic gate. The behavior that we can expect from the Exclusive OR gate goes as follows: Since we have n = 2 inputs, using the 2^(n) formula, we calculate there are a total of 2^(2) = 4 possible combinations or scenarios to handle. A HIGH output (1) results if one and only one of the inputs is HIGH (1). If neither input is high, a LOW output (0) results, if both the inputs are HIGH, a LOW output (0) results. Now, we shall proceed with stating these combinations: The first is when A = 0, B = 0; which makes the output Y = 0; The second is when A = 0, B = 1; which makes the output Y = 1; The third is when A = 1, B = 0; which makes the output Y = 1; And last, but not least, when A = 1, B = 1; which makes the output Y = 0; Now for the comments description: As mentioned previously, the formula which we shall use to verify this statement is the following (A)*(B)' + (A)'*(B) = Y Since both A and B equal 0, Y equals 0: (0)*(0)' + (0)'*(0) = 0 Since only B equals 1, Y equals 1: (0)*(1)' + (0)'*(1) = 1 Since only A equals 1, Y equals 1: (1)*(0)' + (1)'*(0) = 1 Since both A and B equal 1, Y equals 0: (1)*(1)' + (1)'*(1) = 0 It is important to keep in mind that these inputs A and B are Variable, they can represent any Boolean expression's output value. Now I shall open a LogiSim simulation to demonstrate our truth table. This will be a pretty basic example, all we shall need are 2 Pins for the inputs, an XOR Gate, and our LED (or Light Emitting Diode) to display our output value. Now we shall proceed with testing the four cases: As you can observe, when both are 0, the output is LOW, which means the LED remain OFF. When A = 0, and B = 1, the output shall be HIGH, because only one of the inputs was HIGH. When A = 1, and B = 0, the output remains HIGH, as you can see it is bright red, since it met the condition. And when both inputs are HIGH, the output shall be LOW. IF YOU ENJOYED THIS VIDEO, PLEASE GIVE IT A THUMBS UP, AND SUBSCRIBE TO MY CHANNEL FOR MORE VIDEOS LIKE THIS. FOR ANY COMMENTS OR QUESTIONS, YOU MAY CONTACT ME ON THE FOLLOWING PLACES AT TWITTER: @GIOVANNI0918 AT GOOGLE+: +GIOVANNIORLANDOi7 THANK YOU FOR WATCHING, HAVE A NICE DAY!

• published: 02 Jul 2014
• views: 9