Gates, Circuits and Boolean Functions
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1 Lecture 2 Gates, Circuits and Boolean Functions DOC 112: Hardware Lecture 2 Slide 1
2 In this lecture we will: Introduce an electronic representation of Boolean operators called digital gates. Define a schematic representation for digital gates. Use digital gates to create practical circuits. Define two more basic Boolean operators: XOR and XNOR. Introduce the concept of the Control Variable in a digital circuit. DOC 112: Hardware Lecture 2 Slide 2
3 Digital Gates In the last lecture we looked at Boolean algebra from a mathematical point of view, but we also introduced a diagramatic representation of Boolean equations, for example: This diagramatic representation describes an electronic circuit that will implement the equation, and in the circuit the individual Boolean operators are called digital gates. DOC 112: Hardware Lecture 2 Slide 3
4 Schematic Representation of Boolean Operators Instead of labelled boxes, a standard set of easy-to-recognise symbols is normally used to represent boolean functions. A circle is all that is required to indicate NOT. The triangle is provided to indicate the input/output direction. DOC 112: Hardware Lecture 2 Slide 4
5 Inverting functions A circle can be added to the AND and OR symbol outputs to create NAND and NOR gates. Circles can also be placed at the inputs to these gates, but this does not create a recognised Boolean function. DOC 112: Hardware Lecture 2 Slide 5
6 Building blocks for practical circuits Note that NAND/NOR are commonly used building blocks for most circuits: NAND/NOR can easily be constructed from transistors as we will see in lecture 5. NAND is complete (A set of Boolean functions f1,f2, is complete if and only if any Boolean function can be generated by a combination these functions.) DOC 112: Hardware Lecture 2 Slide 6
7 The NAND gate is all we need It is possible to build all other gates out of NAND gates. We can create a NOT gate using the Indempotent law: A A = A therefore (A A) = A DOC 112: Hardware Lecture 2 Slide 7
8 The NAND gate is all we need To create an AND gate we apply the Involution law: (A ) = A making use of our newly designed inverter: DOC 112: Hardware Lecture 2 Slide 8
9 The NAND gate is all we need To make an OR gate we need to apply de Morgan s theorem: A+B = (A. B ) We can just invert the output to make a NOR gate. Logicians call NAND the Sheffer Stroke DOC 112: Hardware Lecture 2 Slide 9
10 Problem Time Can you build every other gate using just the NOR gate? DOC 112: Hardware Lecture 2 Slide 10
11 Building more complex circuits What happens if we cascade two NAND gates? We can analyse this circuit using Boolean Algebra: I = (B.C) X = (A.I) = (A.(B.C) ) Applying de Morgan we get: X = A + B.C DOC 112: Hardware Lecture 2 Slide 11
12 Building more complex circuits An alternative way of analysing the circuit is to build a truth table: A B C I = (B C) X = (A I) DOC 112: Hardware Lecture 2 Slide 12
13 Building a three input NAND gate Cascading two 2-input NAND gates does not do the job, but we can design a 3 input NAND gate using Boolean algebra: X = (A.B.C) = (A.(B.C)) DOC 112: Hardware Lecture 2 Slide 13
14 Two new gates Here we define two new gates which can be very useful: A B XOR A B XNOR DOC 112: Hardware Lecture 2 Slide 14
15 Building an XOR gate from NAND gates Since the NAND gate is complete we should be able to construct an XOR gate using only NANDs. We can start with the Boolean equation: A XOR B = A.B + A.B and use de Morgan s theorem A XOR B = ((A.B). (A. B ) ) This does the job, but it it the best circuit? DOC 112: Hardware Lecture 2 Slide 15
16 No: this circuit is smaller Problem - How do we prove that this circuit implements the correct Boolean function? DOC 112: Hardware Lecture 2 Slide 16
17 The Logician s Solution R = (D.E) = ((A.C).(B.C) ) =((A.(A.B) ).(B.(A.B) ) ) Apply de Morgan R = (A.(A.B) ) + (B.(A.B) ) Apply de Morgan R = A.(A +B ) + B.(A +B ) Distributivity R = A.A +A.B + B.A +B.B Simplify R = A.B + B.A DOC 112: Hardware Lecture 2 Slide 17
18 The Hardware Engineer s Solution Work round the circuit to construct the input/output truth table, and verify that it is the same as the XOR gate. DOC 112: Hardware Lecture 2 Slide 18
19 How many possible gates are there? So far we have seen ONE one-input gate and SIX two-input gates, but in theory there are FOUR possible one-input gates and SIXTEEN possible two-input gates. Are there any more useful one-input gates? DOC 112: Hardware Lecture 2 Slide 19
20 We enumerate all one-input gate possibilities A G0 G1 G2 G Clearly G2 in the above table is the inverter, and the other three possible gates are useless. We can also define the gate function using algebra with the following table: G0 G1 G2 G3 R = 0 A A 1 DOC 112: Hardware Lecture 2 Slide 20
21 What are the possible 2 input gates? For two inputs we have 4 possible input values and therefore sixteen possible two input gates. As before we can enumerate all the possibilities and see what they do. DOC 112: Hardware Lecture 2 Slide 21
22 The sixteen possible two-input gates AB G0 G1 G2 G3 G4 G5 G6 G R = 0 AND A B XOR OR AB G8 G9 G10 G11 G12 G13 G14 G R = NOR XNOR B A NAND 1 DOC 112: Hardware Lecture 2 Slide 22
23 Designing circuits using just NAND and NOR Of all the useful gates, NAND and NOR are simpler and smaller to build and faster in operation. When designing a circuit to calculate a Boolean function, it makes sense to try and engineer it by using these gates only. To do that it is helpful to be able to manipulate the schematic circuit diagram. DOC 112: Hardware Lecture 2 Slide 23
24 Applying de Morgan s Theorem Grapically de Morgan s theorem can be applied graphically. A+B = (A.B ) has the circuit equivalent: For simplicity we can use just circles for inversion: DOC 112: Hardware Lecture 2 Slide 24
25 Manipulating Circuit Diagrams Often circuits can be simplified through symbolic manipulation The transformation increases the number of NAND/NOR gates by 1. DOC 112: Hardware Lecture 2 Slide 25
26 Control and Data Variables Suppose we want to design a controlled function circuit. For example when C=0 the output is 0, and when C=1 the output is A: This is the same as a 2 input 1 output gate BUT functionally we interpret it to have: 1 data input, 1 control input and 1 output. DOC 112: Hardware Lecture 2 Slide 26
27 Control and Data Variables If we construct a truth table from the specification, we find that the above circuit can be implemented with just one AND gate: This example shows why the name gate is used. DOC 112: Hardware Lecture 2 Slide 27
28 An Important Control Circuit Consider a circuit with two data inputs A and B and once control input C and one output defined as follows: if C=0 the output R=A if C=1 the output R=B This is in essence a digital switch and is called a multiplexer. DOC 112: Hardware Lecture 2 Slide 28
29 Implementing a Multiplexer We use the Boolean AND function to gate the signals from A and B individually, and combine the result with an OR gate. We are using the rule that A+0=A and 0+B=B. The complete circuit is: DOC 112: Hardware Lecture 2 Slide 29
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