Exclusive-NOR

The exclusive-NOR circuit (abbreviated XNOR) operates completely oppo- site to the XOR circuit. Figure 4-21(a) shows an XNOR circuit and its accom- panying truth table. The output expression is

which indicates along with the truth table that xwill be 1 for two cases: A B1 (the ABterm) and AB0 (the term). In other words:

The XNOR produces a HIGH output whenever the two inputs are at the same level.

It should be apparent that the output of the XNOR circuit is the exact in- verse of the output of the XOR circuit. The traditional symbol for an XNOR

AB x = AB + AB x = AB + AB = A{B {

x = A{B

x = AB + AB.

AB AB

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FIGURE 4-21

(a) Exclusive-NOR circuit;

(b) traditional symbol for XNOR gate; (c) IEEE/ANSI symbol.

XNOR gate symbols

A 0 0 1 1

B 0 1 0 1

x 1 0 0 1 x = AB +AB

A B

AB A

B

B A A

B

A B

(b)

x = A ⊕ B = AB + AB

(c)

x = A ⊕ B A

B

= 1 (a)

gate is obtained by simply adding a small circle at the output of the XOR symbol [Figure 4-21(b)]. The IEEE/ANSI symbol adds the small triangle on the output of the XOR symbol. Both symbols indicate an output that goes to its active-LOW state when only oneinput is HIGH.

The XNOR gate also has only twoinputs, and it combines them so that its output is

A shorthand way to indicate the output expression of the XNOR is

which is simply the inverse of the XOR operation. The XNOR gate is sum- marized as follows:

1. It has only two inputs and its output is

2. Its output is HIGH only when the two inputs are at the samelevel.

Several ICs are available that contain XNOR gates. Those listed below are quad XNOR chips containing four XNOR gates.

74LS266 Quad XNOR (TTL family) 74C266 Quad XNOR (CMOS)

74HC266 Quad XNOR (high-speed CMOS) x = AB + AB = A{B

x = A{B x = AB + AB

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Determine the output waveform for the input waveforms given in Figure 4-22.

EXAMPLE 4-16

Solution

The output waveform is obtained using the fact that the XOR output will go HIGH only when its inputs are at different levels. The resulting output wave- form reveals several interesting points:

1. The x waveform matches the A input waveform during those time in- tervals when B0. This occurs during the time intervals t0to t1and t2

to t3.

2. The xwaveform is the inverseof the Ainput waveform during those time intervals when B1. This occurs during the interval t₁to t₂.

3. These observations show that an XOR gate can be used as a controlled IN- VERTER; that is, one of its inputs can be used to control whether or not the signal at the other input will be inverted. This property will be use- ful in certain applications.

A B

A

B

x

t₀ t₁ t₂ t₃

x FIGURE 4-22

Example 4-16.

EXAMPLE 4-17 The notation x1x0 represents a two-bit binary number that can have any value (00, 01, 10, or 11); for example, when x11 and x00, the binary num- ber is 10, and so on. Similarly,y1y0represents another two-bit binary num- ber. Design a logic circuit, using x1,x0,y1, and y0inputs, whose output will be HIGH only when the two binary numbers x1x0and y1y0are equal.

Solution

The first step is to construct a truth table for the 16 input conditions (Table 4-4). The output zmust be HIGH whenever the x₁x₀ values match the y₁y₀ values; that is, whenever x₁y₁and x₀y₀. The table shows that there are four such cases. We could now continue with the normal procedure, which would be to obtain a sum-of-products expression for z, attempt to simplify it, and then implement the result. However, the nature of this problem makes it ideally suited for implementation using XNOR gates, and a little thought Each of these XNOR chips, however, has special output circuitry that limits its use to special types of applications. Very often, a logic designer will obtain the XNOR function simply by connecting the output of an XOR to an INVERTER.

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FIGURE 4-23 Circuit for detecting equality of two two-bit binary numbers.

x₁ x₀

y₁ y₀ Binary number

Binary number

z TABLE 4-4

x1 x0 y1 y0 z(Output)

0 0 0 0 1

0 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 0

0 1 0 1 1

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 0

1 0 1 0 1

1 0 1 1 0

1 1 0 0 0

1 1 0 1 0

1 1 1 0 0

1 1 1 1 1

When simplifying the expression for the output of a combinational logic cir- cuit, you may encounter the XOR or XNOR operations as you are factoring.

This will often lead to the use of XOR or XNOR gates in the implementation of the final circuit. To illustrate, simplify the circuit of Figure 4-24(a).

Solution

The unsimplified expression for the circuit is obtained as

We can factor ADfrom the first two terms:

z = AD(BC + BC) + AD z = ABCD + ABC D + AD EXAMPLE 4-18

will produce a simple solution with minimum work. Refer to Figure 4-23; in this logic diagram,x₁and y₁are fed to one XNOR gate, and x₀and y₀are fed to another XNOR gate. The output of each XNOR will be HIGH only when its inputs are equal. Thus, for x₀ y₀and x₁ y₁, both XNOR outputs will be HIGH. This is the condition we are looking for because it means that the two two-bit numbers are equal. The AND gate output will be HIGH only for this case, thereby producing the desired output.

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FIGURE 4-24 Example 4-18, showing how an XNOR gate may be used to simplify circuit implementation.

A

B

C D

ABCD

ABCD

AD

z = ABCD + ABCD +AD

(a)

B C

A D

(b) A + D = AD

B ⊕ C AD(B ⊕ C)

z = AD (B ⊕ C) + AD

At first glance, you might think that the expression in parentheses can be replaced by 1. But that would be true only if it were You should recognize the expression in parentheses as the XNOR combination of B and C.This fact can be used to reimplement the circuit as shown in Figure 4-24(b). This circuit is much simpler than the original because it uses gates with fewer inputs and two INVERTERs have been eliminated.

BC + BC.

REVIEW QUESTIONS 1. Use Boolean algebra to prove that the XNOR output expression is the ex- act inverse of the XOR output expression.

2. What is the output of an XNOR gate when a logic signal and its exact in- verse are connected to its inputs?

3. A logic designer needs an INVERTER, and all that is available is one XOR gate from a 74HC86 chip. Does he need another chip?