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Dueck R.Digital design with CPLD applications and VHDL.2000

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500 C H A P T E R 1 1 • Logic Gate Circuitry

QUAD 2-INPUT NAND GATE

ESD > 3500 Volts

VCC

 

 

 

 

 

 

14

13

12

11

10

9

8

1

2

3

4

5

6

7

GND

SN54/74LS00

QUAD 2-INPUT NAND GATE

LOW POWER SCHOTTKY

 

 

 

J SUFFIX

 

 

 

CERAMIC

 

 

 

CASE 632-08

14

1

 

 

 

 

 

 

 

 

N SUFFIX

 

 

 

PLASTIC

14

 

 

CASE 646-06

 

 

 

 

1

 

 

 

 

 

D SUFFIX

 

14

 

SOIC

 

 

CASE 751A-02

 

 

1

 

 

 

ORDERING INFORMATION

SN54LSXXJ Ceramic

SN74LSXXN Plastic

SN74LSXXD SOIC

GUARANTEED OPERATING RANGES

Symbol

Parameter

 

Min

Typ

Max

Unit

 

 

 

 

 

 

 

VCC

Supply Voltage

54

4.5

5.0

5.5

V

 

 

74

4.75

5.0

5.25

 

 

 

 

 

 

 

 

TA

Operating Ambient Temperature Range

54

– 55

25

125

°C

 

 

74

0

25

70

 

 

 

 

 

 

 

 

IOH

Output Current — High

54, 74

 

 

– 0.4

mA

IOL

Output Current — Low

54

 

 

4.0

mA

 

 

74

 

 

8.0

 

 

 

 

 

 

 

 

FIGURE 11.3

74LS00 Data (1 of 2) Reprinted with permission of Motorola.

11.1 • Electrical Characteristics of Logic Gates

501

SN54/74LS00

DC CHARACTERISTICS OVER OPERATING TEMPERATURE RANGE (unless otherwise specified)

 

 

 

 

Limits

 

 

 

 

 

 

 

 

 

 

 

 

 

Symbol

Parameter

Min

Typ

Max

Unit

Test Conditions

 

 

 

 

 

 

 

 

 

VIH

Input HIGH Voltage

2.0

 

 

V

Guaranteed Input HIGH Voltage for

 

 

All Inputs

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VIL

Input LOW Voltage

54

 

 

0.7

V

Guaranteed Input LOW Voltage for

74

 

 

0.8

All Inputs

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VIK

Input Clamp Diode Voltage

 

– 0.65

– 1.5

V

VCC = MIN, IIN = – 18 mA

VOH

Output HIGH Voltage

54

2.5

3.5

 

V

VCC = MIN, IOH = MAX, VIN = VIH

74

2.7

3.5

 

V

or VIL per Truth Table

 

 

 

 

 

54, 74

 

0.25

0.4

V

I = 4.0 mA

VCC = VCC MIN,

VOL

Output LOW Voltage

 

 

 

 

 

OL

VIN = VIL or VIH

74

 

0.35

0.5

V

IOL = 8.0 mA

 

 

 

per Truth Table

IIH

Input HIGH Current

 

 

20

A

VCC = MAX, VIN = 2.7 V

 

 

0.1

mA

VCC = MAX, VIN = 7.0 V

 

 

 

 

 

IIL

Input LOW Current

 

 

– 0.4

mA

VCC = MAX, VIN = 0.4 V

IOS

Short Circuit Current (Note 1)

– 20

 

–100

mA

VCC = MAX

 

 

Power Supply Current

 

 

 

 

 

 

ICC

Total, Output HIGH

 

 

1.6

mA

VCC = MAX

 

 

 

 

 

 

 

 

Total, Output LOW

 

 

4.4

 

 

 

 

 

 

 

 

 

 

 

 

Note 1: Not more than one output should be shorted at a time, nor for more than 1 second.

AC CHARACTERISTICS (TA = 25°C)

 

 

 

Limits

 

 

 

 

 

 

 

 

 

 

Symbol

Parameter

Min

Typ

Max

Unit

Test Conditions

 

 

 

 

 

 

 

tPLH

Turn-Off Delay, Input to Output

 

9.0

15

ns

VCC = 5.0 V

tPHL

Turn-On Delay, Input to Output

 

10

15

ns

CL = 15 pF

FIGURE 11.3

74LS00 Data (2 of 2) Reprinted with permission of Motorola.

 

do not guarantee these values. An exception to this would be the supply voltage, VCC, whose

 

typical value is simply indicated as the average of maximum and minimum values.

 

Note that IIH and IIL are shown in Figure 11.2 as flowing in opposite directions, as are

 

IOH and IOL. On a data sheet, a current entering a gate is indicated as positive and a current

 

leaving the gate is shown as having a negative value. The reason for these current directions

 

will become apparent when we examine the internal circuits of the gates later in the chapter.

 

 

EXAMPLE 11.1

What is the maximum value of VOL for a 74LS00 NAND gate when the output current is at

 

its maximum value?

Solution When the output is in the LOW state, the output current is given by IOL, which has a maximum value of 8 mA. The output voltage, VOL, is specified for a value of 4 mA and for 8 mA. Since the output condition is specified for maximum IOL (8 mA),

then VOL 0.5 V.

 

 

502 C H A P T E R 1 1 • Logic Gate Circuitry

The 74XX00 NAND gate data is sufficient to represent any logic functions having “normal” output current within its particular logic family. This data can be used for most gate or flip-flop circuits within the family. Some specialized devices with higher-current outputs (e.g., 74XX244 octal tristate buffers) have a different set of electrical characteristics within their family.

In the following sections of the chapter, we will use a NAND gate from each of three device families (74LS00, 74HC00A, and 74HCT00A) for illustrating the general principles of the various electrical characteristics. Devices from other families will also be used in examples and problems. Data sheets for the various devices are included in Appendix C.

SECTION 11.1 REVIEW PROBLEM

11.1What are the maximum values of voltage and current we can expect at the output of a 74LS00 NAND gate when both inputs are LOW?

11.2Propagation Delay

K E Y T E R M S

tpHL

Propagation delay when the device output is changing from HIGH to LOW.

tpLH

Propagation delay when the device output is changing from LOW to HIGH.

Propagation delay occurs because the output of a logic gate or flip-flop cannot respond instantaneously to changes at its input. There is a short delay, on the order of several nanoseconds, between input change and output response. This is largely due to the charging and discharging of capacitances inherent in the switching transistors of the gate or flipflop.

Figure 11.4 shows propagation delay in two gates: a 74XX00 NAND gate and a 74XX08 AND gate. Each gate has an identical input waveform, a LOW-HIGH-LOW pulse. After each input transition, the output changes after a short delay, tp.

FIGURE 11.4

Propagation Delay in NAND and AND Gates

Two delays are shown for each gate: tpLH and tpHL. The LH and HL subscripts show the direction of change at the gate output; LH indicates that the output goes from LOW to

HIGH, and HL shows the output changing from HIGH to LOW.

Propagation delay is the time between input and output voltages passing through a standard reference value. The reference voltage for standard TTL is 1.5 V. LSTTL and CMOS have different reference voltages, as follows.

11.2 • Propagation Delay

503

N O T E

Propagation Delay for Various Logic Families:

LSTTL: Time from 1.3 V at input to 1.3 V at output.

Other TTL: Time from 1.5 V at input to 1.5 V at output.

CMOS: Time from 50% of maximum input to 50% of maximum output.

 

EXAMPLE 11.2

Use the data sheet in Figure 11.3, as well as those in Appendix C, to find the maximum

 

 

propagation delays for each of the following gates: 74LS00 (quadruple 2-input NAND),

 

 

74LS02 (quadruple 2-input NOR), 74LS08 (quadruple 2-input AND), and 74LS32

 

 

(quadruple 2-input OR).

 

 

 

 

 

 

 

 

 

Solution

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Table 11.2 Propagation Delays of 74LS Gates

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

74LS00

74LS02

74LS08

74LS32

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

tpLH

15 ns

15 ns

15 ns

 

22 ns

 

 

 

 

 

tpHL

15 ns

15 ns

20 ns

 

22 ns

 

 

 

Table 11.2 shows the variation of propagation delay among logic gates of the same

 

 

family (74LS TTL). Since each logic function has a different circuit, its propagation delay

 

 

will differ from those of gates with different functions.

 

 

 

 

 

 

 

 

 

 

EXAMPLE 11.3

Use data sheets to find the maximum propagation delays for each of the following logic

 

 

gates: 74F00, 74AS00, 74ALS00, 74HC00, and 74HCT00.

 

 

 

 

 

 

Solution

 

 

 

 

 

 

 

 

 

 

 

 

Table 11.3 Propagation Delays of 74LS Gates

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

74F00*

74AS00

74ALS00

74HC00**

74HCT00***

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

tpLH

 

6 ns

4.5 ns

11 ns

15 ns

 

 

19 ns

 

 

 

tpHL

5.3 ns

4 ns

8 ns

15 ns

 

 

19 ns

 

*Temperature range (74F00): 0°C to 70°C.

**VCC 4.5 V, temperature range (74HC00): 55°C to 25°C.

***VCC 5 V, temperature range (74HCT00): 55°C to 25°C.

As indicated by the notes for Table 11.3, propagation delay (and other parameters) vary with certain operating conditions, such as ambient temperature and power supply voltage. Always make sure that the operating conditions are correctly specified when looking up a data sheet parameter.

All gates in Example 11.3 have the same logic function (2-input NAND), but different propagation delay times. We might ask, “Why not always use the advanced Schottky TTL gate (74AS00), since it is the fastest?” The main reason is that it has the highest power dissipation of the gates shown. We wouldn’t know this without looking up other specs on the data sheet. (We will learn how to do this later in the chapter.) Thus, it is important to make design decisions based on complete information, not just one parameter.

504 C H A P T E R 1 1 • Logic Gate Circuitry

Propagation Delay in Logic Circuits

A circuit consisting of two or more gates or flip-flops has a propagation delay that is the sum of delays in the input-to-output path. Delays in gates that do not affect the circuit output are disregarded. Figure 11.5 shows how propagation delay works in a simple logic circuit consisting of a 74HC08 AND gate and a 74HC32 OR gate. Changes at inputs A and B must propagate through both gates to affect the output. The total delay in such a case is the sum of tp1 and tp2. A change at input C must pass only through gate 2. The circuit delay resulting from this change is only tp2.

FIGURE 11.5

Propagation Delays in a Logic Gate Circuit

The timing diagram in Figure 11.5 shows the changes at inputs A, B, and C and the resulting transitions at all gate outputs.

Assume VCC 4.5 V and temperature range is 55°C to 25°C.

1.When A goes LOW, AB, the output of gate 1, also goes LOW after a maximum delay of

tpHL 15 ns. This makes Y go LOW after a further delay of up to tpHL 15 ns. Total delay: tp tpHL1 tpHL2 15 ns 15 ns 30 ns, max.

2.The HIGH-to-LOW transition at input B has no effect, since there is no difference between 0 1 and 0 0. AB is already LOW.

3.The LOW-to-HIGH transition at input C makes Y go HIGH after a maximum delay of tpLH2 15 ns.

SECTION 11.2 REVIEW PROBLEM

11.2Assume the gates in Figure 11.5 are replaced by a 74LS08 AND gate and a 74LS32 OR gate. Repeat the calculations of for the propagation delays if the waveforms of Figure 11.5 are applied to the circuit. The data sheets for the 74LS08 and 74LS32 are found in Appendix C.

11.3 • Fanout

505

11.3 Fanout

K E Y T E R M S

Fanout The number of load gates that a logic gate output is capable of driving without possible logic errors.

Driving gate A gate whose output supplies current to the inputs of other gates.

Load gate A gate whose input current is supplied by the output of another gate.

Sourcing A terminal on a gate or flip-flop is sourcing current when the current flows out of the terminal.

Sinking A terminal on a gate or flip-flop is sinking current when the current flows into the terminal.

IOL

Current measured at a device output when the output is LOW.

IOH

Current measured at a device output when the output is HIGH.

IIL

Current measured at a device input when the input is LOW.

IIH

Current measured at a device input when the input is HIGH.

We have assumed that logic gates are able to drive any number of other logic gates. Since gates are electrical devices with finite current-driving capabilities, this is obviously not the case. The number of gates (“loads”) a logic gate can drive is referred to as its fanout.

N O T E

Fanout is simply an application of Kirchhoff’s current law: The algebraic sum of currents at a node must be zero. Thus, the fanout of a logic gate is limited by:

a.The maximum current its output can supply safely in a given logic state (IOH or IOL), and

b.The current requirements of the load to which it is connected (IIH or IIL).

Figure 11.6 shows the fanout of an AND gate when its output is in the HIGH and LOW states. The AND gate, or driving gate, supplies current to the inputs of the other four gates, which are called the load gates.

Each load gate requires a fixed amount of input current, depending on which state it is in. The sum of these input currents equals the current supplied by the driving gate. The

FIGURE 11.6

Driving Gates and Load Gates

506 C H A P T E R 1 1 • Logic Gate Circuitry

fanout is determined by the amount of current the driving gate can supply without damaging its output circuit.

The input and output currents of a gate are established by its internal circuitry. These values are usually the same for two gates in the same family, since the input and output circuitry of a gate is common to all members of the family. Exceptions may occur when the output of a particular gate, such as the 74XX244 octal three-state buffer, has additional output buffering or an input of a gate such as a 74LS86 Exclusive OR is equivalent to more than one input load.

EXAMPLE 11.4

FIGURE 11.7

Example 11.4

Output Current due to One Load Gate

The gates in Figure 11.7a and b are 74LS00 NAND gates. Determine the output current of the driving gate in each figure.

H

L

H

IOL IIL

 

a. Low output on driving gate

L

IOH IIH

L

H

 

b. High output on driving gate

Solution From the 74LS00 data sheet, IIL 0.4 mA and IIH 20 A. (There are two values of IIH given in the data sheet. Choose the one for the condition VIN 2.7 V, which is the minimum output voltage of a driving gate in the HIGH state (VOH). The other value is not appropriate since a gate will never have a 7 V output, as specified in the condition, if its supply voltage is 5 V.)

Since the driving gate is driving one load, its output current is the same as the input current of the load gate. Therefore, the driving gate output currents are given by IOL 0.4 mA (positive, since it is entering the driving gate output) and IOH 20 A (negative, since it is leaving the driving gate output).

EXAMPLE 11.5

FIGURE 11.8

Example 11.5

Output Current due to Two Load Gates

Determine the output current of the driving gate in each of Figures 11.8a and b if the gates are all 74LS00 NAND gates.

H

IOL IIL

 

 

 

 

 

H

L

IIL

a. Low output on driving gate

L

IOH

IIH

 

 

 

 

 

L

H

IIH

b. High output on driving gate

11.3 • Fanout

507

Solution Since there are two identical load gates in the circuits of Figure 11.8, the driving gate output current will be twice the load gate input current.

IOL 2 0.4 mA 0.8 mA.

IOH 2 ( 20 A) 40 A.

Figure 11.9 shows the extension of the circuits in Figures 11.7 and 11.8, where the number of load gates is the maximum that can be driven by the driving gate. This is the condition used to calculate fanout.

H

IOL

IIL

H

L

1

 

 

 

IIL

 

 

2

 

 

IIL

 

 

nL

 

a. Low output on driving gate

L

IOH

IIH

L

H

1

 

 

 

IIH

 

 

2

 

 

IIH

 

 

nH

b. High output on driving gate

FIGURE 11.9

Output Current to Fanout Calculation

If the load gates each represent the same load, then by Kirchhoff’s current law (KCL):

IOL IIL1 IIL2 IILnL nL IIL

and IOH IIH1 IIH2 IIHnH nH IIH

The fanout of the driving gate in the LOW and HIGH states can be calculated as:

nL

IOL

IIL

 

 

and nH

IOH

IIH

 

 

By convention, current entering a gate (IIH, IOL) is denoted as positive, and current leaving a gate (IIL, IOH) is denoted as negative. When current is leaving a gate, we say the gate is sourcing current. When current is entering a gate, we say the gate is sinking current.

Note that the output of a gate does not always source current, nor does an input always sink current. The current direction changes for the HIGH and LOW states at the same terminal. The reason for this will become apparent when we study the circuitry of logic gate inputs and outputs.

508

C H A P T E R

1 1 • Logic Gate Circuitry

 

 

EXAMPLE 11.6

How many 74LS00 inputs can a 74LS00 NAND gate drive? (that is, what is the fanout of

 

 

a 74LS00 NAND gate?)

Solution We must consider the following cases:

a.When the output of the driving gate is LOW

b.When the output of the driving gate is HIGH

Output LOW:

IOL 8 mA (sinking)

IIL 0.4 mA (sourcing) nL 8 mA/0.4 mA 20

Output HIGH:

IOH 0.4 mA (sourcing) IIH 20 A (sinking)

nH 0.4 mA/20 A 20

Since nL nH, fanout is 20.

We disregard the negative sign in our calculations, since the input current of the load gate and output current of the driving gate are actually in the same direction. For example, even though IOH is leaving the driving gate (negative), IIH is entering the load gates (positive). These currents flow in the same direction. If we include the minus sign in our calculation, we get a negative value of fanout, which is meaningless.

The fanout in both HIGH and LOW states is the same in this case, but that is not always so. If the values of HIGHand LOW-state fanout are different, the smallest value must be used. For example, if a gate can drive four loads in the HIGH state or eight in the LOW state, the fanout of the driving gate is four loads. If we attempt to drive eight loads, we can’t guarantee enough driving current to supply all loads in both states.

If a gate from one logic family is used to drive gates from another logic family, we must use the output parameters (IOL, IOH) for the driving gate and the input parameters (IIL, IIH) for the load gates.

EXAMPLE 11.7

Calculate the maximum number of Schottky TTL loads (74SXX series) that a 74LS86

 

XOR gate can drive.

 

Solution

Driving gate:

74LS86

IOH 0.4 mA,

 

 

IOL 8 mA

Load gates:

74SXX

IIH 50 A,

 

 

IIL 2 mA

Output LOW:

IOL 8 mA (sinking) IIL 2 mA (sourcing) nL 8 mA/2 mA 4

Output HIGH:

IOH 0.4 mA (sourcing) IIH 50 A (sinking)

nH 0.4 mA/50 A 8

Since nL nH, fanout nL 4.

 

11.3 • Fanout

509

 

What happens if we load a gate output beyond its rated fanout? Adding more load

 

gates will do this by increasing the value of IOL beyond its maximum rating. If enough load

www.electronictech.com

is added, the output of the driving gate might be destroyed by the heat generated by the ex-

 

cess current. More likely, the performance of the driving gate will be degraded.

 

 

Figure 11.10 shows the relationship between output voltage and current for a 74LS00

 

and a 74F00 NAND gate. Figure 11.10a shows that the output voltage (LOW state) in-

 

creases with increasing sink current. Figure 11.10b indicates a decrease in HIGH state out-

 

put voltage with an increase of source current.

 

 

1

VOLTAGE (VOLTS)

0.5

, OUTPUT

 

OL

 

V

 

 

0

0

4

(VOLTS)

3

VOLTAGE

2

, OUTPUT

1

OH

 

V

 

 

0

FIGURE 11.10

TA = 25°C

VCC = 4.5 V

LS00

F00

20

40

60

IOL, OUTPUT CURRENT (mA)

a. Output low characteristic

TA = 25°C

VCC = 5.5 V

LS00

F00

–50

–100

–150

IOH, OUTPUT CURRENT (mA)

b. Output high characteristic

Output Characteristics of 74LS00 and 74F00 Gate. Reprinted with permission of Motorola

In other words, a greater load in either state takes the output voltage further away from its nominal value. This has an effect on other performance factors, such as noise margin, which we will examine in a later section of the chapter.

N O T E

The output voltage of a logic gate is defined in a datasheet for a particular value of output current.

We will examine the fanout of CMOS devices in a later section on interfacing between CMOS and TTL.