## DIGITAL ELECTRONICS

INTRODUCTION: BOOLEAN ALGEBRA AND DE MORGAN'S LAW

There are certain statements that can only be either true or false. Example: The statement today is Friday" is either true, (when made on a Friday), or false, (when made on any other day of the week). A British mathematician, George Boole (1815-1864) created the field that is today known as "Boolean Algebra" by showing that statements such as the one above can be manipulated in a formal, algebraic, manner. While Boolean Algebra has nothing directly to do with physics it is the basis of a field often referred to as electronic logic that has widespread applications in electronic engineering, computer design and experimental nuclear and particle physics. This makes it worthwhile for us to discuss it briefly in the context of some experiments about digital electronics.

Returning to the statement today is Friday, we represent it symbolically with a logical variable A that has the value A = 1 when the statement is true or the value A = 0 when it is false. Now let us consider another statement: today is the 13th. This statement we represent with the variable B which has the value 1 on the 13th of each month and the value zero on any other day.

Next let us consider the conjunctions OR and AND. If we connect two logical statements with an AND that means that both are true. Thus, if you are superstitious you might say: if today is Friday the 13th (A = 1 AND B = 1) I will not travel. If you are very superstitious you might even say: if today is either a Friday or the 13th (A = 1 OR B = 1) I will not travel. In Boolean algebra one formally manipulates statements such as the ones above by representing the OR with an addition and the AND with a multiplication. Let Q be the result of such an algebraic manipulation. Then, on any Friday the 13th the statement.

Q = A × B = 1 × 1 = 1 ------------------------Eq. (1)

is true and if you are superstitious you might resolve that if Q = 1 you will not travel. If you are very superstitious you might even decide not to travel if

A + B \$ 1 -----------------Eq. (2)

 A B A+B A@B 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 1

No statement can be truer than true. There is, therefore, no reason that a logical variable should ever have a value greater than unity and the paradoxical looking statement

1 + 1 = 1 ----------------------Eq. (3)

holds true in Boolean Algebra. All the above can be summed up in the form of the Truth Table, Table 1.
The logical complement of the statement today is Friday is the statement today is not Friday. If A represents the original statement then its complement is customarily represented by , pronounced 'A bar'. Obviously, if A has the value 1, has the value 0 and vice versa. Based on this notation one can easily verify de Morgan's laws, (A. de Morgan 1806 ­ 1871):
Convince yourself that these laws hold by studying Table 2.

Table 2: Truth table for de Morgan's Laws

 A B A + B AB A'+B' (A.B)' A'+B' A'.B' 0 0 0 0 1 1 1 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 0 0 0 0

All of this would, at best, be of interest to some logicians, were it not for the fact that there are circuit elements that one can identify with the statements and operators of Boolean algebra. This means that one can apply Boolean algebra to the design of logic circuits such as the ones that are used in computers. A simple toggle switch has only two states, open or closed, as shown in Fig.1. Figure 1: a) open switch = 0 b) Closed switch =1

If one represents a logic 1 with a closed switch and a logic 0 with an open switch one can build circuits that represent logical operations. Fig. 2a shows an AND circuit in which a current flows only if the first AND the second switch are closed. Fig. 2b shows an OR circuit through which a current will flow if one switch OR the other is closed. Figure 2: a) AND circuit. b) OR circuit.

In computers, electronic circuits take the places of mechanical toggle switches and AND and OR circuits, often called AND gates and OR gates, are represented by the circuit symbols shown in Fig. 3.

Figure 3: a) Symbol of an AND gate. B) Symbol of an OR gate.

Table 3: Truth table for exclusive OR

 A B A(+)B 0 0 0 0 1 1 1 0 1 1 1 0

In circuit elements such as these a logic 1 is usually represented by one voltage level and a logic 0 by another. The conventions vary, but in a typical example (e.g. Transistor Transistor Logic, or TTL) a voltage level of V \$ 2.4 V represents a "1" and V # 0.8 V represents a "0". In a TTL AND gate, Fig. 3a, the voltage at the output lead C will thus be 0 unless there is an input voltage of \$ 2.4V applied to both input A AND input B, in which case it will go to 2.4 V. In an OR gate, such as shown in Fig. 3b, an input level of \$ 2.4 V at either input A or input B (or both) will produce an output level of 2.4 V at the output C. If one employs switches that do not switch from off to on but that toggle between two different current paths, as shown in Fig. 4a, one can build a circuit that switches from one state to the other whenever either of the switches is actuated; see Fig. 4b. You can readily see that a circuit like this allows one to turn a light on and off from two ends of a hallway. In electronics this kind of circuit is known as an Exclusive OR, its circuit symbol and truth table are, respectively, shown in Fig. 5 and Table 3.

In electronic circuitry it is often convenient to have circuit elements that invert the levels and thus give the logic complements of the output signals. Fig. 6a shows the circuit symbol for a NAND gate, (Not AND), a circuit that represents the operation i.e. that produces an output level only when there is no signal present at either input A or input B. Fig. 6b shows a NOR gate representing the operation, see Table 2.

APPLICATIONS
With the basic logic elements that we have introduced above one can build a large variety of logic circuits, many of them very ingenious (and quite complicated). In this lab we can only give you a glimpse of the many possible applications that exist. By far the most important application of digital electronic circuitry is in the design of computers. We have already seen how logical decisions based on the presence of individual (OR) or simultaneous (AND) signals can be implemented. Other important capabilities of a computer are the ability to remember things that it has been told, and, of course, the ability to calculate, i.e. to add. (Subtraction is simply an addition with the opposite sign, whereas multiplication and division can be reduced to repeated additions and subtractions.)

THE FLIP­FLOP
Consider a circuit consisting of two NAND gates, as shown in Fig. 7. Let there be a positive input signal, a logic 1, present at both terminals S and R. What will be the voltage level of the output terminals Q and ? (Assuming that we are using TTL elements.) Could both outputs be at the zero level? Let us investigate. The outputs are connected crosswise to the second inputs, SN and RN, of the two NAND gates. With S and SN at the levels 1 and 0 respectively, Q must be at the level 1, in contradiction with our assumption. Could both outputs be at the level 1? With at the level 1, SN must be 1 and Q = 0 which, again, contradicts the assumption. Q = 0 and  = 1 leads to SN = 1, RN = 0, which would indeed lead to Q = 0 and  = 1, but how can that be? The circuit and its inputs S and R are symmetrical, yet the outputs are not. If Q = 0, = 1 is possible so must be the converse: Q = 1,  = 0. We have thus a bi-stable circuit, i.e. one that can exist in two mutually exclusive but stable states. Which of these two states the circuit occupies must depend on its prior history. We investigate further. If we change the level of S from 1 to 0 Q will flip to 1 which, in turn, will force to flop to 0. Changing the input level of S will thus cause the circuit to change from one of the two stable states to the other. Restoring S to its original state 1 will not make any difference, the circuit will remain in the second state. The circuit can thus be permanently switched, or Set by the application of a short pulse to the input S. Once S has been restored to the 1 state further pulses to the S input will have no effect. We can, however Reset the circuit by the application of a short pulse to the R input. A flip­flop, such as this, can be used as a memory device: It will remember the first pulse applied to the S input and it can be made to forget by the application of a pulse to the R input.
A more mundane but important application of the simple flip­flop is the following: In a mechanical switch two metal contacts are brought together when the switch is closed; it is almost unavoidable that these contacts will bounce several times before they come to rest upon each other. Each bounce will cause a short pulse to enter the circuit. If the switch is used to turn on a light that will not matter. If the switch is used in a computer keyboard and enters "33333333" instead of "3" that is intolerable. We have seen above that once a flip­flop has been flipped, further pulses will have no effect. It is, therefore, a common practice in the design of electronic devices to follow a mechanical switch with a flip­flop that will debounce it.

COMPUTING
We have already mentioned that subtraction, multiplication, and division can all be reduced to addition. It will, therefore, be sufficient for us to discuss the latter operation. The decimal system, in which we normally carry out calculations, is not very well suited to the use in computers. The fact that computers use elements that are capable of only two states that we can identify with zero and one suggests that we employ the binary system.

Table 4: Decimal to binary conversion table
 Dec. 0 1 2 3 4 5 6 7 8 9 10 Bin. 0 1 10 11 100 101 110 111 1000 1001 1010

In the binary system only the numerals 0 and 1 exist. Note: these zeros and ones are numerals, not logical variables. The rules of ordinary, not Boolean, algebra apply. In case you have forgotten, Table 4 gives the first 10 integer numbers in the decimal and the binary system.
 A B S C 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1

The rules of addition for binary numbers look strange due to the fact that one has to carry an overflow whenever a digit is larger than 1 e.g.:

1 + 1 = 10; 10 + 10 = 100; 11 + 11 = 110;

101 + 111 = 1100; etc.

Next we construct a circuit that will allow us to add binary numbers. Consider the parallel connection of an AND gate and an exclusive OR gate as shown in Fig. 8.

This circuit is known as a Half adder for reasons that will soon become clear.

We already know the truth tables for the individual elements and we combine them in Table 5.
Next consider Fig. 9 in which several half adders are combined. Imagine that the three digit binary numbers 110 were presented to the inputs A0, A1, and A2, (i.e. a 1 at A2, and A1, and a 0 at A0). Now go back to Fig. 8 and assume that it represents the rightmost half adder in Fig. 9. According to the truth table, Table 4, the circuit is symmetrical in its response to input signals. That means that even though it contains two different circuit elements it does not matter which of the inputs A or B we identify with A0 or B0. The outputs, however, behave differently and we must identify S0 with the S(um) output. With the number 110 present at the A inputs a 0 would be present at input A0. If a 0 were also present at B0 both S0 and the carry output (which goes into the lower half adder) would remain at zero. Entering a 1 at B0 would produce a 1 at S0 but would leave the carry output at zero. In other words the last stage of this adding circuit gives the correct answers to the problems 1 + 0, 0 + 1, and 0 + 0. Had there been a 1 present at the A0 input, a zero at B0 would have produced a 1 at S0, whereas a 1 at B0 would have produced a zero at S0 and a 1 in the carry output which would have been fed into the next half adder. This, of course, is exactly what the binary addition 1 + 1 = 10 requires: a zero in the last place and a 1 that is carried to the next place. As an exercise investigate how the circuit shown in Fig. 9 will respond if 111 is presented to the A inputs and 110 to the B inputs.

BEFORE YOU START

You will assemble your circuits on the "Op­Amp Designer" breadboard, shown in Fig. 10, which has a fixed 5 V power supply that you can use to power the integrated circuits. If you need any help, please ask your instructor.

PLEASE NOTE: TURN THE 5 V POWER SUPPLY OFF BEFORE YOU PLUG IN OR UNPLUG ANY I.C's. BEFORE YOU TURN ON THE POWER CHECK YOUR CIRCUITS CAREFULLY. WHEN YOU ARE FINISHED WITH ONE EXPERIMENT TURN THE POWER OFF BEFORE YOU CHANGE OVER TO THE NEXT.(A common mistake is to connect power backwards. This will cause the I.C. to get hot. If this happens, turn off the power, check the connection and test the I.C. before proceeding.)

Figure 10

The small circles represent sockets. The lines show which of the sockets are interconnected under the board. Features that are not used in this experiment are crossed out. The integrated circuits should be powered from the 5 V power supply on the left side of the board.

Logic Levels. The following statements apply to the signal levels at the input as well as the output pins:

! FALSE, or logic 0, is represented by a low potential (0­0.8 volts above the ground). Connecting an input terminal to ground puts it in the zero state.

! TRUE, or logic 1, is represented by a voltage of + 2.4 volts or more with respect to ground. Connecting an input terminal to such a potential puts it in the 1 state. An input can be set to the 1 state by connecting it to +5v through a 1000 W resistor, (do not connect directly to +5V!!!). If an input is left unconnected ("floating"), it will usually assume the logic 1 state. However, it is safer to force the input by applying a positive voltage or by connecting it to an output that is in the 1 state. This will avoid spurious signals by electrical noise.

! FAN­OUT. As many as 10 inputs may be tied to one output.

! NEVER connect two or more outputs together; the results are unpredictable and the I.C.'s can burn out. Connect outputs only to I.C. inputs.

! TIMING. Typical TTL gates switch from one state to the other in 15 nanosec (1 nanosec = 10-9 sec). This is the secret of a computer's high speed.

The integrated circuits, chips or I.C's for short, that you will use are all of the TTL type. They usually contain four identical basic circuit elements. The pin assignments are shown in Fig. 11. This time the top view is given. The I.C's should be plugged into the two middle rows of sockets, straddling the center gap, so that each I.C. pin is connected to the four other sockets in its column. The two top and the two bottom rows should be connected to the 5 V power supply; both top rows to the two + 5 V connectors and both bottom rows to the ground connectors.

The input signals should be applied to the I.C's by connecting their input pins by means of the two slide switches in the lower right hand corner of the board to either 5 V (through a 1000 W resistor) or to ground. To find out how the I.C's respond to these input signals you must be able to read out the output signals. This could be done by measuring the output levels with a voltmeter. Instead you should use the Light Emitting Diodes or LED's provided for this purpose. Since not all I.C's provide enough power to light up the LED's you should first set up a Driver circuit, using a 7406 I.C. as shown in Fig. 12.

WHAT TO DO

1). LED Driver. Plug a 7406 chip straddling the middle row near the right end of the board. Connect one LED through a 330 Ohm resistor to 5 V on one side and to one of the even numbered pins (other than 14) on the other, as shown in Fig. 12. Build a second, identical, circuit with one of the other 5 drivers. To test your two circuits connect the corresponding odd numbered pins through a 1 kW resistor to either + 5 V or ground. For the pin assignments see Fig. 11. Leave both circuits on the board, they will be used in the following experiments as indicators.

2). AND gate. Select a 7408 I.C and check its pin connections in Fig. 11. Insert it in the board and connect pin 7 to ground and pin 14 to +5V. Connect the center contacts of the two switches to the two inputs of the gate and their upper contacts through 1 kW resistors to +5 V. Connect the lower contacts of the switches to ground. Connect the output to the LED driver. AFTER ALL CONNECTIONS HAVE BEEN MADE AND CHECKED TURN ON THE POWER. Test the truth table for this gate. Turn off the power.

3). OR, NAND, and NOR gates. Replace the 7408 I.C. successively with a 7432, a 7400 and a 7402 I.C. (Note that the 7402 NOR gates are reversed compared to the 3 other I.C's). Test each of these gates and record their truth tables together with their wiring diagrams in your notebook. Always give the I.C. pin numbers on your wiring diagram as shown in Fig. 8. Any point in the circuit diagram can then be found immediately on the breadboard. Troubleshooting a complicated circuit is almost impossible without proper labeling. (Don't forget to turn off the power before you switch I.Cs.)

4). Flip­Flop. Construct the flip­flop shown in Fig. 7 from two of the NAND gates in a 7400 I.C. Connect one LED driver each to the Q and Q Test it and see if it behaves as described above.

5). Exclusive OR and Half Adder. Test one of the exclusive OR gates on a 7486 I.C. and then connect it with one of the AND gates in a 7408 I.C. to form the half adder shown in Fig. 8. Record its wiring diagram and truth table in your notebook.