There is another important kind of gate,
the NAND gate. Actually, the way to start thinking
about a NAND gate is to think of it as an AND gate with an inverter on the
output. That's shown below.
Actually, however, the symbol for a NAND gate
compresses the inverter down to a dot at the output of the NAND gate as shown
below.
Here is a simulated NAND gate. Check it out.
A
B
C
0
0
1
0
1
1
1
0
1
1
1
0
Wiring a Quad-NAND
Chip
If you want to use gates, you will need to
learn something about their physical characteristics. In this section we'll
walk you through wiring a simple gate circuit using one specific integrated
circuit (IC) the 7400 chip. It's a good introduction to some of the more
complex logic chips that you'll probably be using later.
Here's a picture of the
7400 chip in a circuit board. This chip is
actually an N74LS00P. The LS tells you that it is
a low power Schottky chip. Every manufacturer will embed the 7400 or 74LS00 in
other part numbers.
Notice that this chip has fourteen pins.
If you want to use an IC chip, then you
will always need to know the pinout. That's electrical engineering lingo for
describing the way the pins are connected to the internal circuitry of the
chip. You need to know where the power supply is connected and where the gate
inputs and outputs are connected. Here's the pinout for a 7400 chip.
The first step in wiring the 7400 is to
connect the positive power supply. Use a five volt (5v) power supply and don't
turn it on yet. Connect a lead to pin 14 as shown below, and connect the other
end of that lead to a 5v supply. Keep the power supply turned off until you
have everything connected. Here's what that looks like when the positive supply
voltage to the chip is wired.
The next step in wiring the 7400 is to
connect the ground connection. Connect a lead to pin 7 as shown below, and
connect the other end of that lead to ground.
Notice the pattern to this connection.
The power to this digital logic chip goes to the corners. Remember,
power to the cornersfor logic chips.
Now you can connect the two inputs to one
of the gates on the chip. You're going to put 5v on either of these inputs for
a 1 and ground the input for a
0. There are two wires in the picture below
that connect to pins 1 and 2 on the chip. Those pins are the inputs for one of
the NAND gates on the chip.
Now you can connect the output of the
gate. You will need to connect this output to something like a voltmeter or an
oscilloscope so that you can measure and observe the output of the gate. (And
the voltmeter or oscilloscope will also have to be connected to the ground. You
will measure output voltage with respect to ground.) The output will be near 5v
when the output is a 1 and near 0v when the
output is a 0.
Actually, you can often connect LEDs to
give a visual indication of a 1 (LED
lighted) or a 0 (LED dark). Here some LEDs
are shown, together with 1kW current limiting resistors. If you connect LED
indicators to your circuit remember that an LED is not the same in both
directions, and you have to get the correct end connected to the resistor. The
other end of each LED is connected to ground (or just "grounded"). Here's the
circuit to show the output of a NAND gate:
When the output of the gate is a
1, the output voltage will be five (5) volts.
Current will flow through the series combination of the resistor and the LED, so
the LED will light. When the output of the gate is a 0,
the output voltage will be zero (0) volts and the LED will not be lit. Thus,
the LED lights up when the output is a 1,
and doesn't light when the output is a 0.
You can use this indication scheme to show the status for any signal. (It
doesn't have to be the output of a gate.)
Question
Q1 In the
picture above, (shown again here) is the power turned on for the chip power
supply?
A NAND Gate
Here is a photo of a NAND gate wired to
display the input signals and output signals. In this simulation you can
manipulate the inputs and see the inputs and outputs. Note the following.
The input voltage can be set
to either 5v or 0v (ground) for each input to one of the NAND gates on the
chip. Five volts is a logical 1, and zero volts is a logical zero.
Note how the push buttons
move a connection from 5v to ground when the button is pushed.
When an signal is a 1, there
is an LED that lights to show that the input is 1. When the LED is not lit,
the signal is 0.
Note that there is
a current limiting resistor in series with each LED. If the voltage at
the output becomes 5v and the LED "saturates" around 1.8v, you need a
current limiting resistor. These resistors look to be 1kW.
The power supply connection
and the ground connection to the chip are both shown. The vertical line of
connection points on the circuit board is ground.
Check out how the circuit works and note all of
the connections that you need to make to ensure that the chip works as it is
supposed to work.
Example Problem
Let's reconsider the pump problem. What
happens if there are times when you don't want either pump to pump? Assume you
have a digital signal that is 1 when one of the two
pumps is to pump, and 0 when neither pump is to
pump. For example, if the pH was very close to desired you wouldn't want to do
anything at all so you wouldn't want either pump to turn on..
You still have the other signal that
determines which pump is to pump whenever one of the pumps should pump.
Devise a circuit that will ensure that
both pumps are OFF when the Pumpsignal is
0 and that the correct pump pumps when the
Pump signal is 1.
The circuit you devise in this section
will be simple enough that you can probably implement it with a few chips
although you will need to look for chips with AND gates and inverters. You
should be able to handle that now. Work through the solution in this lesson and
try it out in lab if you can.
Example Solution
Let's look at this problem with a truth
table. Here's the truth table.
Pumps On
1 = ON
Pump Choice
0 = S
1 = W
Pump S
Pump W
0
0
0
0
0
1
0
1
0
0
2
1
0
0
1
3
1
1
1
0
In English, we say to turn Pump S (Strong
reactant) ON when the pumps are ON, and the strong reactant is chosen (Choice 3)
and to turn Pump W (Weak reactant) ON when the pumps are ON and the weak
reactant is chosen (Choice 2). Otherwise, do nothing.
If we examine it closely we see that there
is exactly one term in each function. S is
1 only for choice 3,
that is when you want PUMPS ON and you want the
strong reactant. Similarly, W is
1 only for choice 2.
Here's the truth table again. Note the following:
We have defined Boolean
variables here for the various signals, P, C, S, and W.
We have indicated the
inputs by shading them green,
and the outputs by shading them orange.
P
C
S
W
0
0
0
0
0
1
0
1
0
0
2
1
0
0
1
3
1
1
1
0
Looking at the statement "S
is 1 . . . when you want Pumps ON AND
you want the strong reactant" then you can generate a logic expression
directly from the statement.
and also:
Finally, realize that it doesn't take much
to implement these functions. Note you only need one inverter and two
AND gates. Here's the circuit that turns the pumps
on at the proper time. This is an interactive simulation of the circuit, so you
can toggle the switches with the push buttons. Check it out.
Question
Q2
To check out the circuit you should what?
A QUICK QUESTION
Within the simulated
circuit, determine the part of the circuit that genrates a
1 when the pumps are
ON, and a
0 when they both are
OFF.
What If
The Problem Isn't So Simple?
Not all functions are as simple as this
one, and certainly not all can be implemented with just a few gates. However,
implementing this simple function gives us a clue how to implement more complex
functions.
In the next lesson we'll look at a more
general method for implementing functions - a method that uses only AND and OR
gates and inverters - but a method which can also be implemented with only NAND
gates. We hope that sounds intriguiging to you and that you are looking forward
to the next lesson.
Boolean
Algebra
Clearly at this point we are entering a
realm of a different kind of algebra. We have encountered some example terms in
this algebra.
and:
The algebra is unusual because the
variables in the algebra (S, P, C and W in the example) can take only two
values, 0 and 1. In this section we will examine some of the properties of this
algebra, and the implications of what we have already learned.
There are some simple things we need to
establish before we can proceed.
An
AND gate has this truth
table when the inputs are A
and B, and the
output is C:
A
B
C
0
0
0
0
1
0
1
0
0
1
1
1
So, clearly we have:
0�0 = 0,
and
1�1 = 1,
and
0�1 = 0
Which may be exactly what you expected.
We also need to consider an
OR gate.
An
OR gate has this truth
table when the inputs are A
and B, and the
output is C:
A
B
C
0
0
0
0
1
1
1
0
1
1
1
1
So, clearly we have:
0 + 0 = 0,
and
1 + 1 = 1,
and
0 + 1 = 1
Now, if you are taking a college course, and
you write home that 1 + 1 = 1 is what you just learned, your parents may want
your tuition refunded.
Now, if you accepted what was claimed
above, then you also have to accept the following:
A�A = A
Just let A be either zero or one and remember the
truth table for an AND.
We also have:
A + A = A
Again, just let A be either zero or one and
remember the truth table for an OR.
And - - - believe it or not, this result
for A + A is very useful because it is a fundamental result that will let us
build circuits with fewer gates. We'll come back to that later.
There are some interesting theorems that
can be proved. Note the following:
These two little theorems will prove to be
useful. To prove these theorems you only need to know about the properties of
AND, OR and NOT gates. That is left as an exercise for you.
When we want to prove a
theorem we will take the approach that we can prove the theorem by examining
all possible combinations of the appropriate variables. We can do that
because the possible combinations are finite.
Here is a truth table. It lists all possible
combinations for two variables.
A
B
0
0
1
1
1
1
0
1
1
0
0
0
1
0
0
1
0
0
1
1
0
0
0
0
This truth table proves the following
theorem.
Theorem (de Morgan)
Proof
The proof of this theorem is contained in the
truth table above which lists every possible combination of A and B, and shows
that this result is true.
One final note. There are some further
simple facts that come in useful. Note the following:
and:
Boolean Algebra can be a
confusing and misleading business. De Morgan's theorem above seems almost
trivial. However, there is a very interesting consequence of this theorem.
Here it is:
If you have a Boolean
function that is a sum-of-products form it can be implemented using a two
layer circuit with the first layer composed of AND gates, and the second
layer composed of OR gates.
Applying deMorgan's theorem
to the function the circuit can be built using the same structure, but
replacing every AND and OR gate with a NAND gate.
