Basic Circuit

        Here's a basic circuit that involves just two NAND gates.  There are two inputs to this circuit, X and Y.  Can you generate a truth table for this circuit?

• Note that there is no connection where the two wires running from the output back to the input cross.

        Let's address that issue of the truth table.  Here is a truth table for you to fill in.  (Print this web page if you want to work on it.)

        Let's review what you know about this circuit.  We can focus on what happens when X= 0, and Y = 0, the first entry in the truth table above.

• If X = 0, then P = 1.  We know that if either input to a NAND gate is 0, the output is 1.
• Now. try to take advantage of the knowledge that P = 1.  If P = 1 AND Y = 0, then Q = 1.  It doesn't matter what P is, as long as Y = 0, Q will be 1.
• That gives us the first entry in the truth table above.  Here's the truth table with what we have figured out so far.

  X	Y	P	Q
  0	0	1	1
  0	1
  1	0
  1	1

        Now, let's address the second entry in the truth table.  In that situation, X = 0, and Y = 1.

• If X = 0, then P = 1.  We know that if either input to a NAND gate is 0, the output is 1.  That's the same as before.
• Now. try to take advantage of the knowledge that P = 1.  If P = 1 AND Y = 1, then Q = 0.
• That gives us the second entry in the truth table above.  Here's the truth table with what we have figured out so far.

  X	Y	P	Q
  0	0	1	1
  0	1	1	0
  1	0
  1	1

        Now, let's address the second entry in the truth table.  In that situation, X = 0, and Y = 1.

• If X = 0, then P = 1.  We know that if either input to a NAND gate is 0, the output is 1.  That's the same as before.
• Now. try to take advantage of the knowledge that P = 1.  If P = 1 AND Y = 1, then Q = 0.
• That gives us the second entry in the truth table above.  Here's the truth table with what we have figured out so far.

  X	Y	P	Q
  0	0	1	1
  0	1	1	0
  1	0
  1	1

        Next, let's address the third entry in the truth table.  In that situation, X = 1, and Y = 0.

• If Y = 0, then Q = 1.  We know that if either input to a NAND gate is 0, the output is 1.  That's the same as before.  However, this time, we take advantage of knowing that Y = 0.  Before, it was X that was equal to zero, but X = 1 here and it doesn't help us get started.
• Now. try to take advantage of the knowledge that Q = 1.  If Q = 1 AND X = 1, then P = 0.
• That gives us the third entry in the truth table above.  Here's the truth table with what we have figured out so far.

  X	Y	P	Q
  0	0	1	1
  0	1	1	0
  1	0	0	1
  1	1

        Finally, we have the case where X = 1 and Y = 1.

• If X = 1, then we need to know Q to determine P.
• If Y = 1, then we need to know P to determine Q.
• We are stuck!  There is no obvious way to proceed!

        There is one way to proceed.  We could just assume that P = 1, for example.  Let's try that.

• If Y = 1, and P = 1, then we know that Q = 0.
• If we know that Q = 0, it is an input  to the top NAND gate, so the output there is 1.  Thus, P = 1.  That's what we assumed to start with, so we don't get a contradiction.
• All is copasetic!  Or is it?
• Here is the truth table.

   

  X	Y	P	Q
  0	0	1	1
  0	1	1	0
  1	0	0	1
  1	1	1	0

        This is interesting, but we need to note the following.

• This result is not at all intuitive.
  • The circuit is symmetric.
  • If we have symmetric inputs (X = 1, Y = 1), we should have symmetric outputs.
  • Instead, we seem to have shown that P = 1 and Q = 0.  That's not at all symmetric, and it it bothersome.
• We got to the result by making an unsymmetrical assumption.  We assumed P = 1.

        What if we made the opposite assumption?  Let's assume P = 0.

• If P = 0, then Q = 1 since P is an input to the bottom NAND gate.
• If we know that Q = 1, it is an input  to the top NAND gate, so the output there is P = 0 since the other input to the top gate, X, is also 1.
• But, P = 0 is what we assumed to start all this.  Again, we do not get a contradiction, so everything seems OK.
• Here is the truth table for this situation.

   

  X	Y	P	Q
  0	0	1	1
  0	1	1	0
  1	0	0	1
  1	1	0	1

        Now, we can't have it both ways.  This truth table summarizes what we have found.
 

        There's something decidedly peculiar here.  The output has to be one or the other of these two cases.  Which is it?

        But, there's a related question.

• When you were little and went to the playground, when you walked up to the see-saw, which direction did the see-saw tilt?  Or did you call it a teeter-totter?
• Like the see-saw/teeter-totter, this circuit depends upon what went on before.
• It can exist in either state.  The see-saw can tilt in either direction.  Which state you find depends upon history.
• This circuit has a name with that same-consonant-different-vowel property.  It's called a flip-flop.

        Now, let's imagine that we do the following.

• Let us assume that we input X = 1 and Y = 0.  Then, the output will be P = 0, Q = 1.  Here is the flip-flop in that state

• Then, consider what happens when Y changes from 0 to 1.  Since P = 0, Q will not change when Y changes.  It only takes one zero input to keep the output of a NAND gate at one.  That means that the output does not change when Y changes to 1.

        We can reverse the initial inputs.

• Let us assume that we input X = 0 and Y = 1.  Then, the output will be P = 1, Q = 0.  Here is the flip-flop in that state

• Now, consider what happens when X changes from 1 to 0.  Since Q = 0, P will not change when X changes.  It only takes one zero input to keep the output of a NAND gate at one.  That means that the output does not change when X changes to 1.

        Some observations:

• This circuit is a prototype of a memory element.  It can store and remember one bit of information.
• Before it's a good memory element, it will need a little work.
• The method we used to determine possible states in the flip-flop, including especially the case of inputs with two possible stable states, is the method of contradiction.