Electronic – Truth table to transistor diagram and Boolean experssion to transistor diagram

boolean-algebracmostransistors

I'm having a little trouble, making transistor level diagrams based off truth tables and Boolean expressions.

I was doing a problem to which I understand the first part, but I am unsure on how to do the second part.

Given the truth table

ABC| Output

0 0 0 | 0
0 0 1 | 0
0 1 0 | 1
0 1 1 | 1
1 0 0 | 0
1 0 1 | 0
1 1 0 | 1
1 1 1 | 0

I constructed the transistor level diagram

I feel fairly comfortable with this, however the next part I am perplexed

I am given the expression AB'C'+A'B'C+A'BC'+AB'C to which I simplified to

AB'+B'C+A'BC'
Now how would I construct the transistor level diagram from here? the only thing I can think of is constructing a truth table, but is there an easier/quicker way?

Best Answer

As you know, you use the upper part of the structure (from Vout to Vcc) to set the 1's and the lower part of the structure (from Vout to Vss) to set the 0's.

You can do this directly from the function expression. For example, let say your function is F=A·B*+C. I've chosen a different (and shorter function) for a simpler illustration of the method.

The upper part is easy (A and B* in series for AND) and C in parallel (for the OR part). The variables enter negated into the PMOS gate, because the PMOS will turn ON for 0 V (and not 5 V).

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Then the lower part must implement the 0's. Recalling the De Morgan Theorem, by negating F, AND becomes OR and viceversa. Applying to our function: we'll get F*=(A*+B)·C*

Therefore, in the lower part we have A* and B that are in parallel while C is in series. The variables doesn't enter negated into the NMOS gates because NMOS turn ON for 5V.

This method can be applied also to your original function, it just needs much more space!

Related Solutions

Definition of XOR, and how to remember its expansion in disjunctive normal form for N variables

For 2 input we got:

a XOR b = a'b + ab'

For 3 input it's:

a XOR b XOR c = (a XOR b) XOR c = 
(ab' + a'b) XOR c = 
(ab'+a'b)'c + (ab'+a'b)c' =
(a + b')(a' + b)c + ab'c'+ a'bc' = 
aa'c + abc + b'a'c + b'bc + ab'c' + a'bc' =
abc + b'a'c + ab'c' + a'bc'

Electrical – Making a circuit entirely with XOR 2-input gates

Actally, it is very simple:

f = (y XOR z) XOR x

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This just by looking at the thruth table and realzing that the bottom half is the inverse of thetop half.

Here's the math using boolean calculus.

We know x ^ y = x'y + xy' (^ is the XOR operator). Then:

g = a'b'c + abc + ab'c' + a'bc'
g = (a'b' + ab)c + (ab' + a'b)c'
     ^using a for X and b' for Y
g = (a ^ b')c + (a ^ b)c'

Now we have (a ^ b') and (a ^ b), which may seem not be possible to match, unless you realise that due to the nature of the XOR, (a ^ b') = (a ^ b)' [I'll leave that as an exercise to the reader]. Thus

g = (a ^ b)'c + (a ^ b)c'
g = (a ^ b) ^ c

This is only possible due to the special nature of the original equation. If you need an AND or OR you will find it impossible to reduce the function like this.

ABC| Output

Best Answer

Related Solutions

Definition of XOR, and how to remember its expansion in disjunctive normal form for N variables

Electrical – Making a circuit entirely with XOR 2-input gates

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