Electronic – Making 2-input AND gate, 2-input OR gate and NOT gate using only the “SAND” gate

digital-logic

Suppose there is a special gate called a SAND gate (Single-inversion AND) that looks like this:

enter image description here

How can I make 2-input AND gate, 2-input OR gate and NOT gate using only the SAND gate?

The truth table for this gate is:

$$\begin{array}{|c|c|c|}
\hline
A & B & \overline{A}\ {B} \\
\hline
0 & 0 & 0\\
\hline
0 & 1 & 1\\
\hline
1 & 0 & 0\\
\hline
1 & 1 & 0\\
\hline
\end{array}$$

To make a 2-input AND gate using only SAND gates, I would have to put 2 SAND gates in a row. Is this correct?

I don't know how to make 2-input OR gate and NOT gate using SAND gates. Can you give me a hint please?

Edit: I am only allowed to use SAND gates.

enter image description here

Best Answer

Let \$\text{SAND}(A,B) = \overline{A}B\$

NOT gate
\$\overline{A} = \overline{A}.1 = \text{SAND}(A,1)\$
Connect B to '1' and feed input to A.

AND gate
\$AB = \overline{(\overline{A})}B= \overline{(\text{SAND}(A,1))}B = \text{SAND}(\text{SAND}(A,1),B)\$
Invert the 1st input (using NOT gate implemented above) before feeding to SAND.

OR gate
\$\begin{align}A+B & = \overline{(\overline{A}\ \overline{B})}\\ & = \text{SAND}(\overline{A}\ \overline{B},1)\\ & = \text{SAND}(\overline{A}. \text{SAND}(B,1),1)\\ & = \text{SAND}(\text{SAND}(A,\text{SAND}(B,1)),1)\end{align}\$
Invert the 2nd input before feeding to SAND and feed its output to another inverter.

Related Solutions

Electronic – OR gate from one AND, three NOT

Whoever told you that was simply wrong. In fact, many logic families use just one kind of gate, especially in the early days. RTL logic families were basically all NOR gates, and DTL and TTL families are basically NAND gates. In either case, you can think of NOT as a single-input gate.

You can build any logic function at all from just NOR gates or just NAND gates. Entire computers have been built this way, including the earlier Cray supercomputers.

And don't forget that !(A + B) == !A & !B and that !(A & B) == !A + !B. Take the first equation and negate both sides, and you end up with your example: A + B == !(!A & !B).

Electronic – Implementing 3 to 8 decoder using 4 input NOR Gate

The fact that you were given a NOR gate rather than an OR gate is a significant hint: Look for the patterns of zeros, rather than ones, in your K-map. And remember that don't-cares can be assigned the value zero or one.

Here's the K-map I came up with, based on your truth table:

    A0  0  0  1  1
    A1  0  1  1  0
A3 A2 +------------
0  0  | 0  x  0  1
0  1  | x  0  1  1
1  1  | 0  x  0  1
1  0  | x  x  1  0

If you make all of the don't cares zero, you get this:

    A0  0  0  1  1
    A1  0  1  1  0
A3 A2 +------------
0  0  | 0  0  0  1
0  1  | 0  0  1  1
1  1  | 0  0  0  1
1  0  | 0  0  1  0

Clearly, the left-hand side of the table can be taken care of by feeding not-A0 (using the inverter you were given) into one input of the NOR gate.

The remaining three zeros (Aha!) can be taken from individual outputs of the 3-to-8 decoder, whose A, B and C inputs are connected to A1, A2 and A3, respectively. Specifically, the outputs for "1", "4", and "7" should be connected to the three remaining inputs of the NOR gate.

Best Answer

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Electronic – OR gate from one AND, three NOT

Electronic – Implementing 3 to 8 decoder using 4 input NOR Gate

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