Page 67 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Digital Circuits and Logic Design

Notes
Figure 4.13: Product-of-Sums

We show both the Product-Of-Sums solution, from the previous example, and the Sum-Of-
Products solution from the current problem for comparison. Which is the simpler solution?
The POS uses 3-OR gates and 1-AND gate, while the SOP uses 3-AND gates and 1-OR
gate. Both use four gates each. Taking a closer look, we count the number of gate inputs.
The POS uses 8-inputs; the SOP uses 7-inputs. By the definition of minimal cost solution,
the SOP solution is simpler. This is an example of a technically correct answer that is of
little use in the real world.

Draw the circuit diagram of Sum-Of-Products.

Let us revisit a previous problem involving an SOP minimization. Produce a Product-Of-
Sums solution. Compare the POS solution to the previous SOP.
Figure 4.14: SOP Minimization

Out =A BCD + A BCD + A BCD
+A BCD + A BCD + A BCD
+A BCD + A BCD + A BCD

CD CD CD
A 00 01 11 10 A 00 01 11 10 A 00 01 11 10
B B B
00 1 1 1 00 1 1 1 0 00 1 1 1 0
01 1 1 1 01 1 1 1 0 01 1 1 1 0
11 1 1 1 11 1 1 1 0 11 1 1 1 1 0
10 10 0 0 0 0 10 0 0 0 0

Out =AC +A D+BC+BD Out=(A+ B) (C+D)
Solution:
Above left we have the original problem starting with a 9-minterm Boolean unsimplified
expression. Reviewing, we formed four groups of 4-cells to yield a 4-product-term SOP
result, lower left.

In the middle Figure 4.14, we fill in the empty spaces with the implied 0s. The 0s form
two groups of 4-cells. The solid blue group is (A′+B), the dashed red group is (C′+D). This
yields two sum-terms in the Product-Of-Sums result, above right Out = (A′+B)(C′+D).

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