Seven-segment decoder "b" segment

A BCD-to-7-segment decoder converts a 4-bit BCD digit $ABCD$ (A = MSB) into the seven signals a–g that light the segments of a display. Assume a common-cathode display, so a segment glows when its signal is 1.

     a
   ─────
  │     │
 f│     │b
  │  g  │
   ─────
  │     │
 e│     │c
  │     │
   ─────
     d

Truth table for segment b

Segment b (upper right) is ON for digits 0, 1, 2, 3, 4, 7, 8, 9 and OFF for 5 and 6. Inputs 1010–1111 are not valid BCD, so they are don't-cares.

$$b = \Sigma m(0,1,2,3,4,7,8,9) + d(10,11,12,13,14,15)$$

K-map

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | 1 | 1 | 1 | 1 |
      +---+---+---+---+
 01   | 1 | 0 | 1 | 0 |
      +---+---+---+---+
 11   | X | X | X | X |
      +---+---+---+---+
 10   | 1 | 1 | X | X |
      +---+---+---+---+

• Octet m(0,1,2,3,8,9,10,11) → $\overline{B}$
• Quad m(0,4,8,12) → $\overline{C}\,\overline{D}$
• Quad m(3,7,11,15) → $CD$

$$b = \overline{B} + \overline{C}\,\overline{D} + CD = \overline{B} + (C \odot D)$$

Check: digit 5 ($B=1$, $C=0$, $D=1$): $\overline{B}=0$, $\overline{C}\,\overline{D}=0$, $CD=0$, so b = 0. Digit 6 ($B=1$, $C=1$, $D=0$): b = 0. All other digits give 1.

Simplest logic circuit

Since $\overline{C}\,\overline{D} + CD = \overline{C \oplus D}$, segment b needs only one X-NOR gate, one NOT gate and one OR gate:

B ──[NOT]── B' ──────────┐
                         [OR]── b
C ──┐                    │
    [XNOR]── C⊙D ────────┘
D ──┘

(If X-NOR is not allowed, use three NOT gates, two AND gates and a 3-input OR for $\overline{B} + \overline{C}\,\overline{D} + CD$.)