EOB127-Sayısal Elektronik

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• Tasarlanan bileşik mantık devresinde gerekli olan durum sayısı ikinin

tam kuvveti olmayabilir.

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anlamını taşır.

• Bu durumu bir örnek uygulama üzerinde anlatalım. Örneğimizde 4 bit

iki tabanındaki sayının değerini ortak anotlu displayde gösteren BCD

kod çözücü devresini tasarlayalım.

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D + B.C.D’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D + B.C.D’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fc(A,B,C,D)=B’.C.D’ Fb(A,B,C,D)= B.C’.D + B.C.D’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

AB\CD 00 01 11 10 Fd(A,B,C,D)= A’.B’.C’.D

Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

AB\CD 00 01 11 10 Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

AB\CD 00 01 11 10 Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fe(A,B,C,D)=D Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fe(A,B,C,D)=D + B.C’ Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D Fe(A,B,C,D)=D + B.C’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Ff(A,B,C,D)=C.D Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D Fe(A,B,C,D)=D + B.C’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Ff(A,B,C,D)=C.D + B’.C Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D Fe(A,B,C,D)=D + B.C’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

AB\CD 00 01 11 10 Ff(A,B,C,D)=C.D + B’.C + A’.B’.D

Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D Fe(A,B,C,D)=D + B.C’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X AB\CD 00 01 11 10 Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D Fe(A,B,C,D)=D + B.C’ Ff(A,B,C,D)=C.D + B’.C + A’.B’.D

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

AB\CD 00 01 11 10 Fg(A,B,C,D)= A’.B’.C’

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5 0 1 0 1 0 1 0 0 1 0 0 6 0 1 1 0 0 1 0 0 0 0 0 7 0 1 1 1 0 0 0 1 1 1 1 8 1 0 0 0 0 0 0 0 0 0 0 9 1 0 0 1 0 0 0 0 1 0 0 A 1 0 1 0 X X X X X X X

AB\CD 00 01 11 10 Fg(A,B,C,D)= A’.B’.C’ + B.C.D

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Fg(A,B,C,D)= A’.B’.C’ + B.C.D Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= A’.B’.C’.D + B.C’.D’ + B.C.D Fe(A,B,C,D)=D + B.C’ Ff(A,B,C,D)=C.D + B’.C + A’.B’.D

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Fg(A,B,C,D)= A’.B’.C’ + B.C.D Fb(A,B,C,D)= B.C’.D + B.C.D’ Fc(A,B,C,D)=B’.C.D’ Fd(A,B,C,D)= Fa(A,B,C,D) + B.C.D Fe(A,B,C,D)=D + B.C’ Ff(A,B,C,D)=C.D + B’.C + A’.B’.D

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