Digitale Schaltungstechnik/ Binär nach BCD/ 5 bit
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Vollständige Wahrheitstabelle Bearbeiten
Wert | E | D | C | B | A | Fout | Eout | Dout | Cout | Bout | Aout | |
---|---|---|---|---|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |
1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | |
2 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | |
3 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | |
4 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | |
5 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | |
6 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | |
7 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | |
8 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | |
9 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | |
10 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | |
11 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | |
12 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | |
13 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | |
14 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | |
15 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | |
16 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | |
17 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | |
18 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | |
19 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | |
20 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | |
21 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | |
22 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | |
23 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | |
24 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | |
25 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | |
26 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | |
27 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | |
28 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | |
29 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | |
30 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | |
31 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
Gekürzte Wahrheitstabelle Bearbeiten
Wie wir beim 4 Bit Coderwandler festgestellt haben, kommt die Eingangsvariabel A in keiner Gleichung (ausser Aout=A) vor, weshalb wir sie an dieser Stellen streichen können und so die Wahrheitstabelle kürzen:
E | D | C | B | Fout | Eout | Dout | Cout | Bout | |
---|---|---|---|---|---|---|---|---|---|
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | |
0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | |
0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | |
0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | |
1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | |
1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | |
1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | |
1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | |
1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | |
1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
KV-Diagramme Bearbeiten
Um nun die Gleichungen zu bekommen, verwenden wir das KV-Diagramm:
Aout Bearbeiten
Bout Bearbeiten
B | C B | C B | C B | C B |
---|---|---|---|---|
E D | 0 | 1 | 1 | 0 |
E D | 0 | 0 | 0 | 1 |
E D | 0 | 1 | 0 | 0 |
E D | 1 | 0 | 1 | 0 |
Cout Bearbeiten
C | C B | C B | C B | C B |
---|---|---|---|---|
E D | 0 | 0 | 1 | 1 |
E D | 0 | 0 | 1 | 0 |
E D | 1 | 1 | 0 | 0 |
E D | 1 | 0 | 0 | 0 |
Dout Bearbeiten
D | C B | C B | C B | C B |
---|---|---|---|---|
E D | 0 | 0 | 0 | 0 |
E D | 1 | 0 | 0 | 0 |
E D | 0 | 0 | 0 | 1 |
E D | 0 | 1 | 0 | 0 |
Eout Bearbeiten
E | C B | C B | C B | C B |
---|---|---|---|---|
E D | 0 | 0 | 0 | 0 |
E D | 0 | 1 | 1 | 1 |
E D | 0 | 0 | 1 | 0 |
E D | 1 | 1 | 0 | 0 |
Fout Bearbeiten
F | C B | C B | C B | C B |
---|---|---|---|---|
E D | 0 | 0 | 0 | 0 |
E D | 0 | 0 | 0 | 0 |
E D | 1 | 1 | 1 | 1 |
E D | 0 | 0 | 1 | 1 |
Implementierung Bearbeiten
Hier nun das vollständige Schema:
Da es selbst in der Vergrößerung nur schwer lesbar ist, sind die einzelnen Baugruppen unten nochmal einzeln aufgeführt.
Vergrösserungen Bearbeiten
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Aout
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Bout
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Cout
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Dout
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Eout
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Fout
weiteres Bearbeiten
Da die Schaltung in einem Schaltungsimulator gezeichnet wurde, gibt es noch ein paar weitere Bausteine, die einfach dem Test dienen.