Question

Required to construct counters using synchronous sequential logic. Use one hex digit to display the result. ONLY AND/OR/NOT/XOR gates and flip flops allowed.BCD counters count from 0 to 9.

LOGISIM - not code

1. A 4-bit binary BCD counter with T flip flops WITH enable

if possible please use Logisim to build it and answer with picture of how its made.

Answer 1

He has asked mod 10 (sequential Synchro Goundle countes). It means it will count lo inputs. It has lo stales so we require minimum 4 flip flops to build it. But 4 inputs (or) 4 flip flops means it will count 16 states So after counting 10 states we have to Reset it so that it will again starte counting from 2010. The state diagram box this counter is look cocokon ]-cn010) Now we can use JK flip flops for this counter mostly Il flip flops are used as Counter

Now the excitatation table for the Тк (up klop is ... Poresent state Nent state Required Excitata Qt. Q3 Q. е. д. 9, 0, 4, 5, КТ, КТ Now the excitation table for the JK flip flop Poresent state Next state Required Excitation 4934, 9, 24 43 4, 9, Т.К Е, 3, 5, Оооооооток ох ох 1x oo o oo o ox ox txx | рото оо гоу оу.хо х оот тогоо рухх! * Огоо oт от сухоох (х о( от огтоох хотху) от го отток хотр (х о т т т т р об тxxтxiy( Тооо Tool косхо х1 x Гоо оооо хтохохх? Ik oo On Outl J K о - А -

of Encitati Now the Kmaps for the outputs of Gre table are here for Tifki we do not require kmap as inputs are high and done Carnes Iz knap, Here one quad is ' and h e obtained Q&A - Ka-knap Q.Q1 Die Q&Q9Q1 Q2Q1 Q2 24 x x d One octare K22Q1 - J - Konge 00012 Que ,@, 250, C2Q1 Q2O J3 2QzQ, Qif Q3 x4 FlyHx5 QuQx2 x 5 x 14

K3 Kmq pa 0, Cullen cih, QQ4, Q, Q, QAQ Qua x x x x One Quad el Joth 4 st 12.131.5 K3 cQ, Q1 14 XX Xth I knap Q, QI AQ3 Qom, 0,Q, Q2Q; 840 x x x x One octave Frog zB K4zQ; Que xos 20.0,Q1 27 one dual T 3 Q403 J4 kmap Quelles , QQ, Q2Q, QQy one deal stato J4 2 0322Q, ?, 12 13 14 . . . Qual x 8 x 9 to

The logic diagram for minimal eapo obtained are it Now we have to design mod -16, a bit counter so we use 4 flip flops Here it counte 16 states pow. 30001 6010-2010-0000-000bene (1100 Now here we can use JK flip flop.

Now the excitatation table for the Тк (up klop is ... Poresent state Nent state Required Excitata Qt. Q3 Q. е. д. 9, 0, 4, 5, КТ, КТ Now the excitation table for the JK flip flop Poresent state Next state Required Excitation 4934, 9, 24 43 4, 9, Т.К Е, 3, 5, Оооооооток ох ох 1x oo o oo o ox ox txx | рото оо гоу оу.хо х оот тогоо рухх! * Огоо oт от сухоох (х о( от огтоох хотху) от го отток хотр (х о т т т т р об тxxтxiy( Тооо Tool косхо х1 x Гоо оооо хтохохх? Ik oo On Outl J K о - А -

1 kmap for outputs Here also uk, need not to be deaw as inputs are high. Jakmap and co ll 10 Quta la o 01.11.10 TE I knap 0001.312 solo I Xxx Cu 13 13 14 SYWYYAT J4 2Q 2Q, Q3 J3 kmap Ky2 Q, Q,Q3 T K₃ kmap I | Que Bay 13 2QQ, 2 Cmap Quoi genap 5, 20, all go 632 Q2Q, Q C2 Cmap, Къ, Q) 312 x 161 ho 11 XIX TX7

Expression for minimal inputs The logic diageam using JK. flip flop NOS Hi all I cant to al 1994 : 947

Keno
Thank you..