[help] countdown 9-0 then repeating using jk flip flop sa 7 segment display

[Bell]

i wasn't spying on you,sir marce...
nagkataon lang po na nung tiningnan ko po ang Who's Online, sa K-mapping thread kayo.and matagal kayong nag-stay sa thread na yun kaya i assumed na pinag-aaralan nyo ang k-map but the question is what for? kasi po hindi nyo naman kailangan ang k-map sa ginagawa niyo.hehe dito pala...

[marcelino]

i wasn't spying on you,sir marce...
nagkataon lang po na nung tiningnan ko po ang Who's Online, sa K-mapping thread kayo.and matagal kayong nag-stay sa thread na yun kaya i assumed na pinag-aaralan nyo ang k-map but the question is what for? kasi po hindi nyo naman kailangan ang k-map sa ginagawa niyo.hehe dito pala...

hmmm.... your spying on me!

I GOT AN ADMIRER HERE!
heheheh

pumunta ako dun to get the link for the k-mapping software. i ended using google though. bigla akong tinamad mag sulat sa paper. heheh

[Bell]

I'm not and I wasn't...though I really admire your personality,sir marce.nyahaha   

OT na...baka pagalitan ako...

[rdpzycho]

Bell - no. 1 hijacker.

[marcelino]

^ ay oo nga ano! hehehe nagpost lang si bell, iba na ang usapan...

[ultrasonic™]

Bell - no. 1 hijacker.

hahahaha   (^_^)

[rdpzycho]

  peace bell. back to topic.

[carlsfootprints]

hehehehe

sorry. ngayon ko lang din nakita... mali pala ang connection ko ng outputs ko... nabaliktad ko yung connection sa BCD to sevensegment decoder.

here is the corrected.

sir marcelino.. tama po ba ung gagawin ko? kukunin ko lng ung output dun sa Q(underlined).. then, magiging synchronous(down) counter na xa?

tia

[carlsfootprints]

+1 to you sir marce..

para dun sa reference with state machines.. medyo dinudugo pa ako dun eh.. hehe..

[marcelino]

sir marcelino.. tama po ba ung gagawin ko? kukunin ko lng ung output dun sa Q(underlined).. then, magiging synchronous(down) counter na xa?

tia

basta gawin mo din yung katulad ng table (yung count up), may input(present state), output (next state), then from there, check with the excitation table ng J-K F/F

Code: [Select]

JK flip flop Excitation Table
Transition JK
=========================
0 → 0 0X
0 → 1 1X
1 → 0 X1
1 → 1 X0
=========================
and as an example look at this (count-up)

Code: [Select]

Present         Next JK[3]         JK[2]      JK[1] JK[0]
================================================================
0 0000 0001 0X 0X 0X 1X
1 0001 0010 0X 0X 1X X1
for '0' output, J(3) and k(3) the transition will be, 0 to 0. 0 to 0 transition for JK is 0X.
for '1' output (0001), J(1) and K(1), transition is 0 to 1... kaya naging 1X.
isa pa...
for output '1' again. yung last bit, (J0 and K0), transition is 1 to 0. sa JK 1 to 0 is X1.

since it is 4 bits going down, you shall start from F (or 1111) to 0 (or 0000). values from F to A are don't care kaya X na yan lahat. but you shall put those on the table that will be produce. makakatulong sa pagreduce through K-mapping.

after completing the table, isa-isahin mo nang isolve (kmap or boolean). solve for J3, K3, J2, K2, J1, K1, J0 and K0... i use kmap para madali. just get all the values of J(X) in its column. put the values 1,0 and X to its appropriate location then reduce it.

then draw the circuit.

simulating on proteus helps... kakatamad yan sa breadboarding.

nasolve ko na to... kaso di ko alam kung saan ko sinulat... baka nadelete ko na. sa text file ko yun inilagay.
eto yata yung resulted schematic sa proteus. sana eto na yun. heheh

[carlsfootprints]

tnx tnx sir marce.. hehe.. ang haba ng reply..

utang muna ung +1.. 1 hour pa daw po eh.. hehe..

after completing the table, isa-isahin mo nang isolve (kmap or boolean). solve for J3, K3, J2, K2, J1, K1, J0 and K0... i use kmap para madali. just get all the values of J(X) in its column. put the values 1,0 and X to its appropriate location then reduce it.

sir marce,

dito ako nalilito.. hmm.. weak po kasi ako sa sequential circuits.. don't know state reductions, state machines, etc.. bsta design with sequential.. nag-iiba na panlasa ko..

i only use kmapping sa combinational circuits.. hmm.. i just dont know what will happen after ko mareduced.. let say po may equation na ako for j3.. pano ko malalaman ung input at output nung mga gates na involve? sorry po.. medyo noob.. hehe.. kahit reference na lang po na medyo simplified ung approach..

binasa ko ung link na binigay nyo sir marce.. ok nmn pla.. hehe.. basa-basa muna..

[carlsfootprints]

wala na akong utang

[marcelino]

later, post ko ang complete design... heheh

si misis kasi, panatulog na ako ng maaga kagabi!














kasi daw, maaga pa ako gigising.

[carlsfootprints]

hehe

tnx sir marce

[marcelino]

Code: [Select]

JK flip flop Excitation Table
Transition JK
=========================
0 > 0 0X
0 > 1 1X
1 > 0 X1
1 > 1 X0
=========================

Present Next JK[3] JK[2] JK[1] JK[0]
===================================================
F 1111 XXXX XX XX XX XX
E 1110 XXXX XX XX XX XX
D 1101 XXXX XX XX XX XX
C 1100 XXXX XX XX XX XX
B 1011 XXXX XX XX XX XX
A 1010 XXXX XX XX XX XX
===================================================
9 1001 1000 X0 0X 0X X1
8 1000 0111 X1 1X 1X 1X
7 0111 0110 0X X0 X0 X1
6 0110 0101 0X X0 X1 1X
5 0101 0100 0X X0 0X X1
4 0100 0011 0X X1 1X 1X
3 0011 0010 0X 0X X0 X1
2 0010 0001 0X 0X X1 1X
1 0001 0000 0X 0X 0X X1
0 0000 1001 1X 0X 0X 1X
===================================================

AFTER K-MAPPING:

J[3] = B'C'D'
K[3] = D'

J[2] = AD'
K[2] = C'D'

J[1] = BD' + AD'
k[1] = D'

J[0] = 1;
K[0] = 1;


kung di mo naiintindihan kung papano yung pagK-map nyan. ganito lang naman... i suppose you know how to k-map ha. heheh
kahit boolean solution pwede din. pero tinamad ako kaya gumamit nalang ako ng software. I am using Karnaugh Map Minimizer. libra naman yan... search mo nalang.

for example for J[3].. yung column lang na J[3] ha...

Code: [Select]

Present Next J[3]
=========================
F 1111 XXXX X
E 1110 XXXX X
D 1101 XXXX X
C 1100 XXXX X
B 1011 XXXX X
A 1010 XXXX X
=========================
9 1001 1000 X
8 1000 0111 X
7 0111 0110 0
6 0110 0101 0
5 0101 0100 0
4 0100 0011 0
3 0011 0010 0
2 0010 0001 0
1 0001 0000 0
0 0000 1001 1
=========================
makikita mo ang from F to 8, "don't care" lahat.... then from 7 down to 1, '0'. naging one lang sa 0.

we simply plot that to out map. then get the result.

eto yung mga result ng k-mapping ko.

for J3:


for K3:


for J2:


for K2:


for J1:


for K1:



for J0 and for K0 (same result):


and simply construct the circuit.

[carlsfootprints]

ahh.. hehe.. ganun pala..

tnx sir marce.. +1 ulit

wait lang.. hmm.. ung D natin, nakakabit sa A ng 7448? tama po ba? ung U1:A sa schem un ang D sa reduce equation natin? tama po ba?

[marcelino]

ahh.. hehe.. ganun pala..

tnx sir marce.. +1 ulit

wait lang.. hmm.. ung D natin, nakakabit sa A ng 7448? tama po ba? ung U1:A sa schem un ang D sa reduce equation natin? tama po ba?

nice observation... heheh tama yan!

yan yung naging pagkakamali ko nung una kong try using proteus. i did not bother to look at the datasheet of the BCD to 7 segment decoder. 2 days ko bago nakita na baliktad pala yan. heheheh

ang least significant bit ng BCD to seven segment decoder ay ang A. MSb naman ang D. Whereas sa derivation natin, MSB yung A. ikaw na ang bahala magRename ng mga variables... hehehe

Nga pala, dun sa table (state machine), obserbahan mo yung present and next state... pagdating ng 0 ang next state na nya ay 9. binibigyan ko lang emphasis kasi kahit anong sequences, pwede mong gawin yung same process. i first tried this solution with gray code, but using D-F/F lang.

Ang di ko naiintindihan ay kung papano iimplement sa MCU programming ang state machine.

salamat sa points BTW.

[carlsfootprints]

ang least significant bit ng BCD to seven segment decoder ay ang A. MSb naman ang D. Whereas sa derivation natin, MSB yung A. ikaw na ang bahala magRename ng mga variables... hehehe

so kelangan lang po silang baligtarin? hehe.. nagrun po ba ito sa proteus?

[marcelino]

yeah... narun yan.
sa proteus ako actually nagtest.