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Flip flop mod counter 
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#1
Apr1412, 07:52 PM

P: 5

1. The problem statement, all variables and given/known data
To create a mod counter using flip flops that counts from 5 down to 2 and repeats this cycle...example 5, 4, 3, 2, 5, 4, 3, 2, 5, 4, 3, 2.... 3. The attempt at a solution I currently can get it to count from 7 down to 2 no problems...the issue is...I dont know how to skip states 7 and 6 right now. Using 3 jk flip flops...allows 2^3 states...or 07. Once the state reaches 1...I use an and gate to clear all the flip flops so that it jumps back to 7. Here is my schematic so far...but I'm assuming I have to somehow manipulate the j and k inputs to skip states 7 and 6; basically not hardwire a 1 to j and k. Any help would be appreciated...thanks. 


#2
Apr1512, 02:33 AM

P: 7

Remove the inverter on the output hexD(1)



#3
Apr1512, 05:31 PM

P: 5




#4
Apr1612, 04:21 AM

P: 7

Flip flop mod counter
You're right.
But I have another quick idea  you have to check it, sure. What I was thinking that you are resetting with event "1", it means in fact that you will have this value on the output for a short time, right? So I don't know if it's critical for you, but I'd rather reset on event "2" with a proper delay. This method has an advantage you can use to get your 5,4,3,2. because as you have "2" at the output it means "101" directly at the triggers. Now if you drop the first and the third trigger outputs with reset  the second will turn high on the negative slope of the first. it means after the reset you get "010", inverted into "101" at the output (dear value 5). If you need a scheme, I can add a sketch of how it looks in my imagination. Manipulating JK makes the scheme more flexible, but also much more complicated, let's try to avoid that. 


#5
Apr1612, 02:14 PM

HW Helper
Thanks
PF Gold
P: 7,715

Why don't you do it properly? Implement the next state equations with appropriate JK inputs instead of setting them all 1. Using appropriate "don't cares" for the disallowed states results in very simple JK equations requiring no additional logic gates at all.



#6
Apr1612, 09:59 PM

P: 5




#7
Apr1612, 10:38 PM

HW Helper
Thanks
PF Gold
P: 7,715

Q^+ = J\bar Q +\bar K Q$$so you can calculate ##J## by picking the 1's off the ##\bar Q## portion of the Kmap and ##K## by picking the 0's off the ##Q## portion of the Kmap of your next state equation for each bit. If you don't know what I'm talking about, you probably need more help than I can give you here. 


#8
Apr1712, 01:51 AM

P: 7




#9
Apr1712, 12:36 PM

P: 5

Here are my final next state equations derived from my kmaps Q0 next state = Q0 NOT Q1 next state = Q0 Q2 next state = Q1 NOT * Q0 + Q1 * Q0 NOT Using Q+ = J QNOT + KNOT Q J2 = Q1 NOT K2 = Q1 I dont now how to calculate j0/k0/j1/k1 since their equations are not in the format of Q+ = J QNOT + KNOT Q. It is probably something easy that i'm not seeing...this is the first time i've attempted something like this so bear with me. 


#10
Apr1712, 01:36 PM

HW Helper
Thanks
PF Gold
P: 7,715

Now, turning to ##Q_2##, the most significant bit, I agree again that ##Q_2^+=Q_0\bar Q_1+\bar Q_0Q_1##. You could do the same thing here, but it is better to go to the Kmap, because you can use the don't cares. What you want to do to get ##J_2## is just look at the ##Q_2=0## part of the Kmap and pick the 1's plus any good don't cares. To get ##K_2##, pick the 0's from the ##Q_2=1## part of the Kmap, again using any useful don't cares. See what you get. Finally, for ##Q_1##, check your tables and see if you don't get ##Q_1^+=\bar Q_0\bar Q_1+Q_0Q_1##. Then do either of the above techniques to get ##J_1## and ##K_1##. 


#11
Apr1712, 02:01 PM

P: 5




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