Design a 4-Bit Truncated Sequence Counter (Using JK Flip Flops)
HTML-код
- Опубликовано: 5 фев 2025
- In this video I go through the process of designing a counter that counts from 0-9. With four bits, you could make a counter that counts from 0-15, so this is a counter with a truncated sequence. The design for this counter is done using JK flip flops and is a fairly involved process. The design itself is not extremely difficult, it can just take a while because there are several steps and there are 4 bits, so the J and K inputs to four flip flops must be determined.
you made it so easy,that multiple doubts got cleared all at a time.
This video helped me a lot. I had to design a 4-bit counter that counts up to 0~12, and I had no idea how to deal with 13/14/15th signals. Thank you so much
That is so great to hear. I'm glad to have helped
Excellent explanation, big thanks here from Mexico sir.
this video helped me understand my practice hw. thank you so much!
justtttttttt Excellent, beautifully explained
Thank you this is very clear
thank you, watched a couple of video's on this topic but this one made me get it. great work :D
thank u man u made it all clear!
Glad it helped!
Is that the final counter at the end or do we have to add extra logic gates for the equations
For the final connection among flipflops, you haven't used any of the "not" outputs of the flipflop, are you sure its the doagram you have shown in the beginning of the video? For instance for JA shouldn't you be using QDnot QCnot and QB not?
Another great video david, thanks for the help
tnx for the bit change chart was stuck as had notes but no explanation. great help
Do WE decide which flip flops to use? Because I have to create a 4bit sequencial circuit and I am not sure if I am the one to choose the flip flop, but then again the assignment does not tell me which one to use.
Very helpful. I can use this method to create any counter. thanks a lot
in hte end don't you need the use and or or gates??
don't know if i'm wrong, but how come the next state of 1010 is 0000
Hi. I have a problem with my project.
Design a hardware architecture, using Matlab's Simulink, which works in the following way. There is a 2-4 decoder whose output is loaded (in parallel) in a counter 0-15. After loading, the counter starts its increasing count starting from the loaded value. how should i do?
Thanks sir this was a great help.
i want to desgin a counter that counts from 0-12 With four bits could you help me please ?
Thank you William that was very useful. However my problem is with the connection. I mean there are lots of Boolean expression that I can't put them together. So could you kindly post a video of how to do that.
Thank you
same problem
Next state thik hay kya???
What happens if you change your values past nine to don't care terms?
How about if I want it to count from 9 to 0, asynchronous, with FFs, like the old infamous 7476 FFs?
Can you solve a Kanraugh map that has no aces? Because my assignment has a map that has only 0s and Xs.
I love you doctor
thanks, this helped me a lot.
I tried to do the circuit in digital works and it doesn't work at all. It's counting in disorder.
Very Useful! thank u ! :)
well done :)
are both presets and clears on high?
Preset and clear can be clear on high or clear on low - depends on the implementation. If you are looking at the symbol of a JK flip flop, look for a "bubble" or circle at the input to the preset or clear. If there is a bubble, then it's clear on low, otherwise it's clear on high. The JK flip flops that I drew in this video do not have a preset or clear
ıs that 4 bits and 16 counter ?
Very Nice video I didnt understand the Karnought tho :/
Construct a Synchronous counter circuit that will count the random number 0,2,4,6
repeatedly by using positive edge triggered JK flip flop.
Help me, i not understand
the count 0,2,4,6 has 4 states in it, and so does the count the 0,1,2,3. In binary 0,2,4,6 is 000->010->100->110 while 0,1,2,3 is 00->01->10->11. You might notice that they are identical sequences except that 0,2,4,6 has an extra 0 on the end that never changes. So to count 0->2->4->6, just count 0->1->2->3 and add an extra, unchanging zero on the end.
hello can you provide me the last draw how you connected the flip flops
it same as the one show at the start of video
little bit confused about how to make grouping
Last part of the video is missing. Not set
please, mod 9 johnson counter
Can you add the 10-15 please
hey please tell me how can consist from 1khz square wave to 100hz with FF , How much do I have to use FF
?
A counter, like the one described in the video, that counts 0-9 has 10 states and so can be considered a divide by 10 counter. It rolls over (i.e., restarts) after every 10 counts (or input clock cycles). This makes it a divide by 10 counter, so if your input clock is 1 kHz, the counter would roll over at a rate of 1kHz/10 = 100Hz.
Please send me 4 bit synchronous up down counter using jk flip flop
Also if you can guide how we can convert this into a 0 to 99 counter.
A 0-99 would have 100 states, therefore you would need a 7 bit counter. A simple 7 bit counter would count 0-127, so you would need circuitry to reset to 0 after 99.
100 states? Wouldnt that be impossible. Cant we somehow connect two of the above circuits in a way that when first circuit completes 0-9 it would send a clock signal to the next circuit.
Yes, you could connect two 0-9 counters together where the second counter only counts after the first one counts 0-9. This would require two 4 bit counters which would mean you need a total of 8 flip flops instead of the 7 I mentioned in my previous response. The two 0-9 counter system will still go through 100 different states to count from 0-99 also (the number of values you are counting through = the number of states of your system).
Yes thats right. Thanks man.
Thank you so much.
help me (asynchronous 4 bit jk flip flops
counting from (1 2 3 4 5 6 )) design
You skipped the most important thing for me.
exactly :)
Thank you
can you prove why Ka =1 19:05
No matter what value you have for QA, QB, QC, and QD, KA will either be a 1 or an 'X' - since we can treat the X's as 1's KA is 1
You lost me at the tables with the expresions of JA, JB, JC and JD, also for KA, KB, KC and Kd
+Cristian Marius It's a JK flip-flop. Each FF has 2 input values dependent on the present state variables. Present-state is every possible combination of 4-bits BCD and the next-state is a count up so an increment of 1. From the rules of a JK FF (that little table that was superimposed) you can find the inputs necessary to change each FF from the present-state to the next-state. Then you just simplify with K-maps
Anyone from nutech electric 23
Jassy and kassy hhhhh it's Magnet B*** Mr white hahahaha
Curse you engineering