I just clapped with joy at the end. 👏👏 You cannot imagine how fast I understood everything after watching this video...I just enlightened my class knowledge 100 times! The best Electronic course tutorials are here my fellow engineers.
Wow! With this knowledge, the possibilities are endless. Imagine, instead of clearing each flip flop's state to zero, you use the same NAND gate but connect it to PST and CLR on each FF in a different way. You could manipulate this to use counters to skip to different values
Can't appreciate your amazing explanation enough! Just a simple correction- I think (at 7:55) the signs of the NOT gates should be mirrored, for Qc and Qa. Correct me if m wrong here...!
Sir, you should also mention that we are observing Q as the output...(in the neg-pos edge triggered section in the beginning of the video). And thank you very much for such great videos..appreciate your effort!
Good morning sir. You have explained that we have to clear the counter after 1001 (9). So upon the arrival of 1010 we clear the counter. My question is when Qd Qc Qb Qa (1010) has arrived, we have that count (which was to be eliminated ), since Qd Qc Qb Qa are our outputs. Is that count momentary bef the FFs are cleared? Or the action of the clear is so fast that the FFs are cleared immediately? Am I missing something? Thank you for the lectures.
You have explained that we have to clear the counter after 1001 (9). So upon the arrival of 1010 we clear the counter. My question is when Qd Qc Qb Qa (1010) has arrived, we have that count (which was to be eliminated ), since Qd Qc Qb Qa are our outputs.
It is indeed a fabulous lecture ..which content clear ...but only one correction at the en d..for the two input nand gate you must take qc and qa as 1 nand 1 is 0 but in the lecture the two inputs are qd and qb whose nand is 1 and clr would be 1 ..And sorry for poiniting out such a silly mistake for a great lecture . And shouldnt we stop at 1001 and not at 1010 , because 1010 might also be printed , becoz we are print qd,qc , qb, qa? Correct me if I am wrong
The first part of the video created a great confusion for me As you had used negative edge triggered FF to make a down counter where Q was the input and Q' was used as clock But as you did not mention here in this video that you considering Q as the output for all....It made me think for a very long while. Hope this comment helps someone else too.
You really work hard, I'm worried because I clear concept here but what about semester exam we've to write more and how will we write because we don't have content?
Thank you so much sir. Your videos have be very helpful in my studies. It's just that we cannot find any serial order in your videos which makes it difficult to find which video is next. If you could add a serial order in your videos it would be very helpful.Thank you.
he give 2 clock inputs a and b ,,output of 1st flip flop not given to second as clock,,he gives 2nd input to second flip flop and then ckt is similar,,an logic used that is also different
I am 62 years old and I follow digital design as a hobby.I want make a JK FF counter that gives the sequence 10,22, 5.4.3. I have seen a lot of your interesting videos but I cannot figure out how to do it. Can you give me a hand ? Theodore from Greece.
sir pls i have doubt about it,,is it ic 7490 ripple counter,,because when I reffer ic 7490 ripple counter internal diagram it's seems like little different from u mentioned in video pls suggest fast i have exams 2 days after
sir, if we use +ve edge triggered clock then what about race around condition ?? you taught that to overcome race around condition we use -ve edge triggering .
As clear is active low signal so it will work when clr=1 because when clr=1 then input to ff from clr =0 and output=0 So according to this concept the output of the nand gate should =1 to active the clr input but you are doing opposite to this? Please clear my doubt Am I wrong or you taught wrong?
in 2:09 when you are discussing positive edge trigger you are telling about the clock but not about the output clear from where you are taking the output when positive edge triggring
When to take output in counter with respect to clock pulse in negative edge triggered flip flop?? 1)At the start of clock pulse or 2)At the end of clock pulse
Hello sir, i am a really big fan of your work. I really have a question, how can you make the counter starts from a specific number? like If i.m willing to start counting from 12 to 63 for example
sir in case we have 1100 as input to nand gate ... Qd=1 and Qb=0. by taking there nand gate we have output equal to 1 so our clr input goes to state 1 thus our counter is not goingto be reset in this case..... plz clear this confusion as soon as possible bcz i have exam in next week
So for ripple counters, all we need to do is determine the number of states we have in the count sequence and then attach the clocks according to the type of edge trigger? Do we use synchronous counters to implement the circuitry for sequences that skip states like 0, 1, 2, 4, 6?
Sir, since we do not want 1010 to appear as the output, shouldnt we feed 1001 as the input of the NAND gate? If we reset after seeing 1010 output, dont we somehow count from 0000 to 1010?
Yes we will in fact see for a short period of time 1010 as the output. To solve this problem we can connect the Clock-Enable input of the 4th Flip Flop with the output Qa of the first flip flop. Then we need to connect the J input of the 4th Flip Flop to an AND gate of Qb and Qc. As a result the Qd changes state only when Qa drops from 1 to 0 and Qb=Qc=1and resets to 0 when Qa drops from 1 to 0 every other time (J=0,K=1) .
in that case you have to use Q complement as an output if you use Q an clock in down counter. its more convenient to stick with one output(Q) and just change the clock form Q to Q complement to switch between up and down counter.
the outputs have to be 1010 for not gate to reset the contour so since the outputs are 1010 does that not mean we are already showing number 10 on the display before resting??? please help Im really confused about this. Thank you all.
Hello, if I have a asynchronous/ripple D flip-flop counter with a count sequence of 10, 9, 8, 7, 6, 5, 4, 3 and repeat but with provition for a logic 0 output on 8 and a logic output on 6, how exactly would I go about with this kind of a question. I understand that I will use the positive egde triggered and Q would be the clock. I would the connect the NAND gate with input 3 to clear to ensure that it will repeat the sequence but what of the provitions and would the sequence start at 15 and go down or would I also need to connect a NAND gate to PRE so it starts at 10. Any help would be much appreciated.
It's not that we can just pass our semester exams with ur lectures ..... it's all GATE level content that u r teaching for free👏👏👏 Hat's off to u sir
I just clapped with joy at the end. 👏👏 You cannot imagine how fast I understood everything after watching this video...I just enlightened my class knowledge 100 times! The best Electronic course tutorials are here my fellow engineers.
sir its only because of ur videos i cleared my exam..thnx a lot..keep it up
hopefully +1
i learnd more stuff in a 9min video than a 90min lecture
rather 4.5 min at 2x speed
I feel you..
same man this is excuse i gave to my family when they ask why you arent going university
@@owaiswasim8911 🤣
I wish this guy could cover all of my undergrad degree because his videos are much better than 90 min lectures
Wow! With this knowledge, the possibilities are endless. Imagine, instead of clearing each flip flop's state to zero, you use the same NAND gate but connect it to PST and CLR on each FF in a different way. You could manipulate this to use counters to skip to different values
Thank you so much for all the DE videos...they helped me score really good in my finals...
tommorow is my dld exam.hoping to get good marks because of your videos.
Heroes fight for their people from behind their masks.
Neso Academy fights for students from behind an electronic board.
not a suitable dialogue
good try 🤣
😁 benki🔥
U JUST SAVED MY LIFE THIS SEMESTER
Your videos are really great. Thank you so much. I cleared my exams with good grade.
Thank You So much Sir...it's just because of you i am able to understand Digital Logic circuits course which is in my present semester.
Thanks Neso i learnt better than 1 hr lecture in college #from #Nepal
sir ur lectures are truly worth it gave me great understanding .
sir i also want a lecture on how to design a counter which have only the even states.
Can't appreciate your amazing explanation enough!
Just a simple correction- I think (at 7:55) the signs of the NOT gates should be mirrored, for Qc and Qa. Correct me if m wrong here...!
Yup that's correct!
@@shubhranginidas6818 ohh wow u r clearing his doubt after 1 year
Your lectures are really amazing sir ! I used to hate this subject before I watched ur videos ! Atleast now I am comfortable with it !
its my exam today and at 3:00 am I am listening to your lecture ......... thnks sir
And me at 5:00 am
M1 at 237
I wish I found your videos sooner. This helped me so much thank you so much for making these videos
all your lectures are superb sir.
thanks a lot!!
Woww my favorite channel
Thank you so much❤
I was struggling for 3 hours with a problem. Thank God I watched this video!
One minute into the video and ta-da ! Problem solved. :)
All the lecturers r awesome...I had cleared all my doubts.... specially I liked the flip flops lectures the most.... thanks a lot to neso academy
Sir, you should also mention that we are observing Q as the output...(in the neg-pos edge triggered section in the beginning of the video).
And thank you very much for such great videos..appreciate your effort!
man you are awsummm.. thanx alott fr videos.. bhot duaa milega humlog ka.. God bless
Thank you sir, you have exceptional teaching skills!
you drew the NOT gate symbol in the wrong direction
thank you so much sir these videos are so much helpful
i don't have any words to thank u
Pakka explanation..... awesome.thank you sir
Thanks sir .Watching from Nepal
Now I'm confident about my interns. Thank you
At time 7:11Not gate symbol is reversed and also 7:21 is also reversed. Explanation is very good.
the lectures are very interesting, THANKYOU very much for the videos
Good morning sir.
You have explained that we have to clear the counter after 1001 (9).
So upon the arrival of 1010 we clear the counter.
My question is when
Qd Qc Qb Qa (1010) has arrived, we have that count (which was to be eliminated ), since
Qd Qc Qb Qa are our outputs.
Is that count momentary bef the FFs are cleared? Or the action of the clear is so fast that the FFs are cleared immediately?
Am I missing something?
Thank you for the lectures.
same question
They are asynchronous so output will not wait for next trigger and went off
bache ki jaan bachali apne bauji ,dhanyawad :))
In my book it says to draw kmap for output states then draw circuit for clear input. Here we are taking just the first invalid state
thanks this qs came in exam but with a twist qs was
design a ripple decade counter using jk ff without clr and preset signals
A small Mistake is that when using CLEAR (Qd,Qc,Qb,Qa)=0 then when you are using NAND gate inputs should not be Qd and Qb instead Qd NOT And Qb NOt.
You have explained that we have to clear the counter after 1001 (9).
So upon the arrival of 1010 we clear the counter.
My question is when
Qd Qc Qb Qa (1010) has arrived, we have that count (which was to be eliminated ), since
Qd Qc Qb Qa are our outputs.
That is exactly my question i think the number 10 would also be displayed since we have 1010 before resat???!!
It is indeed a fabulous lecture ..which content clear ...but only one correction at the en d..for the two input nand gate you must take qc and qa as 1 nand 1 is 0 but in the lecture the two inputs are qd and qb whose nand is 1 and clr would be 1 ..And sorry for poiniting out such a silly mistake for a great lecture . And shouldnt we stop at 1001 and not at 1010 , because 1010 might also be printed , becoz we are print qd,qc , qb, qa?
Correct me if I am wrong
Thank you for best teachings...
1 question : why not to use and gate instead of nand gate?
Thank You Very Much! Your videos have helped me alot.
Tqs bro u helped me my exam will start 10:00 clock
Excellent logic bhayyaa
Nice explanation
Thank you so much sir
Your videos helps me alot
The first part of the video created a great confusion for me
As you had used negative edge triggered FF to make a down counter where Q was the input and Q' was used as clock
But as you did not mention here in this video that you considering Q as the output for all....It made me think for a very long while.
Hope this comment helps someone else too.
i suppose he did not mention it here because he's mentioned it in the previous 3/4 bit async UP counter videos.
earlier in down counter video, you used negative edge triggered and Q as clock, you have not mentioned that in this lecture?
You really work hard, I'm worried because I clear concept here but what about semester exam we've to write more and how will we write because we don't have content?
thanks, now i am understanding the working of computer
Thank you so much sir. Your videos have be very helpful in my studies. It's just that we cannot find any serial order in your videos which makes it difficult to find which video is next. If you could add a serial order in your videos it would be very helpful.Thank you.
Thank you for the important notes in the beginning of the video. That's what the lecturer did not explain which confused everyone LOL
This is very helpful.Thank you sir
The besttt teacher ever
awesome lectures sir
he give 2 clock inputs a and b ,,output of 1st flip flop not given to second as clock,,he gives 2nd input to second flip flop and then ckt is similar,,an logic used that is also different
Excellent 👍
you are a gift from the god to us
These videos are really appreciating but in BCD counter we have to use truth table and k - maps while making the mod 10 counters.
I am 62 years old and I follow digital design as a hobby.I want make a JK FF counter that gives the sequence 10,22, 5.4.3. I have seen a lot of your interesting videos but I cannot figure out how to do it. Can you give me a hand ? Theodore from Greece.
What type of counter do we get if flip flops include both negative and positive triggered flip flops
it is very easy method for all ece students
number 1 content.
sir pls i have doubt about it,,is it ic 7490 ripple counter,,because when I reffer ic 7490 ripple counter internal diagram it's seems like little different from u mentioned in video pls suggest fast i have exams 2 days after
you are very good sir thanks
sir, if we use +ve edge triggered clock then what about race around condition ??
you taught that to overcome race around condition we use -ve edge triggering .
+sumit shekhar LPC
It takes place in level triggering
To overcome it we use 3 methods
1)(T/2)>FF propagation delay
2) edge triggering and
3) master slave JK flip flop
As clear is active low signal so it will work when clr=1 because when clr=1 then input to ff from clr =0 and output=0
So according to this concept the output of the nand gate should =1 to active the clr input but you are doing opposite to this?
Please clear my doubt
Am I wrong or you taught wrong?
In our college they are using and gate ....pls reply
in 2:09 when you are discussing positive edge trigger you are telling about the clock but not about the output clear from where you are taking the output when positive edge triggring
Not gate symbol is inverted here in NAND gate. I think you must reverse the symbol
Thanks for your help
Excellent.
When to take output in counter with respect to clock pulse in negative edge triggered flip flop??
1)At the start of clock pulse or
2)At the end of clock pulse
great jop sir that's amazing
U r great sir....
you are great master
sir u r awessome ..
Sir i think in the diagram...direction of not gate will be opposite
At 7:13 minute of the video why did you put the value of Qc and Qa by using a inverted NOT gate and taking the value to NAND Gate
if possible make videos on ptsp (probability theory and stochastic process) subject
Great
Can you explain any explain in which clear is applied to particular filp flop.. Eg if 4ff are there and in only 2ff clear is used.
Hi , can u please also explain us the applications of counters ,registers n other flip flops in our daily life
AMIT KUMAR Meena they are used in computers or various devices like watches
+AMIT KUMAR Meena bro i can bulit traffic light signal by using counters
Hello sir, i am a really big fan of your work. I really have a question, how can you make the counter starts from a specific number? like If i.m willing to start counting from 12 to 63 for example
you code it!
"verilog"
good explination man..
i understand why we removed Qa and Qc but how did we know they are not gate?
sir in case we have 1100 as input to nand gate ... Qd=1 and Qb=0. by taking there nand gate we have output equal to 1 so our clr input goes to state 1 thus our counter is not goingto be reset in this case..... plz clear this confusion as soon as possible bcz i have exam in next week
Someone - Hei. What is your campus?
My mind - Neso academy
So for ripple counters, all we need to do is determine the number of states we have in the count sequence and then attach the clocks according to the type of edge trigger? Do we use synchronous counters to implement the circuitry for sequences that skip states like 0, 1, 2, 4, 6?
Yes we use synchronous counters to implement the circuitry for sequence that skips states
Please release video about timing diagram & truth table for 3 bit synchronous counter
Sir, since we do not want 1010 to appear as the output, shouldnt we feed 1001 as the input of the NAND gate? If we reset after seeing 1010 output, dont we somehow count from 0000 to 1010?
Yes we will in fact see for a short period of time 1010 as the output. To solve this problem we can connect the Clock-Enable input of the 4th Flip Flop with the output Qa of the first flip flop. Then we need to connect the J input of the 4th Flip Flop to an AND gate of Qb and Qc. As a result the Qd changes state only when Qa drops from 1 to 0 and Qb=Qc=1and resets to 0 when Qa drops from 1 to 0 every other time (J=0,K=1) .
is that up/down? or is it just up counter?
Why all ff circuites are in negative edge triggering? Any reason
Sir, but you used Q as clock in the down counter too. Then why did you say to use Q complement as clock.
in that case you have to use Q complement as an output if you use Q an clock in down counter.
its more convenient to stick with one output(Q) and just change the clock form Q to Q complement to switch between up and down counter.
If the preset is not connected then will it affect the circuit??
Why we are restricting to 10 when we are having chance up to count 16.
for positive edge triggered FF the race around condition will appear.How could it be used as counter?could you please explain.
What is the equations for j, k input? Because if i apply k-map the answers are different then =1
to which ff's we should give input to nand gates...you forgot to draw
the outputs have to be 1010 for not gate to reset the contour so since the outputs are 1010 does that not mean we are already showing number 10 on the display before resting???
please help Im really confused about this. Thank you all.
Hello, if I have a asynchronous/ripple D flip-flop counter with a count sequence of 10, 9, 8, 7, 6, 5, 4, 3 and repeat but with provition for a logic 0 output on 8 and a logic output on 6, how exactly would I go about with this kind of a question. I understand that I will use the positive egde triggered and Q would be the clock. I would the connect the NAND gate with input 3 to clear to ensure that it will repeat the sequence but what of the provitions and would the sequence start at 15 and go down or would I also need to connect a NAND gate to PRE so it starts at 10. Any help would be much appreciated.