How do you make a digital counter circuit which counts to 10 minutes, it will rings a buzzer and when it continues to 15 minutes, it will rings a buzz

Thread Starter

fahadnajed

Joined Oct 25, 2019
8
I need to make a digital counter circuit which it counts to 15 minutes for my big project in my campus . When the circuit counts to 10 minutes, it will rings a buzzer. Same when it continues to 15 minutes, it will ring a buzzer again. (the circuit is for public speaking, so the speaker knows that he or she already went speak for 10 minutes and 15 minutes). We weren't allowed to use an Arduino, Microcontroller, PLCs, IC NE555, and a 4000 series CMOS IC but we were allowed to use breadboard, flip flop, decoder and 7 segments display. I'm kinda stuck here and I'm thinking it's impossible to make such a thing. Do you have any idea guys?
 

dl324

Joined Mar 30, 2015
9,150
Welcome to AAC!

I gather that this project is something for which you'll receive a grade. On AAC, we consider any school work for which you will be graded to be homework. Forum guidelines in that case frown on members handing out solutions because that's cheating.

Show your ideas, either in the form of a block diagram or schematic. What were you thinking about using for the time base? Can you use other logic gates? Can you use discrete components?

What happens when the timer gets to 15 minutes?

Are there restrictions on what flip flops, decoders, and displays you can use? You must be able to use other logic gates because you can't decode the time without them.

It might be helpful if you posted the full text of the problem so we don't have to play 20 questions.
 

Papabravo

Joined Feb 24, 2006
12,526
You first need to decide what time units you are going to count. Will it be jiffies (1/60th sec)? Will it be seconds?
Then you need to figure out how many of those basic unit you are going to count.
So if it is seconds you need to count 600 to get to 10 minutes and you need another 300 seconds to get to 15 minutes. Let's round that up to 1024, a power of 2, and say that you need a ten bit counter. Why ten bits? So you can represent any number from 0 to 1023. Then you need a binary decoder to tell you when the counter gets to 600 and another decoder to tell you when it gets to 900. You probably also want a reset to start the whole process over again.

Is all of that clear?
 

absf

Joined Dec 29, 2010
1,942
No 555 timer - fine
No 4000 series logic - fine
Are you allowed to use 74LSxx series logic chips?

Without using 555, I would start to design a 60 seconds timer using TTL logic chips or transistor astable oscillator. When the basic timer is done I'd make a divide by 10 and divide by 15 circuit and use their output to trigger the buzzer via a transistor.

The display of time would be done last.

Allen
 
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Thread Starter

fahadnajed

Joined Oct 25, 2019
8
No 555 timer - fine
No 4000 series logic - fine
Are you allowed to use 74LSxx series logic chips?

Without using 555, I would start to design a 60 seconds timer using TTL logic chips or transistor astable oscillator. When the basic timer is done I'd make a divide by 10 and divide by 15 circuit and use their output to trigger the buzzer via a transistor.

The display of time would be done last.

Allen
How do you explain these? I'm pretty sure I got a little bit brightened. How do you divide the 10 and 15?
 

Thread Starter

fahadnajed

Joined Oct 25, 2019
8
You first need to decide what time units you are going to count. Will it be jiffies (1/60th sec)? Will it be seconds?
Then you need to figure out how many of those basic unit you are going to count.
So if it is seconds you need to count 600 to get to 10 minutes and you need another 300 seconds to get to 15 minutes. Let's round that up to 1024, a power of 2, and say that you need a ten bit counter. Why ten bits? So you can represent any number from 0 to 1023. Then you need a binary decoder to tell you when the counter gets to 600 and another decoder to tell you when it gets to 900. You probably also want a reset to start the whole process over again.

Is all of that clear?
Yes, it's clear. So basicallly, it will be seconds. A 10 bit counter, huh? I see now. Is it seperate the counter and the decoder? Because we get restricted if the IC is built-in Decoder and Counter. I see how that works. I may get brightened by these ideas.
 

Thread Starter

fahadnajed

Joined Oct 25, 2019
8
Welcome to AAC!

I gather that this project is something for which you'll receive a grade. On AAC, we consider any school work for which you will be graded to be homework. Forum guidelines in that case frown on members handing out solutions because that's cheating.

Show your ideas, either in the form of a block diagram or schematic. What were you thinking about using for the time base? Can you use other logic gates? Can you use discrete components?

What happens when the timer gets to 15 minutes?

Are there restrictions on what flip flops, decoders, and displays you can use? You must be able to use other logic gates because you can't decode the time without them.

It might be helpful if you posted the full text of the problem so we don't have to play 20 questions.
I haven't think that yet. Yes! Sure, I can use logic gates and discrete components. So basically, when the circuit goes to 10 minutes, it will give a signs a buzzer (say an LED which acts as a signs), and when it continues to 15 minutes, it will do the same like how the circuit goes to 10 minutes. In this case, the unit minutes is in seconds. Now, I didn't think about will it be stop at 15 minutes or will it be continue? Well, let's say it will continue, because I don't know how to stop exact at 15 minutes.

Any type of flip-flop (SR/D/JK) wether it's counter UP or DOWN, no restrictions on decoders and 7segments display. Mainly, the decoder and counter must be seperate, not built-in IC counter and decoder, but it must seperate.

Yes, you are right. I probably should post the requirements.Capture7.PNG
 

absf

Joined Dec 29, 2010
1,942
How do you explain these? I'm pretty sure I got a little bit brightened. How do you divide the 10 and 15?
Use 4 flip flops (either D or JK) of course. when the outputs are "1010" and "1111", that would be 10 min and 15 min respectively. Use 74LS74 for DFF and 74LS76 for JKFF. I used 74LS393 for simplicity.

There are plenty of this type of circuits in google image. Just search for them. Use AND and OR to decode the outputs you want.

Allen
 

Papabravo

Joined Feb 24, 2006
12,526
The counter and decoder will be separate. If you were to implement this in an FPGA, then they would still be separate blocks inside, exactly as if you were building them with discrete components.
BTW 600 = 0b1001001000 = 0x248 & 900 = 0b1110000100 = 0x384
 

ci139

Joined Jul 11, 2016
806
it depends on your timer output

for example if you use the minute timer -- does not matter (your own choice) whether it's a comparator- or a flip flop type or ther one achieved by frequency divider (counter)) -- then you can use a binary decade counter and SN7474-s or random binary modulus 2 counter might be well another decade counter
___________
the "program" (assuming using SN7490-s or SN74192-s)
_____
7490 (bypass 1 stage use MOD5 counter only -- note : it will be counting from 000 to 100 binary ) using a minute interval clock
note : with 74192 you can't have distinct LSB function but you can have MOD5 output by "2-ble stroke" clock action --or-- tile up the
mod5 from 74x107example of // circuit layout

Digital - Logic - Test - S-7111.png





(-1-) count to 5 ← the first¹ 7490 - the second² 7490 → 2 times // outputs ²010x¹000x
. . . . // ² ¹ mark CT outp.-s /// the bits x must be set to 1by the rising edge of Q.D i must verify it can be done so as i never used 7490 ...
. . . . // ...
according to schematic the 7490's upper CT5 works just fine alone and does not iterfere wit it's LSB function !!
. . . . // you actually ring the bell here when 10' has just passed e.g. when bit 5 has first rizen

(-2-) count to 5 ← the first¹ 7490 - the second² 7490 → 1 more times // outputs ²011x¹000x
. . . . // you now ring the bell here when 15' has just passed e.g. when bit 5 was previously rizen and at that condition bit-4 was risen

for alarm control an additional logic is required -- as it comes out the lowwer bit-s / MOD2-s can be used to this . . .

(-3-) you need to implement the buzzer/ring circuitry ↑preferrably using signals of what you already have↑
__________
maybe there're some hints : video about 74LS90
 
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dl324

Joined Mar 30, 2015
9,150
Any type of flip-flop (SR/D/JK) wether it's counter UP or DOWN, no restrictions on decoders and 7segments display. Mainly, the decoder and counter must be seperate, not built-in IC counter and decoder, but it must seperate.
Your stated requirements in the original post were misleading and the instructions aren't exactly clear.

From #2, you can't use Arduino, microcontrollers, NE555 (or equiv), or counter IC's. In #3, the use of "exemptionally" for up/down counter is syntactically incorrect (and it isn't a word). Exemptionally would be an adverb and you need a verb for it to modify.

The instructions are vague, but I think the intent is for you to design a counter using flip flops and logic gates.

Since this is homework, you need to decide how you want to implement the circuit. We can only offer guidance. Whether you want to count seconds or minutes is up to you, but your choice will affect the amount of logic you need to implement your solution.

You didn't include the description of the desired operation. That should tell you what you're supposed to do after 15 minutes.
 

absf

Joined Dec 29, 2010
1,942
That's why I suggested flip flops instead of counters. But without the 7 seg decoder chip it would be tedious to construct it from discrete logic gates.

The 7 seg decoder should be a lesson on its own. To make the 0.5S time base, I used 74LS14 schmitt trigger inverter gates.
 

ci139

Joined Jul 11, 2016
806
you can double the clock to get Mod5 from decade counter by picking the pre-last bit from previous counter instead of last
or
then add unnecessary complexity as what i first attempted before realizing (such being interesting but not required at this time >.< )
this actually quadruples the clock (?making a MOD 2.5 ??? ↓confirmed below↓)
if you leave the lowermost inverted C (clk)
off then it doubles (making a MOD 5 counter)
Digital - Logic - Test - S-7110.png(!"¤!%"¤& i'm still confused about)
Code:
0 1 2 3 4 5 6 7 8 9 A B C D E F
_______|¯|_______|¯|_______|¯|_
___|¯¯¯|_____|¯¯¯|_____|¯¯¯|___
_|¯|_|¯|_|¯|_|¯|_|¯|_|¯|_|¯|_|¯
_|¯|_|¯|_|¯|_|¯|_|¯|_|¯|_|¯|_|¯
0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7

MOD = N.cyc.in / N.cyc.out = 7.5 / 3 = 2.5
 
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Thread Starter

fahadnajed

Joined Oct 25, 2019
8
it depends on your timer output

for example if you use the minute timer -- does not matter (your own choice) whether it's a comparator- or a flip flop type or ther one achieved by frequency divider (counter)) -- then you can use a binary decade counter and SN7474-s or random binary modulus 2 counter might be well another decade counter
___________
the "program" (assuming using SN7490-s or SN74192-s)
_____
7490 (bypass 1 stage use MOD5 counter only -- note : it will be counting from 000 to 100 binary ) using a minute interval clock
note : with 74192 you can't have distinct LSB function but you can have MOD5 output by "2-ble stroke" clock action --or-- tile up the
mod5 from 74x107example of // circuit layout

View attachment 189753





(-1-) count to 5 ← the first¹ 7490 - the second² 7490 → 2 times // outputs ²010x¹000x
. . . . // ² ¹ mark CT outp.-s /// the bits x must be set to 1by the rising edge of Q.D i must verify it can be done so as i never used 7490 ...
. . . . // ...
according to schematic the 7490's upper CT5 works just fine alone and does not iterfere wit it's LSB function !!
. . . . // you actually ring the bell here when 10' has just passed e.g. when bit 5 has first rizen

(-2-) count to 5 ← the first¹ 7490 - the second² 7490 → 1 more times // outputs ²011x¹000x
. . . . // you now ring the bell here when 15' has just passed e.g. when bit 5 was previously rizen and at that condition bit-4 was risen

for alarm control an additional logic is required -- as it comes out the lowwer bit-s / MOD2-s can be used to this . . .

(-3-) you need to implement the buzzer/ring circuitry ↑preferrably using signals of what you already have↑
__________
maybe there're some hints : video about 74LS90
This is so great. But, I'm still trying to understand, so in the schematics you draw in LTSpice, that three flip-flops, which IC you are using? is it 74LS90 you saying that SN7490? so what you mean by SN7490 and SN74192 is 74LS90 and 74LS192? Isn't? (I'm really sorry for asking too many questions, I'm trying to learn these ideas from all of you but I feel like I'm so stupid)
 

dl324

Joined Mar 30, 2015
9,150
(I'm really sorry for asking too many questions, I'm trying to learn these ideas from all of you but I feel like I'm so stupid)
Since this is homework, you should be trying to come up with a design on your own. Members can give guidance, but they're not allowed to give you solutions.

Show us your attempt and we can tell you if you're headed in the right direction.
 

ci139

Joined Jul 11, 2016
806
Since this is homework, you should be trying to come up with a design on your own
I agree, but we should also realize, that for the "first-timer" getting around with the poor specifications and often even the typos or miss-references - the task is near impossible even for more experienced people . . .
___________
that three flip-flops
if you mean the post #10 -- try the last link from 74x107example of // circuit layout ◄◄ ← if you zoom in they got 74HC73 printed on it

but they both seem to have the identical function - the SN74x107 and 74HC73


The 74HC73 is a dual negative edge triggered JK flip-flop with individual J, K, clock (nCP) and reset (nR) inputs and complementary nQ and nQ outputs.
The SN74107 contain two independent J_K flip-flops with individual J_K clock, and direct clear inputs. The 107 is a positive pulse-triggered flip-flop (= falling edge triggered = negative edge triggered . . . ← but that's where the fun begins because you need to verify that this is the case in practical experiment).


 
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dl324

Joined Mar 30, 2015
9,150
getting around with the poor specifications and often even the typos or miss-references
Students should ask their instructor questions when instructions are unclear.

The student has made no attempt to show us his/her attempts. Until then, we shouldn't start designing things for him/her.
the task is near impossible even for more experienced people
Seems like a fairly simple problem to me.
 

ci139

Joined Jul 11, 2016
806
Seems like a fairly simple problem to me.
i mean in general -- as with analog designs -- as and with the digital or mixed -- if you try to suck out the sense from carelessly written datasheets.. (when the only reason for the existence of such seems to be that they must exist -- but not in order to provide any consistent technical information . . . well that's the worst case) .. -- then you'll soon learn that random amateur project site in www provides a better reference . . . unfortunately - - - You can try your luck with LM308 equivalent circuits -- very interesting zoo you can find from d/s-s of the different manufacturers of it
 
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ci139

Joined Jul 11, 2016
806
I'm really sorry for asking too many questions, I'm trying to learn these ideas from all of you but I feel like I'm so stupid
Don't be sorry to ask questions
__

If you feel "stupid" insecure about the area in question ← then that is where you estimate/hypotize about your capabilities -- in simple terms you tell yourself to be cautious about your decisions = you simply need to verify the work you've done ← that until you get more acquainted with the field

but still i always multiple check my work -- you have a one viewpoint to it while your "in it" (doing it) -- then , after a while your mind will tile it up from the different perspective revealing perhaps the detail it couldn't notice before -- etc. ...
 
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