Course Work Question

[tarasmuz]

Hello,

I'm really confused with my 555 astable. I want it to count 3600 seconds (1 hour), but it can't do it. So I decided to use the 10 seconds time period (Resistor R1 = 10000KΩ, Resistor R2 = 22KΩ) and to have a frequency of 360. But I'm not sure how to 'manage' the frequency of my 555 and I'm not sure if my plan is possible at all. So can some explain it to me ?

Thanks to everyone who will help me

 

[MikeMl]

Have you read and understood the data sheet? All of the limitations (max resistance, Capacitor leakage) are discussed in the "applications" section. Also, look into the CMOS version of the 555 to get longer delays.

 

[tarasmuz]

Have you read and understood the data sheet? All of the limitations (max resistance, Capacitor leakage) are discussed in the "applications" section. Also, look into the CMOS version of the 555 to get longer delays.

The data sheet I have doesn't say anything about frequency and I'm not allowed to use CMOS version of 555

 

[MikeMl]

To get longer delays, use the 555 as a clock to a binary counter. Or use a CMOS 4060 chip.

Read this.

 

[tarasmuz]

Can I not use what I suggested?

 

[MikeMl]

You didn't explain how you were going to get from a period of 360 seconds (not frequency) to a period of 3600 seconds.

 

[tarasmuz]

Well, I thought that I will just make a time period of 10 sec and then change my frequency to 360 somehow, but I was not sure how to change the frequency so I posted the question here

 

[MikeMl]

Frequency = 1 /Period

Period= 1/Frequency

They are not independent.

The "frequency" of an astable multivibrator running at a period of 360 seconds (6 min) is 1/360s = 0.0027777 Hz

What you need to do is "count" 10ea. 360s periods (multiply the period by ten, same as dividing the frequency by ten).

 

[meowth08]

Hi,

Maybe what tarasmuz is saying is that he would make a 360 Hz which is 360 cycles per second for 10 seconds.
That would give him 3600 cycles in total since (360 cycles/ second) * (10 seconds) = 3600 cycles.
Maybe he's also confused with period and frequency.
Period is the time it takes to complete a positive and negative alternation.
Frequency is the number of cycles that can be measured in one second time.

 

[tarasmuz]

Well, yeah. That's what I'm saying, but I'm really confused as I don't really know how to make it

 

[meowth08]

Doing what you are thinking would not count for 3600 seconds.
It would give you 3600 cycles.

But if it is what you need, you should use two 555- one astable and one monostable.
Make a monostable that would have an on time of 10 seconds.
Then, make an astable with frequency of 360 Hz.

Figure out first how to do this from the application note of the datasheet providded by sir MikeMl.

 

[MikeMl]

What about a clock generating pulses, and then counting them with a digital counter don't you understand?

 

[tarasmuz]

Thanks a lot, this helps.

 

[tarasmuz]

I don't really get it and for me meowth08's method looks easier. Anyway - Thank for help!

 

[meowth08]

I don't really get it and for me meowth08's method looks easier. Anyway - Thank for help!

It's not my method. I just tried to explain what you were thinking.
From the beginning of the thread, you were unable to explain to us the actual requirement.

From your first post, you mentioned you wanted to count 3600 seconds but the solution that you have given will not give you 3600 seconds.
It will only give you 3600 cycles for 10 seconds. If you get the period (time to complete a cycle), it would be (10 divided by 3600).
For the frequency (number of cycles in one second), it would be (3600 divided by 10).

You will not realize that your method of solving your problem is wrong unless you understand concept of period and frequency.
If you have time, please provide drawings of the circuit on your mind so we know how to help you.
Sir MikeMl has already given you the starting procedures.

Tell us which part you don't understand.

 

[ClydeCrashKop]

There are a couple 555 timer calculators here http://www.csgnetwork.com/electronicsconverters.html

 

[alec_t]

If your 555 is set up to provide output pulses with a repetition period of 10 secs then you will have to use multiple counters to count 360 pulses from the 555 if you want a 1 hour timeout. Because 360 is not a power of 2 you would not be able do the counting with a single readily-available binary counter. It would be more convenient to set the 555 to give a pulse repetition period of, say, 3600/2^14 = ~0.22 sec, then use a readily-available 14-stage binary counter (e.g.CD4020).