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4017 Decade counter

jackorocko

Apr 4, 2010
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Ok..It is not necessary but it is one of the ways to have the clock pulse given to the 4017 right ?:)


No, re-read what I said and what steve said. YOU MUST HAVE A CLOCK PULSE for the 4017 to work, but a 555 is most certainly not the only way to provide a clock pulse.

edit: I ain't sure how to take your comment. But the 4017 needs a clock pulse as steve said. Sorry if I misinterpreted what you meant
 
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vick5821

Jan 22, 2012
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No, re-read what I said and what steve said. YOU MUST HAVE A CLOCK PULSE for the 4017 to work, but a 555 is most certainly not the only way to provide a clock pulse.

edit: I ain't sure how to take your comment. But the 4017 needs a clock pulse as steve said. Sorry if I misinterpreted what you meant

What I means is that there are many ways to give clock pulse to 4017, and ic 555 is one of the ways, right ?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Yes, that is correct.

If you have a 555 that will do.

There's an even easier way with Schmitt trigger inverter, but even though you only use a single 2 pin function, the chip has 6 of them and is consequently a 14 pin chip. 555 is an 8 pin chip and seems simpler.
 

vick5821

Jan 22, 2012
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Hey, whats for we connect diode to combine two or three output Q?
 
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vick5821

Jan 22, 2012
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Same here.:D

Vick post your diagram. I hope we won't be discussing internal circuit of IC.:D

4017-1.png


The circle part in the diagram is what I referring to..why do we connect diode to combine the output pins? Can I just connect them using normal wire ?
 

Rleo6965

Jan 22, 2012
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4017-1.png


The circle part in the diagram is what I referring to..why do we connect diode to combine the output pins? Can I just connect them using normal wire ?

This was already answered by duke37.
The diodes are necessary since if one output is high and another is low, they will fight to determine the output voltage. The diodes give a high output when any output is high.

Think and use your imagination.

What will happen if you replaced 4 diodes with jumper wire.? You will be shorting Q5,Q6,Q7,Q8 and damage the 4017 ic.
 
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Rleo6965

Jan 22, 2012
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Oh you want detailed or step by step explanation. Let's wait for other members to post. They have better english and knowledge than me.:)
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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The circle part in the diagram is what I referring to..why do we connect diode to combine the output pins? Can I just connect them using normal wire ?

Good question Vick5821.

Remember that CMOS outputs can source and sink current.

If you connect 2 of them together and they have different logic levels, you will force current from one to the other. This is almost always *not* what you want.

The diode ensures that current cannot flow back into the other outputs, and effectively wire-or's the outputs.

You should also look at open collector (or open drain) outputs as they provide another solution to this problem, although it soed not apply here because the 4017 has normal outputs.
 

vick5821

Jan 22, 2012
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Good question Vick5821.

Remember that CMOS outputs can source and sink current.

If you connect 2 of them together and they have different logic levels, you will force current from one to the other. This is almost always *not* what you want.

The diode ensures that current cannot flow back into the other outputs, and effectively wire-or's the outputs.

You should also look at open collector (or open drain) outputs as they provide another solution to this problem, although it soed not apply here because the 4017 has normal outputs.

Perhaps intially the current will flow to Q0, but if I didnt use a diode to ensure in one direction flow, it will cause the current to be output through like for example Q2 ? Is this what you meant by steve?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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In a 4017, only a single output is high -- think of it as connected to V+

All the other outputs are low -- think of them as connected to V-

Imagine what happens when you connect V+ to V- -- because that's what you do when you connect two logic outputs together that have different levels.

Fortunately for your power source, the outputs have significant resistance, but you should still be able to see what happens.
 

vick5821

Jan 22, 2012
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In a 4017, only a single output is high -- think of it as connected to V+

All the other outputs are low -- think of them as connected to V-

Imagine what happens when you connect V+ to V- -- because that's what you do when you connect two logic outputs together that have different levels.

Fortunately for your power source, the outputs have significant resistance, but you should still be able to see what happens.

Ohhh...so you mean if I use wire to connect both output together, I will get the +V and -V connected together ? If I use a diode, it will avoid this from happens by ensure the current flows in one direction ? Correct steve ?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Pretty much. You ensure that the only valid path is the one from Vcc, through the diodes, then the load, to ground.
 

vick5821

Jan 22, 2012
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Untitled-2.png


Can anyone explain to me reason for this type of connection for PIN 15(RESET pin) ?

FYI, I connected one LED at Q0 and another LED at Q1 of counter 4017. So when I did not connect the pin 15 in that way, I found out that randon number of clock pulse will lights up LED at Q0 and follow by LED at Q1. Suppose the first clock pulse will lights up the LED at Q0. But when I off my power souce and ON again, sometimes, 5th clock pulse will only lights up the LED at Q0 and sometimes 2nd clock pulse will lights up LED at Q0.

Any reason for that ?

Thank you :)

Image source : http://www.kpsec.freeuk.com/projects/trafficlight.htm
 

CocaCola

Apr 7, 2012
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The cap/resistor forces a single reset with every power up, without this forced reset the counter will not always start at 'zero', you will instead (may) get odd behavior with each power up and a random counting start point...
 
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