4017 Decade counter

[vick5821]

Hi,

This is a 4017 decade counter.

&

This is a simple traffic light project using 4017 decade :

I would like to ask,
What does these mean :
1.For some functions (such as flash sequences) outputs may be combined using diodes.
2.why the reset is connected to the negative of a capacitor　?
3.so it provides an output at 1/10 of the clock frequency
4. The ÷10 output is high for counts 0-4 and low for 5-9, so if I connect one LED to this output, then I connect another 5 LED to Q0 - Q4, then then one LED will be high from Q0 to Q4, then the 5 LED will take turn to be high from Q0 , then Q1, then Q2, then Q3, then Q4 ? Am I correct ?
5. The 4017 counter is connected from the output of IC 555, means that when the output is HIGH, it will make Q0 HIGH ? then when low, Q0 low ? then when next 2nd round HIGH, it will make Q1 HIGH ? Is it like this ?

Thank you

 

[jackorocko]

why the reset is connected to the negative of a capacitor

It is an RC circuit, when first powered on the capacitor will look like a short and so the reset pin will be held high until the capacitor charges up and then it will be low. Google RC circuit if you want to learn more. With no certainty I would assume it is meant to wait for the 555 output to stabilize and set all outputs to LOW before it starts counting.

The ÷10 output is high for counts 0-4 and low for 5-9, so if I connect one LED to this output, then I connect another 5 LED to Q0 - Q4, then then one LED will be high from Q0 to Q4, then the 5 LED will take turn to be high from Q0 , then Q1, then Q2, then Q3, then Q4 ? Am I correct ?

I think that sounds right, look here for some visuals. http://www.doctronics.co.uk/4017.htm

The 4017 counter is connected from the output of IC 555, means that when the output is HIGH, it will make Q0 HIGH ? then when low, Q0 low ? then when next 2nd round HIGH, it will make Q1 HIGH ? Is it like this ?

No, look at the above link. It must count to 10 before pin Q0 is HIGH again.

so it provides an output at 1/10 of the clock frequency

Yes, it is what they call a frequency divider. In this case it is dividing by 10. If you have a 10Hz input clock freq you will have a 1Hz freq output

 

[vick5821]

It is an RC circuit, when first powered on the capacitor will look like a short and so the reset pin will be held high until the capacitor charges up and then it will be low. Google RC circuit if you want to learn more.



I think that sounds right, look here for some visuals. http://www.doctronics.co.uk/4017.htm

I got some knowledge on RC circuit

When it is connected to DC power supply, the capacitor will be charge up, after a long time, the capacitor will acts like a open circuit to DC current ?

 

[Rleo6965]

The purpose of diodes connected to Q output was to Turn On LED in the event that any Q output goes High.

 

[vick5821]

The purpose of diodes connected to Q output was to Turn On LED in the event that any Q output goes High.

You mean how ? Why I must connect them ? cant I connect resistor ?

 

[vick5821]

Yes, it is what they call a frequency divider. In this case it is dividing by 10. If you have a 10Hz input clock freq you will have a 1Hz freq output

So, the input clock is from the output of IC 555 ?

 

[jackorocko]

When it is connected to DC power supply, the capacitor will be charge up, after a long time, the capacitor will acts like a open circuit to DC current ?

Yes and if you think about this a little more that would mean the pin is connected to the ground through a resistor that would be acting like a pull-down resistor.

 

[duke37]

The reset pin needs to be high to reset the counter so that all outputs are at 0 and the counter starts at zero. A time delay on this pin means that the power is up before the reset goes off.

In this application, a reset is not needed. The counter will start with maybe one or more outputs high but will reset itself once it has been round once.

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.

I have built a similar circuit and showed the circuit on another post yesterday

 

[Rleo6965]

Vick.

Use the Circuit Wizard from the link I gave you.

 

[vick5821]

The reset pin needs to be high to reset the counter so that all outputs are at 0 and the counter starts at zero. A time delay on this pin means that the power is up before the reset goes off.

In this application, a reset is not needed. The counter will start with maybe one or more outputs high but will reset itself once it has been round once.

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.

I have built a similar circuit and showed the circuit on another post yesterday

I cant figure out this ><

 

[Rleo6965]

I cant figure out this ><

You won't. Unless you simulate it w/ the Circuit Wizard.
Use it. It took me 2 days to find crack version for you without the virus,

 

[vick5821]

Yea, I simulated them..Still figuring.The connection got a little bit different from mine..

So need time to study it

 

[vick5821]

can decade counter count and LIGHT up LED on its own without combine with IC555 ?

Thank you

 

[(*steve*)]

can decade counter count and LIGHT up LED on its own without combine with IC555

Well, strictly speaking, yes. But the LED which is lit won't change without a source of clock pulses. The 555 is *a* method of supplying a series of clock pulses. The 4017 can't do it by itself.

 

[Rleo6965]

Any clock source can be use to the input of 4017. Even a mechanical switch can be use.
But I'm not sure if 4017 High Output current can drive the LED. Never tried it before.

 

[vick5821]

Well, strictly speaking, yes. But the LED which is lit won't change without a source of clock pulses. The 555 is *a* method of supplying a series of clock pulses. The 4017 can't do it by itself.

Oh, so you mean, if I do not have the clock pulse, the LED wont take turn to be HIGH ? It will be like just lights up ? Means all output HIGH when switch ON ?

 

[Rleo6965]

"Know your enemy before you conquer".

Read the datasheet to get more information about cd4017.
http://www.alldatasheet.com/datasheet-pdf/pdf/50846/FAIRCHILD/CD4017.html

 

[(*steve*)]

No.

*An* output will be high on power on. It won't change without a clock pulse (or a reset pulse)

 

[vick5821]

No.

*An* output will be high on power on. It won't change without a clock pulse (or a reset pulse)

Ok..so it is necessary to connect an IC555 in astable mode to giv the pulse.

Thank you

 

[jackorocko]

Ok..so it is necessary to connect an IC555 in astable mode to giv the pulse.

absolutely not! But it is probably one of the easiest ways to attached a clock pulse.