please help with my crossroad traffic light project...

[DerStrom8]

i know i read this but can you make understand in a very not technical language

There is not much technical language there, if you read past the sophisticated way of saying it. All it's saying is the following:

1. The 4017 has 10 outputs that you cycle through
2. The input pins include a clock, a reset, and an inhibit
3. The output transitions to the next on the positive clock signal change, assuming Inhibit is low (0v). If the inhibit signal is high (5v), however, it does nothing.
4. Connecting the Reset pin to +V resets the output back to the first one
5. A carry-out pin allows multiple 4017s to be "cascaded" one right after another, to allow more outputs to be controlled by one clock.

Now that's not too complicated, is it? Just read the text carefully, and don't be daunted by the fancy speech

Regards,
Matt

 

[D@rkL0rd]

Here is a quick simulation I just ran to show you how the 4017 works. Obviously in the real one you'll want resistors to limit LED current and whatnot, but this should show you how it works overall. Notice that the LED changes when the input (clock) goes HIGH (the small square next to the clock input turns red).

thnx

 

[D@rkL0rd]

I guess I am a geek because I talked in a technical language with my peers for most of my life.
Maybe someone else can patiently teach you the basics of electronics.

i know i just a newbie here...but never mind i understood it...now i know that i need a high time pulse for .5 sec and low time pulse for 0.5...thnx for your time...

 

[D@rkL0rd]

There is not much technical language there, if you read past the sophisticated way of saying it. All it's saying is the following:

1. The 4017 has 10 outputs that you cycle through
2. The input pins include a clock, a reset, and an inhibit
3. The output transitions to the next on the positive clock signal change, assuming Inhibit is low (0v). If the inhibit signal is high (5v), however, it does nothing.
4. Connecting the Reset pin to +V resets the output back to the first one
5. A carry-out pin allows multiple 4017s to be "cascaded" one right after another, to allow more outputs to be controlled by one clock.

Now that's not too complicated, is it? Just read the text carefully, and don't be daunted by the fancy speech

Regards,
Matt

thnx...can you please telll what the inhibit pin does i am really not clear on that... and when we connect reset to +V it rests output after full cycle of ten pins or after first pin and moves in a loop...

 

[DerStrom8]

thnx...can you please telll what the inhibit pin does i am really not clear on that... and when we connect reset to +V it rests output after full cycle of ten pins or after first pin and moves in a loop...

When high, the inhibit pin prevents the output from advancing even if it receives a clock signal.

Connecting reset to +V will reset it back to the first output regardless of which output is set at that particular moment. The way my circuit is set up in the simulation video, I have the last output connected to reset. This way, it will reset whenever the last output is reached, allowing it to restart the loop. You can also connect the reset pin to +V at any time (perhaps via a push-button) to manually reset the chip to the first output.

 

[audioguru]

i know i just a newbie here...but never mind i understood it...now i know that i need a high time pulse for .5 sec and low time pulse for 0.5...thnx for your time...

NO you don't need the clock to have a "high time pulse for 0.5 sec and low time pulse for 0.5 sec". You just need the clock to go high each second so the timing can be high for 0.9 sec and low for 0.1 sec., or high for 0.7 sec and low for 0.3 sec.
It will even advance each second if the clock was high for 0.1 sec and low for 0.9 sec but it is difficult to get a 555 oscillator to Doo Dat.

 

[D@rkL0rd]

NO you don't need the clock to have a "high time pulse for 0.5 sec and low time pulse for 0.5 sec". You just need the clock to go high each second so the timing can be high for 0.9 sec and low for 0.1 sec., or high for 0.7 sec and low for 0.3 sec.
It will even advance each second if the clock was high for 0.1 sec and low for 0.9 sec but it is difficult to get a 555 oscillator to Doo Dat.

thnx ...

 

[D@rkL0rd]

When high, the inhibit pin prevents the output from advancing even if it receives a clock signal.

Connecting reset to +V will reset it back to the first output regardless of which output is set at that particular moment. The way my circuit is set up in the simulation video, I have the last output connected to reset. This way, it will reset whenever the last output is reached, allowing it to restart the loop. You can also connect the reset pin to +V at any time (perhaps via a push-button) to manually reset the chip to the first output.

thnx

 

[alec_t]

Post deleted

 

[D@rkL0rd]

NO you don't need the clock to have a "high time pulse for 0.5 sec and low time pulse for 0.5 sec". You just need the clock to go high each second so the timing can be high for 0.9 sec and low for 0.1 sec., or high for 0.7 sec and low for 0.3 sec.
It will even advance each second if the clock was high for 0.1 sec and low for 0.9 sec but it is difficult to get a 555 oscillator to Doo Dat.

well i finally got the required pulse from the timer ....but now in the counter the pin 0 is not going high in the first loop it goes high int he second loop...please help

 

[DerStrom8]

well i finally got the required pulse from the timer ....but now in the counter the pin 0 is not going high in the first loop it goes high int he second loop...please help

First of all you should connect the EN pin to ground. Right now you aren't even enabling the 4017!

 

[D@rkL0rd]

First of all you should connect the EN pin to ground. Right now you aren't even enabling the 4017!

so how is it oscillating and connecting it also doesnt solve the problem...

 

[DerStrom8]

so how is it oscillating and connecting it also doesnt solve the problem...

I guess I misunderstood the question. You should still have it connected to ground, and you should have Q9 connected to reset so that it loops without extra delays. This may solve the issue with the LED not turning on "on the first loop". It IS turning on in the first loop, but there is an extra delay when Q9 is on and none of the other pins are. That's why you need to reset the 4017 when Q9 is reached (meaning the light cycle is complete).

 

[D@rkL0rd]

I guess I misunderstood the question. You should still have it connected to ground, and you should have Q9 connected to reset so that it loops without extra delays. This may solve the issue with the LED not turning on "on the first loop". It IS turning on in the first loop, but there is an extra delay when Q9 is on and none of the other pins are. That's why you need to reset the 4017 when Q9 is reached (meaning the light cycle is complete).

why isnt the led turning on what am i doing wrong here...

 

[DerStrom8]

why isnt the led turning on what am i doing wrong here...

Didn't you read my post? Connect Q9 to Reset and try again.

 

[D@rkL0rd]

Didn't you read my post? Connect Q9 to Reset and try again.

i did connect it but after taht the 9 pin is not going high its just starting from 1 ... i need all ten pins to go high for 1 sec each...i have uploaded the logic table for which i need to create the circuit...

 

[DerStrom8]

Oh, so you wanted pin 9 to go high but not do anything? Seems a little odd, but ok.

So what is the exact problem you're having? You aren't making much sense. Can you take a screen recording of the simulation running? Try a program like Camstudio if you don't have screen recording software already.

 

[D@rkL0rd]

Oh, so you wanted pin 9 to go high but not do anything? Seems a little odd, but ok.

So what is the exact problem you're having? You aren't making much sense. Can you take a screen recording of the simulation running? Try a program like Camstudio if you don't have screen recording software already.

what i want is every pin of the deacade counter to go high on every loop for 1 sec ...i.e, any pin form 0-3 should for first 4 sec ...but in the first loop the first pin(pin 0)is not going high but in the next loop it is...

 

[DerStrom8]

Oh, I see what you're saying now.

Could you post the schematic you're following so that I can get straight which LED is R1, G2, etc?

 

[DerStrom8]

By the way, I have simulated this myself and I, too, only see Q0 on for a fraction of a second before going to the next output. I'm not sure why this is happening, it may be a SPICE glitch....