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cascading 4017 troubles...

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Hi all,
Im building a running light circuit. Im working form a this diagram for cascaded 4017 chips...;
Running Lights
its sort of working a bit but a kind of strangely; it counts through the outputs from the first chip, then for each cycle of the this it advances the second chip one step. WIERD!
Can anyone see any glaring errors with it? or have i just made some mistake?
I tried turning the diode round but that didnt work. Argh.

Cheers!
 

audioguru

Well-Known Member
Most Helpful Member
The 555 oscillator is horrible as an oscillator for logic ICs because it has a huge supply current spike each time its output switches. It needs the two supply bypass capacitors recommended in the datasheet for the LM555, or use an oscillator that does not have a huge supply current spike like a Cmos oscillator or a Cmos 555.

Why don't you cascade the CD4017 ICs like the datasheet shows?
 
hmmm

sadly attaching capacitors (a ceramic 0.1uf and electrolytical 10uf one) between + and gnd didnt sort it. Perhaps i'll try a different circuit altogether, or buy one. Anyone know where i could get a 16 step running light circuit from, internet searching is proving fruitless.
 

audioguru

Well-Known Member
Most Helpful Member
oh yeah and would resistors before leds help at all? if so what sort of value would be good.
The circuit you copied has no supply voltage shown. If the supply voltage is more than about 9V and if you have 2V red LEDs then the 4017 ICs are straining to supply enough current and the one output that also feeds the other 4017 will not have enough voltage so then the circuit will not work.
Try a 1k current-limiting resistor that feeds all the LEDs.
 
Thanks $crooge
Im powering it with 4AA bats (6v). So do you recon not having a resistor before each led could be another factor in its erratic behavior? And if I put them in and decouple it it might work?...

Mp
 

audioguru

Well-Known Member
Most Helpful Member
There is only one output on one CD4017 that drives an LED and also must have a logic output so it needs a resistor to limit its output from being shorted by the LED.
 
Sorry scrooge im confused, I thought the majority of the pins are outputs on a 4017 (on the diagram im working from it looks like 1,2,3,4,5,6,7,9 and 10 are outputs).

Just to clarify im not torpedoing your response, you seem to be alot more clued up than me so im sure you are correct, im just confused!
 

audioguru

Well-Known Member
Most Helpful Member
Pin 3 on the first CD4017 drives LED#0. But it also must provide a logic high (at least +4V) to pin 15 of the second CD4017. It cannot provide +4V if the LED holds its voltage at 2V. So it needs a current-limiting resistor in series with its LED so that it can go as high as +4V.

Your 6V supply will drop as low as 4V so the CD4017 can supply only a very small current. Then the LEDs will be very dim.
 

sheldonstv

New Member
that schematic is a load of poo!!! pin 15 is the reset for the 4017 which when taken hi resets the second 4017-also there is no turn on reset for either 4017....both clock ip pins(pin 14) on each 4017 must be joined together-why is there a 33k resistor connected to the second clock ip???
i have uploaded you a 2 stage counter schematic that when the first counter reaches end of count it inhibits itself by taking its inh pin(pin 13)hi,and at the same time this signal is inverted so the inh pin of the second 4017 counter goes lo so that ic is now counting while the first counter is inhibited....both clock inputs are joined together on each 4017 ic(pin 14) so both ics are parralell clocked-when the second 4017 reaches end of count both ics get reset so the 1st counter is now free to count with the second ic inhibited and the whole cycle will repeat until you switch it off or your batterys go flat........
 

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  • 2 stage 4017 counter.pdf
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sheldonstv

New Member
let me know how you get on....ou can use 1 gate as the clock and the remaining gates for your 4017gating and the last could be used as your 4017 reset.......
 
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