CMOS CD4060 timer

[atollp]

Hey forum,

I'm going to build this circuit, but after the output goes high (after the delay) I'm assuming it latches. I want it to go high and then low after a period of time. It is possible to add on a resistor-capacitor arrangement so that it won't latch but will instead go low?

Again, I don't know if it latches for sure.

I've seen what I want on other circuits but don't know how to implement it onto this one.

Thanks in advance.

http://www.coolcircuit.com/circuit/timer_4060/timer.GIF

 

[(*steve*)]

No, it seems to turn on when you press the start button, then stops when the time delay runs out *or* if you press the stop button.

 

[Rleo6965]

May I ask why did you choose this circuit? This is not a monostable. Its a long delay oscillator output. Q3 output have the fastest change of transition and q13 was the longest delay transition.
I think S2 should be SPDT. So that you can turn off the relay while the cd4060 still continue running.

With original S2 circuit. Pushing on the S2 will turn off the relay. But the counter timer still running. You have to turn off S1 to stop the cd4060 which also therefore turn off the relay.
This means. Why do we need S2 if we can use S1 to turn off CD4060 and the relay.

Question also with the original circuit was how long do you push S2 to stop relay. Because when you release S2 the relay will turn on again.

Is this the function of circuit that you needed?

 

[(*steve*)]

I think S2 should be SPDT. So that you can turn off the relay while the cd4060 still continue running.

I thought that at first too.

With original S2 circuit. Pushing on the S2 will turn off the relay

Note that the relay has contacts in parallel with S1. Pressing S1 momentarily pulls the relay in which keeps power applied.

When the timer gets to the appropriate count it turns on the transistor which causes the relay to drop out (s2 does the same thing). This also removes power so the relay stays off.

You really need another set of contacts on the relay to allow it to do anything useful though.

 

[Rleo6965]

Those Switches and relay contacts was very confusing.

Contact switch relay parallel with S1 was NC or normally closed or already switch on. Then why do we need S1 to turn the circuit on while it was already " Turn On".

EDIT:

I think this circuit will automatically turn on if you plug it into ac outlet and delay will start for a certain period. Depends on selection switch and turn off the relay when delay period was reached. The turning off of the relay was very fast i.e. few milliseconds and turn on again. WE really don't need S1 and S2 and it can be out of the circuit.
This was the problem of Atollp. He wants the fast turning off of relay remain off. I can't figure out the solution now. I'm sleepy. It's already 1:30am here in Philippines.

 

[atollp]

Thanks for the replies,

I assumed, because of the separate diagrams, that this circuit normally operates on a 12v DC source, but could be modified to work off 120AC, and the modification is shown in the diagram.

If that's the case, and you only look at the top diagram, as soon as power is applied the circuit should begin the delay and after if expires it will either turn on or off the relay. This isn't my circuit I just found it online.

 

[(*steve*)]

Rleo6965, good point. Yes, it starts a timing cycle when power is applied.

I would prefer to see S1 and the relay contacts in the secondary rather than mucking about with the mains like that.

atollp, the bottom part of the circuit is required (or art least the function is required if you want to be able to manually start the timer.

I understand it's not your circuit.

 

[Rleo6965]

I also saw the circuit when I google CD4060 yesterday.
But I think we can help improve the circuit to suit your requirement.

 

[atollp]

I think what I'm going to do is leave off the mains and step down transformer, and rectification unit off of it, run it off two AAs in series with a 9v, and add a switch in between the power source and circuit. Does anyone know if it energizes the relay after the delay is expired? Or does it energize it as soon as power supply is connected and then de-energize after the delay expires? If the latter is the case, is it possible to reverse it so that after the delay expires it will energize the relay?

Thanks for the replies...sorry for all the follow up questions

 

[(*steve*)]

The relay will energise on startup, or when you press the start button.

It will release after the time period expires (or earlier if you press the stop button)

 

[Rleo6965]

I suggest that relay should be not energise during startup to conserve battery current if required delay was too long.

 

[(*steve*)]

Maybe you do, but what does the OP want?

I still don't really know what function he requires, but this looks pretty close.

You could probably do a reset via an RC delay which would prevent the circuit from starting when the power is applied... maybe

 

[atollp]

Damn, I was hoping for it to do the opposite. Is that possible?

 

[Rleo6965]

If you want that relay should be energize after the delay period expire. Remove R5 10k resistor and place relay in its place. Dont't forget to place 1N4001 diode in parallel with Relay for T1 transistor protection. Connect diode Cathode to +Vcc side of relay and Anode side to collector of Q1. It will look diode symbol pointing upward.

 

[atollp]

Thanks! I'll build the circuit when I get a chance!

Thanks for the help guys!!