Looking for ideas on a simple timer. Less components the better.
When a switch is closed the time should start. If the switch is opened before 10 secs the timer should reset back to zero.
The output relay should only become energised if the switch is held closed for 10 secs
LM555 would be ideal.

 

What's the voltage?

Once the output is ON, how do you want to reset it?

 

What are the relay ratings?

There already exist "delay on make relays" you can buy.

You can also use a comparator ( Open loop) where with your switch you would just charge a capacitor via a resistor.
The IC compares a fixed voltage with the voltage level on the capacitor and switches the output when the input voltage polarity inverts (after your delay).
example: LM239 , open collector can typically sink 16mA , single or dual voltage power supply
An extra resistor is needed to discharge the capacitor when you release the switch.

So:
- 1 IC
- 4 resistors
- 1 capacitor
- 1 switch

 

Thank you for your reply.
12v dc supply.
Timer would reset every time the momentary switch is opened.
Small pcb mounted relay would only energise if momentary switch was kept closed for 10 secs.
I guess I could source a ready made delay relay but this is my hobby!
While I have been waiting I have come up with a solution using a capacitor/resistor, zener diode and a pair of transistors in a darlington circuit. This is working fine but I would like to find a solution using an IC.

Many thanks for your help.

 

Small pcb mounted relay would only energise if momentary switch was kept closed for 10 secs.

You're going to hold a switch closed for 10 seconds?

 

1 IC (PIC12HV615)
1 capacitor
1 resistor
1 MOSFET
1 relay (assume suppression diode built in, otherwise it would blow my parts budget!)
oh, and the switch too.

I can build that circuit in 6 parts!

I actually designed a relay timer used by one (but more on the hook) relay manufacturer. Those are some of the parts.

 

Hello ian.blue

You say:
A- When a switch is closed the time should start.
B- If the switch is opened before 10 secs the timer should reset back to zero.
C- Timer would reset every time the momentary switch is opened.
D- Small pcb mounted relay would only energise if momentary switch was kept closed for 10 secs.
Is it True?.

So while SW1 remains closed for more than 10 seconds the relay should remain closed.
It means that if SW1 is opened within 10 seconds the relay should not energized.

I think the circuit contained in the attached image meets those statements.

When SW1 is closed starting time, the 555 output remains high for a time determined by the value of VR1 and C1 components.
When that time runs out the output of 555(PIN 3) will be low.
Q1 change that level.
Q2 energize the relay.

Q2 should be selected according to voltage and current required by the relay.
Time, 10 seconds, can be adjusted more precisely by means of the potentiometer VR1.

regards
at your service

 

Hello ian.blue

You say:
A- When a switch is closed the time should start.
B- If the switch is opened before 10 secs the timer should reset back to zero.
C- Timer would reset every time the momentary switch is opened.
D- Small pcb mounted relay would only energise if momentary switch was kept closed for 10 secs.
Is it True?.

So while SW1 remains closed for more than 10 seconds the relay should remain closed.
It means that if SW1 is opened within 10 seconds the relay should not energized.

I think the circuit contained in the attached image meets those statements.

When SW1 is closed starting time, the 555 output remains high for a time determined by the value of VR1 and C1 components.
When that time runs out the output of 555(PIN 3) will be low.
Q1 change that level.
Q2 energize the relay.

Q2 should be selected according to voltage and current required by the relay.
Time, 10 seconds, can be adjusted more precisely by means of the potentiometer VR1.

regards
at your service

If I press the switch for 8 seconds then release it C1 has to be rapidly discharged. If not, I would press the switch after a few seconds again and it wouldn't take 10 second for the 555 to trigger because C1 wasn't discharged.

 

Yes, the capacitor has to completely discharge each time.
I should explain the (switch) is not a manual switch but a liquid level switch. The function is to indicate when the level is low but since there is turbulence in the liquid, false warnings would constantly register.
That's where the 10 second time lapse comes in.
Rather like the low fuel indicator light in an automobile.
I wonder how that works?
Can anyone come up with a solution?

 

You could use digital counters. The CMOS 4000 series works with 12V. You'll need an oscillator which you could make out of cristal and and inverters. You pass this to a counter like the 4060 (or two of them) and stop the counter when it reaches the 10s output.

Why didn't you like your analog solution?

 

Having just read your post I think I do like my analogue solution.
No point in taking a sledge hammer to crack a nut.
There is of course that reset problem but I think that is not a problem at all since in practice the turbulance is fairly regular so the float switch is bobbing up and down causing the capacitor to be half charged. If I accept that as the norm and increase the time delay to say 20 seconds I should have the answer.
Thanks to eveyone for their help.

 

Why are you bailing out? No one has given you what you asked for yet. Stick around a bit longer and you will get what you asked for.

 

Well, I thought someone would have worked this up for you by now.. Guess not. I believe from the description of your requirements this is what you want. Let me know if it isn't. BTW, you may not need Q1 but since I didn't know the requirements of your relay I opted not to drive it directly from the 555.