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Help wiring a Ne555 Timer delay module (monostable?)

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I bought a ne555 timer delay module which was advertised as being able to be used in a monostable manner, some sellers provide this diagram and I have been using it;

s-l500.png

https://i.ebayimg.com/images/g/7aEAAOSw8HBZNRLy/s-l500.png

The problem I am facing is that when I wire it this way and flip the switch the relay and "load" stay on indefinetly, the adjustable potentiometer does not seem to affect the circuit in this configuration and I bought 4 modules so I think I can rule out a defective one. I am thinking about pulling the ne555's (they are smd tho :[ ) off and making my own monostable modules but I suspect I may be doing something wrong with the wiring.

Here are the exact modules I ordered https://www.amazon.com/gp/product/B06XYF8Y1D/ref=ppx_yo_dt_b_asin_title_o03_s00?ie=UTF8&psc=1

here is a picture of my wiring, I am shorting the common and CK ports with a wire to act as the switch

1vlRg4i.jpg

https://imgur.com/1vlRg4i
 
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The board is not a faulty one. It may bes on the wiring diagram given.

Make an experiment . Just Connect Vcc and Gnd terminal to power supply .Dont connect any other load or switch.

See whether the LED on board gets switchedON for a specific time.
 

The board is not a faulty one. It may bes on the wiring diagram given.

Make an experiment . Just Connect Vcc and Gnd terminal to power supply .Dont connect any other load or switch.

See whether the LED on board gets switchedON for a specific time.

I have done that an the board works fine in the normal configuration.

But I still need a way to send a 'trigger' and get a 10 second output.
 

I have done that an the board works fine in the normal configuration.

But I still need a way to send a 'trigger' and get a 10 second output.

The board does not have the option of separate trigger.
Switching ON of pwr supply generates the trigger internally.
 

The board does not have the option of separate trigger.
Switching ON of pwr supply generates the trigger internally.

yes i realize this, but the board is advertised as being a monostable circuit with the wiring diagram as in the original post. my question is how do i get the module to work in a monostable fashion.
 

if you want the module to have a trigger input in monostable operation ,
trace the circuit in the board and modify it.
 

If I'm right, the original wiring diagram shows the switch across the normally open relay contacts which are in series with the supply to the board. Closing the switch powers the board up and it stays powered up until the monostable times out, the relay opens and it turns itself off again. That seems to be what it was designed to do, there is no external trigger signal.

It may be possible to modify it but as Srizbf stated, you need to trace its schematic to see how it is currently connected.

Brian.
 

Thank you all for your replies, your assessment of the circuit is almost accurate and perhaps even the intention of the designers. while it is true that the switch is connected across the open terminals of the relay and the board receives power when it is pressed, it only the board only receives a pulse as the relay does not switch until AFTER the mono-stable timer runs out. Because of this the sequence of events are as follows; switch pressed (must pressed until the timer triggers the circuit otherwise the board just loses power and nothing will happen) -> relay closes and the board now supplies power to itself -> relay stay on perpetually as the timer has already been triggered and there is no self resetting. The problem is that the board does not stay on until the mono-stable times out, it stays on after the mono-stable times out and only gets there if the witch is held long enough.

Unfortunately It doesn't seem like there is a way to achieve a trigger input with a single board, I have managed to solve my issues using two of these boards incorporating the normally open relay ports on one and the normally closed ports on the other.

I am still interested in this circuit and would like to experiment with the suggestion srizbf made, that is to trace the trigger port and splice a switch into it but i don't know how that would affect the rest of the circuit, now I am no expert and this seems strange to me but as far as I can tell the Trigger pin on the ne555 IC is connected to the minus side of the capacitor and what I think is the ground pin (the pin directly under the screw) of the potentiometer. Would I need to run a switch from the 12v supply to one of those nodes or what, I dont know where to go from here.
 

What is it that you actually want the board to do?
As far as I can tell, the sequence of events is:
- you press the switch which powers up the circuit
- when the circuit is powered, it automatically triggers the monostable action of the 555 but in itself that does NOT power the relay
- when the monostable times out, it will power the relay and so close the relay contacts
- when the relay contacts are closed, it will continue to power the circuit, even if the switch is then released
In effect, this circuit is turning something on but requires the switch to be on for at least a set period of time. There is no way with this circuit to turn it off.
Is that what you want?
If not then tell us the sequence that you intend.
Susan
 

If you have a test meter, do a continuity test between pins 1 and 2 of the NE555. Then do the same between pins 8 and 2. In either case, if there is a direct connection between them you will have to cut it so pin 2 is isolated from everything else. Unfortunately, if they are joined it is likely to be by copper trace under the IC.

If you can isolate pin 2, it is the trigger input to the NE555. Never connect it to a negative voltage or a voltage higher than supply with reference to ground.

In general, I avoid these 'DIY modules'. Some of them are VERY badly designed. I have one here that claims to be four opto-isolated isolated relay drivers. It does have four relays and four opto-couplers but they all share a common ground and transistor driver circuit, meaning the isolation can't exist. In fact to improve performance I remove the opto-isolator and link it's input pins to it's output pins!

Brian.
 

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