Electrolytic is fine.

 

 

What is the source of the 5ms 12 volt pulse?
You can eliminate the mosfet and use the reset pin 4 as the trigger to start the timer.

 

So the reset starts the timer with a high instead of a low? The trigger is for example the output (pin 6) of an optoisolator H11D1M.

 

So the reset starts the timer with a high instead of a low? The trigger is for example the output (pin 6) of an optoisolator H11D1M.

Correct.

 

Yes this also works. Don't forget the diode between out und reset of you do it like that

 

Can you explain why I would need a diode? Pins 3 and 4 are not connected to each other. If pin 2 is not used for the trigger then is that left open? Which way round would the diode go?

 

Can you explain why I would need a diode? Pins 3 and 4 are not connected to each other. If pin 2 is not used for the trigger then is that left open? Which way round would the diode go?

Look at the schematic I posted
Pin2 is connected to pin6-7.
The diode is required to hold pin4 high after the 5ms pulse is gone.
The diode provides a latching function.

 

Got it. I didn’t get to see the graphic as I thought it was a repeat of my earlier one. I will be building a breadboard demo so I can easily try both approaches.

 

I will be building a breadboard demo so I can easily try both approaches.

Are you actually using a IRF840 for Q1? If so that is way overkill. A npn transistor such as a 2N3904 should be fine as well.
Change R1 to 100K.

 

Earlier on a FET was proposed and as I have a few 840s I used that, but as you say it can be done in a ‘lighter’ way.

 

I have now built both versions of this circuit and my preference is to use the one with the 2N3904 to present a low to the 555 trigger input. Here I have used LEDs to see the timed pulses from the relay.

I have one remaining query about the input. If I want to have a higher trigger voltage such as 50V, I would clearly need to bring that down for the input to the base of Q1 using a divider of sorts but would I need to connect the other end of the pot to ground, as in the circuit below, or can it float?

Also how critical is the total input resistance to the base of Q1 such that if the divider R1 was set to 5k, would I need to reduce R2 accordingly or is there a lot of flexibility around that for Q1 to saturate and operate effectively? Given the above post suggestion to use 100k for what is now R2, maybe there is since the present 10k seems to work ok. At the moment Vbe is 0.74V with a 12V input signal.

 

Also how critical is the total input resistance to the base of Q1 such that if the divider R1 was set to 5k, would I need to reduce R2 accordingly or is there a lot of flexibility around that for Q1 to saturate and operate effectively?

Not critical in this circuit.
Setting R2 at 47K should suffice for a 5 to 50 volt trigger.

 

Not critical in this circuit.
Setting R2 at 47K should suffice for a 5 to 50 volt trigger.

Ok so the pot can roam it’s full range from 0-50k. What about having to connect the pot to ground?

 

Ok so the pot can roam it’s full range from 0-50k. What about having to connect the pot to ground?

The pot should not be connected to ground, as it is acting as a rheostat (variable resistance), not a potentiometer.
Often though, the wiper is connected to one end of the pot, so that a failure of the wiper contact will not cause an open circuit (which may not be important in this application).

 

Thanks