0
\$\begingroup\$

I have a 555 monostable set up which is triggered when an output from another circuit goes low. This then remains on for 10s. This all works almost perfectly.

My issue is that when the output from the other circuit goes high again and then back down within the initial 10s timing , this 10s timer does not restart, it is locked in its first cycle.

Example: If i send the trigger pin low, the 10s timer begins and the diode illuminates. Then if I wait 5s and then send it high and low again, it continues the initial timing cycle and the diode goes off after 5s.

Each time the input circuit goes low, i would like to restart the 10s timer.

Anyone know how this can be achieved? I have tried playing around with the reset pin but with no success.

P.S. i know the time constant for this circuit is actually around 100ms, it's just the limitations of the software I am using , can only simulate 10ms/s

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ You need a "retriggerable" timer. The 555 is not retriggerable, although you can fake it by adding external components. \$\endgroup\$ – Dave Tweed May 31 at 11:27
0
\$\begingroup\$

You can do that, with a bit of effort.

However, you're really crossing into regions where it's much easier to simply learn how to use an arbitrary low-cost microcontroller (e.g. an Arduino-compatible) and program exactly what you want. Much more flexible, much less potential for analog headache!


To achieve that, you could catch the falling edge of the other circuit and use it to discharge C1.

Not having tried that, but:

I think it would be possible to achieve using a capacitor to couple in the falling edge to the base of an NPN transistor, which you use as inverter, to another drive another NPN's base whose collector-emitter is in parallel with C1.

\$\endgroup\$
0
\$\begingroup\$

This can be achieved with a retriggerable monostable, such as CD4538. This will need a buffer, such as 2N7000, to drive the relay, but otherwise should be fine running off 12V.

You need a catch diode across the relay coil. You can trigger the CD4538 from either a positive- or a negative-going edge - just tie the other input accordingly.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.