1
\$\begingroup\$

In a monostable configuration of the 555:

http://circuitdigest.com/electronic-circuits/555-timer-monostable-circuit-diagram

I have connected pin2 to the ground all the time. My goal is that once the circuit is powered on it starts the counting down right away then turns off after x seconds but this is not what's happening.

The output load will be turned on this way all the time.

How to do it properly then?

Thanks

\$\endgroup\$

2 Answers 2

Reset to default
3
\$\begingroup\$

You need a second R-C to do the triggering.

schematic

simulate this circuit – Schematic created using CircuitLab

Just make sure the R1-C1 time constant is shorter than the R2-C3 time constant — but longer than the rise time of the power supply.

\$\endgroup\$
2
  • \$\begingroup\$ I would go further and say to make the R1-C1 time constant as short as possible; the counting doesn't start until pin 2 is high (or perhaps "no longer low") thus the total on-time includes both time constants. \$\endgroup\$
    – CharlieHanson
    Commented Mar 24, 2016 at 14:01
  • 1
    \$\begingroup\$ @CharlieHanson: No, that is incorrect. The "counting" starts as soon as the 555's internal flip-flop changes state, which is on the leading edge of the trigger pulse. As long as the trigger pulse ends before the timing interval ends, the length of the interval is not affected. Think about it -- if what you are saying were true, then it would not be possible to connect pins 2 and 6 together to operate the 555 in its astable mode. \$\endgroup\$
    – Dave Tweed
    Commented Mar 24, 2016 at 14:55
0
\$\begingroup\$

Here's what works for me. You can ignore Q3 and R51, who's base is grounded by that remote relay contact, which causes RL6 to be triggered independently. With values shown timing is from 6 seconds to many minutes depending on RP1 setting. Sorry about the two pics! enter image description hereenter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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