0
\$\begingroup\$

This is to trigger a 12v solenoid (acting as a laser shutter) with 5v (JP2) thru an optocoupler. The problem is 5v line have bit fast interruptions on a laser show, so I'd like to add a bit of delay OFF (say 0,5sec) in order the shutter remains active (energised) when this fast OFF interruptions happens (you know, to avoid click-click shutter open-close noise) So I think I need to put a capacitor and maybe resistor+capacitor, but sorry, I don't know where on this circuit...could you help me please? Many thanks enter image description here

maybe this one? enter image description here

\$\endgroup\$
  • \$\begingroup\$ Why not get rid of the "fast interruptions", instead? It sounds like your shutter controller is bad. \$\endgroup\$ – WhatRoughBeast Mar 16 '16 at 16:12
  • \$\begingroup\$ The "fast" interruption is something current in laser show (when beam are momentary off), I can't avoid them. This is why people add a bit of "delay off" on solenoid circuits (acting as a laser shutter). A manual E-stop is also there to cut 12V if necessary. \$\endgroup\$ – Jorsy Mar 16 '16 at 16:22
1
\$\begingroup\$

The biggest problem with your circuit is that it works backwards from what you want. With 5 volts applied to the optoisolator, the OI will turn on, and the MOSFET gate will be pulled low, and the MOSFET will turn off. So the shutter coil will not be energized when 5 volts is present. Is this a problem? I'll assume not.

Then, for a 0.5 second delay, your proposed circuit won't work - both the resistor and the capacitor are too small. A 10k resistor and a 100 to 200 uF will probably do what you want. Just be aware that the delay will vary from MOSFET to MOSFET, over about a 2:1 range.

\$\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.