2
\$\begingroup\$

For my project I have to close to contacts. It's bascially a remote for a Canon camera.

With Canon you can use 3 pins (2.5mm jack) to "remote" controll it, 1 of these 3 is the common pin, 1 is the focus and the last is the shoot contact.

My initial plan was to use 2 NO relays for this, put after a look in the datasheet, I noticed the "minimum contact load" and after some searching I found a post about all of this and "mercury wetted relays".

My problem is that I am pretty sure that there is absolutely no load on this circuit.

Now to my question, could I just use a MOSFET in this case, basically use a MOSFET in a circuit with no current?

\$\endgroup\$
2
\$\begingroup\$

There must be some current flowing or the switch/relay will have no effect. You should measure the current flowing when you close the switch.

If you want to use a MOSFET then you need to check the voltage across the terminals when the switch is open, but I strongly expect that this will be a sufficiently low voltage that almost any MOSFET will meet the voltage range requirement. And, yes, MOSFETs can be used to switch very small currents. That's why we build computer chips with them.

\$\endgroup\$
  • \$\begingroup\$ I did exactly that -- a pair of Canon cameras, triggered with dirt-cheap BSS-138 MOSFETs -- in a series of aerial camera pods I built for a client. Worked just fine. The camera battery voltage is just a bit under 7.5V. \$\endgroup\$ – Dave Tweed Aug 13 '19 at 15:59
  • \$\begingroup\$ @DaveTweed Ok, I was planning to use the BSS138 for the relay anyway. So I could just use an ESP to controll the MOSFET (with a pulldown on gate), and just connect the commen pin to the drains and the other two to the 2 sources? \$\endgroup\$ – Max Aug 13 '19 at 19:04
  • \$\begingroup\$ No, the sources connect to the common pin, and the drains to the control signals. And of course, you also need to connect the sources to the ground of the ESP, to provide a reference for the gate drive. \$\endgroup\$ – Dave Tweed Aug 13 '19 at 19:18

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.