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I understand that typical optoisolators can logically be considered as "normally open". However I have a situation where I'd like the circuit controlled by the opto to be normally closed, mainly for the failure state but also so that the opto's led doesn't have to be activated for 99% of the time.

How can this be done on the transistor side?

Failure states aside, should I be concerned with having the opto activated indefinitely (in a normally open circuit)?

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    \$\begingroup\$ Use an SSR instead, you can get isolated SSR's that are NO or NC \$\endgroup\$
    – Voltage Spike
    Aug 2, 2019 at 18:50
  • \$\begingroup\$ Can you recommend a part number? The circuit that needs to be NC is 12vdc \$\endgroup\$
    – Spammy
    Aug 2, 2019 at 20:06
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    \$\begingroup\$ Try CPC1117N. These normally closed relays are often called Form-B. \$\endgroup\$
    – Steve G
    Aug 2, 2019 at 20:51

3 Answers 3

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Instead of switching the actual load with the opto, use the opto to drive a transistor which switches the actual load.

However, note that if the opto fails short then that will open the transistor.

The risk could be potentially reduced by having two optos wired series on the output and parallel on the input so that both must close (or fail) for the transistor open.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Any thoughts on the second part of my question? Should I even bother - is it considered normal usage to just keep an opto activated for extended periods of time? I've ordered some 2N3904s and 2N3906s to add to my parts box anyway. \$\endgroup\$
    – Spammy
    Aug 2, 2019 at 20:27
  • \$\begingroup\$ LEDs wear out, but then again, so do transistors while conducting so it's six of one or half a dozen of the other. The LED would also consumes mA though that might not matter depending on how much your load current is conducting when normally closed. I guess it matters how fail-safe your thing has to be. \$\endgroup\$
    – DKNguyen
    Aug 2, 2019 at 20:31
  • \$\begingroup\$ I actually have some triacs lying around. Could they be used in the place of the transistors? Or would the current at the gate not become low enough? \$\endgroup\$
    – Spammy
    Aug 4, 2019 at 8:06
  • \$\begingroup\$ Triacs don't turn off until the current through them falls below a holding value. In practice they're only used on AC circuits. \$\endgroup\$
    – Transistor
    Aug 4, 2019 at 8:15
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Yes, there are DIP opto-MOS relays with depletion-mode MOSFETs on the output.

Or you can use a photovoltaic gate-drive optocoupler with a discrete depletion-mode transistor. Just remember to connect it to drive the gate negative.

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This kind of optoisolators are pretty common. They are refereed as form B. These can be easily implemented with depletion mode MOSFETS. However, how suitable are these for your application depends very much on the current and voltage of the output circuit. This one, for example, can handle up to 75mA at a maximum voltage of 60V.

You can easily find optocouplers that can handle currents as high as 2A with blocking voltages up to 600V.

About your second question, the answer is you shouldn't be concerned. As long as you keep both sides of the circuit working within safe limits. For example, in the datasheet I referred above, there is a section called Absolute Maximum Ratings @25 oC As long as you never exceed these values, your circuit will not be stressed. If your circuit will be working at temperatures higher than 25oC, just compensate. A good rule of thumb is to work the output not higher than 80% of the mentioned maximum ratings and drive the input with the least amount of current possible (e.g. from 5 to 10mA.)

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  • \$\begingroup\$ Is there a reason they don't just have "regular" opto-isolators that have a single depletion mode FET on the output? Is the market just not big enough so they all have back-to-back FETs for bidirectional blocking, at which point they become an SSR more than a simple opto? I never gave it much though, but is there even a depletion mode equivalent for BJTs? Most optos in the schematic show a photo-BJT on the output rather than a MOSFET and I never gave it much thought as to why that is. \$\endgroup\$
    – DKNguyen
    Aug 3, 2019 at 17:43
  • \$\begingroup\$ Indeed many manufacturers market optocouplers with FET outputs as SSR. There is no depletion mode BJT (maybe something similar to a IGBTs can be made depletion mode but I haven't seen any implementation.) Most people (including me) would use an opto with BJT outputs without giving much thought but many applications will work as well with optos with FET outputs. Even more, there are some applications that become much simpler with FET outputs (e.g. AC switching.) \$\endgroup\$
    – Krauss
    Aug 4, 2019 at 8:45

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