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--I am a hobbyist, not an engineer.

--I was looking in some datasheets about optocouplers and i cross over this part "off-state output terminal voltage”. I google it and i was redirected to this website and one of its questions about this subject. >> here is the thread: What does "off-state output terminal voltage" in optocouplers mean?

--I want to know more about what Damien said: "the maximum voltage you can have across the detector pin when it's turned off. It is equivalent to the Vce max of a BJT or Vds max of a MOS. "

--In my mind, I imagine it's about the voltage spike. When the voltage applied spikes to 600V, the component can not be damaged when is NOT active, NOT coupled, NOT working. And what is happening when is turned on? (when the component IS working?) What is the difference between turned off and turned on? I mean, that voltage, 600V should be allowed to cross when is turned ON. And not when the junction is turned OFF/innactive. It's how i understand it. Please make some light for me. And thank you!

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An BJT functions (in a really greatly simplification) works like a switch. This also applies to an optokoppler or a MOSFET.

This mean if the device is "off", that the path between some pins get insulating. If the part is "on" the same path gets low-resistance. That means for an BJT if the base-emitter voltage rises above a certain voltage, than the collector-emitter path gets low-resistance. If the the base-emitter voltage falls under this certain voltage, then the collector-emitter path gets insulating. You can only apply high voltages on an insulating path, as you would have to drive insane high currents through an low-resistance path to obtain high voltage over these paths.

Again: This is a very great simplification

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  • \$\begingroup\$ EXCELLENT explanation ! Thank you Very much. Unfortunatly i am not allowed to upvote your answer since im a new member on this site at this moment. but you do deserve a BIG upvote. In a very timid way, i was thinking from your perspective but not very sure on my thought. Your answer really pushed me in the right direction. So you are my hero at this point. Bravo and again, you are Super. \$\endgroup\$ Apr 20, 2020 at 11:18
  • \$\begingroup\$ So, i understood perfectly. When the junction is closed (not current activated), its practically a big resistance in circuit, that supports maximum up to 600V in this case to not get damaged. But what if the junction is "on"(it's Gate/Base is open) and the same 600V are passing through? Is the same thing as when is in "off" mode? It's the resistivity of the metals + some conductivity ? Then i should ask, what is the maximum voltage permitted when is "on" ? Still 600V maxim (in this case) or greater(because of conductivity)? \$\endgroup\$ Apr 20, 2020 at 11:54
  • \$\begingroup\$ I have provided a lot of explanation in my answer (I have updated it). You are not being clear about WHICH 600V are you talking about, WHERE on the component exactly? A standard opto-coupler with a transistor is usually rated for less than 100V across the output pins, but there are optos with TRIACs on the output, capable of controlling higher voltages directly. In your question you haven't posted the exact component either (and this matters, as it helps to provide a useful answer), but the one mentioned in the linked post was MOC3023 (TRIAC-based). \$\endgroup\$ Apr 20, 2020 at 12:12
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Here is a representation of an opto-coupler below.
- It is made out of an LED and a photo-transistor, both contained in a single case which doesn't let any light in or out, but lets the light from the LED reach the photo-transistor through a clear (transparent) plastic channel which at the same time provides an electric isolation between those 2 components.
- This transparent channel provides isolation of a few thousand volts between the LED and the transistor.
That's why an opto-coupler is also called an opto-isolator, as its purpose is to both couple an input signal on the LED with an output on the transistor while providing an electrical isolation between the two. - When there is no voltage/current applied to the LED on the input, the output transistor is off, not conducting current, it has very high resistance.
- When you apply some current to the LED, the transistor will start conducting, its resistance between collector and emitter will be reduced.
- The more current you run through the LED, the more will the output transistor turn on, and its resistance will reduce even further.
- The "output terminals" in this case are the collector and the emitter of the photo-transistor.
- "Off-state output terminal voltage" is basically the maximum voltage the transistor can withstand between its collector and emitter when it is off (when there is no current through the input LED to turn the transistor on).
- If you know the basics about transistors and their ratings, this answer should clarify it for you now. If not, you should read up on bipolar transistors and their ratings, and possibly on some most basic circuits with a bipolar transistor.

schematic

simulate this circuit – Schematic created using CircuitLab

"In my mind, I imagine it's about the voltage spike. When the voltage applied spikes to 600V, the component can not be damaged when is NOT active, NOT coupled, NOT working. And what is happening when is turned on? (when the component IS working?) What is the difference between turned off and turned on? I mean, that voltage, 600V should be allowed to cross when is turned ON. And not when the junction is turned OFF/innactive. It's how i understand it. Please make some light for me. And thank you!"
- If you are assuming that the "Off-state output terminal voltage" is about the "voltage spikes" across the whole opto-coupler (input to output), you are wrong.
THAT rating is called "Isolation voltage", and is much higher than the 600V you're mentioning here, normally in the thousands of volts (normally at least 2000V).
Also, the isolation voltage rating DOES NOT depend on the state of the opto-coupler (whether it is turned on or off), it is a static value.

EDIT: In your question, you asked about opto-couplers in general:
"--I was looking in some datasheets about optocouplers and i cross over this part "off-state output terminal voltage”."
I have checked the component which was mentioned specifically in the post you linked, but not in your own question, and it does make a difference in helping you.
The component in question is MOC3023, and it is a TRIAC driver output opto-coupler (using a DIAC as a photo-receiver). In its datasheet there is: VDRM - Off−State Output Terminal Voltage
specified at 400V, which means that in the off state, the maximum voltage across the output terminals of this opto-coupler before damaging the opto-coupler's output DIAC is 400V, even if it's a short voltage peak.
Looking further, I have come across Toshiba's "Glossary of Photocoupler and Photorelay Terms" which mentions Opto-TRIACs with this rating, and again it is the maximum voltage that you can apply to the output terminals without damaging the output of the opto-coupler (or opto-TRIAC, also known as an SSR - Solid State Relay), when it is off.
Almost ANY opto-coupler device, regardless of the kind of output, and especially the one in your case will have a greatly reduced output resistance when it is turned on, often acting as a short, across which there is a minimal voltage of no more than just a couple volts, so this off-state voltage rating has nothing to do with the on-state voltage across the output terminals.
Your lack of understanding of this most basic concept is why I have explained opto-coupler basics here, because it is appears that you don't know how voltages and currents propagate through a circuit and
The last paragraph of your question, analyzed:

--In my mind, I imagine it's about the voltage spike."

It's about maximum voltage capability, whether it's a steady DC value, regular peak value, or spike.

When the voltage applied spikes to 600V, the component can not be damaged when is NOT active, NOT coupled, NOT working.

I wouldn't say "not coupled" as coupling means the output is ready to show something if there is a signal at the input, regardless of it not being on at the moment, so it is always coupled.

And what is happening when is turned on? (when the component IS working?) What is the difference between turned off and turned on?

That depends on the specific opto-coupler but it normally means that its output resistance is very high (acts like an open switch, not conducting current) and a maximum voltage (the applied voltage) appears across its output pins/terminals when it is off, and its output resistance becomes very low (like a closed switch, like a short-circuit) and a very low or almost no voltage appears across its output pins/terminals.

I mean, that voltage, 600V should be allowed to cross when is turned ON. And not when the junction is turned OFF/innactive. It's how i understand it. Please make some light for me. And thank you!

Voltage allowed to cross when turned on? How does a voltage cross, what do you mean by this?

Based on your lack of basic knowledge about how opto-couplers are connected in a basic circuit, here is a very basic schematic:

schematic

simulate this circuit

A voltage/power source is never connected directly across the output terminals of an opto-coupler, just like they're never connected across a switch. Opto-coupler in this case acts like a switch (in other cases it acts like a current-controlled variable resistance, but don't sweat it right now!).
Just like you would connect a switch in series with the power source and the load it controls, you would in a similar fashion connect an opto-coupler in a simplest circuit.
The whole voltage from the power source appears across the opto-coupler's terminals when the opto-coupler is off because there is no voltage drop through the load when opto-coupler is off, and there is no current through the opto-coupler. If you know the basics about parallel and series connected resistors, you will understand this.
When the opto-coupler is on, almost all of the voltage now appears across the load, and almost no voltage across the opto-coupler output terminals because they are now almost like short-circuited, just like it happens with a switch in a circuit.

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  • \$\begingroup\$ :) It is not what I'm asking ! Your answer is way too basic and simplistic. This is what everyone knows and is very intuitive. \$\endgroup\$ Apr 20, 2020 at 11:12
  • \$\begingroup\$ If you read my whole answer carefully, you will see that it provides an explanation that will help you understand, but I will edit it and include more explanation regarding your last paragraph. \$\endgroup\$ Apr 20, 2020 at 11:16
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In my mind, I imagine it's about the voltage spike. When the voltage applied spikes to 600 V, ...

enter image description here

Figure 1. Image from Edin's answer!

Make sure you are talking about the right parameter. There will be parameters for the LED, parameters for the photo-transistor and there will be parameters for the insulation between the two.

Opto-isolators will be typically rated for 1 to 2 kV. That means that the maximum difference in voltage between pins on the LED and the transistor may be up to 1 or 2 kV. (It does not mean that you can apply 1 kV to the LED or to the transistor.)

... the component can not be damaged when is NOT active, NOT coupled, NOT working.

It could. The LED might be off but if the difference in voltage between the LED and the transistor exceeds the breakdown voltage of the insulation then isolation will be lost.

And what is happening when is turned on? (when the component IS working?) What is the difference between turned off and turned on? I mean, that voltage, 600 V should be allowed to cross when is turned ON.

No voltage is ever allowed to cross. Only light crosses. That's the whole idea!

And not when the junction is turned OFF/inactive. It's how I understand it.

Remember that the LED turns on, light is emitted, the photo-transistor turns on and conducts current from its collector to its emitter.

Please make some light for me.

If I could you wouldn't see it. The opto-coupler LEDs are infrared!

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  • \$\begingroup\$ I like your sense of humor. :D \$\endgroup\$ Apr 20, 2020 at 11:18
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    \$\begingroup\$ You can see that Edin and I both had difficulty understanding your question. I answered it point by point to make sure I addressed what you appeared to be asking. It wasn't at all clear where this 600 V pulse was going to come from. It's also not clear what insight you got from the answer you accepted which discusses base-emitter voltages which aren't directly applicable in opto-transistors which generally don't have the base connected to an external pin (although some do). \$\endgroup\$
    – Transistor
    Apr 20, 2020 at 11:37
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    \$\begingroup\$ "Your explanation is too basic and simplistic(from what i intended) . This is what everyone knows and is very intuitive." But so is basic transistor operation. I assumed you weren't asking about that. "Close but still off point because you were refering to the inside mechanism." What else would we be discussing? \$\endgroup\$
    – Transistor
    Apr 20, 2020 at 11:39
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    \$\begingroup\$ I think that teodoric8 hasn't even grasped the basics of the most basic electrical circuits, but is too embarrassed to get off his high-horse and admit that. NoiseEngineer's answer was just as unclear as teodoric8's question, yet he somehow gained a super-clear understanding from his answer and couldn't praise him enough. Sounds like someone has some issues. \$\endgroup\$ Apr 20, 2020 at 12:30
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    \$\begingroup\$ After reading some comments and the more acurate and more detailed answers of Edin Fifić and Transistor, I think Edin Fifić might be correct with the comment above \$\endgroup\$ Apr 20, 2020 at 14:15

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