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I don't understand how can I enable/disable the 6N137 optocoupler, because in the datasheet a negative current is shown. Is that a mistake in the sheet?

It states that the low/high voltages are positive: 0..0.8 and 2..VCC respectively, but the high/low enable current is negative.

Just to state that I blew up one charger the last time I was trying to enable it with the pin (with a resistor of 10 kΩ connected to VCC, just for experimenting). I know that it's enabled by default and that connecting it to GND will disable it. Also, it seems to consume less power when its explicitly enabled, so at least for that it was worth it.

I can't be sure that this was the cause of the blast, but the optocoupler stopped working after that (only the output side where the enable/disable pin is; the input LED still conducts).

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Reset to default
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Note that the measurements are not done when enable pin is set to VCC, but voltages below VCC.

If you pull the pin down to 0.5V, up to 1.6mA flows out from the pin.

If you pull the pin up to 2.0V, still up to 1.6mA flows out from the pin.

It simply means that there is internal biasing or pull-up circuitry that pulls the enable pin high, to keep it enabled with no external connections, and externally it is needed to pull current out of the pin to bring the voltage down.

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    \$\begingroup\$ I understand the words, but I do not understand how to use this exactly. Can you please give an example with a simple resistor how to make the pin low and high? \$\endgroup\$
    – verbessern
    Commented Sep 9, 2022 at 8:09
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    \$\begingroup\$ No, it means none of that. It means if you short enable pin to GND, there is 0V and chip outputs up to 1.6mA. It also means you can't connect 10k to ground, because it is too weak to pull it low. It means you must put a strong enough resistance like 500 ohms or less, to make sure that the pin voltage is at most 0.8V when 1.6mA flows out of it. You can freely connect VCC to enable. \$\endgroup\$
    – Justme
    Commented Sep 9, 2022 at 8:24
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    \$\begingroup\$ No, I am not saying it is an output pin, that's your thinking. It is an input pin, and current will flow out of it if you try to pull it low. Just imagine the pin has internal pull-up resistor. In real life it might be a TTL logic input or something similar, or CMOS input with current source as a pull-up instead of a resistor. \$\endgroup\$
    – Justme
    Commented Sep 9, 2022 at 8:58
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    \$\begingroup\$ Imagine you have a PCB with CMOS input with say 4k7 pull-up to VCC of 5V so the input pin is at 5V when idle. Would current flow if you pull the pin to ground? Yes. Would any current flow if you connect it to 5V? No. \$\endgroup\$
    – Justme
    Commented Sep 9, 2022 at 9:41
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    \$\begingroup\$ Justme, this seems to have become a discussion/tutorial in comments with @verbessern. Instead, could you edit and improve your brief answer so it explains it all sufficiently, with a diagram for absolute clarity. Otherwise, future readers would have to piece together a com3 answer. Thanks. \$\endgroup\$
    – TonyM
    Commented Sep 9, 2022 at 10:07

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