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In the datasheet it is stated that the maximum VCC voltage is +5.5V.(https://www.vishay.com/docs/84732/6n137.pdf)

Because ive been told to question everything (and i have enought of them to kill one or two) i used it with 12V. It works great as it should. Now what should be limiting the voltage and what could happen if ill use it at 12V?

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    \$\begingroup\$ 7V (not 5.5V) is the absolute maximum voltage for this device. That means the manufacturer does not guarantee that the device will survive any voltage over 7V. It may work, or it may not, but you can't blame Vishay because they told you not to do it. \$\endgroup\$
    – Hearth
    Feb 10 '19 at 20:00
  • \$\begingroup\$ yeah sure, but why is it 7V in the first place? \$\endgroup\$ Feb 10 '19 at 20:01
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    \$\begingroup\$ Because that's the voltage they chose to design the thing around. \$\endgroup\$
    – Hearth
    Feb 10 '19 at 20:01
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    \$\begingroup\$ Because the manufacturer guarantees that each and every device will provide its expected full lifetime if you never exceed 7V for more than one minute. Your circuit is living on borrowed time. \$\endgroup\$ Feb 10 '19 at 20:02
  • \$\begingroup\$ Like at TTL with Schottky transistors these are 5V Logic outputs. \$\endgroup\$ Feb 10 '19 at 20:47
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The original 6N137 (designed by HP over fourty years ago) is specified only for 5 V (± 10 %), but actually regulates its supply:

6N137
(source: Optoelectronics Applications Manual)

Vishay's 6N137 does not copy this circuit (it has hysteresis, and the output transistor is a MOSFET), so it is not known whether it has a regulator. But the use of a MOSFET points to a possible failure more, which is described (for similar transistors in completely different chips) in TI's AHC Designer's Guide:

Input voltages greater than 7 V must be avoided to preclude damage to the gate oxide of the input stage. This damage is not necessarily permanent, but will adversely affect the expected lifetime of the circuit. The gate oxide of AHC devices is only 200 Å thick. An input voltage of 7 V corresponds to a field strength over the gate oxide of 350 kV/cm. Although breakdown of the oxide is expected only at input voltages above 10 V, electrons tunnel increasingly into the gate oxide at field strengths greater than 350 kV/cm, influencing characteristics of the transistors and causing failure.

So it is possible that the 12 V has already done some damage to your chip.

In any case, the presence of the regulator shows that the circuit is likely to be sensitive to changes in the supply voltage, so it is quite possible that the chip runs outside the guaranteed electrical characteristics.

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You are supposed to provide a power supply that limits the voltage to 7V or less. The manufacturer hasn't designed this part to run at 12V and it may not function properly.

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    \$\begingroup\$ It's more correct to say that you're supposed to provide 5.5V or less, as that's the maximum in the recommended operating conditions. It's not guaranteed to work outside of that, but it is guaranteed not to be destroyed by voltages less than 7V, its absolute maximum rating. \$\endgroup\$
    – Hearth
    Feb 11 '19 at 1:31

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