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I am using 6n137 opto-coupler to drop down 5v pulses to 3.3v level pulses. I found in data sheet the minimum input voltage is 4.5V at Vcc. and the circuit given in the data sheet has pulled up Vo to Vcc using RL resister. Is it ok if i supply Vcc with 5V and pull up Vo to another separate 3.3V level,to get 3.3V pulse output? as shown in below figure. data sheet 6n137 enter image description here

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  • \$\begingroup\$ Yes, it is OK to pullup to 3.3V. \$\endgroup\$ – Todor Simeonov Jul 15 '17 at 21:33
  • \$\begingroup\$ @Todor Simeonov the way i am gonna do is in above figure UPDATED ! can you check ? \$\endgroup\$ – oppo Jul 15 '17 at 21:34
  • \$\begingroup\$ Yes, your schematic is correct. You may need a smaller value for R1 to achieve better turn-off (rise) times. \$\endgroup\$ – Todor Simeonov Jul 15 '17 at 21:39
  • \$\begingroup\$ Absolutely correct. You can pull the output up to 3.3 V and have it operate the way you want. The output of the device is a conventional Open Collector. See this data sheet: farnell.com/datasheets/1057512.pdf \$\endgroup\$ – Jack Creasey Jul 15 '17 at 22:36
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    \$\begingroup\$ pullup R and load C determine rise time and 15pF with internal switch control fall time to approx ESR of 600mV max /Iol=13mA = 46 Ohms \$\endgroup\$ – Sunnyskyguy EE75 Jul 15 '17 at 23:00
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As the datasheet says, the device has an open-drain NMOS output.

This means the output of the device looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Which means the output cannot drive itself to a voltage, but can pull the signal down to ground. It does have a small internal leakage of up to 1μA so if it goes into a very high impedance input, even without a pull-up resistor it will eventually float up to a voltage between GND and Vcc.

But, since the device cannot drive a voltage on its output directly, you are free to pull-up to any voltage you require, just keep an eye on the fact that the output will only go to 0.3V to 0.6V if the resistor is small (hundreds of Ohms), so for very low voltage signals, like 1.2V that may not be low enough to be considered "off".

The value of your resistor will depend on how fast you want your signal to go back up. If you have fast signals, in the MHz range, or need fast flanks, you'll probably want less than one kilo Ohm. For slower stuff and slower edges, say 10~100kHz range or even lower, probably 3.3k to 5.1k will work fine. But it does depend on the amount of trace or wire connected to the output as well. Bigger trace makes it slower, because those have more capacitance.

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Correction

6N137 is 5V+/-10% Vcc IC with a CMOS "compatible" output using a pullup R and an sink current equivalent to approx 50 Ohms +/-50% but limited to 50mA max for thermal reasons.

Thus 330R pullup to 3.3V or total series R of 50+330 results in 8.6mA and Vol= 50*8.6= 430mV +/-50%.

Higher R* for given C operating load values can be used according to "need for speed"

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    \$\begingroup\$ The 6n137 is NOT a CMOS output. See this datasheet: farnell.com/datasheets/1057512.pdf .....while there may be slight differences in the architecture for various manufacturers, the output is Open Collector ....so the OP can pull it up to 3.3 V and have it work as desired. You don't need a resistive divider at all. \$\endgroup\$ – Jack Creasey Jul 15 '17 at 22:32
  • \$\begingroup\$ I stand corrected Vishay indicated CMOS compatible, I didn't see an open drain on collector on their spec, but I see the Ioh now. Your 2005 Fairchild spec, however, does show it. My err. going too fast. \$\endgroup\$ – Sunnyskyguy EE75 Jul 15 '17 at 22:54

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