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
As the datasheet says, the device has an open-drain NMOS output.
This means the output of the device looks like this:
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.
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"