Opto-isolator basic question

I have build a standard opto-isolator (4N37) circuit with input coming in on pin 1 &2. The opto is powered by 5V to Pin5 through a 10K resistor, Pin4 is grounded, Pin6 is grounded through a 400K resistor.

When the input side of the opto is low, Pin4 has a voltage, but when the input goes high, Pin4 doesn't have voltage. It is reversed from what I would expect, the other side of the opto would only have voltage if the input was high. I'm sure this is something basic that I am missing and don't understand. Basically, the opto seems to be working as its picking up the changes in one circuit and switching but not in the manner I expect.

hi,
The output is inverted due to the transistor.


low, Pin4 has a voltage, but when the input goes high, Pin4 doesn't have voltage.

Do you mean pin 5 ???


EDIT:
Is it connected as in this image.?

Thank you very much. Yes, I did mean Pin 5 and it is just like the image. How can I invert that signal?

n9rex said:

Thank you very much. Yes, I did mean Pin 5 and it is just like the image. How can I invert that signal?

Click to expand...

hi,
One way you could try, is to connect the collector [pin 5] to +5V and place a 4.7K in the emitter [pin4] to 0V... keep the 400k connected to the Base.

There should be a signal now on pin 4, NOTE: it will not be +5V!...
If the output voltage level is not high enough you will need to add an inverter transistor to your original circuit... OK?

To invert, tie pin 5 to +5 and pin 4 to ground thru the resistor. You inverted signal is on pin 4. Sometimes it is not necessary to have a bias resistor on pin 6 at all.

Awesome, I will try this today. Thank you for helping me with this. I am obviously still learning the basics.

The bias resistor on pin 6 reduces the opto gain but tends to make it faster. So its value determines the tradeoff between speed and sensitivity.