Hello all I'm trying to make a small circuit in which i have to choose the resistance value.. can someone tell me how to choose the resistor value at the input of Optocoupler IC. And my input voltage is 3.3V .. and current 25mili amps ..

 

What is the output of the opto connected to?

 

It's just like choosing the resistor for any other LED. If you want to achieve significant frequency, then it needs a more careful choice - but if you wanted it to work at any speed you wouldn't have chosen an 817.
Choosing the pullup resistor on the output also needs to be considered.

 

Hello, I have this optocoupler reference image i have sheared having four legs .. first pin i want to give input with resistor and second pin i want to connect LED light and third pin I want to connect with resistor and resistor output to npn transistor.. and fourth pin to 5V ..

basically I'm trying to make a optocoupler relay module trigger voltage 3.3V . and power voltage 5V.

IC pin like 1 opposite 4, 2 opposite 3. this is what inside circuit of IC..

 

forward voltage of the optocoupler diode is 1.2v in the datasheet. You can do the math to figure out what size resistor you need to put in series with that to give you 20mA. The datasheet listed 20mA as forward current. Or you can use a calculator like digikey has (led series resistor calculator) to do the math for you.

 

Another way of doing it is to think of it as a transistor with Vbe=1.2V and hfe=0.5

 

Hi @J253
Is this a College or Homework assignment.?

Moderation.

 

Thank you for this suggestion I have also calculated in my case it's 100 ohm.. and on second pin I have connected an LED 2V 20ma.. now when I'm measuring voltage at first pin it's showing 3.26 and on second pin 2.15.. where is the voltage drop across resistor ..