0
\$\begingroup\$

This question already has an answer here:

I have a circuit like below.

I wanted to know how these resistance values were calculated. I wanted to use 12V/24V in place of 9V.

Let me know selection of opto-coupler and resistance value.

enter image description here

\$\endgroup\$

marked as duplicate by CL., JRE, RoyC, DoxyLover, markrages Jun 25 at 7:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ Do also check possible duplicate: Resistor value for Opto-coupler input \$\endgroup\$ – Huisman Jun 20 at 6:18
  • \$\begingroup\$ @CL. What's the additional value of repeating my comment? \$\endgroup\$ – Huisman Jun 20 at 6:27
  • \$\begingroup\$ @CL Ah, thanks for clarification \$\endgroup\$ – Huisman Jun 20 at 6:33
0
\$\begingroup\$

Follow the next steps:

  1. Calculate the value of resistor R1 (for LED) --> U = I x R --> R = U / I We know the following things:
    • Uin = 12 V
    • Uled (opto) = 1,2 V (take a look into your datasheet)
    • ILed (opto) = 4 mA (take a look into your datasheet)

So, the total voltage = 12 - 1,2 = 11,8. R = U / I --> R = (11,8 / 4x10^-3) = 2950 Ω = 3 kΩ

  1. Calculate the value of resistor R2 --> U = I x R --> R = U / I we know the following things:
    • 0.3 V – Saturation voltage of photo transistor (voltage across photo transistor when current flows through it)
    • Ic= Collector current (Current flowing through photo transistor when light falls on it) see datasheet. Let take 1 mA in this case.

So, the total voltage = 5 - 0,3 = 4,7 V . R = (4,7 / 1 x 10^-3) = 4700 Ω = 4,7 kΩ

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.