Resistor value calculation for optocoupler input

Question

I am working on an optocoupler (LTV-845) circuit.

Below is my working circuit:

Forward current flowing to opto input would be

(Vcc-Vf(LED)) / Rf = (24V-2.1V)/470 = 0.046A.

Is my understanding correct?

Also need to understand whether LTV-845 is capable of handling 0.046A, as the forward current is 50mA maximum and collector current is 80mA.

Need clarification.

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This would be my circuit:

What would be the current flowing to the load connected to pin 3 with RL=1K?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Redrawn circuit.

• D1 has a V_f of 2.1 V at 20 mA.
• D2 has a V_f of 1.4 V at 20 mA.
• V_R1 = 24 - 2.1 - 1.4 = 20.5 V.
• At 20 mA, R₁ = V / I = 20.5 / 20m = 1 kΩ.

Also need to understand whether LTV-845 is capable of handling 0.046A, as forward current is 50mA max and collector current is 80mA.

50 mA is the infrared LED's absolute max. You don't want to be operating near that value. Since your LEDs are in series they both have to have the same current. The maximum current will have to be set by whichever has the lower limit.

The current transfer ratio of that device is quite high - over 600 at I_F = 1 mA. This means that you probably don't even need 20 mA in the LED.

Don't forget to calculate the power dissipated in R1. The 1k will have P = I²R > 900 mW. Reducing the current to 5 mA would require a 4.1 kΩ and power would be 100 mW.