I am using an FOD817DSD optoisolator.
According to the datasheet:
- CTR: 300% - 600% @ 5 mA
- VF = 1.2 V (typical) @ If = 20 mA.
We have 10 optoisolators on the PCB which is connected to some other board IO pins. There is no proper information on required IC for the third party board. I assumed max 15 mA is sufficient for any digital IOs.
I have used 12 V supply through 1k SMD resistor (tried with 125 mW, 250 mW, 400 mW) to the optocoupler input. Voltage on input pin is 1.15 V (I can get 1.2 V If the resistor is 540 ohm, but heating > 50°C) and collector voltage is 5 V, everything works fine. But all 1k input resistors are getting heated up to 50°C. Even though I tried with 400 mW resistor.
I found this formula in the datasheet, CTR =(IC / IF) * 100%.
- IF=(12 V - 1.2 V) / 1k = 10.8 mA But am getting 1.15 V instead of 1.2 V. So power dissipation in the resistor is P = 10.8 * 2 mA X 1k = 117 mW. I have used more than 125 mW rated resistors. Why is it getting hot?
As per the formula “IF” should not exceed 17 mA because given IC max is 50 mA. What then is the meaning of "VF is 1.2 V @ IF is 20mA" as mentioned in the datasheet?
The heating is the same for both 125 mW and 400 mW rated resistors. What is the role of resistor wattage here?
If we use higher wattage resistors will they also heat up to the same temperature or will the heating be reduced?
Is there something wrong with the circuit? Will heating break the resistor or reduce its performance? How I can reduce the heat?