# Resistor getting hot - up to 50°C. Is it an issue of the resistor wattage or nature of the resistor?

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%.

1. 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?
2. 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?

3. The heating is the same for both 125 mW and 400 mW rated resistors. What is the role of resistor wattage here?

4. If we use higher wattage resistors will they also heat up to the same temperature or will the heating be reduced?

5. Is there something wrong with the circuit? Will heating break the resistor or reduce its performance? How I can reduce the heat?

• What physical type of resistor are you using? All resistors will heat up with current through them - the only question is how much which depends on the temperature coefficient. If you are using a 125mW device with 117 mW of dissipated power you should expect significant self heating. – Peter Smith Jun 3 '19 at 11:17
• It is quite possible you don't need to drive the optocoupler that hard. If all you are doing is switching a digital signal, you can probably reduce the current through the LED to just 1 mA. That would greatly reduce the power wasted in the resistor, and greatly reduce the heating. But, it depends on the circuit you are driving on the output side of the optoisolator. – JRE Jun 3 '19 at 11:25
• The datasheet for your resistor may indicate what the expected temperature rise is given the load power. You can use that to see if 50°C is about right. Second the datasheet rated power is only valid if the circuit pads are the same as shown in the datasheet. If you have not used the manufacturer's recommended PCB footprint then you will not get the performance on the datasheet. – scorpdaddy Jun 3 '19 at 12:20
• 50 deg C is not that hot, assuming that this board is always operated at reasonable temperatures, you will be fine. Did you use wide traces to conduct the heat away? The vendor should give you some guidelines. – Mattman944 Jun 3 '19 at 14:09

Any 1k resistor will have to dissipate the 118 milliwatts. The resistor rated wattage just tells if the resistor can handle it or not. If all else remains the same, they all will end up at the same temperature. As resistors that can handle more power are larger and contain more mass, they just take longer to reach that temperature. To reduce the temperature, the heat needs to be conducted away from the resistor via PCB or airflow.

• Typically, the higher the power rating the resistor has, the bigger dimensions it has. So, likely also more surface area for convective heat transfer, which may result in a relative lower steady state temperature. – Huisman Jun 3 '19 at 12:11

Point 1:

The Watt rating of a resistor refers to its maximum power dissipation.

The heating you are referring to is joule heating given by P = I^2 R

As you can see the heating is not related to the rating of the resistor. The watt rating is an absolute maximum

To determine the amount of heat the resistor will generate / can take you will need to look at the datasheet

https://en.wikipedia.org/wiki/Joule_heating

Point 2:

Ic max is 50mA but this has no bearing on the led current. If the led is saturating the transistor then 50mA is the max collector current, the collector current is dependant on what load you connect to the transistor

Point 3:

...See point 1

Point 4:

...See point 1

Point 5:

That depends on the resistors ratings for max temp, to reduce the heat: ...See point 1

• If the PCB has sparse regions of copper to provide heat flow paths (no underlying GND planes), then the region will heat up regardless of the resistors size. Standard-thickness copper foil (1 ounce/foot^2, 1.4 mils thick) has thermal resistance of about 70 degree Centigrade/square for any size square. – analogsystemsrf Jun 3 '19 at 11:52
• Maybe i'm missing something but I don't see how this relates to my answer? – LazyMoggy Jun 3 '19 at 12:46
• Thank you all for your answers. I understand, Power will be dissipated as Heat and it is uncontrollable unless the current through the resistor should be low. In Resistor Datasheet specified as, 70°C is the common operating ambient temperature for all the rated power (ex: 1/10,1/8,1/4,1/2 W… etc). So all lower wattage and higher wattage resistors will be Derated once it hits 70°C. – Naveen K Jun 5 '19 at 11:52
• As per the graph upto 70°C is fine for working. In my Circuit 1KΩ/125mW/±100ppm/°C and 1KΩ/400mW/± 200ppm/°C Resistor dissipate 118mW and both Heatup upto 50°C. If I increase the Power dissipation to 300mW, 1KΩ/125mW resistor may heatup upto 150°C and it may breakdown. My question is what will happen to 1KΩ/400mW Resistor when power dissipation is 300mW? In 118mW case it was 50°C,What is the temperature when its dissipate 300mW? We can calculate Power Dissipation by using P=I^2*R, Is there any calculation/Formula for Calculating the Temperature for the above case as example? Thanking You – Naveen K Jun 5 '19 at 11:53