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I'm new to electronics and designed a four-layer SMT board,

I am seeing a temperature rise of around 10°C (17°C ambient to 27°C after a few minutes) of a few components and the PCB, and I'm not sure why or if it's even a problem.

Using a thermal camera I have narrowed it to be in the area surrounding the optoisolators. Using the probe on my multi-meter the opto's are 27°C after a few minutes.

  • DIR_F_L ../.. is 5 V from an Arduino 22 mA.
  • FLOUT ../.. are 5 V with 1k pullup 5 mA.

I believe the resistors could be a potential source of heat generation as they are 0805 1/8 W.

Circuit digram

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    \$\begingroup\$ P = U*I applies regardless of the package of a resistor. Temperature rise depends on the watts dissipated, and the thermal resistance to the ambience. \$\endgroup\$
    – sh-
    Jan 6 '21 at 19:59
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    \$\begingroup\$ Even with conservative worst-case values from here: vishay.com/docs/53048/pprachp.pdf you're still only looking at 6C rise for the 560 ohm resistor and actual rise is probably less than half of that. \$\endgroup\$
    – vir
    Jan 6 '21 at 20:06
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My apologies - in my earlier comment, I used 5mA as your driving current straight from the datasheet and didn't double-check your resistor value. With a typical forward voltage for the LED of 1.2V, your 560 ohm resistor is dropping 10.8V at a current of 19.3mA. Power dissipation inside the opto is 23mW, well within limits, but your resistor is dissipating 209mW, which is closer to the 1/4W rating than is really proper. The Vishay data therefore gives a very conservative (high) temperature rise of 28C on a "normal" copper board with natural circulation. I would recommend bumping the resistor up to 1.07k to drop current to ~10mA which should be enough to pull down your output at worst case current transfer ratio and reduce the power dissipated to 107mW.

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  • \$\begingroup\$ Thanks for the update, This morning I ran the board and put the thermocouple directly onto the 12v resistors, they are immediately getting warm. The 12V resistors are 1/4W The 5V resistors are 1/8W Should I, Increase the package size i.e 1/2W and 1/4W? Would this reduce the heat? Or should i recalculate the CTR of the optos and reduce the current to say 15ma? \$\endgroup\$ Jan 7 '21 at 9:35
  • \$\begingroup\$ Up to you, really. 1/2W resistors would probably be fine with the values you have now. The heat generated would be the same, but you wouldn't be as close to burning up your resistors as you are now. Worst case CTR for that opto is 50% so as long as you have 10mA going through the input you'll pull down your output. You might also see if you could increase the pull-up resistors to say 10k and drop the input current to the opto even more to the 5mA they seem to use as the recommended minimum. \$\endgroup\$
    – vir
    Jan 7 '21 at 17:00
  • \$\begingroup\$ OK, The 5V resistors have all been upgraded to 1/4W resistors, but values left as is. The 12V resistors have been changed to 820R but left at 1/4W, I believe the dissipated power will be 142mw. \$\endgroup\$ Jan 7 '21 at 17:10
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I see a major thermal problem using 12V to drive the LEDs with 560R.

You want to never exceed 50% of power rating of device to operate at 50% of it's 100'C rated temp rise to 125'C from 25'C.

If you want to use 1/8W R's then raise impedance x5

Rin = 2.7 kOhm Rc= 5 kOhm

For same CTR of 4mA input to 1mA output = 25%

enter image description here

You can also go lower with Rin=5.4k and Rc= 10k as this uses 10% of 20mA rating as the reliable minimum operating current of LED for long term.

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If the inputs are active it is more likely the isolators themselves, but in any case it is an unusual amount for the application.

Are you certain the temperature increase applies to the entire PCB ? It is not clear to me if you are reporting a local rise or overall rise.

Furthermore, an IR camera is not a very reliable method for quantifying thermal rise, although it is very effective for finding hot spots, it should not be relied upon to quantify the thermal change with any accuracy. It would not surprise me if the components on the PCB would show up brighter in the it camera after warming up.

5mA pullups at 5V is .025W of power, likely not the cause of the thermal increase of the whole pcb, you may observe some local increase of the component. The 1/8W rating is a maximum rating and you are currently at 20% of that rating.

Local heating of a passive component to a steady state is generally benign, sometimes it may change the value of the component in a circuit in a way that requires consideration but this is not a concern for a pull-up application.

Particularly warm components or small pcb can result in overall heating of the whole assembly. A good ground plane with many solid connections improves the thermal conductivity of the assembly. Mounting screws connected to this ground plane further extract the heat to the enclosure if necessary. A dedicated heatsink for the component can be considered if the plane cannot handle the thermal current.

Sometimes , thermal isolation or positive feedback can lead to a situation where the temperature of a component does not stop increasing (even possibly over course of many hours) leading to thermal runaway and catastrophic failure.

For more thermally sensitive and high power devices there is a universe of detail and detailed modeling available to consider so the above is just a brief overview that is appropriate for a basic digital design like the isolator design you presented.

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  • \$\begingroup\$ Thanks, the input of the opto's are active from an Arduino pin at 22ma, 0.11W. A comment on the thread indicated at worse a 6'c temperature rise. I'm using a thermocouple with my multimeter to measure the component temperature, but the PCB is via the thermal camera, but locally near the opto's feels warm. \$\endgroup\$ Jan 6 '21 at 21:02

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