fighting with heat

Hello All

I`m connecting to the optocoupler ILQ615 throught resistor network 2.2kohn 1/4W ( i know should be 1/2W.

supply 4 x 28V.

with all 4 inputs ON that resistor gets up 70degC.

Instead i would like to put maybe 2 or 3 1/4W resistors in series to split the load and heat generated.

3x 680 ohm 1/4w in series will do? or maybe 2x 1k 1/2w?

Any advice greatly appreciated.

Hi,

Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?

It will make you explanation easier to understand.

Thanks… Tom… :slight_smile:

mOskit:
with all 4 inputs ON that resistor gets up 70degC.

Yeh, 28V / 2.2k = 12.7mA = 0.356 Watts

You should use 4 separate resistors, one for each input. If you have 'em use 2k7 1/2W

If you only have 680ohm 1/4W and 1K 1/2W, then either of these are ok:

3x 680 ohm 1/4w in series will do? or maybe 2x 1k 1/2w?

or 680ohm in series with the 2k2 will be ok.

Yours,
TonyWilk

Sorry.

Screenshot attached

You should use 4 separate resistors, one for each input. If you have 'em use 2k7 1/2W

Can we estimate heat dissipation or temperature is going to reach?

Hi,
Ops circuit.
e2fb51da5bd33ee6341ce0ef7587f02b6b8160a9.jpg
Tom… :slight_smile:

Hi,

If all four optos have to come on together, why not connect all the opto inputs in series.

That will mean you only need one string of resistors, their value will be lower because your volt drop across the optos will be 4 times bigger than it is now.

Any reason you need to opto isolated some LEDs?

You could switch them with a MOSFET instead of optos and not have the current and heating problem you have now.

Thanks.. Tom.. :)

Hi Tom

Sketch is simplified view of all.

Each opto input is independent. if you look closer, there are “D” marks on the outputs. Led is indication only purpose.

I wanted to isolate the circuit from 750VDC if the ark decide to go through the machinery :smiley:

mOskit:
Can we estimate heat dissipation or temperature is going to reach?

Heat dissipation: easy, just calculate resistance * current2.

Temperature reached: this depends on the thermal properties of the part, and how it can dissipate its heat (ambient temperature, forced air flow, heat sink). Much harder to calculate.

I`ve bench tested this just now and as before, with “Resistor Network” circa 70degC.

With 3x 820R in series, barely 30degC.

Hi, Do you have a power supply lower than 28V that you could use to supply the opto input and use a MOSFET on each opto input?

If you do use the resistors in your diagram. When you put them into the PCB, mount them up on their leads off the PCB to allow air flow and minimize damage to the PCB. Make the copper pad you solder the resistors to as large as possible so heat that travels down the leads does not cause the pads to eventually lift.

Tom.... :) (I'm off to bed.. will check in the morning)

or you can use full watt resistor instead of half watt or quarter watt, that means, the biggest resistor in size but the same resistance. they mostly are grey in colour

No wonder - you did go from 356 mW on a 1/4W part, to 320 mW over three 1/4W parts, or just over 100 mW for each.

This is all the reason i`m fighting with the heat.

That 30-40degC is i thin acceptable. i cannot use any method of ventilation and the circuit must be sealed and potted in thermal conductive compound outside.

Potting normally has very poor thermal conductivity. Even just putting it inside a sealed container will make it heat up significantly. Those thermal ratings are typically for when exposed to freely convecting air.

In that case, using a 1 Watt resistor won't make any difference. 356 mW dissipated is 356 mW dissipated. Period. The final temperature in the enclosure will be the same.

mOskit: i`m fighting with the heat.

Whatever you do, there's still the same amount of heat (still the same no. of milliWatts dissipated)

Best you can do is 'spread it about a bit' as you have done with the 3 x 820R

It's sometimes a good idea to really overkill the resistor rating - say a 5W or 10W wirewound ceramic, simply because it is larger with more surface area.

That 30-40degC is i thin acceptable. i cannot use any method of ventilation and the circuit must be sealed and potted in thermal conductive compound outside.

That doesn't sound too bad as long as the potted thing has free air around it or is bolted to the inside of a metal enclosure.

Yours, TonyWilk