SSR cooling requirements (not a wiring question)

[oley55]

I just added a second PID and 25 watt SSR to my previously built setup. Although my project box had vent slots in the back I went with the externally mounted heatsink: https://www.auberins.com/index.php?m...roducts_id=244

The SSR is normally mounted in the center of the much larger heatsink. The heat sink is large enough to mount (2) 25 AMP SSRs side by side after drilling and tapping new SSR mounting holes.

I wonder/worry if the single 25 AMP specific heatsink is adequate for both 25 AMP SSRs operating simultaneously. The 25 watt SSRs seem like overkill for our low watt/AMP lead pots or hot plates. Being on the low end of the SSR's capacity does that mean less heat being generated? Or does the number of on/off cycles generate the heat? Is there a high temp failure point for SSRs? I was thinking I could monitor the SSR's temperature by checking the heatsink with a thermal gun?

I think I'm ok but thought I'd throw the question out there.

edit added: I see the cut in temp for a cooling fan is 113 degrees. I suppose any temps over 113F are approaching component critical temps. https://www.auberins.com/index.php?m...ort=20a&page=2

[Cast10]

My understanding is it is rated for a max of 25 amps. But it functions continuously at a much lower rate.

That’s the reason so many folks go to 40 amps, as my studies revealed.

My first session with my new PID/40 SSR indicated no heating in the box, without heat sink, too.

[JonB_in_Glencoe]

1. Being on the low end of the SSR's capacity does that mean less heat being generated?
Technically no, but it's not that simple. The wattage dictates the heat being generated. If a 25 amp SSR is loaded with 6 amps will generate x amount of heat. If a 40 amp SSR is made in the same package, it will generate the same, x amount of heat if loaded with 6 amps. Often these components are packaged differently to better dissipate heat generated, with or without heat sinking. So if the a 40 Amp SSR is packaged in a larger and more efficient heat dissipating package, while the heat generated is the same, the SSR will remain overall cooler. I've seen 25amp and 40amp Chinese SSR's made in the same package...add to that, amp ratings and other specifications are often fudged by the Chinese.

2. Or does the number of on/off cycles generate the heat?
Electrical current generates heat, so a higher "rate" of on/off cycles will generate more heat.

3. Is there a high temp failure point for SSRs?
Yes, Heat is a major factor in electronics failure. There should be specifications on your particular component.

4. I was thinking I could monitor the SSR's temperature by checking the heatsink with a thermal gun?
That should work.

If you need more specific answers, you need to provide more specific details.

My own anecdotal evidence:
My Homemade PID is in a perforated metal box to run a Lee 4-20 pot that draws 6 amp when it's heating up. The Chinese 25Amp SSR is mounted inside the box directly on the perforated metal. The area of the box will be slightly warm to the touch during the heat up time, but remains cool to touch during normal operating during the casting session. If your heatsink gets hot, I would consider a cooling fan.

[Mal Paso]

SSRs generate a little over 1 watt of heat for every amp controlled so When On a single pot will produce about the same heat as an old fashioned 7 watt night light. Less as it reaches full temperature an is not on as much.

A 40 amp SSR is just overkill, they were probably cheap and the number looks impressive.

[popper]

SSR don't like heat. The on-off time is very small but increases with temp, which increases temp. Called thermal runaway. They are spec'd for a standard voltage loss at rated current/temp so yes, a higher rated device is better. Probably cheaper to get a 40 amp than hooking up a fan.

[BamaNapper]

The heat sink you have states that it provides cooling for a SSR running at 25 amps. It does not matter if that 25 amps is the total load of 1 or 2 SSRs mounted to the heat sink, or 25 small ones passing 1 amp each. The heat generated by a SSR (or multiple SSRs) is based almost solely on the current through it/them. And your heat sink does not seem to require a fan to dissipate that much heat.

If you have 2 SSRs on the heat sink and the combined current is less than 25 amps, you should be fine without a fan. If you're running a couple 700 watt pots, all should be good. If you're running 1200 watt pots you're pushing the limits.

BTW, a PID cycles on/off as needed. If you assume it's on only half the time when the lead's up to temp, that means that once the lead is hot the SSR is only generating half the heat.

A rule of thumb I've used my whole career is that if you can comfortably touch the heat sink, the device is going to be happy. Much hotter than that and the device is probably going to fail prematurely.

One last comment: use thermal paste. A little bit of paste makes a big difference in the amount of heat that will be transferred from an electronic device to a heat sink.

[Half Dog]

^^^^ +1 on the thermal paste.

[oley55]

I used a Thermal Pad under each SSR.

I just ran my pot and hot plate together from cold start, did numerous major cool down and reheat cycles. My ambient heatsink start temp was 86 degrees and the peak temp did hit 98, but only for a minute or less. Average normal heat cycles generated 93-94 and routine operation generated 90-91 degrees. (heatsink temps via infrared thermal meter)

I feel good with the setup and have no worries over heat killing my SSRs.

Many thanks to all for your thoughts and expertise.

[Mike W1]

I like to, and have posted this comment many times and here goes again. I did not originate.Before I retired from Zeiss, at least 2-3 times a year I had to engineer systems
that would measure the exact temperature of machined parts prior to
measuring their dimensions. (Engine blocks, heads, etc. in Big 3 auto plants.)
So I've done more than my share of study of temperature measurement
methods. Let me say categorically that these non-contact "laser" measurement
systems are almost worthless. They measure infrared radiation from the target,
and that radiation varies wildly with what is called the "emissivity" of the
surface. The industrial suppliers of these things claimed that they could tell me
a temperature within 2-3 percent of displayed value, IF we sent a large
number of samples to them so they could measure the emissivity and calibrate
their sensors. Any time a sensor required replacement, they had to go through
the whole rigamarole again. Any time the surface finish changed, ditto.

Without calibration, and given unknown surfaces, they wouldn't guarantee
accuracy to better than 25% of the displayed value! And that's industrial
grade equipment costing thousands of $$. Bottom line, if you calibrate a
Harbor Freight instrument against an accurate contact thermometer, you may
be with 5% for the exact object you calibrated for. Any other object, you
can be wildly off.

BTW the laser has nothing to do with the measurement process. It's just an
aiming aid.

Best method is a low-mass thermocouple or RTD device, (thermistor).
These are not expensive. Some under-$100 Volt-Ohm-Meters come with a
thermocouple probe. I have had two of these, and they worked pretty well.
I've never stuck one into melted lead, but there are versions that are
encapsulated in ceramic which I would.

[jim147]

I've ran an outside walk in 220 volt cooler for years with a SSR mounted in a black metal box with no heatsink. Unless it is just a cheap SSR I don't think you will have a problem.