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Old 02-08-2010, 01:27 AM   #1
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Default SSR Heat Output?

Is there a way to determine the heat put off by an SSR?

I'm using 25 amp SSR to control my March 809. The pump is rated at 1.4 amp @ 115 volt. I also have SSR's controlling my 24 volt gas valves. Same 25 amp SSR's.

I don't think it will put off much heat but figured it would be a good topic for those who run heating elements with bigger loads. And I'd like to know if my plan to mount the SSR's to a piece of 1/4" aluminum is sufficient or overkill.

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Old 02-08-2010, 02:23 AM   #2
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That SSR won't get warm with the pump. No heatsink needed.

There should be resistance value for your SSR. It is probably a curve, Resistance vs. Current (your current is, in this case, 1.4A). Check the datasheet.

Watts = I²R

Now, how hot will it get? No idea. Without a heat sink, you can probably run 5A safely. More than 10 and that thing will get really hot.

I run 23A through a 25A SSR with a big heat sink, no problem. Barely gets warm.

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Old 02-08-2010, 02:29 AM   #3
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Isnt it the switching that makes it warm?

I mean... if I have the SSR on 100% of the time, will it still get hot?

Ive used them for a while, but always on elements that were running 1sec. duty cycles.

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Old 02-08-2010, 02:59 AM   #4
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Originally Posted by The Pol View Post
Isnt it the switching that makes it warm?
No, it's a function of the current passing through the semiconductor switch (due to internal resistance). I've seen figures quoted in 1.1 - 1.3 watt/amp range.

So, assuming 1.3 watt/amp, an SSR will dissipate 30 watts with a 23 amp continuous load (23 x 1.3 = 30). 30 watts of heat dissipation is not insignificant....Now, scale that to the "9K jigawatts" you'll be running on the new rig and you can fry some eggs on the back of your control panel.

But ya, less than 5 amps does not require a heatsink. Ambient air cooling is sufficient.
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Old 02-08-2010, 03:14 AM   #5
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Thanks guys, very good info.

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Old 02-08-2010, 11:22 AM   #6
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Huh, good to know!

My current rig actually sends less current across the SSRs than my last one did, and that heatsink barely got warm. Cool. Thanks.

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Old 02-08-2010, 12:10 PM   #7
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Quote:
Originally Posted by lamarguy View Post
No, it's a function of the current passing through the semiconductor switch (due to internal resistance). I've seen figures quoted in 1.1 - 1.3 watt/amp range.

So, assuming 1.3 watt/amp, an SSR will dissipate 30 watts with a 23 amp continuous load (23 x 1.3 = 30). 30 watts of heat dissipation is not insignificant....Now, scale that to the "9K jigawatts" you'll be running on the new rig and you can fry some eggs on the back of your control panel.

But ya, less than 5 amps does not require a heatsink. Ambient air cooling is sufficient.
Close..it's the current and the voltage drop of the semiconductor, but yes, 1.1-1.3W/A will get you in the ballpark.

The internal resistance of semiconductors is so small it's insignificant at low current. Now on a 10,000A recitifer, it's a differenct story!
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Old 02-08-2010, 01:56 PM   #8
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Quote:
Originally Posted by The Pol View Post
Isnt it the switching that makes it warm?

I mean... if I have the SSR on 100% of the time, will it still get hot?

Ive used them for a while, but always on elements that were running 1sec. duty cycles.
Pol, you are correct, sort of. The time during which the SSR is switching from ON -> OFF (or OFF -> ON) does consume much more power than the constant ON state. However, it switches so fast and (relatively) infrequently that it is insignificant compared to the ON state dissipation.

OFF -> ON time for my SSR is stated as 0.1 msec (0.0001 seconds!).

It is W*t that gives total energy (joules).
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Old 02-08-2010, 03:25 PM   #9
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Default Calculating an SSRs Internal (Junction) Temp

The SSRs typically have a limit of 125C internal junction temp. You must stay below this!.

To figure out what that junction temp will be, you need to 1) add up all the thermal resistances in your system between that junction and air, then 2) multiply that sum times the SSRs internal power dissipation (watts). This wattage is NOT the load tied to the SSR ... more on that below)

1) Thermal Resistances
Without a heat sink, the resistances are:
  • Junction -> Case (case is the metal back of the SSR; see SSR datasheet)
  • Case -> Air (not usually stated in datasheet; use 20 C/W)

With a heatsink,
  • Junction -> Case
  • Case -> Heatsink (with thermal grease, use 0.1 C/W)
  • Heatsink -> Ambient (air) (see Heatsink datasheet)
2) SSR Internal Power Dissipation
The SSR internal power dissipation can be calculated by multiplying the SSR's max voltage drop (datasheeet) times your intended current. This number is the amount of the heat we are trying to dissipate. See example below.
Example
I have a Crydom D2425 SSR with the following specs:
  • Max Load Current: 25A
  • Max Voltage Drop (@ max current): 1.6V
  • Thermal Resistance, Junction -> Case: 1.02 C/W
  • Heat Sink Thermal Resistance: 0.7 C/W
  • My Load Current (5500W heating element): 23A

1) Total Thermal Resistance: (1.02 + 0.1 + 0.7) = 1.82 C/W
2) SSR Power to Dissipate: (1.6 * 23) = 36.8W

So, the increase in the SSR's junction temperature, above ambient, will be

Thermal Resistance * Power = Max Temp Rise
1.82 * 36.8 = 67 C.

If Ambient is 23C, the junction temperature of the SSR will be 90 C during a constant load of 23A.
Example 2
Let's say you know your load, but you need to find a good heatsink for your SSR.

Using my example above, I know I will generate 36.8W of power in my SSR. I want to keep the SSR junction from rising more than 100C above ambient.

Max Temp Rise = Thermal Resistance * Power
or
Thermal Resistance = Max Temp Rise / Power

100 / 36.8 = 2.71 C/W <-- that is for all the resistances, junction to air

Since I already have about 1 C/W from the SSR junction to the case, my heatsink must be less than about 1.7 C/W. I chose one that is 0.7 C/W, so I'm good!
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Last edited by passedpawn; 02-08-2010 at 03:49 PM. Reason: I'm stupid. Thanks Bjorn.
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Old 02-08-2010, 03:35 PM   #10
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Quote:
Originally Posted by passedpawn View Post
2) SSR Power to Dissipate: (1.6 * (23^2)/25) = 33.8W
You total power is V*I + I^2*R

The I^2*R = 23^2 * .0002 (if that) = 0.1W ....it's minuscule

So really it's just V*I = 23 * 1.6 = 36.8W

The rest is spot on, nice.
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