I'm using a 40A 120VAC SSR to control a 10A resistive load. It gets hot during use. Is this heat associated with the switching, with the load current, or with the voltage? And where does it come from?

(Please provide an explanation that can be understood by a trade electrician with little electronic engineering knowledge.)

  • \$\begingroup\$ Define hot. Is it heatsinked? If hot is "above 70C", that's probably not so great. \$\endgroup\$
    – uint128_t
    Feb 6 '16 at 20:00
  • \$\begingroup\$ Yes, it's on a heatsink in a ventilated enclosure. I'd define hot as "hot enough that I wonder if it's worth using over mechanical relays." \$\endgroup\$
    – feetwet
    Feb 6 '16 at 20:08
  • 1
    \$\begingroup\$ What is the model of the SSR you are using? Why are you using a SSR instead of a switch or relay? Is your SSR an SCR type of Triac type? I am a trade Electrician. 20 Years ago when I started using SSRs in response to Contactor failures, and I observed heating of the SSR - I tried to calculate the Heat Losses and started experimenting with Heat Sinks and Different types of SSRs. A good quality SSR will have a datasheet explaining some of the Heat Losses you are describing. \$\endgroup\$
    – Tinkerer
    Feb 6 '16 at 20:09
  • \$\begingroup\$ @Tinkerer I'm using the "Inkbird" SSR that came with this nifty PID temperature control kit. \$\endgroup\$
    – feetwet
    Feb 6 '16 at 20:15
  • \$\begingroup\$ Somewhat telling that Inkbird and part number (SSR-40 DA) come up with no datasheets, just plenty of sales links. Not the sign of a reputable manufacturer, IME. I note also that the kit specifies the physical size of the heatsink, but not its thermal resistance. As it does not appear to be a faithful ripoff of Crydom's heatsinks (crydom.com/en/service/crydom_productlineoverview.pdf) I suspect it's a poor heatsink design - note that all heat has to flow out the center rib, rather than having fins attached directly to the spreader plate... \$\endgroup\$
    – Ecnerwal
    Feb 6 '16 at 20:43

SSRs (solid state relays) are usually SCRs or triacs that are activated via a opto-coupler, all built into a single package. One problem with triacs and SCRs is the voltage drop across them when on. This can be 1.5 V or more. Check the datasheet of the SCR you are using. It should tell you the voltage drop.

Electrical power into a device is the current thru it times the voltage across it. If the SSR drops 1.5 V and you're putting 10 A thru it, then it will dissipate 15 W. Again, check the datasheet to see what kind of heat management might be required of this SSR.

There are SSRs that have lower voltage drop by using FETs, but these usually don't handle much current and are usually more expensive.

If you're not planning to switch often, then perhaps a mechanical relay might be more suitable. These don't have one nice features of SCR and triac SCRs though, which is to turn off at the current zero-crossing. That reduces emissions.

I know of one commercial product that used a SCR and relay in conjunction for switching AC line loads. The SCR did the actual on and off switching, nicely synchronized to the AC zero crossings. The relay did the heavy lifting by being turned on a little later and off a little earlier than the SCR. The SCR therefore only dissipated power for a short time, and the relay never saw high switching stresses. The lifetime tests suggested over 1 M switching cycles, something the relay would otherwise not have been good for.

To be clear, I'm not suggesting you do something like that. I mention it to illustrate some of the properties of these things and how they have been harnessed and worked around in the past.


When the SSR is switched ON, it takes a little bit of power to do the switching, which heats it up a little, but when it's switched ON, the load current has to go through the switch's ON resistance, which causes the bulk of the heating.

Consider: if the SSR looks like a tenth of an ohm when it's ON and there's 10 amperes going through it, into the load, then the relay will be dissipating:

$$P = I^2 R = 10A^2\times 0.1\Omega = 10 \text{ watts} $$

  • \$\begingroup\$ So the vast majority of the heat generated by a SSR is due to the resistance of the relay path, and hence the load current times the duration switched on. \$\endgroup\$
    – feetwet
    Feb 6 '16 at 20:17
  • \$\begingroup\$ @feetwet: Yes; precisely right. \$\endgroup\$
    – EM Fields
    Feb 6 '16 at 20:30

SSRs have resistance when they are on - considerably more than most mechanical relays, actually. Perhaps more resistance than other SSRs if you have some sort of dubious brand which apparently lacks published specifications.

A 40A SSR is only a 40A SSR if it's attached properly to a proper heatsink, or on for VERY short times. Follow the data sheet. Well, assuming there even IS a datasheet.

10 amps of current, squared, times the resistance when on of your SSR = heat. What looks to be a poorly designed heatsink (and is there any sort of thermal pad ensuring excellent contact between the SSR and heatsink?) means they get hot, even with "only" 1/4 load.

  • 3
    \$\begingroup\$ The SCR or triac that is likely in the SSR acts more like a voltage drop than a resistance. Power is therefore closer to linear with current, not the square of the current. \$\endgroup\$ Feb 6 '16 at 20:38
  • \$\begingroup\$ A fixed voltage drop is a resistance that varies with current; looked at as a resistance, the power is still 100X Reff(10A). Since the question does not entail altering the current or voltage and the part has no datasheet from which to extract actual numbers, P = I•V = I•I•Reff(10A) \$\endgroup\$
    – Ecnerwal
    Feb 6 '16 at 21:15

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