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I'm using this Solid-State-Relay to transfer 350 Watt:

Input: 45VDC x 10A
Output: 10VDC x 10A

The following is derived from it's data sheet:

Maximum junction temperature 150 C
Junction to Ambient 30 C/W
Junction to Case 7 C/W

In order to keep it's features from changing because of a high temperature, I'd rather for it to have a maximum temperature of 50 Celsius degrees.
The ambient temperature is 25 C.

A simple calculation shows that a proper thermal resistance of a heat sink for this job is: (50-25)/350 = 0.07 C/W which is crazy!!!

What do you suggest for me to do in order to keep the device below 50 C ?
Would it be unrealistic to use this kind of component for this kind of task?

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  • \$\begingroup\$ Have you considered using multiple devices? \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Sep 1, 2013 at 20:32
  • \$\begingroup\$ @IgnacioVazquez-Abrams: Multiples of the same SSR? Thought of that, but I think it's not reliable. Not all of them would work synchronously (one will pass power before the other etc). Do you think it's a good solution? \$\endgroup\$
    – Dor
    Sep 1, 2013 at 20:37

1 Answer 1

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Your specified operating conditions don't make sense. The maximum voltage drop across one of these relays is 1.4V at 10A, so you can't have 45V in and 10V out. So, the maximum power dissipation in the relay itself is 14W. The temperature drop from junction to case will be 98C. You need a heatsink with a thermal resistance of (150-98)/14=3.7C/W or less.

If you really want to keep the junction temperature below 50C you must reduce the load current. Suppose you add an infinitely good heatsink, with zero thermal resistance. The thermal resistance from junction to case is 7C/W. If the ambient (and case) temperature is 25C then you can only allow a 25C increase from case to junction, so you are limited to 25/7= 3.5A of current.

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  • \$\begingroup\$ Oh my god I can't believe I got confused between relay to regulator !!! Sorry and thank you! BTW I think you meant to 3.5 Watt, not current :) \$\endgroup\$
    – Dor
    Sep 1, 2013 at 20:47
  • \$\begingroup\$ Doh! You are right, 3.5W so at 1.4V across the relay the current is 2.5A. \$\endgroup\$
    – Joe Hass
    Sep 2, 2013 at 2:20

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