Fuse for solid state relay protection - semiconductor I2T or standard

For protection of hockey puck style solid state relays (say 10 A at

125 V), it is suggested by vendors to use semiconductor I2T type fuses because they blow in 2 msec which is much faster than standard fast blow fuses. But since the semiconductor fuse can cost more than the solid state relay, I am wondering how important it really is to use them, specifically for protection against shorted loads where the relay would be subject to very high currents.
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Reply to
Steve
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"Steve"

** I suggest you need to *eliminate * short circuit load events - only sheer carelessness allows them to happen.

A 10amp SSR will typically withstand 100 to 150 amps for one cycle - enough time for a fast HRC fuse or magnetic breaker.

Possibly add series resistor into the circuit or a non-saturating inductor of 1.5 mH to get the short circuit current down.

..... Phil

Reply to
Phil Allison

Perhaps the "relay" would turn into a very nice ultra low R wire instead of opening..

Reply to
Robert Baer

Do consider the consequences of various failures. Blown fuse, down for an hour. Blown semis, down for a day or two. Blown motor, down for weeks. Or maybe completely different. But only you and your customer really know the application and the consequences, that is part of the engineering.

Reply to
JosephKK

Monitor the desired switched line in question. Monitor the voltage on that shunt. If the max value is reached, "OPEN" the low voltage DC excitation line that activates the relay, NOT the line experiencing the high current.

In other words, turn off the relay, instead of attempting to open the line the relay switches with yet another set of contacts of some sort (a breaker).

The relay is already the switch/circuit/line breaker.

Reply to
lurch

correct., assuming its contacts are rated to break the fault current (and it's poor engineering if they aren't)..

Reply to
who where

"lurch"

** Nice idea - but it cannot work with short circuits. Problem being, that one cannot instantly turn off a triac or normal SSRs by removing drive.

It will take unit the end of the half cycle ( ie the current must fall to near zero) for the triac to commutate off - so the full fault current will have to be tolerated for that long at least. Good bye 10 amp triac.

For your idea to work, impedance must be added in series with the load to limit the fault current to a safe value. Plus the circuit must latch off and be manually reset when the fault is cleared.

Much more practical to just use an up-rated SSR, add a bit of series impedance and fit a fast acting HRC fuse.

Works for professional lighting dimmers.

.... Phil

Reply to
Phil Allison

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