Solid state relay burned on first test!

Bought this one, sounds perfect, up to 220V DC 100A!
https://www.ebay.com/itm/SSR-100A-D...e=STRK:MEBIDX:IT&_trksid=p2057872.m2749.l2649

Connected a 98V battery to a 50A resistor dummy load, connected a 12v input to the relay and within about 5 seconds I started hearing hissing and popping sound from the inside of the relay and it died and staid closed.

One thing I don't understand, the relay is just an on off switch, two positive and negative for the 12v input and two terminals that are either opened or closed.
I connected the two high current terminals to the + wire to open or close the circuit but looking at the little diagram on the relay it shows that one is + and one is - while it doesn't make sense to have a positive and negative as it is just opening and closing the connection of a single wire.

Am I missing something?

rg12 said:

Bought this one, sounds perfect, up to 220V DC 100A!
https://www.ebay.com/itm/SSR-100A-D...e=STRK:MEBIDX:IT&_trksid=p2057872.m2749.l2649

Connected a 98V battery to a 50A resistor dummy load, connected a 12v input to the relay and within about 5 seconds I started hearing hissing and popping sound from the inside of the relay and it died and staid closed.

One thing I don't understand, the relay is just an on off switch, two positive and negative for the 12v input and two terminals that are either opened or closed.
I connected the two high current terminals to the + wire to open or close the circuit but looking at the little diagram on the relay it shows that one is + and one is - while it doesn't make sense to have a positive and negative as it is just opening and closing the connection of a single wire.

Click to expand...

The switching device is a FET. In one direction it's a switch. In the other direction it's a diode with a forward voltage of ~.6V. So if you had a 50A load, then you were dissipating (50*50*.6) = 1500 watts inside the device.

Note that they make bidirectional SSR's (two FETs back to back) but if the SSR has a + and - terminal it's not bidirectional.

billvon said:

rg12 said:

Bought this one, sounds perfect, up to 220V DC 100A!
https://www.ebay.com/itm/SSR-100A-D...e=STRK:MEBIDX:IT&_trksid=p2057872.m2749.l2649

Connected a 98V battery to a 50A resistor dummy load, connected a 12v input to the relay and within about 5 seconds I started hearing hissing and popping sound from the inside of the relay and it died and staid closed.

One thing I don't understand, the relay is just an on off switch, two positive and negative for the 12v input and two terminals that are either opened or closed.
I connected the two high current terminals to the + wire to open or close the circuit but looking at the little diagram on the relay it shows that one is + and one is - while it doesn't make sense to have a positive and negative as it is just opening and closing the connection of a single wire.

Click to expand...

The switching device is a FET. In one direction it's a switch. In the other direction it's a diode with a forward voltage of ~.6V. So if you had a 50A load, then you were dissipating (50*50*.6) = 1500 watts inside the device.

Note that they make bidirectional SSR's (two FETs back to back) but if the SSR has a + and - terminal it's not bidirectional.

Click to expand...

So I only connect the + of the battery.
To which side I should have connected it, negative or positive?

rg12 said:

So I only connect the + of the battery.
To which side I should have connected it, negative or positive?

Click to expand...

The current should flow from + to -. So if the battery is your source, the + of the battery should go to the + of the SSR.

billvon said:

rg12 said:

So I only connect the + of the battery.
To which side I should have connected it, negative or positive?

Click to expand...

The current should flow from + to -. So if the battery is your source, the + of the battery should go to the + of the SSR.

Click to expand...

Yep, thats how I connected it...

Maybe you posted the wrong picture, but it shows a 10A capacity for the relay contact.


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E-HP said:

Maybe you posted the wrong picture, but it shows a 10A capacity for the relay contact.


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Click to expand...

I was concerned about it too while ordering and he assured me that it's a 100A and I did receive a relay that says 100A on it.

Were you using a decent size heat sink?


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rg12 said:

Yep, thats how I connected it...

Click to expand...

Hmm.

If you try another one, measure forward voltage under load and calculate heating. Unheatsinked it can handle a few watts of heating. With a good heatsink it can handle tens of watts.

What do you mean tens of watts?
How can it be rated for 220V 100A?
It took it about 5 seconds to die, it wasn't a heating issue, it just isn't built for that and the rating is just BS.

rg12 said:

What do you mean tens of watts?
How can it be rated for 220V 100A?

Click to expand...

OK.

You talked about your load - 98V 50A. That means your load will dissipate about 4900 watts. However, the other stuff in the circuit (the wiring, fuses etc) will not dissipate 4900 watts, even though they are carrying 50 amps and seeing 98V relative to other conductors. For the stuff in line with the current, the formula is I^2R or VI, where I is the current, V is the voltage ACROSS the device/wire/fuse, and R is the internal resistance.

Any switch has some internal resistance. Let's say the MOSFET they use has a 5 milliohm internal resistance (good for a 250V MOSFET.) That means that at 50 amps you will see 50 * 50 * .005=12.5 watts. That is a lot for a small device to dissipate. Think about how hot a small 10 watt incandescent bulb gets - and keep in mind that the FET generally has to stay below a die temp of 150C. (That's die temp, not case temp, which is much lower.)

So for you to use that SSR at 50 amps it would definitely need a heat sink to get rid of that 12.5 watts.

But let's say they didn't use a good FET. Let's say they used a 15 milliohm FET. Now your heating is at about 40 watts. And that would quickly damage the FET.

Now let's say that it was hooked up backwards (i.e. the terminals were mislabeled or something.) Now you use the other formula. Forward voltage could be as high as 1 volt (depending on FET) and current is 50 amps - so heating is 50 watts. Now it blows very quickly.

It took it about 5 seconds to die, it wasn't a heating issue, it just isn't built for that and the rating is just BS.

Click to expand...

I wouldn't assume "it wasn't a heating issue" until you understand the problem better. It may well be a fake part - but in the end most of the damage (even from counterfeit parts) comes from heating.

Billvon is 100% right.

If you do a search on eBay you will see that in order for them to operate at their current rating continuously you will need the heatsink that some of them come with.
https://i.ebayimg.com/images/g/R7UAAOSwCsRbpcON/s-l1600.jpg

But, there are also fakes out there.
Some interesting links.
https://www.te.com/commerce/DocumentDelivery/DDEController?Action=showdoc&DocId=Data+Sheet%7F1308242_SSR%7F0116%7Fpdf%7FEnglish%7FENG_DS_1308242_SSR_0116.pdf%7F1393030-4

https://www.youtube.com/watch?v=DxEhxjvifyY
https://www.youtube.com/watch?v=2UtL2uAYCUA

billvon said:

Any switch has some internal resistance. Let's say the MOSFET they use has a 5 milliohm internal resistance (good for a 250V MOSFET.) That means that at 50 amps you will see 50 * 50 * .005=12.5 watts. That is a lot for a small device to dissipate. Think about how hot a small 10 watt incandescent bulb gets - and keep in mind that the FET generally has to stay below a die temp of 150C. (That's die temp, not case temp, which is much lower.)

Click to expand...

Ya, I was doing a similar guesstimate, which is why I asked about the heat sink. On the other hand, not sure I'd trust a seller that can't even post the correct picture of their product.

rg12 said:

Connected a 98V battery to a 50A resistor dummy load,

Click to expand...

50A is a current, not a resistance. What was the resistance?

Chalo said:

rg12 said:

Connected a 98V battery to a 50A resistor dummy load,

Click to expand...

50A is a current, not a resistance. What was the resistance?

Click to expand...

I meant the total amps.
They are 3 1500w resistors in parallel where each is 5ohm

rg12 said:

Chalo said:

rg12 said:

Connected a 98V battery to a 50A resistor dummy load,

Click to expand...

50A is a current, not a resistance. What was the resistance?

Click to expand...

I meant the total amps.
They are 3 1500w resistors in parallel where each is 5ohm

Click to expand...

Did they survive? Assuming your 98 volt pack was fully charged to 108 volts, then that would be about 65 amps through your 1.67 ohms, or about 7000 watts through resistors rated at 4500 watts. That's like 4 or 5 heat guns worth of heat so those things usually need pretty big heat sinks to dissipate that much energy and survive at double their ratings. Have you measured their resistance after the relay fried? Maybe something shorted on the load side.

E-HP said:

rg12 said:

Chalo said:

rg12 said:

Connected a 98V battery to a 50A resistor dummy load,

Click to expand...

50A is a current, not a resistance. What was the resistance?

Click to expand...

I meant the total amps.
They are 3 1500w resistors in parallel where each is 5ohm

Click to expand...

Did they survive? Assuming your 98 volt pack was fully charged to 108 volts, then that would be about 65 amps through your 1.67 ohms, or about 7000 watts through resistors rated at 4500 watts. That's like 4 or 5 heat guns worth of heat so those things usually need pretty big heat sinks to dissipate that much energy and survive at double their ratings. Have you measured their resistance after the relay fried? Maybe something shorted on the load side.

Click to expand...

The pack full is 100.8V, it was 98V when tested and the measured current with a clamp meter (voltage sags a bit under load) was about 50A.
Thats just a little over their max rating and I use it for short load tests and not for constant discharging.