I am not very electronically knowledgeable. I usually plug things together until they work or the smoke comes out. I recently bought some SSRs from Jameco (LCA110). The datasheet did not come with a basic circuit, so I just hooked up the input side to the 5v rail of a computer power supply and the other side to the same power supply but through a resistor to an LED. My goal was to test my connections before hooking it into a circuit to ensure I was connecting it properly.

I did not use a resistor on the input side and was getting a flicker on the led. I wanted a steady on state. I put a 100uF cap across the power/ground to smooth out the power, but it did not help....now for the really not smart part. Since the first one wasn't working well, I decided to test other SSRs to see if they did the same thing. Most of them did nothing.

My questions are, did I blow out all of the internal LEDs by not using a resistor on the input side and is there a way to test if they are still working?...again, I am getting nothing from the output side in both the DC and the AC/DC setup.

If an oscilloscope is needed, Then I'm out of luck.

Here is a link to the datasheet:
https://www.jameco.com/Jameco/Products/ProdDS/1582997.pdf

Thanks. Hopefully I gave enough information.

 

Well, with electronic parts, you really have to read the datasheets in order to get the best performance out of them, or even to simply scan the "absolute maximum ratings" to make sure you don't let the smoke out.

The absolute maximum for the input control current for continuous operation is 50mA; you can go up to 1A for 10mS, but beyond that - smoke or silent death syndrome.

So, you have to limit the input control current to < 50mA.

It's not obvious from the absolute maximum ratings what you will have to do in order to keep the current below that, but it's safe to say that Ohm's Law does work. 5v/50mA = 100 Ohms; so even if the input LED had a Vf of 0v with 50mA flowing through it, you know that 5v across 100 Ohms will only let 50mA flow.

But, let's look down further in the datasheet.

On page 4, under "Performance Data", the upper left-hand plot, shows the forward voltage of the LED for various temperatures and current.
We can see that with 50mA current at 80°C, the Vf might be as low as around 1.38v.

In the plot to the right of that, we see that the optimum turn-on time occurs when the control forward current is around 20mA-25mA.

So, it would seem prudent to keep the control current somewhere around 25mA. Looking back at the chart to the left, if you're going to be operating the circuit at around room temp (25°C), we can pretty much see that 25mA current would intersect right at about 1.38v too. So, we need to figure out what resistance you need to get 25mA current when the Vf is 1.38v.

R= (Vsupply - 1.38v) / 25mA = (5v-1.38)/0.025 = 3.62/0.025 = 144.8 Ohms.

144.8 Ohms is not a standard value of resistance.
Here is a decade table of standard resistance values: http://www.smpspowersupply.com/resistor/decadetable.html
Bookmark that page.
Use the E12 and E24 columns, as those are usually available.
150 Ohms is the closest value to 144.8 from the E12 and E24 columns.

So, to calculate the current, we'll go 3.62v / 150 Ohms = 24.13mA - close enough.

You either fried your SSR's, or you fried the LED you were using for the test. Try it again using a 150 Ohm resistor to limit the control current with a 5v supply, and try a fresh LED w/resistor for testing the output side.

 

Thank you very much for the very detailed answer. Of the 16 SSRs that I started with, two of them still work. Wasted money, but at least I learned something.

Thanks again.

 

Thanks for the article, Sgt Wookie. Now I can get on with building a remote keyboard that will allow my grand-nephew to play the major chords with his diminished-utility right hand (using a rigged prosthesis)while only hitting one key. It is a switch-selected option on the Casio keyboard but it is on the left-hand end of the keyboard. I'll use 36 LCA110s in the project.