Fried SSR?

I think I know the answer to this, but here is one for the group.

Did a brew Friday night and something odd happened out of the gate, when I threw the contactor switch, I somehow got a short and tripped the breaker. Turned everything off, opened the box examined all wiring, reassembled had no issues....until.

When it came time for the boil, I noted the PID was no longer controlling the output to the element. I have mine wired so I can see when current is going to the element, glad I do. The light for the element stays on even when the PID is NOT sending the signal. This leads me to believe the short may have fried open the SSR. What do you think?

By the way, I believe the reason for the short, I recently installed a volt/amp meter and before the brew I moved the transducer(?) to a different wire, in doing so, the transducer may have come in contact with the neutral. After examination of the transducer, there is a small scortch mark. Since I am using a GFCI, I recognize it would not take much to trip the breaker. I have since removed the transducer...it was for show more than use.

Bumping to hope someone has a thought on this.

1. Completely disconnect the SSR
2. To the output, connect a light bulb
3. To the INput, connect a 9 volt battery

Does the bulb light up as it should? If not, replace SSR.

I am confused...if I connect a light bulb and 9 volt to the output does that not defeat the process. I am wondering if I need to connect the 9 volt to the input side? This would make sense as I believe the SSR is fried in the position where it allows current to pass through non-stop.

Also as more information. I ran a continuity test across the output side with no power and the volt meter is beeping which I think confirms my thinking. Should not the output be open in the absence of a signal telling it to be closed?

Thanks

Jps101 said:

I am confused...if I connect a light bulb and 9 volt to the output does that not defeat the process. I am wondering if I need to connect the 9 volt to the input side? This would make sense as I believe the SSR is fried in the position where it allows current to pass through non-stop.

Also as more information. I ran a continuity test across the output side with no power and the volt meter is beeping which I think confirms my thinking. Should not the output be open in the absence of a signal telling it to be closed?

Thanks

Click to expand...

You are correct, it is the INput that takes the 9v. I've corrected my post above.

According to your continuity test, the bulb should stay lit with or without the 9v battery applied across the input and you'll know the SSR is fried.

Thanks...that is what I was thinking.

Thank you for the feedback.