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Have been testing the new electric brewery set-up  and had some suspicions about the SSR that is in a PID controlled circuit for a heat exchanger. I have run some tests and have found that if I run the 4 kW heating element via the SSR (switched on full time) it takes twice as long to heat a volume of water to boiling than if I bypass the SSR. So it seem that I am losing half the power of the heating element by running it through the SSR. (I am bypassing the SSR right at the location of the SSR, so it is not anything to do with the rest of the circuitry). 

This is a 40A SSR from Jaycar. Is it a case of all SSRs being horribly inefficient, or is it that 'cheap' SSRs are like this, or is it that I have a faulty SSR?

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This is a reply to mattd2's reply - forum software not giving me a reply link to his comment...

Sorry, not fully up to speed on the various labels on a relay's inputs, outputs etc.  What is the 'duty'?

Anyway, to take a punt - If I'm understanding correctly, you are talking about chaining the SSR off another relay by switching it off the output from the relay?  If so, doesn't that kill one of the original reasons for going with a SSR ie. the potentially fast switching time?  I guess though that it means the SSR can still be used for the heavy lifting re. the large currents involved.

Thanks for the explanation Guy,

OK - so what you guys are suggesting is a way of manually and physically isolating the circuit using a relay if it fails shut.  I guess what I was after was some sort of failsafe design that automatically cut power if the SSR failed.

Anyone able to shed any light on why I can't reply to responses to my own posts?  Same for Firefox and Chrome...

Good point...  The things you never think of.

I had a look at the manual for the Love 16B and don't quite understand the outputs. It states that the outputs are:

SPDT relay (rated for 250V 5A), 14V voltage pulse, 4-20mA & Linear voltage 0-5/10V

But it does not tell you (or at least as far as I could understand) how to select the output mode, or did you have to select the output when you purchased it?

If it is fixed and it sounds like you have the relay output (as you have to supply an external power source to the output terminals) it might be easier to get a SSR with a 240V control and then just run 240V through your controller's output to the SSR.

If you can select the output type then set it to 14V pulse and hook it up (with the correct polerisation) to you exsiting SSR.

Yes that is an issue with the manual that has always perplexed me - I rang the supplier and they could not shed much light on that aspect either. I am going to hook up the existing internal 6V AC supply within my control box to a small bridge rectifier that I got yesterday from Jaycar and run that through the PID switch to see if that works. It currently seems to work at half pace on the 6V AC supply, (and the SSR only needs 3V DC), so it may work out. If not, then will try hooking it into a 14v output in the same manner. Then it is off to find a different SSR as you suggest..... (And one day I will actually brew some beer!)

Did a bit more research and found this manual http://www.dwyer-inst.com/PDF_files/E_90_BPC.pdf

Page 3 shows your options = Model number - 16B-XYZ

X (output 1) = 2 Voltage pulse, 3 relay, 4 current, 5 linear voltage

Y (output 2) = 2 Voltage pulse, 3 relay

Z - Options

What model number do you have?

Ah Matt you genuis! Thanks for that. I will probably have to take my control box apart to find out, but I am pretty sure I have a relay model, so now it is all making a lot more sense. If this is the case then I am not limited to the 14V but any DC voltage 3 - 30Vdc will do the trick, and so my previous plan should work. Will pursue over the next few days/week and see how it goes. 

System now rejigged with proper dc through the PID, tested and works fine. Thanks for all the help!

If you put the heat sink in the HX and the HX gets hot then it won't sink much heat from the SSR which is the job its supposed to be doing...

I read somewhere that SSR wastes up to 30W of power at 20A load. No point of trying to use that to heat anything (except the heat sink ;) ).

I'm just testing my new HLT and I measured the power loss at my SSR. I've 2kW heating element. For the Voltage drop over SSR I measured 0.988VAC (true RMS).

I = 2000 / 230 = 8.7A,

P = 0.988VAC * 8.7A Power loss on the SSR:  8.6W = 0.0086kW Not much I would say.

I'm pretty sure that the SSR can handle that much even without a heat sink. I've a tiny one on mine and it's barely over the room temp.

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