PID:SSR issue

[cheese1566]

:?:
I found that my recently purchased 25 amp SSR is allowing full current to pass over the contacts (which is just letting my pot run full like it is plugged into the wall). I disconnected the wiring and found that it works like it should with a low voltage continuity tester (multimeter), but not with 120 volts...

My SSR has a small LED light on it and it goes on/off when the PID sends the signals. I have double checked the wring and tested everything.

Is it possible for a SSR to "test" OK, but not properly function with a load?

(A few weeks ago I installed the same exact components in my espresso machine and it is functioning flawlessly.)

I did email the vendor where I bought it from for some tech help and hopefully a replacement. Bummer!

[cheese1566]

Double Bummer, I should have posted this in the Casting Equipment section. :killingpc

Can a moderator move it there or to Special Projects?

[ANeat]

Its possibe, the SSR is rated for 120 vac right???

[shaune509]

ssr=solid state relay
pid controls the relay function, power contacts are passing full applied voltage when activated, pid turns on/off to maintain tothe hold temp. so it is working as designed. if you dont want full voltage then a resistor will need to be put inline between the ssr and the load.
shaune

[garandsrus]

Moved for you :)

As was suggested, the SSR is nothing more than a switch with no moving parts. When the PID tells it to turn "on", voltage passes through and the pot heats up. When it is turned "off", no voltage goes through and the pot cools down.

A PID works by turning the power on/off while monitoring the temp instead of using a resistor to increase/reduce the amount of power flowing through. The pot temp should be turned all the way up when using a PID as the temp control on the pot is not being used.

One other thing, I think that the control voltage is a low voltage. It is not 120v.

John

[Lee W]

SSRs can be bad. They are just electronic components that allow current to pass when a control voltage is applied. Good ones are expensive, not so good ones are cheap.
Send it back if it is still under warranty if you can.
Nice to see you got the espresso machine going. What temp are you setting it to?

[cheese1566]

We are starting at 232 degrees, just by what we found on the web. Seems to be working good so far.

[cheese1566]

A ha!
I got a response from Auber Instruments already...
3. Can I test a SSR with a Multimeter?
1) Q. When there is no input for the SSR, I can still measure a live voltage on output of the SSR. Does that mean the SSR is defective?

A. Not necessarily. The SSR is a semiconductor device that has a leak current. The leakage is less than 40 mA. If you measure the output voltage at open circuit condition, you will get a live voltage that is the same as the AC input. That does not necessarily mean the SSR is burned through. However, if there is a load (such as a heater) on the output, you should not detect any significant voltage because it is drained by the load.

2) Q. When I measured the resistance of the output of SSR, it changed from mega ohm to 1-20 kilo ohm range when an input trigger is applied. Should it be zero ohm?

A. No. It should not. This is because almost all AC SSR on the market has a zero voltage crossing detection circuit. It will not turn on (change resistant to zero) unless it detects an AC voltage that is change the flow direction (across zero voltage). The zero voltage detection function is for protecting inductive load. Without an AC load, the output resistance will never change to zero. It should be noted that changing the resistance from mega ohm to kilo ohm does not mean the SSR is good. It only tells you it has not burned open and is responding to input.

3) Q. How to test an AC SSR?

A. The most reliable way is connect the output with a 40W light bulb. If the light bulb responds to the input trigger, the SSR is fine.

http://auberins.com/index.php?main_page=page&id=10&chapter=0

[cheese1566]

:oops:That was it!

very good response from Auber Instruments!!!

Lee, here is my Rancilio Silvia with your benefit PID.
26273

[cheese1566]

Here is the finished, tested, and working PID controller for my melters.
I made it portable so I can use it primarily for my 20 pound pot, but can move it to my 10 pounder for occassional use. Built from scrap goods, computer power supply box, SSr from Auber, and the PID & probe from Lee W's benefit a while back.
The red light on the front is set to come on when the temp gets within 5 degrees (+/- 10) of the set point.

26275 26276

[Doby45]

I think I am about to build one of these but I want to add a "by-pass" switch. That way I can bypass the PID on the initial heat up and melt and then when the melt has become liquid I can flip the switch to PID mode. Should be great fun..

[lwknight]

Unless you have the TC mounted in the bottom of the pot , you won't need a by-pass. I can can just lift the TC out of the pot temporarily.

An easy way to do the by-pass is using a plug recepticle like pictured above. Just unplug from the control and plug into the wall outlet.

I use a 24v transformer on my contactor control so an override would be easy. If your using the PID with the ssr control it may be harder.

[Lee W]

Cheese,
Nice to see the finished products. Very nice job.

[Muddy Creek Sam]

cheese1566,

Nice looking unit, now how much you charging?

Sam :D

[Just Duke]

Here is the finished, tested, and working PID controller for my melters.
I made it portable so I can use it primarily for my 20 pound pot, but can move it to my 10 pounder for occassional use. Built from scrap goods, computer power supply box, SSr from Auber, and the PID & probe from Lee W's benefit a while back.
The red light on the front is set to come on when the temp gets within 5 degrees (+/- 10) of the set point.

26275 26276

Do you sell these?

[Frozone]

I think I am about to build one of these but I want to add a "by-pass" switch. That way I can bypass the PID on the initial heat up and melt and then when the melt has become liquid I can flip the switch to PID mode. Should be great fun..

I've got to ask.... WHY?



Duke, I can make you one, but you may not like the price! I was going to produce a number of these but shipping cost are rather high and drive the price up past what I though most would pay.

[Doby45]

I've got to ask.... WHY?

For the simple fact that I wanted a switch that I could switch instead of plugging and unplugging the unit. Then I have a question I've got to ask.... WHY NOT?

[Frozone]

You didn't answer my question.

Why do you think you need to plug/unplug anything?
Why Not? The PID will do the work. It will run the heat wide open same as your switch will do until it's up to the set temp.
Things won't heat up any quicker with your switch in the circuit, and Unlike You IT won't forget to shut the switch off.
As soon as things reach set point It will then begin controlling the temp to keep it at there.

[lwknight]

For the simple fact that I wanted a switch that I could switch instead of plugging and unplugging the unit. Then I have a question I've got to ask.... WHY NOT?
__________________

That is totally unnecesary. The PID will run full power till the setting is reached.
The PID does not somehow lower the current.

The only reason that an override would be useful is in a case of the TC being mounted in the bottom of the pot where the temperature would read high till it actually gets covered with alloy. This could only be the case when initially melting ingots in a dry pot.

Like I said earlier, you could just lift the TC out of the pot and it will not cycle.

[Frozone]

The only reason that an override would be useful is in a case of the TC being mounted in the bottom of the pot where the temperature would read high till it actually gets covered with alloy.

I have the TC mounted in the bottom of the pot; Even empty it doesn't cycle. <edit> I guess I should say, it doesn't read a high temp. It reads low until there is melt in the bottom. The sensor doesn't need to be completely covered however, just 1/4" will get the PID going. </edit>
Now mine is a 'through the bottom' design. It is mounted in a brass compression fitting that goes though the pot bottom. The TC is actually ~ 3/8" above the bottom, but it is connected to it by the brass fittings. A direct read off the pot sheet metal Might give different results.

[Doby45]

I was under the impression from earlier reading on the whole PID topic that using it in PID mode during the initial warm up puts undo extra stress and heat through the SSR. That was the whole reason I was looking at a bypass switch that would run the pot as normal until it melted and then flip it to PID so the PID did not take the brunt of the wattage to get to initial melt.

[cheese1566]

On the $$ questions, no I generally won't build them for people. Then my hobbies become work...plus it would cost too much in the end run by the time I order all the components from various places. I built these from the PID units in the benefit auction a while back. I knew I wanted some, so when they were offered-I jumped! Plus, I like to do things economically by reusing stuff that generally gets thrown away. Like the box, indicator lights, power cords...some see junk-I see treasure. I see how some group buys and co-ops can easily go bad- not worth the risk for my past time here.

But if someone had all the right stuff and no time, drop me a PM. I do some stuff during the winter months for trades.

They are easy to assemble if you get the correct components. Plenty of threads and wiring diagrams on the topics. Plus everyone will help out if you have a question. Just like my thread. (But, if you are uncomfortable doing wiring like household outlets/switches, best not attempt it.) The confusing part sometimes can be the set up and intial programming. Each brand/manufacturer of PID does that a little different.

[lurch]

I was under the impression from earlier reading on the whole PID topic that using it in PID mode during the initial warm up puts undo extra stress and heat through the SSR. That was the whole reason I was looking at a bypass switch that would run the pot as normal until it melted and then flip it to PID so the PID did not take the brunt of the wattage to get to initial melt.

The solid state relay passes the power current to the pot heating element. The current consumption of the typical lead pot element will make the typical SSR get pretty warm during initial heat up unless the SST is mounted on a suitable heatsink. If the heatsink is sufficient then there is not going to be any wear & tear on the SSR and nothing to worry about. (What kills electronic parts is heat primarily. Even in the case where voltage ratings are exceeded, the eventual destroyer of carefully crystallized silicon is still heat since there will be a place where a p-n junction goes into avalanche conduction causing current to flow in a very localized spot which in turn cause a localized hot spot which also in turn destroys the semiconductor device when severe enough.)

A bypass switch is just an extra doo-dad that while might be neat to play with, is not really necessary. If the thermal management of the SSR is such that a bypass switch really helps, then you need to improve the thermal management of the SSR. i.e. use a bigger heatsink.

[Doby45]

Then it is not needed for sure, because I have some really nice CPU heatsinks that I can pick from to use. My only concern was the heat, simple is better. Thanks Lurch.

[lwknight]

The SSRs are cheap and easy to replace.
I use an electro/mechanical contactor myself. It will carry a sustained load indefinately and and should have a cycle life of more 25000 with only a 800 watt element on it.

These are for condenser units and are rated 30 amps at a cost of a whopping $8.95 .

[JIMinPHX]

There are different varieties of SSR. Some are zero crossing type, others are not. Some are for AC load use. Some are for DC load use. There are also different types of trigger signals. You need to get the right type of SSR for the application at hand. Some of the cheap ones actually work quite well when properly applied.

A heating element for a lead melter is an ohmic heating load. The inrush current on it is normally 10 x full load amps when it is cold. After it heats up, the full voltage current load is much less than it is when the element is cold. If the SSR is sized to accept the 10x inrush current, then there is no problem with running it without a start up bypass switch. If the SSR is marginally rated, then each time it experiences a cold start, the output thyristor in the SSR will get a little chunk chopped out of it. After a while, enough little chunks will add up & the SSR will fail. If you have a big chunky SSR, then you don't need a cold start by pass switch. If your SSR is marginally rated (smaller & cheaper) then a cold start by pass switch is a good idea.

The advantage of an SSR over a traditional contactor is that the SSR can switch states very quickly & has a massive capacity for total number of cycles. The SSR is better for high frequency pulse width modulated applications. A traditional contactor has better durability for occasional inrush spikes & voltage spikes.