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Old 07-19-2010, 02:18 PM   #1
onelegout
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Default PID Relay output driving an SSR?

Hi everyone,

Just a quick question.

I've got a PID which is only fitted with a output rated at 2 amps. How can I use this to drive an SSR? I was thinking of using a battery pack with 2x 1.5v AA batteries. My idea is that as the relay isn't rated high enough for the elements, I can use them to switch a small current which will in turn switch the SSR on and off.

Will this work?



Cheers!
H

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Old 07-19-2010, 04:44 PM   #2
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I've seen some SSRs that need at least 4vdc. Which one are you looking at? Perhaps 6-12vDC power supply would be easier?

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Old 07-19-2010, 04:50 PM   #3
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You could have the PID drive a small relay that drives the larger relay...

Kinda overcomplicated, I'd save my pennies for an Auber PID.

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Old 07-19-2010, 07:18 PM   #4
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if the internal pid relay is mechanical, it won't last very long when operated in pid mode..

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Old 07-19-2010, 07:22 PM   #5
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Do you mean it only outputs 2 volts? You said 2 amps, which is a lot for the control current.

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Old 07-19-2010, 07:44 PM   #6
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Old 07-19-2010, 09:06 PM   #7
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Quote:
Originally Posted by klyph View Post
Do you mean it only outputs 2 volts? You said 2 amps, which is a lot for the control current.
I mean it doesn't output anything :P It is the switch side of a relay - so it can't drive an SSR by its self as it doesn't output any current. I think I'd need a control current to switch the SSR. The current would be switched just like in Sawdustguy's picture - will this work?
H
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Old 07-19-2010, 09:55 PM   #8
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You have to find out how much voltage is required by the relay. The voltage you need to find is written on the relay itself. If it says 12V DC, then use a 12V battery or power supply in place of the 9VDC power supply shown in sawdustguy's drawing.

One more thing on that drawing... the schematic shown is in the wired off closed position, meaning when the PID is "adding heat" it will open the switch and turn OFF the heating element. If you chose to wire your PID to the SSR like this, I suppose you could wire the SSR to the heating element in the off/CLOSED configuration.

For the shematic shown, 6 and 7 should go to the solenoid in the SSR, with the appropriate voltage supply somewhere in line with the SSR, and the heating element should wire into the normally open terminals on the SSR.

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Old 07-20-2010, 01:00 AM   #9
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Quote:
Originally Posted by onelegout View Post
I mean it doesn't output anything :P It is the switch side of a relay - so it can't drive an SSR by its self as it doesn't output any current. I think I'd need a control current to switch the SSR. The current would be switched just like in Sawdustguy's picture - will this work?
H
Oh, now I see what you're saying. Ya carry on, listen to the above posts.
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Old 09-01-2010, 03:57 PM   #10
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Sorry to dig up an old thread. I just thought I better close this off with a post confirming that this method will indeed work. It would be far better, however, to use a PID with an SSR driver output, as the relay output might not be able to stand up to the constant on/off action of the PID, depending on your PID's settings.

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