looking for a solution to the ssr leakage / indicator light problem.

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-TH-

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I have a 240v led connected t
o one hot leg and also the ssr-switched leg with the intent for it to be on when the ssr closes. My research here (also verified by personal experience) shows that the ssr will leak current enough to turn on the led at all times, unless the element is plugged in. For most this is ok but I would really like for it to function properly even when the element is unplugged. Is there any way I can make this work the way I want it to?
 
use a 12v LED on the signal side instead of the high voltage side?
Yeah I could do that, but there aren't any 12v LEDs available that match the other two on my panel.

I thought I saw a post on another thread somewhere suggesting the use of a resistor in parallel with the LED.
 
I have a 240v led connected t
o one hot leg and also the ssr-switched leg with the intent for it to be on when the ssr closes. My research here (also verified by personal experience) shows that the ssr will leak current enough to turn on the led at all times, unless the element is plugged in. For most this is ok but I would really like for it to function properly even when the element is unplugged. Is there any way I can make this work the way I want it to?

One possible solution is to not use an SSR. In it's place you could simply use a relay or contactor. It would be a bit more noisy and the relay contacts would eventually wear out (after about a million cycles), but it would achieve your objective.

If you do this, make sure you use a PID capable of relay output like the Auber SYL-2362 or SYL-2342.
 
Strap a PTC fuse in parallel with the LED.

When the current is low (i.e., under the "trip current" of the fuse), the fuse looks like a short and the leakage current bypasses the LED.

When the current exceeds the trip current, the PTC heats up and becomes very resistive. This allows the LED to turn on.

Has anyone measured the leakage current?
 
I wired mine just to show what element I was powering at the time regardless of ssr activity. Wired the indicators to my switch.
 
I had tried a 120v indicator wired from switch to common which worked great until I plugged IN the element - then the element allows current to flow through it to the indicator light.

I really want the light to come on when only when the element is on. I'm using a pwm so I want to see it on with the pulses. It works fine now its just that I unplug the elements often because I switch from one to another, or when I cool, etc. It just bugs me that the "element on" light comes on.

I did find a 12v light that is the same size and style as the ones I have except its incandescent vs LED. I might just order one and run it off the pwm output.
 
I think your indicators are working as designed.

One phase of element has power, other phase is controlled by the PWM. With PWM OFF, one phase goes through element resistor to indicator light attached to the other phase. There is a small voltage drop across element resistor but

If you want the element OFF, then disconnect power from both phases using a contactor or DPST switch and the light will be OFF. Keep the element indicator light and add the 12V light for the PWM On/Off if desired.
 
I was able to get my indicator to go off with a much larger 1/4 watt resistor. I think it was like 500K or a megaohm. Costs like 5 cents.
 
I was able to get my indicator to go off with a much larger 1/4 watt resistor. I think it was like 500K or a megaohm. Costs like 5 cents.

Not sure what you mean by much larger but just so I'm sure, you put a 1/4 watt, 500k or 1 megaohm resistor in parallel with your indicator light?
 
I have a whole set of 1/4 watt resistors and I basically used the smallest ohm value that would dissipate less than that. I can't remember if I put the resistor across the input terminals or the output.
 
Ok after a bit of research this is what I've learned:

My Auberin 40A SSR has a leakage current of 3.0 mA (from spec page)
My LED indicator light is rated at 1.5 mA (from spec page)

So as expected there is enough leakage current from the output of the SSR to turn on the LED - or to put it another way (based on what I’ve read) the load of the LED is not great enough to “release” the SSR.

To solve this problem in general it is common to use a "bleeder" or “bypass” resistor which is simply a resistor placed in parallel with the load (in my case the LED) which will increase the total load enough to release the SSR. The higher the resistance that still works the better, one reason being it will require a less wattage resistor.

So in my case, I need to add another 1.5 mA load (3.0 – 1.5) to my SSR output. To get the additional 1.5 mA, I need to use a resistor equal to (or less than): R = E/I = 240V/.0015A = 160kohms. Less is ok too because it would just increase the load (more amps). So lets say I want to try a 130kohm resistor (I happen to have one). Now I calculate the W rating needed for that resistor. The current used is: I = E/R = 240/130k = 1.85mA. Therefore the power calculates to be: P = E x I = 240V x .00185A = .444 Watts. So my 130k ohm, 1/2 W resistor should do the trick.

I will test my theory tonight and report back…
 
As an engineer and fellow nerd, I appreciate the math and the well thought out solution. Here's hoping the transition from paper to practice works! Let us know!
 
Ok after a bit of research this is what I've learned:

My Auberin 40A SSR has a leakage current of 3.0 mA (from spec page)
My LED indicator light is rated at 1.5 mA (from spec page)

So as expected there is enough leakage current from the output of the SSR to turn on the LED - or to put it another way (based on what I’ve read) the load of the LED is not great enough to “release” the SSR.

To solve this problem in general it is common to use a "bleeder" or “bypass” resistor which is simply a resistor placed in parallel with the load (in my case the LED) which will increase the total load enough to release the SSR. The higher the resistance that still works the better, one reason being it will require a less wattage resistor.

So in my case, I need to add another 1.5 mA load (3.0 – 1.5) to my SSR output. To get the additional 1.5 mA, I need to use a resistor equal to (or less than): R = E/I = 240V/.0015A = 160kohms. Less is ok too because it would just increase the load (more amps). So lets say I want to try a 130kohm resistor (I happen to have one). Now I calculate the W rating needed for that resistor. The current used is: I = E/R = 240/130k = 1.85mA. Therefore the power calculates to be: P = E x I = 240V x .00185A = .444 Watts. So my 130k ohm, 1/2 W resistor should do the trick.

I will test my theory tonight and report back…

1/4W resistors are much more common, so just tie two 330k 1/4W resistors in parallel. Double the power rating, half the resistance.
 
Well I tried a 160 Kohm 1/2W resistor in parallel with the indicator light and no dice (meaning the light still came on when element was unplugged). I also had a 130 Kohm 1/2W resistor so I tried that and still no luck.

Next up is a 7.5 Kohm 8W resistor I found at work which I will try tonight. That one calculates out to draw 32mA and 7.7W @240V, so if that's not a big enough load then I am completely misunderstanding something (no big surprise there) and I will scrap this whole idea and install the 12V lamp I already ordered (due today) on the input side of the ssr. I might do this anyway because then I could put the 230V indicator lamp on my SPA panel box for a visual of when that is switched on. Either way I will report back what happens.
 
I just live with the leakage. Its only a problem when the powered element is not plugged in.

The thought of using a 12v LED as the indicator on the signal wiring is interesting excepting that it only tells the operator if the signal is getting through and not if power is getting through. Having the LED connected to the power side of the circuit tells you if the signal and the SSR are working.
 
I agree with mabrungard. If you put the LED on the SSR control circuit and the SSR fails closed, you'll never know except that your mash will get too hot. The control circuit will still do it's thing and the element will have power the whole time.
 
If you are disabling 220v power properly with a contactor upstream or downstream of the SSR then ther SSR leakage and light are no longer an issue. The other benefits are no power to the element when it is "off" and can successfully turn off power event if the SSR or PWM ckt fail.
 
I have a 40A switch (rotary cam lock-out) in my panel that cuts power to the entire panel. But there are times when I want the power switched on and also have the elements unplugged. What I do not like is that in this instance the "element on" indicator comes on falsely, due to the ssr leakage current. Also, I cannot check to see if the PWM is functioning properly unless I plug in an element which in turn means water in the kettle.

I agree that having a 12V light on the SSR input is not ideal. However I just had an interesting thought - in addition to the 12V light I could put 230V indicator lights on my element junction boxes that are attached to the kettles.
 
If you are disabling 220v power properly with a contactor upstream or downstream of the SSR then ther SSR leakage and light are no longer an issue. The other benefits are no power to the element when it is "off" and can successfully turn off power event if the SSR or PWM ckt fail.

To clarify, are you saying that the relay provides enough resistance so that, when the relay is closed, the led will not light from leakage, even if no element is plugged in?

Or, are you saying that with the relay the led can still light from leakage with no element plugged in, but you should keep the relay switched open unless you have the element plugged in and submerged, and not worry about it?
 
No, I am saying that the 2 phase contactor cuts power to the outlet (and light and element).
No power = no leakage = light OFF.
It sounds like TH has a single contactor at the input to the panel and relies on the SSR to cut power to the element, keeping 1 leg of the element hot at all times. Depending on how the panel is designed, you could enable/disable the main power contactor when one element is enabled and run the PID and pumps off of fused 110v upstream of the contactor.
 
No, I am saying that the 2 phase contactor cuts power to the outlet (and light and element).
No power = no leakage = light OFF.
It sounds like TH has a single contactor at the input to the panel and relies on the SSR to cut power to the element, keeping 1 leg of the element hot at all times. Depending on how the panel is designed, you could enable/disable the main power contactor when one element is enabled and run the PID and pumps off of fused 110v upstream of the contactor.

That's what I thought. Run the SSR hot through one pole of the contactor, run the other hot through the other pole of the contactor, and switch the contactor off unless an element is plugged in and submerged. Contactor off, then light off. Contactor on and element plugged in, with no power through the SSR, then light off. Contactor on and element plugged in, with power through the SSR, then light on. Thanks.
 
No, I am saying that the 2 phase contactor cuts power to the outlet (and light and element).
No power = no leakage = light OFF.
It sounds like TH has a single contactor at the input to the panel and relies on the SSR to cut power to the element, keeping 1 leg of the element hot at all times. Depending on how the panel is designed, you could enable/disable the main power contactor when one element is enabled and run the PID and pumps off of fused 110v upstream of the contactor.

Right now I have a key switch to turn off and on the pwm but if I were to do it again I would also use a contactor that switches both 230 legs. This seems best and most logical. However its too late in the game for me to do that. I'll let you know what I end up doing...
 
Tonight I tried a 7.5 kohm 8W resistor wired in parallel with the LED and BINGO! No more light on when the element is unplugged.
 
wonder what the life expectancy of that resistor is tho...

Well I'm not an electrical guy so I have no idea. But what I do know is that a 7.5 Kohm 8W resistor will draw 32mA @240V and that equates to 7.7 Watts (just under the 8W rating). So I would think with just a few hours a month use it should last a while. But like I said I really don't know for sure.
 
Well I'm not an electrical guy so I have no idea. But what I do know is that a 7.5 Kohm 8W resistor will draw 32mA @240V and that equates to 7.7 Watts (just under the 8W rating). So I would think with just a few hours a month use it should last a while. But like I said I really don't know for sure.

Life depends on several things including wattage, how much use, and temperature. The resistors we use at work have a wattage de-rating down to 50% at our normal operating temperature according to the datasheet.
 
Life depends on several things including wattage, how much use, and temperature. The resistors we use at work have a wattage de-rating down to 50% at our normal operating temperature according to the datasheet.

How long would you guess mine will last?
 
Well I'm not an electrical guy so I have no idea. But what I do know is that a 7.5 Kohm 8W resistor will draw 32mA @240V and that equates to 7.7 Watts (just under the 8W rating). So I would think with just a few hours a month use it should last a while. But like I said I really don't know for sure.


But it's not going to have 240 across it as it is in series with the leakage resistance of the SSR. Measure the voltage across it, then use V*V/R to calculate the power dissipated. It will probably last a long time.

[Edit]

Cancel that thought as it will have 240 across it when the SSR is on.

Have you thought about sensing element current i.e. with a current transformer?
 
Or, maybe a series resistor and varistor in series with an LED.

Nobody liked the varistor in series with the LED? A 120V varistor, then the resistor, then the LED would probably work.

Not sure what's available at Radio Shack, but they probably have them. You could also scrounge one out of an old 120V power strip with surge suppression. Or, just buy from Digikey.
 
Um, no. If the control side was operating properly, WHY would it ever command the SSR on, if the SSR was 'failed closed'?

OK, you're right if the PID is in auto mode, where the temp sensor is the feedback channel.

However, in manual mode, where the PID ignores the feedback, I'm right.

I'll split the difference with you.
 
Nobody liked the varistor in series with the LED? A 120V varistor, then the resistor, then the LED would probably work.

Not sure what's available at Radio Shack, but they probably have them. You could also scrounge one out of an old 120V power strip with surge suppression. Or, just buy from Digikey.

Since I know nothing about varistors you would have to explain to me how/why it would work, and then also why it would be better than a single resistor in parallel with the LED that I've already got working.
 
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