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Old 12-25-2012, 12:44 AM   #11
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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.

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Old 12-26-2012, 10:32 PM   #12
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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.

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Old 12-26-2012, 11:32 PM   #13
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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.

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Old 12-27-2012, 01:35 PM   #14
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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?
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Old 12-27-2012, 03:36 PM   #15
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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.

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Old 12-27-2012, 08:09 PM   #16
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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…

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Old 12-27-2012, 08:22 PM   #17
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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!

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Old 12-27-2012, 08:27 PM   #18
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Quote:
Originally Posted by -TH- View Post
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.
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Old 12-28-2012, 03:18 PM   #19
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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.

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Old 12-28-2012, 04:25 PM   #20
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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.

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