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1 year, 8 months ago via answers.hackaday.com

I have an AC adapter that puts out 12VDC at 5 Amps. This runs a type of lighting system made from LED's. When the adapter is initially power

ed up, it surges to almost 18vdc for a split second. This is enough to shorten the life span of the LED's considerably because it's turned on an off frequently. My question is what spec of zener diode can I put in the mix to regulate this? I know its 12v, does the wattage have to equal the adapter i'm using or how exactly do I figure this? OR is a zener not the right answer?
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jkreed | 1 year, 8 months ago
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I'm assuming here that the power supply is a switcher.

If the supply is custom-made, the root cause of the problem is likely a controller issue. Specifically, the "compensator" portion of the PWM circuit, which forms the PI controller (proportional & integral), was either not designed properly or not implemented properly. Another possible explanation is that a capacitor should be connected to the 12V output but for some reason isn't. Check any documentation that came with it - you may have to supply this externally.

Regarding how to fix the issue, if you have knowledge of controller circuits you can try modifying the compensator directly. That's the most difficult but the "proper" way to do it.

A better way is to add a low-pass filter to the output, either with an RL or RC circuit. Caps are generally easier and cheaper than inductors. Figure out how long the transient (spike) lasts in seconds. Let's say it lasts 0.1s, so we want the filter to have a time constant of 0.2s for safety. This means that at 0.2s after turn-on, the output voltage will be 63% of 12V, or 7.6V, gradually increasing to the final 12V. If R is in ohms and C is in farads, the time constant is equal to R*C, thus multiplying the two numbers must give you 0.2.

When choosing your R and C, there is a trade-off. You can use a large R and small C (cheap), or small R and large C (perhaps more expensive). When 5A is running through the output, the 12V output is decreased by 5*R in volts, so a smaller R is best for staying close to 12V output. For example, R=0.1 ohms leads to Vout=11.5V at 5A assuming constant 12V from the power supply. This also leads to C=2F, which isn't reasonable at all!

Play around with the numbers and try to estimate the duration of the voltage spike as best you can - the shorter the time the smaller the C is needed. Another responder said to use an RL filter, which may be easier, but still (in the above example) requires a 20mH, 5A, dc inductor which is not exactly a cheap part.

Once you have the duration nailed down (use a scope if you have one), you can also try making a very large capacitor out of super/ultra caps. Most have low voltages - say 6.3 max, so to get 15V (incl a safety margin) use 3 in series but remember that total capacitance for that string decreases to 1/3 of the individual capacitor values. Then you have to add more of those series strings in parallel to increase the total capacitance.

GL!

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jkreed | 1 year, 8 months ago Report

Okay, one more approach that is way easier. You say the PS is a switcher, then it should be current regulated. Can you verify that if you short the output, that 5A can flow without causing a fuse to blow?

If this does indeed work, the capacitor can be made much smaller than what I said above. Try adding a 47,000 uF, 20V electrolytic cap to the output. They cost about $2 each.

With a cap on the output at 0V, it's like the power supply is shorted and will therefore be in current regulation. Treating it as a current source, the cap on the output will have its voltage increase linearly with time. Starting with the capacitor equation Q=CV, we can make derive the equation C = 5A*0.1s/12V, which gives 0.042F or 42,000uF. In other words, it'll linearly ramp from 0V to 12V in 0.1s when supplied with a 5A source.

I hope this works. You mention that you're turning the LEDs on and off frequently - how frequently? Every time you charge and discharge the cap it will heat up and decrease its life. If this is an issue, using several smaller caps in parallel can distribute the heating and extend the overall lifetime.

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rangerx52 | 1 year, 8 months ago
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what kind of lighting system is it?

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facefusion | 1 year, 8 months ago Report

You know the Malibu yard lights that you stick in the ground around your landscaping? Pretty much like that but without the timer and transformer they use. I use the AC adapter mentioned. Each fixture has 2 - 10mm LED's with a 510ohm resistor at each positive leg allowing me to run 12vdc to them. The daily on/off with the jump in voltage is degrading them far to soon. This is why I am looking for a simple solution to keep the voltage from leaving the 12vdc range during power up.

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ejp1087 | 1 year, 8 months ago
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You say it only surges to 18V for a split second? You could add a low pass filter in order to filter out the high frequency surge to 18V. This might also cause your LEDs to appear to fade on and off when you turn them on and off, but if the frequency of the surge is high enough your LEDs could still look just like normal. Do you know the time it takes for the power to go from 0 to 18V and the time it takes to go from 18V back to 12V? If you know these times, then I'm sure you could design a simple low pass filter to protect your LEDs. It might mess with the current to the LEDs, but only a very tiny bit.

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st2000 | 1 year, 8 months ago
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I'm thinking you are going about this wrong.

1) Maybe this power supply is not so good. Maybe better supplies do a better job at regulating voltages. Including the initial voltage! What voltage does the supply put out w/o a load. If it's still over 12VDC I would plan on getting and testing other supplies. BTW, is this a switcher?

2) If you are not going to replace the power supply, I would toss a relay on the output to keep the LEDs isolated until the power stabilized. You could probably use a simple cap-resistor R/C time constant to keep it activated for a second or how ever long it might take. My TV does this. My stereo does this. However, those are probably done under microprocessor control.

-good luck

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facefusion | 1 year, 8 months ago Report

It is a switching supply. When powered on, voltage sits idle at 12.32 vdc. After a load of 25 fixtures (see my reply above) the drop is to about 11.8 vdc. Both of these voltages are within spec for the way I currently have it configured. The reason I can't go with a different supply is that I had this one custom made to run at 5 amps to accommodate the the length of the run that varies from time to time. With this supply I can run over 100 fixtures and still have a good result. I searched for over a month to find one pre-made. Best I could find was 2.5 amps. Which is fine for some cases, but I wanted to be able to use one supply for every scenario versus having different supplies for different quantity configurations.

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st2000 | 1 year, 8 months ago Report

I wouldn't have thought a switcher would have start up problems. If really custom made - maybe you can go back and have it fixed. If not possible, pull the specs on the switcher chip and see if there are allowances for adjusting the initial conditions. If none of that is possible, I still like the idea of a simple timing device (one shot) controlling a relay such that power is not allowed out of the box until the supply has stabilized.

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crkrjkhd | 1 year, 8 months ago
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you would want a zener diode with a power rating over 60w and a 1 ohm resister comming from + before the fork to the zener.. search google for zener diodge voltage regulator

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ryzic | 1 year, 8 months ago
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Have you considered using a RL circuit? Consider that the inductor is under DC conditions a short circuit, but under transients will attenuate a signal (low pass filter). You might not need to worry about a zener diode if this is only a startup thing.

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opossum | 1 year, 8 months ago
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Get a 12V 5A switching power supply that works properly. It is a standard stocked component at DigiKey, Mouser, Jameco and many others.

Here is one for $24 at Jameco: http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_123335_-1

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facefusion | 1 year, 8 months ago Report

What I am using is a sealed AC adapter. The power supplies you listed, I am aware of, but being that I need an AC adapter, I cant use any of them. I also have about 200 of these adapters left in stock that I need to use up before finding a better alternative. They ALL do the same thing. The way I NEED to fix this issue is to put a single component at the + output of the adapter and then put my system after the component. I may be hoping for something that might not be possible, until I know for sure that it isn't possible, I have to keep looking. All the suggestions I have heard are excellent, if I were just dealing with a one time problem. This is a lighting system that I install for a living. So I have to go back to every job i've done and correct the problem. This is why I am hoping it can be a cheap, one component fix to just add in. I wish in the beginning I had noticed a spike, but I had no reason to be looking for one. The output was 12v and that was that. Something custom like this, I would have expected it to be on the money. That's what I get for assuming. I spoke to the people that manufactured my adapters and they say I'm crazy. I am in the process of making a short video of the adapter hooked to a scope so they can see I'm not making it up. Either way, I know they aren't going to do anything to correct the ones I already have, but at this point I just want to show them that the product I got from them is crap.

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ryzic | 1 year, 8 months ago Report

He didn't ask for a new supply, he asked for how to adapt his current one.

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opossum | 1 year, 8 months ago Report

The current one is clearly defective.

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