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repairing corn cob led bulbs

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  • repairing corn cob led bulbs

    one of the arrays is faulty. lamp works if i bridged it as pictured. i want to bridge it with a resistor. how do i calculate the value? there are three groups in series. a group consists of five parallel leds. i know how to calculate resistance but not sure what resistance one led has. they supposedly are some kind of 5730 type (measure 5.3 x 3.0) and the array will work on 9v, but that might be overdriving them. if i look up 5730, they are 2.8-3.6v, 150ma. so does that mean i can calculate with 20 ohms? 3 x 4 = 12 ohms for the array? Click image for larger version

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  • #2
    can you figure out the drive current by by plugging a voltmeter into either end of that duff array? That would give you current for that array so resistance would be whatever you need to drop 9V at that amperage.

    Looks like the lights are run of rectified line voltage (unless there are some components inside the tube under the driver boards), so there may also be a largish resistor somewhere on the driver board that is the current limiting resistor. Or maybe it's on the back side of the board each array is on. Eitherway, that would also get you the drive current value.

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    • #3
      The resistance needed to bridge one LED or to bridge one of those strips of 15 LEDs can be found. But more details are needed. Please fill in the table below and I will tell you how to proceed to get an accurate value for that resistance. When I am asking about the "supply Voltage", I want to know what Voltage exists at the input terminals of the LED array itself, not what is used, like a battery or the AC from a house outlet.

      Supply Voltage: __________
      Supply Type (AC/DC): __________
      If DC, what type and number of batteries: __________
      Number of strips of 15 LEDs: __________
      You say the jumper allows the lamp to work.
      Number of 15 LED strips that light up WITHOUT the jumper: __________
      Number of 15 LED strips that light up WITH the jumper: __________
      What test equipment do you have access to: __________

      If you can, I would like to know exactly how the 15 LED strips are wired together. I would suspect that they are in series. But there could be several series groups of them that are then connected in parallel.

      Two facts that you can easily get immediately if you have a VOM are the DC Voltage across one of the 15 LED strips and the current that flows through them. For the Voltage, leave your jumper in place and with a DC Voltage scale selected, connect the meter across one of the working 15 LED strips. For the current, remove your jumper and set your VOM to a high current scale, usually around 5 or 10 Amps on most meters. Then connect the meter where you had the jumper connected. The remaining LEDs should light up and you can get a current reading. You will probably need to switch the VOM to lower, more sensitive current scales, but it is best to start at the highest in order to protect the meter from damage. Take the current reading at the lowest scale where the needle or digital display does not go over the top. I may be able to give you a resistance value from these two, but it will help if you can provide the other answers above. Or at least some of them.

      PS: LEDs are not linear devices. That means that their current vs. Voltage graph is not a straight line. A resistor, by way of contrast, does have a straight line graph (note below). And that further means that they do not display any SINGLE resistance value. Their effective resistance depends on how much current is flowing through them. And that current is determined by other things in the circuit. This is unlike a conventional incandescent light bulb which does display a somewhat linear behavior around their operating point on that current vs. Voltage curve. So, talking about "the" resistance for an LED is not really a good way to envision them.

      Note: if you are wondering about why a straight line graph indicates a constant resistance, remember Ohms law: I = V/R. If you turn that equation around you see that R = V/I. And the slope of a graph of the current (I) vs. the Voltage (V) is also given by ΔI/ΔV. So the slope of the current/Voltage graphic plot is the inverse of the resistance. If the inverse slope of the graph's line is constant (a straight line) then the (ΔI/ΔV) is also constant and that means that the resistance is constant. And conversely, if the plot of those values does not produce a straight line (LEDs or any diode are examples) then the resistance is not constant.
      Paul A.
      SE Texas

      Make it fit.
      You can't win and there IS a penalty for trying!

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      • #4
        With that many leds in total, it's unclear what the full connection is. But if you get it to work with a short as you show, then a 20 ohm resistor should be safe, and you could potentially go with about 68 ohms.

        There will be a point where the power dropped across the resistor is at a maximum, and it may require a 2 watt size. Bridging it like you show must be making other strings of leds glow brighter than they should be- the trick would be to select a resistor value that lets them all glow with the same brightness. The 'trick' with that is trying to see how bright they are without burning your retinas. One way I deal with this is to operate the lamp from a variable voltage so I can hit a point where they all glow but are not too bright to look at. Some arrays don't like a variable voltage, and you most often can get around this by using a single series resistor instead of trying to supply a critical voltage level. Even a 10k resistor will allow enough current to make the leds glow. This is just a test condition to allow you to find the correct value for the permanent shunt resistor.

        You might find that a few resistors in series would be better than a single resistor- this would make for a lower profile, plus spread the heat out- possibly allowing you to use 1/2 watt resistors, or maybe even the 1/4 watt size.
        I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-

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        • #5
          I wonder why the string has gone out. If you have some fine probes and a power supply (or even a partially discharge li-ion battery) you can test each LED. In fact, you can even use the diode function on most voltmeters. You may find that one of the strings of 5 or even a single LED has gone - bypassing that one will be easier and have less effect on the light overall. I've even bypassed those types of LEDs using old 5mm LEDs, works a treat.

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          • #6
            What is the cost of replacement?

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            • #7
              Cost vs price....What's the price of tossing something easily fixable with stuff on hand and a bit of time into a landfill? Not every hour of every day can be a billable hour...

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              • #8
                Most lamps of that type use a current regulated source that will apply the proper current to all LEDs in the series circuit. If your information on the individual LEDs is correct, each group of five in parallel will draw 5 * 150mA = 750 mA. Each column of 15 LEDs will drop about 9 volts, so the power for each column is about 7 watts. It looks like you have 10 columns so the total voltage will be about 90 volts. Thus, the lamp should work from about 100 VAC (or DC) and probably up to 250 VAC (or DC). It will have a switching regulator which will provide the proper current over that entire range, and will not be affected by shorting out one of the columns as you have done. Replacing the jumper with a resistor will serve no purpose other than burning up watts as a little space heater.

                If you can get to the LED terminals, you should be able to find one group of five with full rectified voltage across them (possibly 300 VDC or more on 220 VAC), and you might be able jumper just that group. You may also find that there is just a bad connection between the groups, although it's quite possible that one of the group of five failed open, which applied 190 mA through the remaining four. The weakest of that bunch then failed open, causing 250 mA to flow through the rest. They would blow out in short order until all were open. If the LEDs in the parallel groups are not well matched, or if they are not at the same temperature, current imbalance would cause individual overloads and the sequential failures as noted. Lamps like these often push the LEDs up to and beyond their best ratings, in order to squeeze more lumens at the expense of longevity. They do want you to buy new lamps pretty often, and would rather not have them last 50,000 hours or more as they would if driven more conservatively.
                http://pauleschoen.com/pix/PM08_P76_P54.png
                Paul , P S Technology, Inc. and MrTibbs
                USA Maryland 21030

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                • #9
                  Originally posted by Dan Dubeau View Post
                  Cost vs price....What's the price of tossing something easily fixable with stuff on hand and a bit of time into a landfill? Not every hour of every day can be a billable hour...
                  Merely a question, does not appear to be a typical Edison base component from the pictures. May quite likely be a high end industrial lamp costing many hundreds of dollars.

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                  • #10
                    Even the higher Wattage equivalent bulbs with Edison bases can cost a significant amount. I purchased a couple of 150 Watt equivalent LED bulbs with standard Edison bases and they weren't cheap.
                    Paul A.
                    SE Texas

                    Make it fit.
                    You can't win and there IS a penalty for trying!

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                    • #11
                      Most industrial lighting uses Edison 39 MM threads (mogul), most common consumer lighting products use Edison 26 MM threads.
                      A 100 w (light out put equal to a an incandescent lamp) LED mogul bulb may cost well over $200.00 each. At that price fixing such a lamp may be useful.

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                      • #12
                        This appears to be something like a 100 watt actual power bulb (about 300 watt halogen equivalent) which costs nearly $200:

                        https://normanlamps.com/cob-high-out...ite-5000k.html

                        Click image for larger version

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                        Paul , P S Technology, Inc. and MrTibbs
                        USA Maryland 21030

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                        • #13
                          Actually, it COULD be helpful to know the voltage range for the "bulb". It depends on how the current regulation is done. Presumably that is for 230V, so a nominal range of 200 to 240V is reasonable. If it were 100-277V the result might be different. The OP is in a 220V country.

                          Some LED bulbs use analog current regulators, which obviously dissipate more heat, but are cheap and effective. They tend to be made for specific narrow voltage ranges, because that minimizes the heat dissipation. With such a regulator, cutting out some LEDS will heat the regulator more, due to the increase in voltage across it.

                          Others use SMPS type regulators, which are more efficient, and can have a much larger range of voltage. They will not be bothered by cutting out some LEDs, in general, although that is not a guarantee..... if several sets of LEDs are parallelled, then a set with some LEDs cut out might hog most of the current, probably damaging it.



                          1601

                          Keep eye on ball.
                          Hashim Khan

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                          • #14
                            What you could do is temporarily put a resistor in place of the test lead, then check the voltages on the other strings. Chances are the voltage across your resistor would be different, so you would raise or lower the test resistor value until it more or less matches. I'd start with 68 ohms. Make sure to monitor how hot that resistor gets so you don't smoke it. You might well be right at 20 ohms- if your short isn't cooking anything you could pretty much start with any value resistor.

                            You'll obviously be eliminating one complete string and any potential light that could give. Alternatively, you figure out which section of that string has a higher voltage across it, then you add the resistor there. The higher voltage would mean the associated leds are open- bridging them with a resistor would allow the other leds in that string to light. This might be difficult to do though, seeing how compact that all is.
                            I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-

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                            • #15
                              to be honest LED bulbs that have that many LEDs are typically mains voltage rectified to DC with a simple diode bridge and a resistor (or resistors in each string) to limit current. I don't see much by way of the typical circuitry on the boards in the pic.

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