V Reg not working (I don't get it)

Discussion in 'General Electronics Chat' started by reconchris, Jan 11, 2019.

  1. reconchris

    Thread Starter New Member

    Aug 2, 2017
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    Ok so I have a standard 7.5V output voltage regulator. The max current it can deliver is 2A. I have a LED array that will operate at 7.5V at 3A. This means that when the array is connected to my power supply the voltage stabilizes at 7.5V once it draws 3A and goes into constant voltage mode.


    So the solution is simple right! Just run two 7.5v V Regs in parallel so they can both share the current while supplying 7.5V and your done right. The the V-Regs output 7.5V and the pannell will draw 3A at that voltage.


    Nope! the V-regs operate normally when they are being tested on potentiometers and other loads. However when I connect it to the led panel the V-Regs stop regulating completely and I see the input voltage across the LED panel which means the V-Regs act like they are not there.


    The circuit is so simple that this problem is making me feel dumb.

    the V-Reg I am using is a L78S75C.

    Moderators note : reduced font size. Large fonts are like shouting.
     
    Last edited by a moderator: Jan 11, 2019
  2. dl324

    AAC Fanatic!

    Mar 30, 2015
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    Please post a schematic and use a smaller font.
     
  3. reconchris

    Thread Starter New Member

    Aug 2, 2017
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    Here is the diagram.
    KLR light diagram.JPG
     
    Last edited by a moderator: Jan 11, 2019
  4. Sensacell

    Moderator

    Jun 19, 2012
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    Not going to work.

    Whichever regulator has the higher output voltage (even slightly) will hog the current and overheat.
    The regulator has only a short circuit limit of 3 A, it will just overheat and shutdown.
     
  5. Reloadron

    Distinguished Member

    Jan 15, 2015
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    This is your L78 series regulator data sheet. I do not see any input capacitors on the inputs or outputs? Next, trying to use series regulators in parallel is not a good idea and seldom works.

    Paralleling Voltage Reference-Based Linear Regulators

    "Current sharing with linear regulators is traditionally not as simple as connecting the parts in parallel. Two voltage reference-based linear regulators set to the same output voltage and with the outputs tied together will not share current equally. An LDO's output voltage is determined by the reference voltage multiplied by a gain factor based on the feedback resistors. Due to tolerance errors in the voltage reference and feedback resistors, the output voltages will be mismatched. With unmatched outputs, the LDOs will not share current; one LDO will provide the majority of the current until it hits current limit, thermal limiting or its output droops low enough for the other LDO to begin supplementing its current. These three situations present circuit operation challenges and can pose reliability concerns, leading to possible premature failure of the overstressed LDO".

    I see Sensacell posted and I am sure you get the idea.

    Ron
     
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  6. crutschow

    Expert

    Mar 14, 2008
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    What is the input voltage?
    Are the regulators on a heat-sink?
    A small resistor (say 0.2Ω) in series with each output will help them share the load.
     
    Last edited: Jan 11, 2019
  7. MrSoftware

    Senior Member

    Oct 29, 2013
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  8. MisterBill2

    Well-Known Member

    Jan 23, 2018
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    In post #5 Ron is right!! Those regulators will oscillate very well at some high frequency and the function will not be correct. Those bypass connectors need to be installed like the application note says. Not only are those capacitors important, they are VITAL!!
     
  9. spinnaker

    AAC Fanatic!

    Oct 29, 2009
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    You don't need to regulate voltage for LEDs. You just need to limit current. Just use the appropriate value and wattage of resistors. Or better use and switching LED driver to be most efficient.


    What are the specs on your LEDs? What is your supply volatge.
     
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  10. MisterBill2

    Well-Known Member

    Jan 23, 2018
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    There is an alternative that works quite well, and is simple to add. That is to use a PNP power transistor "wrapped around" the 3-terminal regulator. This will raise the current capability of the circuit to the sum of the limits of the two devices. So 20 amp transistor and a 1 amp regulator can provide almost 21 amps. I can't provide the circuit drawing right now, but it is fairly common and presented by a number of sources. the emitter of the PNP device connects to the supply, the base is connected to the input of the regulator, and the collector of the PNP is the output, with the output of the regulator also tied to the collector. So as the load draws more current the regulator increases the base drive to the PNP device, turning it on more. There are also series resistors from the supply input to both the regulator input and the PNP emitter. Those are well covered in the write-ups. AND the whole thing costs less than adding a second regulator.
     
  11. LesJones

    Well-Known Member

    Jan 8, 2017
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    This is the method that "MisterBill2" was suggesting.

    Screen Shot 01-12-19 at 10.14 AM.PNG

    Les.
     
  12. MisterBill2

    Well-Known Member

    Jan 23, 2018
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    Many thanks to Les for posting the circuit I was talking about. My computer with cad has not worked since a lightning hit a while back, and I have not installed the cad program on the replacement one yet. Lightning is hard on computers.
    Some versions put a second small resistor in series with the emitter of the power transistor, about 0R1 ohms, to avoid any tendency toward instability.. And the ratio of current division is then related to the ratio of the resistors.
    R1 is needed to assure that the transistor does turn off when the load is zero or very small, since the leakage current of power transistors may be a bit larger than the smaller ones.
     
  13. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Or you could go for a different voltage regulator like the LT1083 or LT1084:

    LT1083_84_85_comparisson.png

    Bertus
     
  14. spinnaker

    AAC Fanatic!

    Oct 29, 2009
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    Again the TS is just trying to light some LEDs. As long as the supply voltage is reasonably regulated, why does the TS need to regulate to 7.5V? It is the current that needs to be limited not the voltage.
     
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  15. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    There is no datasheet of the led array posted, so we do not know if the array has some internal constant current arrangement.
    @reconchris , do you have a datasheet of the led array that you can post here?

    Bertus
     
  16. MisterBill2

    Well-Known Member

    Jan 23, 2018
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    Upon occasion we need to go by the labels that come with things. Some of the time they are correct, some times not. An input of 7.5 volts allows for a small series resistor and two 3.6 volt LEDs all in series, and any number of those. And while a simple series resistor is the least efficient constant current method, it can serve quite well if maximum efficiency is not the primary parameter to be optimized. I have a very nice 12 volt light bar that is built that way.. A regulated current supply would double the price.
    Dropping 0.3 volts at 100mA would be wasting 30 milliwatts by my calculation. For a 3 Amp array having 30 sets like that, the power wasted would be 900 milliwatts. Not that bad, really
     
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