Paralleling LM317T's

Discussion in 'The Projects Forum' started by iONic, Nov 20, 2008.

  1. iONic

    Thread Starter AAC Fanatic!

    Nov 16, 2007
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    Quick question.

    I have read that placing voltage regulators in parallel can increase the current.
    For example, a couple of 317's with max of 1A each could produce 2A. But am I right to assume that if the input voltage source can not provide the 2A, let alone the 1A, then this is not possible? Would it increase the output current at all?
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    If your input voltage source cant provide a current more than one Amp then you cant increase this by using a regulator, you have to change the voltage source itself.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    LM317T's have a rated max continuous current of 1.5A, not 1A. You're confusing them with the 78xx series, which ARE limited to 1A (and have poorer regulation specs to boot).

    You can't get more current out of a regulator than you put into it. If your voltage source is limited to 0.9A, that's all you'll be able to get out of the LM317T, no matter how many you put in parallel.

    Don't forget that when used as a voltage regulator, some of the current will be consumed in the resistor network (R1,R2) used in establishing the output voltage. The standard value for R1 is 120 Ohms, which guarantees regulation within specifications, as Vref (voltage between OUT and ADJ) is specified to be 1.2v to 1.3v inclusive; 1.25v nominal.

    There's also the LM150/LM350 (3A) and the LM138/LM338 (5A) which are similar to the LM317, but they aren't current generators either.
     
  4. BradofCanada

    New Member

    Jul 10, 2014
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    Hi!

    Just a few things to be aware of when attempting to parallel voltage regulators to increase output current. Yes it can be done, however the best way to go is to use a single regulator capable of delivering the output current you need.

    You need to keep in mind that as precise as manufacturers try to make devices identical: there will always be variances, and these variances will create potential differences between the two (or more) semiconductors you are attempting to parallel. The potential differences will cause voltage and current losses, as the devices get into a heated argument with each other.

    Yes, the differences will generate extra heat - not tremendous, but a definite reflection of the inefficiency of the total circuit. If you are stuck with a surplus of regulators, and are determined to parallel them: you can add external rectifier diodes to help minimize some of the variances between the two or more regulators.

    A good rule of thumb I use when creating a power supply is to design it to deliver at least 50% more current than the device you want to power will ever need. More commonly I go a full 100% more than required, as this helps minimize output voltage drop when starting an inductive load.

    Also I suggest a minimum of 1000uf input capacitance (before the regulator) per 1000mA of output required. This will help compensate for voltage dips the front end of the supply that may occur, especially if you have heavy inductive loads running from the same supply line.

    Hope that helps!
     
  5. Lestraveled

    Well-Known Member

    May 19, 2014
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    There is a very simple way to increase the current output of a LM317 with an external transistor. The circuit is in one or two of the common data sheets (TI/Fairchild). I have used it with two 2n3055 to make a 10 amp supply.

    Mark
     
  6. Lestraveled

    Well-Known Member

    May 19, 2014
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    Here is the basic schematic
     
  7. BradofCanada

    New Member

    Jul 10, 2014
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    Nice circuit Lestraveled! It should work well for most applications.

    I still prefer using a higher current regulator, and minimizing the parts and the bulk, but this makes it look like we're really achieving something! Lol
     
  8. crutschow

    Expert

    Mar 14, 2008
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    No linear regulator can output more current than is available at the input.

    Of course, you can increase the output current as compared to the input current with a switching regulator. It acts similar to a transformer to convert a higher voltage at a low current to a lower voltage at a higher current. But you still can't get more power out than power in (VA out can never be more than VA in). And you will lose some power in the conversion process (switching regulators are typically 80-90% efficient) so output power is always some less than input power.
     
  9. Lestraveled

    Well-Known Member

    May 19, 2014
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    True. You give up something, to gain something else. My personal experience with this circuit is that it is very robust.

    Mark
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Sorry I deleted this message before I realized someone had already responded to it. I deleted it because, after doing some simulations, I realized that LM317s don't parallel well due to their significant reference voltage tolerance.

    Prior Deleted Post:
     
  11. crutschow

    Expert

    Mar 14, 2008
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    I did find something interesting in my simulation of the boost circuit (below), similar to that shown in post #6. If you use a value of about 0.8Ω for R3 and .04Ω as an emitter resistor (R5) than the LM317 current limit will also limit the current of the D45H1 10A transistor to <9A (with a 10V source). That should be sufficient to protect the transistor against momentary short circuits at least.

    Of course how well that reflects the real circuit depends upon how close the LM317 and transistor models are to the real devices. Some tweaking of R3 and R5 will likely be needed to get the desired current limit.

    LM317 Boost.gif
     
  12. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Did you see this schematic in the LM317 datasheet?

    [​IMG]

    As you can see they are using some small resistors at the outputs of the LM317's to balance the currents.

    Bertus
     
  13. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    One of the regulars here recently mentioned making a PSU with 3 LM317's paralleled, that he said had worked fine for many years.

    Probably the wiring and internal resistance of the devices was enough to share the load well enough.
     
  14. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Using a Buck regulator you can get more current out than the current in, but you need to have more voltage in than you have out then.

    For example, if you have a DC wall wart that can put out 1 amp at 10 volts and your regulator puts out 5 volts, then with a buck you can get almost twice as much current out. It wont be twice as much because the buck converter is not 100 percent efficient, but you get the idea. If you want twice the output current then you need more than twice the input voltage. If you want three times the current out, then you need more than three times the input voltage.

    So the formula then is:
    Vin=(Vout/eff)*Iout/Iin

    where 'eff' is the efficiency of the converter, and Iin is less than Iout.

    Using the above, an 80% efficient buck with a 5v output at 2 amps requires 1 amp input at a voltage of:
    Vin=(5/0.8)*2/1=12.5 volts input at 1 amp.
     
    Last edited: Jul 12, 2014
  15. Patrick Joseph

    New Member

    Feb 23, 2015
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    I like your circuit. Can I ask you about the diodes? Are they just standard 1n4007 diodes? And the values for the resistors? I'm used to seeing 100 Ω and such, not 2R2.
     
  16. Lestraveled

    Well-Known Member

    May 19, 2014
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    D1 and D3 are protection diodes so 1N4007 (1 amp) diodes are the smallest I would go. D2 can also be a 1N4007.

    2R2 is a notation for low value resistors. 2R2 = 2.2 ohms. The R is a decimal point.
     
  17. Patrick Joseph

    New Member

    Feb 23, 2015
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    I finally built this regulator circuit. It works great! Thanks for the suggestion.
     
  18. RodneyB

    Active Member

    Apr 28, 2012
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    How do you get the resistor power rating
     
  19. Lestraveled

    Well-Known Member

    May 19, 2014
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    I assume you mean the wattage rating of R4 and R5. Each will see half of the output current, so use the equation, W = I * I * R. Use the next higher wattage resistor.
     
    RodneyB likes this.
  20. beic

    New Member

    Jun 13, 2016
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    Can I get that schematics of 2N3055, please?! and can it be done to be regulated with 5k pot. trough R2?
     
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