Linear voltage regulator question - input amp vs out amp

Discussion in 'General Electronics Chat' started by RogueRose, Jan 20, 2016.

  1. RogueRose

    Thread Starter Member

    Oct 10, 2014
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    I've been reading some spec sheets on the LM317 and I'm confused at what it can do. Here is the PDF I've been looking at.
    www.ti.com/lit/ds/symlink/lm317.pdf

    Now on the first page it says under features "output current greater than 1.5a" further down it says max amperage is 1.5 and in another place it says 2.2 A.

    From what I have come to believe about how these transistors work, is if a voltage of 7.5v is given to it and 5v is the output, then it wastes the 2.5v as heat. So if it was drawing 1.5 A for output, does that mean it would be drawing 1.5A * 7.5v = 11.25watts so output is 5v * 1.5A = 7.5watts Heat would = 11.25 - 7.5 = 3.75watts. Is this correct?

    If I desire the same output of 5v @ 1.5A and I give it a 3.5v input would it draw 3.5v @ 2.15A and have no waste heat? I'm confused as to how much input current is allowed.
     
  2. Lestraveled

    Well-Known Member

    May 19, 2014
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    Your heat calculation is correct.

    The input voltage must always be greater than the output voltage by at least 2.5 volts. (There is a chart for "Drop Out Voltage" verses current in the data sheet.)
     
    Last edited: Jan 20, 2016
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  3. #12

    Expert

    Nov 30, 2010
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    The 1.5 amp number is the guaranteed capability. The 2.2 amp number is, "as close as they could get" to a proper shut-down (with the safety circuits) without degrading the quality of the output voltage at 1.5 amps. The output current is exactly the same as the input current within less than 1 milliamp which is wasted in the ground leg of the chip. In your 3.5 volt example, the chip will waste some of the power simply because it can not pass current without a little bit of voltage across the chip. In essence, it can not become, "invisible" when supplied with insufficient voltage, even though that is the best one could hope for.
     
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  4. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    It must have a minimum of 2.5v input higher than the output, and they make a 5Amp version.
     
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  5. crutschow

    Expert

    Mar 14, 2008
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    There are "low drop-out" type regulators made that require less voltage across them for proper operation.
    They are useful when you want a regulated output voltage close to the input voltage.
    The LM317 is used otherwise since it is common, cheap, and rugged.
     
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  6. dl324

    Distinguished Member

    Mar 30, 2015
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    This won't work with a linear regulator like LM317. For the output voltage to be higher than the input, you require a step up switching (boost) regulator.

    If you operated a linear regulator with an input voltage lower than the desired output voltage plus the drop out voltage, the regulator would not regulate.

    A more appropriate term for what you're calling "heat" is power dissipation.
     
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  7. RogueRose

    Thread Starter Member

    Oct 10, 2014
    189
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    I greatly appreciate your very helpful responses! I'm very new to this more detailed electronics (have some experience with 12v automotive, computer/networking equipment & some minor 120vac work).

    I thought that the LM317 could take any Vin (within range of like 2v-36v) and give you any range from between that 2-36 but with the 2.5v difference requirement - so I thought it could take 5v and give 25v @ 1.5A - which is why I was so confused as to how it was done. A per what DodgyDave said, I can't get a higher voltage than what the input voltage is? I thought these chips were used in boost converters.?

    What I need to figure out is the best way to get a 5v output from either a single 3.7v source or multiple 3.7's in series (up to 5 in series). I think 2.2A would be sufficient but I would certainly not mind the 5A version (LM338?). I need a travel battery for my electronics. So if you were able to build the circuit with any of the 5 voltages (3.7, 7.4, 11.1, 14.8, 18.5 - but I think they actually are 4.0-4.1 when the batt is topped off, does that make much of a difference?) which would you choose for this application (let's assume I'm going to run 2 chips either 2 317's, 2 338's or a combo)?

    Thanks again for your help!!
     
  8. #12

    Expert

    Nov 30, 2010
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    317's and 338's are not boost regulators. (You can't get there from here.:D)
     
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  9. RogueRose

    Thread Starter Member

    Oct 10, 2014
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    Ok, good, so they may work for what I need. Now IDK if this will work. If I need the 5v and I have 20v source, Can I get 4A output from drawing 1A input on the LM338 or would it draw 4A and "waste" 15w / A as heat? Is this type of conversion possible with these or would this be something like a switching regulator?
     
  10. #12

    Expert

    Nov 30, 2010
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    Absolutely not.

    These chips are not wattage converters, they are voltage regulators.
     
  11. Lestraveled

    Well-Known Member

    May 19, 2014
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    @RogueRose
    Why don't you talk about what you want to do so we can stop wasting time and effort.
     
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