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LM1117 linear regulator.

Discussion in 'General Electronics Discussion' started by HellasTechn, Aug 12, 2020.

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  1. HellasTechn

    HellasTechn

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    Apr 14, 2013
    Greetings dear forum members.
    Today i am considering a power supply for a microcontroller. In particular one that will implement an LM1117 regulator the TO-252 package.
    In the past i have used the 7805 regulator many times and i am pretty familiar with its cooling requirements but when it comes to the LM1117 i have no idea.
    I need to step down 12V to 5V. My microcontroller will consume 70ma at startup, 150ma at full load for a few seconds and around 120 during normal operation. So it will have to dissipate around 1 W

    How can i calculate how hot it might get and if i need some short of cooling for it e.g. heatsink or fan?

    Thank you so much !
     
  2. bertus

    bertus Moderator

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  3. HellasTechn

    HellasTechn

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    Apr 14, 2013
    Thank you bertus. If i understand correctly for the TO-252 1 watt will pull its temp 45.1 deg above ambient, am i correct ?
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Right, if no extra cooling is applied.
    Observe chapter 11 in the datasheet. The value for RTja is valid only for a layout that follows the guidelines in this chapter. You need to supply enough copper area to serve as a heatsink.
    Max. junction temperature is 150 °C, so the chip will work in theory up to ~ 100 °C ambient temperature. The lower tamb is, the better for the endurance of the chip.
     
  5. HellasTechn

    HellasTechn

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    Apr 14, 2013
    Yes that makes perfect sense.
    Now i keep thinking what are the real advantages of using an LM1117 over an LM7805 ? Is it only its small size and SMD package or anything else ?
    In my particular case i now use the 7805 with only two ceramic disc capacitors and a small heatsink. If i switch to the LM1117 TO-252
    i will no longer need the heatsing since the PCB will act as such but i am going to need tantalium caps instead of ceramic.
    Though on the arduino uno boards i have i can clearly see that they do not use tantalium capacitors but ceramic SMD.
    I am a little confused here. Is it that they just do not follow the datasheet ?
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

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    When stepping down from 12 V it is size and max. power dissipation only.

    The true advantage of the LM1117 comes into play when the input voltage is barely above the output voltage. The LM1117 is a low drop regulator with a drop out voltage of max. 1.25 V. Meaning it can regulate e.g. a 5 V output from as little as 6.25 V input voltage. The 7805 has a drop out voltage of typ. 2 V (can't find a worst case value on the fly). So it requires at least 7 V input voltage.

    At the time the 7805 was developed, no ceramic capacitors of the required capacity were available, that is a more recent development. Small size at high capacity was available in the form of tanalum capacitors only. That's why the datasheet refers to tantalum (even never revisions, as these sections tend to be not updated).
    However, there is one concern with loop stability. The datasheet states that the output capacitor should have an ESR in the range 0.3 Ω ... 22 Ω. Ceramic capacitors typically have way too low an ESR and additional compensation (resistance) may be required. You can read more about this effect here and here.
    On the other hand, the official schematic of the arduino UNO shows a "big" electrolytic capacitor parallel to a small ceramic one in the power suppl. That's usually o.k.:

    upload_2020-8-13_9-51-57.png
     
  7. HellasTechn

    HellasTechn

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    Very interesting ! I will keep that in mind. Thanks !
    In the end of the 2nd pdf it states that it is a good idea to test the actual board (with the chosen caps installed) under various loads and see how it performs.
    So as for my prev question 7805 for input voltages of 8V or above and 1117 for voltages lower than 8V. Then Heat dissipation and filter caps consideration.
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    The second part is correct, but the first part not necessarily: you can use an LM1117 for input voltages above 8 V, too, if you take care of the thermal conditions.
     
  9. HellasTechn

    HellasTechn

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    Apr 14, 2013
    I am going to check both the datasheets for their thermal characteristics but after readig the above i suspect the 7805 is betterfor larger input voltages.
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    The TO220 case can dissipate more waste energy, so it is "better" for higher input voltages than the LM1117. However, considering the loss in usefully available energy my first choice for high input voltages would be a step-down regulator. Complete small modules based on the LM2596 are readily available.
     
  11. HellasTechn

    HellasTechn

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    Apr 14, 2013
    Yes that is also true but for the specific project i want to have a power supply as smooth as possible so i am not thinking to use a buck converter now.
     
  12. bertus

    bertus Moderator

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    Nov 8, 2019
    Hello,

    A linear regulator will make some heat thet must be dissipated.
    A TO220 version has the advantage that a heatsink can be mounted.

    A switching regulator will make some noise, but be much more efficient.
    Recom makes a switching version of the 7805:
    https://uk.rs-online.com/web/p/switching-regulators/0163451/

    Bertus
     
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