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When going from 14.5V to 5v with linear regulator

Discussion in 'General Electronics Discussion' started by adamjohnson, Oct 14, 2016.

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  1. adamjohnson


    Oct 14, 2016
    Hello all,
    I was wondering if its ok to use a voltage divider to drop voltage and feed that to the Vin of a linear regulator lm2940 for example. IE: Voltage divide from 14.5 down to 9v then go from 9v to 5v with the LM2940. I'm going to go out and assume that this a janky way of getting a
    linear regulator to work with out over heating it.

    I'm wanting to draw at most 100ma and right now doing it 14.5v to 5v its getting up to 170f at room temperature of 71f yikes..! I know a lot of that comes from shedding voltage difference in heat.

    I also have a second question if i reduce my current pull down to 10ma the LM2940 gets around 105f is that pretty safe for a automotive environment.

    I'm confined to a package and can't get a heat sink really in there to help with the regulator.
    I could possible lay it down and solder it to the board if that would help for longevity.

    I appreciate any help, kind of been something I have been trying to figure out. I was originally using a buck converter but it didn't have revers polarity protection, or load dump protections like the LM2940 does So I went that route as anything else buck converter wise was way way over price point I was trying to stay in.

    I'm also open to suggestions.
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    That's not the way to go. The voltage divider's output voltage will vary wildly with changing output current. It will also dissipate additional power due to the current through the divider which is not sed to power the load.
    A linear voltage regulator is explicitly designed to regulate a lower voltage from a higher input voltage.

    The lm2940 is spec's for a max. input voltage of 26 V, so it can easily cope with the 14.5 V you're going to supply.
    At 5 V output voltage the voltage differential across the lm2940 is ~ 10 V (o.k., 9.5 V to be exact ;)). 10 V * 100 mA equals 1 W. Depending on the case style you use and the mounting conditions, this can create up to 60 ° C overtemperature for the junction. The chip will get hot but will operate at ambient temperatures up to ~ 90 °C without exceeding limits (max. junction temperature is 150 °C). Even at higher temperatures the chip will internally limit the power dissipation to protect itself.

    If you really need to reduce the power dissipation of the regulator, insert a zener diode in series between the 14.5 V power source and the input of the regulator. When using e.g. a 6.2 V zener, the power will be distributed as follows:
    • zener diode: 6.2 V * 100 mA = 0.62 W -> chose a 1 W type
    • regulator: (14.5 V - 6.2 V -5 V) * 100 mA = 3.3 V * 100 mA = 0.33 W
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