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I just built a small 5.1V regulator with LM317, using 220 and 680 ohm resistors. However, I measure 8.6V at the output without a load connected. Is this normal, i.e. under a load (about 120 ohm) it would drop to 5.1V? The input is a 9V/300mA unregulated wall wart, which reads about 13V without a load.

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    \$\begingroup\$ As a note, 220 and 680 are fairly large/strong resistors to use for a regulator. While it helps in this situation as the answerer Hans has pointed out, most of the time you would want a smaller pair, 2.2k/6.8k or even 22k/68k, to reduce the quiescent current draw from mA to µA. \$\endgroup\$
    – Passerby
    Nov 4, 2013 at 3:03

3 Answers 3

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You need a minimum load of 10mA for it to be stable. A tip is to add a LED and run it on 5mA. You'll know whether the supply is on, and it will add additional load to make it stable.

Note that the adjustment resistors of 220+680 ohm also give some load, of about 5.6mA. Together with a LED you get more than the required 10mA.

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  • \$\begingroup\$ National's datasheet says 5 mA; where are you getting 10 mA from? \$\endgroup\$ Mar 23, 2011 at 22:25
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    \$\begingroup\$ @reemrevnivek: LM317 parts manufactured by other companies like ON Semiconductor and Texas Instruments have the minimum required load current specified at 3.5 to 10 mA, so that's probably where Hans is getting the 10 mA figure from. \$\endgroup\$
    – In silico
    Mar 24, 2011 at 2:22
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    \$\begingroup\$ Because the specifications are measured at Iout = 10mA. The reference voltage is specified between 10mA and Imax. Even more important, 5mA is typical, 10mA is worst case. If you do get a bad batch, or a different supplier you want to make it work as well. Some values are also 'ensured by design but not tested'. In other words, don't fixate on exact precise figures in datasheets. Go for a robust solution. \$\endgroup\$
    – Hans
    Mar 24, 2011 at 18:28
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One way to ensure a minimum load is to make sure the feedback network draws the minimum load. For example 240 ohms and 75 ohms gives a 5.25V output and draws 16mA. Tweak the resistors to get it perfect.

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The regulator works mostly like a resistor which is automatically adjusting itself so if there is too little current the voltage drop on the regulator will be too small to regulate the output.

It is possible to make regulators that work without load but they tend to suffer from high quiescent current.

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  • \$\begingroup\$ This is very important knowledge, but really could be used in an addendum to an answer to the actual question. \$\endgroup\$
    – Kortuk
    Mar 24, 2011 at 1:13
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    \$\begingroup\$ But the question is "Is it normal?" and not "How to fix it?" ;] \$\endgroup\$
    – jpc
    Mar 24, 2011 at 1:16

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