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I tried to make variable voltage regulator using LM317T IC.enter image description here I constructed the above circuit with input of 12V DC supply. Since I currently don't get 240ohm for R1 (as recommended by datasheet), I used 110 ohm and 470ohm on two trials. On both trials, the output only varies between 0.5V to 9.5V (approx.). Now, my doubts are,

  1. Why it's does not reach nearly 12V ?
  2. So , Is 240ohm at R1 mandatory to exact functioning ?

Edit: there is no load at output - I only measured it using DMM


After changing some values with the same input, i.e., R1 = 470 ohm, R2 (variable resistor) = 100K trimpot

The differential voltages become varies between 0.06V to 10.32V (which the answers suggested). Now,

  1. Is these high value of R2 recommended to increase output?
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  • \$\begingroup\$ Are you sure about the voltages you measured? By what the LM317 does and its datasheet specs, I would expect 1.25V up to 10.5 volts (maybe slightly less, but 9.5V seems too low). Why do I expect 10.5 volts - see the "dropout voltage" specification in the datasheet. \$\endgroup\$ Commented Sep 4 at 13:45
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    \$\begingroup\$ Do you have C1 and C3 in your circuit? \$\endgroup\$ Commented Sep 4 at 13:50
  • \$\begingroup\$ @anrieff that's my doubt \$\endgroup\$ Commented Sep 4 at 14:20
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    \$\begingroup\$ You do know about series and parallel resistor combinations: if you have two of each value of resistor : series 110 + 110 = 220 ohms or parallel pair of 470 ohms gives (470 / 2) ohms = 235 ohms : so you could have reached almost the ideal value. \$\endgroup\$ Commented Sep 4 at 14:37
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    \$\begingroup\$ @RajeshR There IS a load on the output; R1 MUST draw a few milliamperes, for proper regulation. The 470 ohm R1 isn't enough load according to the specifications (which assume worst case scenario) \$\endgroup\$ Commented Sep 5 at 0:10

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On both trials, the output only varies between 0.5V to 9.5V

What you may not be aware of is that the device is not "a low drop-out" type. "Drop-out" is the voltage difference required (for the device to function correctly) between input and output voltage. Here's a graph from the ON Semi data sheet: -

enter image description here

At 25 °C and, a low output current, the drop-out is about 1.5 volts. As load current rises the drop-out becomes worse. With a drop-out of 2.5 volts it seems you might have nearly a full-load current.

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  • \$\begingroup\$ But there is no load at output - I only measured it's using DMM \$\endgroup\$ Commented Sep 4 at 14:14
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    \$\begingroup\$ @RajeshR nevertheless, the point I'm making is that you cannot expect better than 10.5 volts on a light load and, if your input supply is maybe only 11 volts, you'll get 9.5 volts out but, you may have a fake part or some partially broken part or some other thing like lack of decoupling capacitors. Also have you checked the value of R2? Measure input voltage with the circuit connected is something you have to do. \$\endgroup\$ Commented Sep 4 at 14:29
  • \$\begingroup\$ Its correct - input voltage is 11.99V \$\endgroup\$ Commented Sep 4 at 17:30
  • \$\begingroup\$ @Whit3rd the claim of no loading is made by the OP under other answers so why are you targeting my answer for your comment? \$\endgroup\$ Commented Sep 4 at 23:51
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    \$\begingroup\$ @Andyaka - Hi, I don't see a problem here. Your answer mentioned load current. The OP commented. If another site member sees an apparent misunderstanding in a comment (here, it's a comment by the OP) it's fine for that other site member to respond below where that mistaken comment was written. There's no need for site members to specifically avoid replying to other comments below your answers. This is just how comments go sometimes. You likely see this more simply because of the number of answers you write :) And what if the OP had only made that comment here? Responses would be here! TY \$\endgroup\$ Commented Sep 5 at 0:22
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The LM317 regulator has a dropout voltage of about 1.5V under light load (suggesting about 10.5V output with 12V input), and your numbers are considerably less, so there seems to be an issue if your load is indeed light.


Possible causes: You should have the capacitors shown in place to prevent possible oscillation. If your input has some ripple that will affect the readings since the regulator would drop out on the troughs of the input waveform. Or possibly measurement error. Since the minimum output voltage has to be approximately 1.25V (not 0.5V) that might be a possibility. Or maybe the schematic you are showing us is not complete (maybe there's a series diode or something like that).


The 240Ω is not a mandatory value, lower is fine, in fact 120Ω is the maximum that meets worst-case Vin-Vout and temperature minimum loading for the LM317. The purpose of having it that particular value is to draw the minimum current required by the regulator when the load is open, while not wasting excessive power.

enter image description here

With the nominal 1.25V Vref it will draw 5.2mA which is typically adequate (especially for Vin-Vout \$\le\$ 20V, but is not really guaranteed. Using 120Ω meets all conditions for the LM317.

The consequence of using too low a value is wasted power. The consequence of using too high a value is that the output voltage may rise above the set value with a light load (usually considered very undesirable behavior). In extreme cases, the voltage may be less stable because the adjustment pin current and current change will affect the voltage more.

If you are making an adjustable supply you would vary the rheostat value along with the resistor in question to get the desired voltage range, based on the formula in the datasheet.

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  • \$\begingroup\$ But, There is no load at all \$\endgroup\$ Commented Sep 4 at 14:24
  • \$\begingroup\$ Yes (other than the 5-10mA load provided by the R1/R2), so it should be more like 10.5V maximum/1.25V minimum with steady 12V input and the circuit shown. \$\endgroup\$ Commented Sep 4 at 14:27
  • \$\begingroup\$ The LM317 specified max output current is 10ma so R1 should be no greater than 120Ω for it to stay in regulation. The 5mA/240Ω value is for the LM117 mil temp range version. \$\endgroup\$ Commented Sep 4 at 14:51
  • \$\begingroup\$ @CarlRutschow Yes, mil version only. The 240Ω is okay for limited Vin - Vout only. Will edit. \$\endgroup\$ Commented Sep 4 at 15:01
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    \$\begingroup\$ @CarlRutschow The typical curve in my answer above shows that the minimum current is typically 2mA for Vin-Vout 20V or less. That's why most of the published circuits for LM317 can get away with 240Ω- a >2.5:1 margin which is deemed adequate. \$\endgroup\$ Commented Sep 4 at 16:15
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  1. it is normal that output does not reach 12 V. Your expectations about the regulator operation are unreaistic. A linear regulator with 12 V on input cannot output 12 V.

    LM317 recommended operation is 3 V drop between input and output, so you can expect only 9 V out, maybe 9.5 V out with little or no current.

  2. 240 Ω resistor does not have anything to do with output not reaching 12 V. It simply provides the minimum load of about 5 mA to keep the regulator output at the set voltage.

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  • \$\begingroup\$ Is it true that, it can drop 3V ? \$\endgroup\$ Commented Sep 4 at 14:23
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    \$\begingroup\$ @RajeshR Yes. It is true. Check from the datasheet if in doubt, it reads there. It's a regulator. Under normal use, it will drop at least 3V and it can drop up to 40V. \$\endgroup\$ Commented Sep 4 at 15:00
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  1. There is minimum voltage between input and output pins required for proper operation of any regulator. It is called "dropout voltage". It is in the datasheet but I would assume at least 3V for something with bipolar Darlington pair as a main regulating element.

Low dropout voltage (LDO) regulators do exist, but 317 is rather at the high side instead.

  1. 110 or 470 ohm are both good enough for most purposes. The exact value is a compromise between the circuit stability and the wasted power. The datasheet contains the minimum load current that you may be interested in.
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