I know that ratio R1 to R2 determine the output voltage of LM317. E.g. R1 = 200, R2=330 ohms will produce about 3.3V. My question is, what if I use 2K and 3.3K for R1 and R2? What is the impact of increasing the resistors values but keeping the ratio the same?

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    \$\begingroup\$ to reduce error due to incorrect datasheet lookup can you give us a link to the datasheet for your chip? \$\endgroup\$ – Kortuk Aug 22 '11 at 20:45
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    \$\begingroup\$ The National LM317 datasheet is at national.com/ds/LM/LM117.pdf. The adjust pin current has a maximum of 100 uA. \$\endgroup\$ – Kevin Vermeer Aug 22 '11 at 21:35
  • \$\begingroup\$ @JGord - Did your comment end up on the wrong post? This doesn't have much of anything to do with an op-amp. If this was a mistake, please flag the comments so a mod can delete both. \$\endgroup\$ – Kevin Vermeer Aug 22 '11 at 21:45

The output voltage is determined not by the ratio of R1 to R2. It's given by the following equation:

\$ V_{OUT} = 1.25 \left(1+\frac{R_1}{R_2}\right) + I_{ADJ}R_2\$

For ordinary purposes, the \$I_{ADJ}R_2\$ term can be discarded, because \$I_{ADJ}\$ is on the order of \$100\mbox{ }{\mu}A\$.

You've multiplied your resistors by 10, so this error term will also be multiplied by 10, going from 33 mV to 330 mV, or 0.33 V.

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    \$\begingroup\$ Thanks Kevin. But wouldn't I_adj also go down by factor of 10, thus compensating in the equation above? \$\endgroup\$ – lyassa Aug 22 '11 at 21:55
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    \$\begingroup\$ @lyassa - No, \$I_{ADJ}\$ is a function of the internals of the device, and is largely independent of the resistors. \$\endgroup\$ – Kevin Vermeer Aug 22 '11 at 22:07

Several people have correctly pointed out that the LM317 output voltage is affected by the Iadj current which flows in R2 (see example circuit below).

Two factors are potentially relevant for Iadj - its absolute values of 50 uA typical, 100 uA maximum, and its variation across the load range of 0.2 uA typical, 5 uA maximum. As others have noted, R2 need to be small enough such that Iadj voltage drop in R2 can be ignored or it must be allowed for. If R2 is large then the change in Iadj through R2 under load may be significant. For example, if Iadj changed by it's maximum value of 5 uA across load and if R2 was 100k (much larger than usual) then the change in Vout would be V=I.R = 5 uA.100k = 0.5Volt! Even a 20k here would cause a change of 0.1 Volt, which may be of concern in some cases. (If it was then you should probably not be using a simple 3 terminal regulator, but that's another story).

Typical LM317 voltage regulator

Less subtle issue: There is a second less subtle but sometimes overlooked factor. The LM317 internal electronics are "operated" by the dropout voltage across the regulator and a minimim current MUST flow through the regulator to achieve regulation.

The LM317 datasheet specifies 10 mA max, 3.5 mA typical as the minimum load current (on page 4 of the referenced datasheet). (A maximum minimum is a nice concept :-)). 'Proper' design requires that 10 mA worst case be allowed for. IF the external load always draws 10 mA or more then all is well. However, if external load current can fall to below 10 mA then the design must provide a load to provide this 10 mA. Worst case, with no load, R1 provides a convenient way of providing the 10 mA while also providing a nicely "stiff" divider. R1 will always have 1.25V across it in normal operation. Using R1 = 240 ohms as shown in the datasheet example gives I = V/R = 1.25/240 = 5.2 mA which is more than the 3.5 mA typical minimum load needed but less than the 10 mA worst case minimum load needed. If there can be zero external load then you need no more than R=V/I = 1.25V/10 mA = 125 ohm for R1 if this is the way you obtain your minimum load current. SO the 240 ohm resistor shown for R1 would not meet the LM317 minimum load requirement worst case. Either a lower value of R1 must be used or a minimum external load suitable to bring the total up to at least 10 mA must always be present.

With R1 set, R2 can now be dimensioned to achieve the desired output voltage. With 10 mA flowing in R1 + R2, Iadj is insignificantly small in all except critical cases.

When 'designing" a circuit (rather than just "making it work") it is essential that the worst case parameters are used. What constitutes 'worst" will vary with the parameter and, in some cases, you may have to use the mimimum value of a parameter for one design calculation and the maximum value of the same parameter for another calculation.

Efficiency issues:

"For interest" - the LM317 has a minimum dropout voltage of about 1.5V to 2V for most of the range of conditions that would typically apply. (25C, 20 mA to 1A. ) Dropout can be as low as 1V at 20 mA at 150 C (!!!) and as high as 2.5V at 1.5A at either -50C or +150C (!). 2V is a safish value for dropout for scoping calculations. Worst case for your design needs to be established when doing final design.

At say 5V out then efficiency = <= Vout/Vin = 5/(5+2) =~ 71%.

At very low currents the minimum load current of 10 mA may be significant. eg at 1 mA out efficiency = 1ma_load /10_ x 71% = mA_min = 7.1% ! :-) :-(.

At 5 mA out its 5/10 x 71% =~ 35%.

Maximum efficiency rises towards typically 70% with increasing loads.

BUT all the above is what happens when the regulator is just at the point of "dropout". Where Vin is more than about 2V above Vout it is the regulators job to drop the excess voltage. So efficiency must be lower than the maximum possible in most cases.

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    \$\begingroup\$ This is a great detailed description. +1 \$\endgroup\$ – Al Kepp Dec 9 '12 at 23:46
  • \$\begingroup\$ "V=I.R = 5 uA.100k = 0.5Volt" Why 5uA? is it not 100uA? \$\endgroup\$ – Adrian Maire Sep 23 '16 at 12:40
  • \$\begingroup\$ @AdrianMaire 5 uA is correct in this context. What I am saying is that the specification allows a maximum change of 5 uA in Iadj for changes of load from minimum to maximum. If the voltage is set correctly by the resistors when there is no load then changing the load CAN worst case cause Iadj to VARY by up to 5 uA and IF R2 was 100k, then the voltage drop across R2 would vary by 0.5V. In practice the change in Iadj is typically much smaller. and R2 is unable to be as large as 100k due to the other points discussed. \$\endgroup\$ – Russell McMahon Sep 23 '16 at 15:36

Others have already pointed to the equation

\$ V_{OUT} = 1.25V \left(1+\dfrac{R_1}{R_2}\right) + I_{ADJ}R_2\$

which also can be found in the datasheet. Rearranging for \$R_1\$ gives us

\$ R_1 = R_2 \left( \dfrac{V_{OUT} - I_{ADJ} R_2}{1.25V} - 1\right)\$

If \$V_{OUT}\$ is in the order of volts (most likely) and \$R_2\$ is in the hundreds of \$\Omega\$ the term \$I_{ADJ} R_2 << V_{OUT}\$ and can be ignored, since \$I_{ADJ}\$ is maximum 100\$\mu\$A. We get a simplified equation then:

\$ R_1 = R_2 \left( \dfrac{V_{OUT}}{1.25V} - 1\right)\$

For example for \$V_{OUT}\$ = 5V and \$R_2\$ = 100\$\Omega\$ the first equation gives us a value of 299.2\$\Omega\$, while the second gives us 300\$\Omega\$, an error of only 0.3%.
On the other hand, if you would pick 10k\$\Omega\$ for \$R_2\$ you would get values of 22k\$\Omega\$ and 30k\$\Omega\$ resp. for \$R_1\$. Using the 30k\$\Omega\$ would result in 6V out instead of 5V, an error of 20%!

There's another good reason to pick low values for \$R_1\$ and \$R_2\$. The datasheet mentions a minimum load of 3.5mA typical, 10mA maximum. It's better to choose 10mA, not only because you always have to calculate for worst case, but also because the 10mA is given as a minimum condition for the other parameters.
For 5V out you'll want \$R_1\$ + \$R_2\$ < 500\$\Omega\$ then.

  • \$\begingroup\$ @stevenh - Noted that you said Imin_max was 10 mA. Recalled that I'd said 5 mA. Looked at data sheet - you were correct :-). Changed my answer to accomodate. No real differences of note except that typical value given for R1 in about every example circuit I've seen violates data sheet spec for minimum current at no load. Interesting. \$\endgroup\$ – Russell McMahon Aug 23 '11 at 11:41
  • \$\begingroup\$ @Russell - Personally I would only let the 10mA flow through R1 + R2 if it's a very low power application, but then I might use a low ground current regulator as well, like the Seiko S-812C (1\$\mu\$A!). In other situations your load may include a LED or so which already draws twice the required current. \$\endgroup\$ – stevenvh Aug 23 '11 at 16:20
  • \$\begingroup\$ Indeed. ie we are both agreeing that " 'Proper' design requires that 10 mA worst case be allowed for. IF the external load always draws 10 mA or more then all is well. However, if external load current can fall to below 10 mA then the design must provide a load to provide this 10 mA. Worst case, with no load, R1 provides a convenient way of providing the 10 mA while also providing a nicely "stiff" divider." deja vu :-) \$\endgroup\$ – Russell McMahon Aug 23 '11 at 22:50
  • \$\begingroup\$ I didn't understand how did you write the condition R1+R2<500 for 5 v. \$\endgroup\$ – user46714 Aug 8 '14 at 7:36

You also have to take into account Iadj, which is around 100uA. As this remains constant at all times, but the I through R1 changes depending on it's resistance, you need to make sure that the 100uA is not a large part of the program current.

So the higher you have R1, the more "error" Iadj will cause, as it starts to become a significant part of the overall current.

With your example:

(1.25 * (1 + (330/200))) + (100e-6 * 330) = 3.3455V

With resistor x10:

(1.25 * (1 + (3300/2000))) + (100e-6 * 3300) = 3.6425V


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