I want to calculate the resistors required with:

Input voltage = 9.0 volts

Output voltage = 3.0 volts

All the datasheets omit to give the input voltage in the formula:

Vout=1.25*(1+R2/R1)+(near zero term in R2)

Where R1 is the fixed resistance (about 240R) and R1 is the variable resistor.

  • \$\begingroup\$ Please edit your question to add a link to the datasheet you are using and a page or figure reference for the circuit you are using. Better still, add a copy of the schematic into your question. Make it easy for those you are asking for help. \$\endgroup\$
    – Transistor
    Commented Mar 3, 2019 at 11:58
  • 1
    \$\begingroup\$ The output is independent of the input voltage (provided the input voltage is larger than the desired output voltage) \$\endgroup\$
    – Huisman
    Commented Mar 3, 2019 at 11:58
  • \$\begingroup\$ 3-V headroom is recommended (VI – VO) to support maximum current and lowest temperature. \$\endgroup\$
    – Huisman
    Commented Mar 3, 2019 at 12:03
  • 3
    \$\begingroup\$ All the datasheets omit to give the input voltage in the formula And why do you think that the input voltage needs to be in the formula? Isn't the function of a regulator to make the output voltage independent of the input voltage? Perhaps all you need to do is take care that the input voltage is high enough for the required output voltage and not too high such that it would damage the LM317 (and also power dissipation is relevant btw). \$\endgroup\$ Commented Mar 3, 2019 at 12:10
  • \$\begingroup\$ \$\Large R_2 = R_1 \cdot \frac{V_{OUT} - V_{REF}}{V_{REF} + I_{ADJ} \cdot R_1}\$ \$\endgroup\$
    – G36
    Commented Mar 3, 2019 at 12:14

2 Answers 2


The LM317 datasheet does not neglect to include the input voltage when calculating the resistors for the desired output voltage.

The input voltage is not needed. The entire point of a voltage regulator is for the output voltage to be independent of the input voltage.

If the resistor values depended on the input voltage, then you would have to change the resistors every time the input changes. That wouldn't be a very practical regulator.

If you are thinking of the common case of reducing a fixed, regulated voltage to a lower voltage, then I can see where the misunderstanding comes from.

But, linear regulators are intended to take a varying voltage and lower it to a fixed, stable voltage. Linear regulators have an input range they can work in.

Your LM317 will provide a stable output voltage as long as the input voltage is below the maximum allowed (no more than 37V between input and output voltage) and the input is above the minimum (about 3V between input and output.)

So for an output voltage of 3V, you have a maximum input voltage of 40V and a minimum of 6V. For any voltage in that range, the output will be fixed at 3V. No resistor changes needed.

The above ignores overload conditions and the minimum load conditions. Those are described in the datasheet. If you have trouble with that, ask in a separate question.


Adding something to the already good answer by JRE.

You must take into account also power dissipation in your design. The LM317 is a linear regulator IC, this means that it lowers the input voltage by dropping the excess voltage as a "automatic regulating resistor". In other words, by dissipating electrical power as heat.

The exact value of the dissipated power is given by the formula \$P=(V_{input}-V_{output}) \cdot I_{load} \$.

If your load draws, say, \$500mA\$, your chip is going to dissipate \$(9V-3V)\times 0.5A=3W\$ of power. This will turn your LM317 very hot if you don't add a suitable heat sink to it.

Assuming an (optimistic) thermal resistance of \$ 20K/W \$ for a TO220 package, the chip will reach a temperature which is \$20K/W \times 3W = 60 °C\$ hotter than ambient. Without an heat sink you will probably reach a point where the internal thermal protection of the chip will kick in and stop its operations.

I don't know what you are going to build, i.e. why you need such a circuit to lower that voltage. If you don't need a variable output, probably you'll better off by using one of those very cheap switching buck converter modules you find on Amazon or eBay.

Nowadays there are little reasons to use a linear regulator to lower a voltage from a fixed one if substantial load current is involved.

Principal reasons to use linear regulators instead of switching regulators are:

  • you need lower output ripple/noise (linear regulators are quieter);
  • the differential input/output voltage is tiny (say a 1-2V max), so you use a LDO (low drop-out) linear regulator;
  • the current of the load is tiny and power dissipation is not an issue, so you choose a simpler circuit (no inductors; less critical design);
  • you need a precise voltage reference, with very low drift and thermal variation (almost no current is delivered to the load in this case, and specialty linear regulators are built for the purpose).
  • \$\begingroup\$ Many thanks for your detailed response. \$\endgroup\$
    – Gsy Niall
    Commented Mar 4, 2019 at 10:08
  • \$\begingroup\$ Many thanks for your detailed response. I wished to use the LM317 because it is so small - but I see the problems with it now. (Please tell me how to select a new paragraph - when I hit return it sends the comment!) \$\endgroup\$
    – Gsy Niall
    Commented Mar 4, 2019 at 10:14

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