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The following LM317 adjustable voltage regulator example uses R1 & R2 to adjust the output voltage:

enter image description here

And the datasheet for the output formula is given as:

enter image description here

Now instead of using R1=4.2k and R2= 10k one can also use:

R1 = 420 and R2 = 1k or

R1 = 42k and R2 = 40k ect.

So we can obtain the same output voltage when R1 and R2 are as small as hundreds of ohms or very big as tens of kOhms or even more. I can see that Iadj changes with resistance values.

Regarding the options of R1 and R2 values, is there a reason to chose one option to the other? From power perspective one might chose tens of kOhms. But that makes the Iadj very small. Is there any relation between the output voltage stability and the choice of resistors here? What could be the correct way of reasoning?

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    \$\begingroup\$ Not mentioned...this regulator requires a minimum load current to regulate (somewhere between 3.5mA and 10mA). These resistors can serve two purposes: (1) draw minimum load current, (2) select output voltage. So unless you can assure that load current never goes to zero, keep R1,R2 small-valued. \$\endgroup\$
    – glen_geek
    Feb 12, 2020 at 14:29
  • \$\begingroup\$ What kind of stability? Temp, Input ? Load, Step Load? Aging drift? Iadj is constant ! \$\endgroup\$ Feb 12, 2020 at 20:30
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 I meant output voltage stability with a varying load. \$\endgroup\$
    – GNZ
    Feb 13, 2020 at 9:36
  • \$\begingroup\$ We call this Load Regulation error and not stability. It is purely a resistance ratio. Stability would be an overshoot and settling time effect which is at least 2nd order.. bias feedback resistance will only affect an offset error and not step load error as it is constant current bias. \$\endgroup\$ Feb 13, 2020 at 9:45

2 Answers 2

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The simplified calculation for \$V_{out}\$:

\$V_{out} = 1.25 (1+\frac{R2}{R1} )\$

Is only accurate when the current through \$R_1\$ and \$R_2\$ are the same.

When that's not the case (when \$I_{Adj}\$ is not much smaller than the current through \$R_1\$) then the calculation will be slightly wrong, we would need to use the formula form the datasheet:

\$V_{out} = 1.25 (1+\frac{R2}{R1} )+I_{Adj}R_2\$

Note that that formula needs \$I_{Adj}\$ as an input value!

As \$I_{Adj}\$ can vary per device (each LM317 will have a slightly different \$I_{Adj}\$) you would need to measure it for each device. That's a hassle!

If we simply make the current through \$R_1\$ and \$R_2\$ so large that the value of \$I_{Adj}\$ doesn't matter, that's much simpler!

To get that (the large current through \$R_1\$ and \$R_2\$) we need to use low values for \$R_1\$ and \$R_2\$.

So you can use large values for \$R_1\$ and \$R_2\$ but that will make the output voltage vary more between each LM317 (unless you measure \$I_{Adj}\$ for each and adjust the values of \$R_1\$ and \$R_2\$ to compensate).

It is much easier to use low value resistors.

If that takes too much power for your use case then you should not be using an LM317. There are other voltage regulators that can work with high value resistors for the feedback network.

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Choosing a higher value for R2 does not make Iadj any smaller.
Iadj is the regulator's quiescent operating current (and could vary from one part to the next) - it is not determined by the value of R2, but by the regulator IC.
As you can see in the formula for the output voltage, you need to account for Iadj in your calculations, but you need to remember that it's not necessarily going to be exactly the same value for every unit you build.
As such, choosing a lower value for R2 means that the potentially different Iadj values have less of an effect on the output voltage.
However, you're also fighting power consumption - because obviously lower resistor values there lead to more power lost in the resistor divider.
You need to weight the trade-off between output accuracy & power consumption when choosing the resistor values you use.

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