Does Voltage Divider Impedance Matter with LM317?

[audiogeekess]

I need to make a voltage divider for my LM317 so I can have the output voltage be 5v. I saw some data sheet somewhere for it that used 2 resisters each under 1k ohm. I have a bunch of resisters that are 1k to 20k ohm and was wondering if it would be ok to use a combination of them instead, increasing the impedance. Is this ok? Does it matter what the power supply impedance is?

[gmarsh]

It'll work fine, just be aware of the Iadj current.

[martin clark]

It will work fine providing your prospective load draws at least 5mA - otherwise the output voltage will drift a bit high (see the LM317 datasheet).

Of course, a little DC imprecision might not matter at all for the application you have in mind.

Either way I'd recommend you include the optional cap across the lower voltage set resistor (from Vadj to 0v) for best results.

[sofaspud]

You could use four of the 1k resistors in parallel connected from output to adjust pins to get close to the 240 ohm used in the datasheet circuit, and the needed 5mA minimum current. Regulation suffers and the output voltage may rise without that minimum current.

[ticknpop]

If you read Walt Jung's super regulator articles you'll see it's better to use 1k rather than 120 or 240 ohms for the uppper resistor and increase to 100 uf the resistor bypass capacitors with 317/337 reulators for better ripple rejection. Even better is using LM329 as shown in the Linear Technology LT1033 article. Yes you need 10 ma or more output load to get them to regulate with the higher value resistors, but worth it for the higher ripple rejection.

[AndrewT]

you can get the 317 to work very well @ low output voltages by replacing the lower resistor with a LED or LED string.
Two 1.9V LEDs + 1.25V of the load/adj pins gives 5.05V.
The only resistor required is that between the load/adj pins.
Changing the value of this resistor changes the current through the LEDs. That in turn allows one to trim very precisely the output voltage.
But do ensure that the 317 passes more than minimum operating current at all modes of operation.

[Soundminded]

The rule of thumb is that compared to the load impedance of what you are trying to power, in addition to the proper resistance value ratios the resistance across the load from the divider should be at least an order of magnitude lower. If it isn't, the load itself will become a significant element of the divider and the voltage across it will drop lower than you want it to.

[paulb]

Originally Posted by Soundminded

... the load itself will become a significant element of the divider and the voltage across it will drop lower than you want it to.

Are you sure? Two loads paralleled across a voltage source.

[Soundminded]

Originally Posted by paulb

Are you sure? Two loads paralleled across a voltage source.

Yes, a voltage divider consists of to or more resistors in series, the voltage dividing in proportion to each resistor compared to the total resistance. When you connect a load, you are connecting it across one of the resistors. If the load resistance is not an order of magnitude larger than the resistor it is being connected across, it will reduce the parallel combination of itself and that resistor and will therefore reduce the proportion of voltage across them. This is basic DC electricity.

[Conrad Hoffman]

From what you can see above, it appears they chose the resistors they did just to improve the odds of having the minimum load on the regulator. Higher impedance dividers work just fine, though I've never pushed the limits to see how high.