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I'm using a 317 Voltage regulator in order to reduce a 12v into 3.7v output.

Using the mathematical relation Vout = 1.25*(1+ R2/R1), A ratio of 2 between R1 and R2 will be suffice for my needs.

Does it matter if I'm using R1=200 [ohm] or R1=10K [ohm] as long as I keep the mentioned ratio ?

I used R1=10K, and I get an error ( Vout ~11.6v ).

Guy

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  • \$\begingroup\$ Was your 11.6 V measured with a load present? If you check the datasheet you'll find that there's a minimum load (10 mA or so?) for stability. That's what Elliot is discussing. Hopefully you haven't omitted the decoupling capacitors. Have you calculated the power dissipation in the LM317 and what sort of heatsink you'll require? \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 18:28
  • \$\begingroup\$ without load. Load is at max- 200mA \$\endgroup\$
    – guyd
    Commented Apr 13, 2020 at 18:49
  • \$\begingroup\$ (1) So your reading is unreliable. Add a resistor to provide adequate load as per the datasheet. (2) That's a current estimate. You need to calculate the power dissipated in the LM317. \$ P = V_{317} I \$ where \$ V_{317} \$ is the voltage across the 317, not the load. \$\endgroup\$
    – Transistor
    Commented Apr 13, 2020 at 18:51
  • \$\begingroup\$ Tried with a 300mA load - same effect \$\endgroup\$
    – guyd
    Commented Apr 14, 2020 at 5:18

2 Answers 2

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You want to select resistor values such that the current through them is much greater (at least 10X) the quiescent current of the regulator itself. Resistor R2 passes the quiescent current of the regulator as well as the current coming from R1.

The quiescent current may also be called the "adjustment pin current". A typical value for R1 is \$240\,\Omega\$ which gives a current of about 5mA through the resistor divider. For the TI LM317 the adjustment current has a maximum value of 0.1mA, so the divider current is about 50X bigger than the adjustment current.

The formula provided to set \$V_{out}\$ is based on the notion of a resistive voltage divider, where you have 2 resistors and by selecting the ratios of the resistors you can set the voltage at the point where they are connected together. Connecting the adjustment pin of the LM317 to the resistor divider injects an additional current into the bottom resistor so the divider formula isn't quite right. However, if you choose resistors so that the current from the adjustment pin is small compared to the current that would normally flow through the divider then the error can be minimized.

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  • \$\begingroup\$ Can you please explain more ? typical current consumption is around 200mA \$\endgroup\$
    – guyd
    Commented Apr 13, 2020 at 18:22
  • \$\begingroup\$ your last paragraph, is the essence of my lack of knowing: can you say that 10K resistor causes that ? \$\endgroup\$
    – guyd
    Commented Apr 13, 2020 at 20:18
  • \$\begingroup\$ Having thought about it, I'm not sure that the problem I am describing would make the output voltage too low. Try adding a load as Transistor suggests. \$\endgroup\$ Commented Apr 13, 2020 at 21:12
  • \$\begingroup\$ Tried with 300mA - same effect \$\endgroup\$
    – guyd
    Commented Apr 14, 2020 at 5:17
  • \$\begingroup\$ You may have purchased defective parts. Next time, buy from a more reputable supplier. It is a very bad sign that you don't know who the manufacturer is. \$\endgroup\$ Commented Apr 14, 2020 at 12:00
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You want to select the resistance based on what the datasheet of your exact component tells you. There are so many different LM317 manufacturers that their implementation will vary, and you can select the resistance that fits for all LM317 type regulators.

LM317 type regulators must have a minimum load current to keep their output well regulated. This current is stated in the datasheets. But if it is 5mA, then R1 should be 240 ohms in most datasheets. Some datasheets say 10mA, so R1 would be 120 ohms.

Since you used 10k resistor, it does not draw enough load current, and the regulator output voltage is not what you expect.

And you seem to want 3.7V; I hope you are not using it to charge lithium batteries.

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  • \$\begingroup\$ a) no, it is for powering up an ESP8266-01, b) I do not know what manufacturer it is, since it was bought on eBay ( manufacturer is not specified ) \$\endgroup\$
    – guyd
    Commented Apr 13, 2020 at 20:11
  • \$\begingroup\$ The manufacturer logo should be on the LM317. Also in the future, you might want to ask the Ebay seller for more info before buying unknown stuff. Or buy known stuff from somewhere else. Anyway best to have 120 ohm resistor then, just to be sure. \$\endgroup\$
    – Justme
    Commented Apr 14, 2020 at 7:26

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