I've bought an LM39302R for my GSM module (SIM808):

enter image description here

The datasheet of the module recommends a Microchip regulator which is not available in my location so I've searched and found next best option, LM39302R, although I would be happy to find a cheaper replacement with acceptable voltage dropout.

Anyway, the LM39302R datasheet said it follows the famous rule of adjustable LDO regulator:

enter image description here

I built the PCB, and I've put 1K resistor on the top and 2.2K for R2 (I know VOUT trace of the regulator is too thin!)

enter image description here

According to the formulation which you can see in this online calculator it should be 4V on the white arrow on the PCB picture, but I read only 2 volts.

Why does this happen?

Almost all implementations use 100K resistor for R1 and 47K for R2. I don't see how it could output 4 volts, it should be 1.8375 volts.

The formulation has made me think, what is the difference if someone puts 1Kohm for R1 and 2Kohm for R2 or 1-ohm for R1 and 2-ohm for R2 or 1-Mega-ohm for R1 and 2-Mohm? Does it make any difference?

Any help would be appreciated!

Update 1:

lm39302 datasheet.

Sim808 datasheet

Update 2:

It is a shame but the low voltage that I've read was because of a damage capacitor in the circuit! the value in the PCB picture which is recommended (100k and 47k) works fine and lm317 calculator values is almost the same for this chip.

  • \$\begingroup\$ Please hyper link to the data sheet and show your schematic (your schematic please). \$\endgroup\$
    – Andy aka
    Commented Sep 11, 2020 at 7:28
  • \$\begingroup\$ What voltage are you supplying it with? \$\endgroup\$
    – user16324
    Commented Sep 11, 2020 at 11:36
  • \$\begingroup\$ @BrianDrummond I use 5V 1A power adapter. \$\endgroup\$
    – malloc
    Commented Sep 11, 2020 at 11:49

2 Answers 2


I don't know this part specifically, but regulators usually require decoupling (bypass) capacitors on their inputs and outputs. Check the data sheet and, if available, an application note for details of what is required. A quick look at the data sheet shows that these are required.

On your last part about the absolute value of the resistors, yes there are some very practical points that the simple equation hides. The device itself will place a small load on the circuit, so check for any specification of leakage current, or internal resistance. If your resistor values are too high then the leakage will become significant an cause errors in the voltage divider. Too low and you will at a minimum be wasting power.


The calculator you are using is good for LM317 devices. There, the REF voltage is referenced relative to OUT. Here, the REF voltage is referenced relative to GND (pdf 185.), so by exchanging the values ​​calculated with the calculator, you get the desired voltage. (With a small error, because the ADJ input current is very small here compared to the LM317.) Because of the small ADJ current, you can also use larger resistors, which means lower power consumption. It gives exactly 4V if the resistance on the GND side is 1k and the resistance on the OUT is 2.2k. Or multiples of these (100k / 220k).


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