Long shot, I know...

I have this Boost convert 1-5V in out 5.1-5.2V and I hope to modify it for a different output voltage.

Boost convert 1-5V in out 5.1-5.2V

So I'm trying to identify and find the datasheet for the main chip.

I think the chip reads 31WOC, but I can't find a datasheet for that number. Anyone familiar what the type is of the chip?

There is no more datasheet than the following:

IN 1-5V
OUT 5.1-5.2 
Current: Rated 600mA (single lithium input), the maximum 700mA (single lithium input)

€ 1,50
  • \$\begingroup\$ Where is the data sheet for the board? Schematics? \$\endgroup\$ Jun 27, 2012 at 16:25
  • \$\begingroup\$ Added a few details to my question, that's all I have available. \$\endgroup\$
    – jippie
    Jun 27, 2012 at 17:52

2 Answers 2


The 31WOC is probably not the actual part number, rather some type of code described in the datasheet for different variants of the chip.

I think I would forget about what the actual part number is, and just locate the feedback pin and adjust the resistor divider as necessary (assuming it's an adjustable regulator). Most use an internal reference of ~1.2V. Look for the pin with this voltage on, attached to the centre of a resistor divider.
If it's a fixed regulator then it's a little more tricky, you will probably have to experiment with a few resistor values from the feedback pin to ground (start with 1-10k?). You will only be able to raise the output voltage this way.

EDIT - on the picture you have added, it looks like the divider is the two resistors at the top right marked "18C" (15k?) and "473" (47k). Measure these resistors, and also the voltage at the pin they are connected to (top right pin of IC). The output voltage formula should be 1.24V * ((Rtop/Rbottom) + 1) which in your case is 1.24V * ((47k / 15k) + 1) = 5.12V, which is around what you are seeing.
Adjust as desired, but be aware if you are increasing the output, that the chip may not be rated for voltages much above it's current level.

  • \$\begingroup\$ The left-top two resistors have 1V26 and 3V96 across them :o) \$\endgroup\$
    – jippie
    Jun 27, 2012 at 15:16
  • \$\begingroup\$ Yep, that sounds about right. \$\endgroup\$
    – Oli Glaser
    Jun 27, 2012 at 15:21
  • \$\begingroup\$ Good point about the overvoltage, didn't think of that. But I want to reduce the voltage to 3V. Thinking of glue'ing on a 50k trim pot. Of course set it to an 'about-right' position before power up, so i won't blow up the device and just do the last bit of trimming. \$\endgroup\$
    – jippie
    Jun 27, 2012 at 15:21
  • 1
    \$\begingroup\$ If you place the 50k in parallel with the 47k resistor then you should be able to adjust from 3.3V down to 1.24V. \$\endgroup\$
    – Oli Glaser
    Jun 27, 2012 at 15:30
  • \$\begingroup\$ I decided to go for a 33k resistor, because the cost of a multiturn potmeter would nearly double the cost of the device ;o) Anyways, it powers a laser module from a penlight at just under 3V, as intended/designed. \$\endgroup\$
    – jippie
    Jun 27, 2012 at 20:11

What Oli said looks very good.
Comment on voltage may not fully apply - see below - but is irrelevant if you are reducing voltage.

The SOT23? device labelled 141M is almost certainly a MOSFET which does the actual inductor drive and which is driven by the IC.
Lower pin as shown is source, upper is gate - connected to IC and middle pin in drain - ie std MOSFET pinout. It looks like they MAY have a high side current sense resistor - the R100 = 0.1 Ohm? just to right of MOSFET. This is unusual. It just may be in series with output to limit FET current but that would also be unusual.

The SS22 labelled device is almost certainly a Schottky diode acting as the main output rectifier. SS22 data sheet here - 2A 14V RMS. Usually they use an SS12 1A diode. They may have had a few zillion 2A devices to use up - 1A is usually ample for this sort of device.

The 31WoC is probably a batch code. Alibaba sellers often list these as if they were part numbers but I could not see it listed.

The diagram below is not for your specific IC but shows a typical arrangement which will be close to what you have. Yours seems to use all 6 pins which is, again, unusual, as most functions can be accomodated with 5 pins. Top circuit is with MOSFET (as in your example) and the bottom circuit is for internal main-FET versions.

enter image description here

Normal practice is to feed the IC Vdd from the output (so supply voltage is bootstrapped) if Vin is low and from Vin if Vin_min is say 3 Volts. Startup is usually down to 0.8V to 1.0V range but they tend not to pull the skin off a rice pudding at these voltages and may not start under very heavy static load.

As you are reducing Vout and not changing Vin the arrangement of Vdd that exists now should be appropriate. If you had been changing to a much higher Vout it may have been desirable to change where Vdd is driven from.

The high side resistor from Vout to feedback pin usually benefits from a small capacitor in parallel with it. This can be thought of either as coupling Vout changes to the feedback pin directly without division or as wreaking dark magic on the overall transfer function to assist stability. Either view works. It is not obvious from the diagram whether such a capacitor is present and if not it may not be needed but bear it in mind.

The 4R7 labelled inductor MAY be 4.7 uH but these ICs tend to be not terribly high frequency (100 kHz range typically) and a value of 47 uH would be more normal.

  • \$\begingroup\$ About V(DD) bootstrapping: I noticed the other day that the device was able to drain the battery to under half a volt. But when I applied such a low voltage a bit later, the converter refused to start until input voltage rose to about 1V. \$\endgroup\$
    – jippie
    Jun 27, 2012 at 18:02
  • \$\begingroup\$ Managed to pick up the magnetic field from the coil and the device oscillates at 20kHz. \$\endgroup\$
    – jippie
    Jun 27, 2012 at 18:26
  • \$\begingroup\$ @Russell - FWIW with the overvoltage comment, I noticed the external FET, but as I see it the output seems to connect to the bottom left pin of the IC (and the top of the divider) \$\endgroup\$
    – Oli Glaser
    Jun 27, 2012 at 22:15
  • 1
    \$\begingroup\$ @OliGlaser - I'm (so far) happy with my version :-). Pin connecting to output is a good match for Vdd pin. Vdd = output connected for Vout up to a certain voltage and low Vin. In several places a number of vias side by side suggest major tracks under board - Vout can easily be transferred from Schottky SS22 to output. Component at top left is cap coupling Vout from SS22 to Vout - at top left. I think Vout+ is aat middle left. \$\endgroup\$
    – Russell McMahon
    Jun 28, 2012 at 2:11
  • 1
    \$\begingroup\$ @jippie - I'd guesstimate that A2 was a smaller but still capable diode connected in reverse bia across Vin+ - Vin-. If the supply is reversed it shorts te supply and attemots to produce magic smoke or blown fuse or whatever - anything better than blowing up IC etc. \$\endgroup\$
    – Russell McMahon
    Jun 28, 2012 at 9:19

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