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I bought this kind of module for my led light project. Unfortunately it did not come with instructions.

It can be found from eBay with "3.5-28V to 1.25-26V DC-DC Converter Boost Buck Step Up Step Down Voltage Module"

My plan is to set a load on the "OUT" side, set up a multimeter and turn that screw in the down middle until I get the correct reading. Is that the way to fry the board?

Voltage module

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  • \$\begingroup\$ Can you provide a link to the ebay page you purchased it from? \$\endgroup\$
    – sptrks
    Commented Apr 30, 2012 at 6:42
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    \$\begingroup\$ My advice: don't buy things which don't come with or for which you cannot find the specifications. \$\endgroup\$
    – Telaclavo
    Commented Apr 30, 2012 at 10:53
  • \$\begingroup\$ This module and circuit has ben discussed on SE befor - a search may find it. LM2577 is a fixed boost to 28V and then LM2596 bucks this to from 1.25V to 26V variable with the pot. Rather overkill in many cases but quite a neat idea. They seem to not know about SEPIC converters which do this in one pass. Still fun though. \$\endgroup\$
    – Russell McMahon ♦
    Commented Apr 30, 2012 at 12:47

3 Answers 3

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Strange. This is a single supply module, yet it has two switchers on board.

Anyway. Yes, connect a voltage between 3.5V and 28V to the input and measure the output voltage, which you indeed should be able to adjust with the multiturn potmeter. Check the switchers' datasheets for the maximum output current. The right IC seems to be an LM25005, that can deliver 2.5A. May be a bit high for the coil though. The left one I can't read.

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  • \$\begingroup\$ I am guessing that they may have used a comparator like technique to turn one ICs at a time, depending on the input and output voltage? Edit: It just occurred to me that they may have used a buck-boost converter by connecting them in series? \$\endgroup\$
    – abdullah kahraman
    Commented Apr 30, 2012 at 7:12
  • \$\begingroup\$ @abdullah - It's possible that they first boost to say 28V and then buck to the variable output voltage, but that would be Very Bad Design. Efficiency will be the product of both regulators' efficiencies, and there exist buck-boost regulators! \$\endgroup\$
    – stevenvh
    Commented Apr 30, 2012 at 7:19
  • \$\begingroup\$ There are buck-boost converters, as wikipedia says: "A buck (step-down) converter followed by a boost (step-up) converter". Since the output of a boost converter is almost the same as the input when the input voltage is higher than or equal to the set output voltage, it sounds reasonable to me when the price is an issue. \$\endgroup\$
    – abdullah kahraman
    Commented Apr 30, 2012 at 7:38
  • \$\begingroup\$ @abdullah - wikipedia sucks! That board OP bought is probably designed by that wikipedia author. A real buck-boost switcher may achieve efficiencies up to 95%, which you can't by cascading. \$\endgroup\$
    – stevenvh
    Commented Apr 30, 2012 at 7:44
  • \$\begingroup\$ Ah, yes, I love LTC3789. However, you will need a four layer PCB to design a good end-product with it. The IC is really pricy, too! By the way, how can I search the Digi-Key for buck-boost topology? \$\endgroup\$
    – abdullah kahraman
    Commented Apr 30, 2012 at 7:49
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I found the datasheets for both the LM25775 and the LM2596S.

Here is the typical application of the LM25775enter image description here

Here is the typical application of the LM2596Senter image description here

If this is all one module it looks like the resistor values are changed by the small pot you were referencing.

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  • \$\begingroup\$ In an LED-driving application, I would be tempted to connect the load to ground through an appropriate resistor, then feed that back to pin 4, thus making a constant-current regulator. R = 1.23 / Iout \$\endgroup\$
    – markrages
    Commented Apr 30, 2012 at 19:57
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From the picture, it seems like You assume the circuit is a "black box" and connect the input and you should get a voltage at the output. The output voltage seems to be controllable via the small (blue) pot near the bottom of the picture.
What you have described is correct, connect the input and measure with a multimeter at the output and adjust till desired voltage is achieved. Though while connecting your actual load you might want to ensure the current drawn is within the acceptable range of the board :)

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