Pocket-sized Power Supply




Introduction: Pocket-sized Power Supply

Here is a miniaturized adjustable power supply,

you can adjust the output from 1,2V to 16,8V (DC).

Step 1: Material :

You will need what there is on the picture plus a power cable 19V DC


LM317 (2pcs)

10K potentiometer

prototype board (20pcs)

Alligators clamps


These links are mostly amazon products so you can find cheaper on other websites (banggood, aliexpress, ebay, ...).

(the other pieces are from recycling so there is no link to find the exact same)

Step 2: Circuit :

The power supply is based on the LM317.

I modified the basic circuit of the LM317 to add switches and a voltmeter,

folow this schematic and weld the components to the board.

The "switch 1" is the black one, it is the power switch.
The "switch 2" is the lever when it's in "OFF" position the voltmeter displays the voltage and you still can adjust it, but no current flows. However, when it's "ON" current flows and the led turns on.

The voltmeter has 3 wires: red, black and yellow (or white),
red and black are the wires which gives current to the LEDs of the display. So black goes to ground and red goes to VCC (after the power switch).

The Yellow one is the wire which measure voltage it's why I place it at pin 2 of the LM317 (OUT)

You can add a 10µF capacitor between out and ground for a better stability

Step 3: Drilling :

Draw gide-lines on the box and drill some little holes around the emplacements of the voltmeter and the switch with a drill-press, then with a hammer, break the center like on the picture and sand the edges to allows voltmeter to fit in.

Step 4: Close the Box :

it's time to close the box!

I secure the potentiometer and the lever in place with nuts.

The voltmeter and the power switch are clipped.

For the led and the DC jack, I resort to use hot glue to keep them in place.

I also put paper to insulate the electronic circuit from the metallic box.

however you can screw the regulator to the box in order to use it as an heat sink because the LM317 will heat up !

Step 5: End :

This project is now ended ! let a comment, and vote for this project in the "pocket-sized" contest !!!



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    20 Discussions


    Question 3 months ago

    For Vi=9v I have trouble with the voltage output, the potentiometer is really sensitive and it's very hard to have 5v or 3.3v.

    So I searched in the datasheet and find this equation:

    Vo=1.25 (1+R/240)
    Where R is the potentiometer.
    So I drew the curve (see the picture or https://ggbm.at/TpG4NV6r )

    Then, this is my question, why using a 10k potentiometer to adjust Vo from 1.5v to 8v?
    A 2k potentiometer seems better for this application.

    (ps: maybe I am the only one to have this problem of sensitivity)

    No way you have more than 9V out, and why the diode on the output?

    Data-sheet has voltage-drop of ~3V from input to output, and no diode on output.

    2 replies

    hey! my input is 19V DC so it's allow me to adjust it from 1.2 to 16V and I placed a diode beacause when I opened the circuit (with the second switch) it supplied me -16V I don't know why so I placed a diode and it works

    So your guide is misleading: "You will need what there is on the picture plus a power cable 12V DC"

    Besides that it would be best practice to put the switch on the positive.

    And the circuit is waaay obsolete, you can buy switch-mode regulators ready-made on eBay at $1.20, about the same price Amazon wants for a LM317, way more efficient and even buck-boost from 1.25-60V at any input from 5-35V.

    Nice job.What are the sizes of the aluminum box?

    1 reply

    Hi! the box is 89.1x35x30.3 mm :)

    Great job! Mount the LM317 to the enclosure, and the enclosure can be used as a heatsink for the LM317.

    Very nice and superbly built! Question: why not connect the voltmeter to measure the output instead of the input?

    Again, well done.

    1 reply

    it is measuring the output :) it's connected to the output pin of the LM317 red: pin3 (input) black: ground yellow: pin2 (output) the yellow wire measure ;) .


    Tip 5 months ago

    This is a regulator, not a supply. The voltmeter doesn't show the real output voltage which is 0.6V lower. A single resistor of 1k to 4k is enough for the Led. An output capacitor is always better for voltage stability, count 1000µF / A. As the LM317 is a linear regulator, input voltage must be at least 1.5V higher than output voltage and it generates heat depending on input output voltage difference and intensity delivered, it's not luxury but needed to mount it with a heatsink.

    nooo I had the "mini power supply" idea too for this contest ahaha

    I can't compete with yours..it's so awesome!!

    well...congrats you've got my vote :D

    1 reply

    haha you can still publish it ;) no problem and thanks for the vote :D


    5 months ago

    Nice one, thanks for sharing and keep up the good work :)

    3 replies

    thank you, it's good to see that what we share is liked

    For sure, it's a good instructable!
    I love to see when people take time to share their knowledge :)