Introduction: Make the Simplest Adjustable Power Supply in 10-15 Minutes!

For about $5-$10 you can have your own adjustable power supply to power various electronics!

Step 1: The 3 "main" Parts You Will Need

1 - LM317 (I used LM338, however any adjustable regulator with "LM" in it should work fine)
1 - 5k ohm variable resistor (potentiometer)
1 - 220-240 ohm resistor

Step 2: Identify the Pin Order on Your Regulator

LM adjustable regulators should have 3 total pins, one for voltage in, voltage out (ground in), and adjust. 

From left to right they should be voltage in, adjust, and voltage out (ground in). 

Or you can look at the picture, pin 1: voltage in, pin 2: adjust, pin 3: voltage out (ground in)

Notice that pin 3 is both your voltage output for ground and input for ground. 

Step 3: 220-240 Ohm Resistor

Now we have to solder the resistor to the regulator's pin 1 and 2.

This step should be easy since there is no difference on which side you solder the resistor (any side will work). 

Step 4: Add in the Variable Resistor (potentiometer)

Now we have to add in our 5k ohm variable resistor (allows us to adjust voltage accordingly)

Solder the variable resistor to pin 1 of the regulator

The other end of the variable resistor will be now be our new "voltage in". So we would connect the main power source to that pin on the variable resistor.

Do not get confused with pin 1 and the pin on the variable resistor, if you give power to pin 1 on the regulator you will not be able to adjust your voltage output. 

Step 5: You're Done!

Now you can solder two wires onto the regulator so you can connect it to a power supply (24v max), you will then be able to adjust it from about 1.2v~23v. 

Please keep in mind that this is just the bare basics of making an adjustable power supply. Most diagrams would include capacitors in their design to help smooth out voltage fluctuations, although they are not completely necessary, they will help with more consistent voltage output.

If you feel comfortable with adding capacitors to the design, all you have to do is connect the capacitors in parallel with the voltage in (+) and voltage out (-) of the regulator. Make sure that the capacitor is rated for at least 25v or higher, any lower and it may explode/leak. 

You may want to double check your voltage output to make sure that it is working before trying it on different devices. 
You may also need to attach a heatsink to the voltage regulator if you are going to be running heavy loads, as it does get quite hot under those conditions. 

Lastly, after you have done all the correct connections and verified that it is working properly, you may put it in a case so everything is protected.


Zhuna87 (author)2017-02-24


Can I connect 12 VDC 1 Amper power brick to this?

LeeZ10 (author)2017-02-03

Can I use a 9v battery to power it

Lectric Wizard (author)2013-04-05

Voltage regulators are more of an integrated circuit than just a transistor, so the ratings are different. The power rating of transistors themselves vary considerably using the same case. You have to look up each type to be sure. Cheers!

-max- (author)Lectric Wizard2015-11-26

To truly figure out how much power a regulator can dissipate, you need to do some basic thermal calculations. Different case styles have different thermal resistances. There are generally 3 thermal resistances that need to be considered: silicon junction to the package/case (found on datasheets for transistors & regulators and stuff. 5 C/W for the 7805), then one between the package/case and the heatsink (depends on how good you mount the transistor to the heatsink), and finally the one from the heat sink to the air. (are you using a small baby heatsink, or a CPU cooling heatsink, or a bigass car radiator!?)

You can simply add all these thermal resistances up to get an idea of what θJA (junction to ambient). As a matter of fact, you can treat thermal resistance just like electrical resistance! Anyways generally speaking the heatsink will have the largest/most significant thermal resistance. Smaller heatsinks are less effective at getting the heat from the transistor into the air because there is less surface area

To figure out the maximum power you can dissipate, use the formula PD = (TJ − TA)/θJA. PD is power dissipation in watts, TJ is the temperature of the silicon junction, Ta is the temperature of the air, and θJA is thermal resistance.

When it comes to any device, the power dissipated is nothing more than the voltage dropped across it multiplied to the current through it. So that formula can help you solve for any device. For linear regulators specifically, the power dissipation will be the difference in voltage, times the current. (Vin -Vout)*A. You can work backwards from that to figure out what voltages/currents can be pulled safely from a power supply without causing the regulator to overheat.

Lectric Wizard (author)-max-2016-12-15

Yes but for most applications you can ballpark it using the datasheet ratings and your voltage & current

-max- (author)Lectric Wizard2016-12-15

The datasheet can give you the maximum power dissipation, and that is generally assuming you can keep the case of the package at room temperature, and the junctions will be at the maximum temperature (150*C) Of course in the real world it is almost impossible to achieve like 100W of dissipation while keeping the transistor at 25*C w/o an active heat pump.

Lectric Wizard (author)-max-2016-12-17

Again your bringing transistors into this. Linear Voltage regulators have a finite max. current they can supply. Exceed that and if your lucky it will shut down on overcurrent, unlucky smoke ..

-max- (author)Lectric Wizard2016-12-17

Many voltage regulators have protection mechinisms in place, sure. That does not mean you should be careless about how you treat them, I have had many regulators die from abuse from heat and short circuits. They are not fool proof.

By "junction", I am referring to the primary pass transistor inside that is the culprit of all the heat.

Lectric Wizard (author)-max-2017-01-11

That is my point. Follow manufacturers specs, not what someone without the necessary understanding tells you. I never suggested abusing them !!!!

-max- (author)Lectric Wizard2017-01-11

So then my original point holds true. To truly figure out how much power a regulator can dissipate, you need to do some basic thermal calculations. Different case styles have different thermal resistances. (goto comment 1)

TO220 is typically just under 1 *C/W, from the silicon chip inside to the case. If you let the package stand in free air its about 35 *C/W from junction to ambient, depending on air flow.

Exceed the thermal and you ARE abusing the part. Thermal cutoff typically occurs at 150*C which is too hot. 100*C is the max for reliable operation

Lectric Wizard (author)-max-2017-01-11


-max- (author)Lectric Wizard2017-01-11

My original comment was simply adding to what you said (apologies if that was mistaken to be a correction), then you argued that you can ballpark it for many applications, which I agree with for typical non-critical DIY projects (particularly when breadboarding and having access to a reasonable size heatsink), but that does not make it good practice. A linear regulator is most often (almost always) limited by thermal dissipation capability, at least in my experience. Sometimes a power derating curve is offered by the datasheet, but this is basically just power dissipation limits calculated for you.

I tried to stick a 7805 inside of a small LCD module which was drawing 350mA at 5V. Powered from a 12V RC battery. While on the breadboard mounted on a small heatsink it got a little warm but not bad. However once assembled the regulator overheated, melted the inside of the case, cause LCD clouding (thermal heat damage) and the regulator died. It was not even close to the rated 5A but I neglected power dissipation inside a plastic case with no ventilation. It was dissipating over 2W of heat.

I know linear regulators inside out (Literally!!!), as I have built many linear regs from BJTs on multiple occasions. I am working on a really nice 14 bit microcontroller lab bench power supply capable of 0 -- 5A; 0 --15V. I not have a "beta" build working on a perf-board to improve performance and reduce the changes of marginal instability. I plan to use LT1007's in the final design for the low noise, lower voltage offset, higher precision, and MUCH higher slew rate. Only thing that needs to be done is to layout a proper PCB and figure out frequency compensation components.


I have probably replaced a thousand of them over the years.


(when a shorted component overloaded them)

lazar33 (author)2016-11-08

Basics, add an old transformer, 4 diodes (grec), and one bigger cap and you have power supply that can be powered by 110V (230V in my country).

Gelfling6 (author)2013-04-03

The key, any "Adjustable" regulator.. the fixed LM7805, as key example.. but You're right, this is as simple as it needs to be.. but, also keep amperage rated to the regulator as well.. I.E. TO-220 (your example), has a wattage of 150W (22V input, never push past 5-Amps, or you'll be looking at smoke!) T0-3's (metal can) rate 300W (can push 10-12amp.).. an image on , gives a quick run-down on Transistor cases, same applies to regulators as well.

Lee Wilkerson (author)Gelfling62014-10-22

If you're going to try to coax 1.5A. through a TO-220 case or 3A. through a TO-3, you'd better be adding a substantial heatsink with some heatsink grease.

Lectric Wizard (author)Gelfling62013-04-04

78xx regulators are only good for 1.5A current in the to-220 package & 3-5 A in the TO-3 case.

Gelfling6 (author)Lectric Wizard2013-04-05

Thanks for that correction.. the rating I gave, was going by the detail on the webpage I noted.. (transistor rating).. I imagine the LM317 or LM338 would have about the same rating in TO-220 form.

pfred2 (author)2013-04-03

Perhaps you could share a schematic diagram of your circuit?