Introduction: Power LED's - Simplest Light With Constant-current Circuit

Picture of Power LED's - Simplest Light With Constant-current Circuit

Here's a really simple and inexpensive ($1) LED driver circuit. The circuit is a "constant current source", which means that it keeps the LED brightness constant no matter what power supply you use or surrounding environmental conditions you subject the LED's to.

Or to put in another way: "this is better than using a resistor". It's more consistent, more efficient, and more flexible. It's ideal for High-power LED's especially, and can be used for any number and configuration of normal or high-power LED's with any type of power supply.

As a simple project, i've built the driver circuit and connected it to a high-power LED and a power-brick, making a plug-in light. Power LED's are now around $3, so this is a very inexpensive project with many uses, and you can easily change it to use more LED's, batteries, etc.

i've got several other power-LED instructables too, check those out for other notes & ideas

This article is brought to you by MonkeyLectric and the Monkey Light bike light.

Step 1: What You Need

Picture of What You Need

Circuit parts (refer to the schematic diagram)

R1: approximately 100k-ohm resistor (such as: Yageo CFR-25JB series)
R3: current set resistor - see below
Q1: small NPN transistor (such as: Fairchild 2N5088BU)
Q2: large N-channel FET (such as: Fairchild FQP50N06L)
LED: power LED (such as: Luxeon 1-watt white star LXHL-MWEC)

Other parts:

power source: I used an old "wall wart" transformer, or you could use batteries. to power a single LED anything between 4 and 6 volts with enough current will be fine. that's why this circuit is convenient! you can use a wide variety of power sources and it will always light up exactly the same.

heat sinks: here i'm building a simple light with no heatsink at all. that limits us to about 200mA LED current. for more current you need to put the LED and Q2 on a heatsink (see my notes in other power-led instructables i've done).

prototyping-boards: i didn't use a proto-board initially, but i built a second one after on a proto-board, there's some photos of that at the end if you want to use a proto-board.

selecting R3:

The circuit is a constant-current source, the value of R3 sets the current.

- LED current is set by R3, it is approximately equal to: 0.5 / R3
- R3 power: the power dissipated by the resistor is approximately: 0.25 / R3

I set the LED current to 225mA by using R3 of 2.2 ohms. R3 power is 0.1 watt, so a standard 1/4 watt resistor is fine.

where to get the parts:
all the parts except the LED's are available from, you can search for the part numbers given. the LED's are from Future electronics, their pricing ($3 per LED) is far better than anyone else currently.

Step 2: Specs & Function

Picture of Specs & Function

Here i'll explain how the circuit works, and what the maximum limits are, you can skip this if you want.


input voltage: 2V to 18V
output voltage: up to 0.5V less than the input voltage (0.5V dropout)
current: 20 amps + with a large heatsink

Maximum limits:

the only real limit to the current source is Q2, and the power source used. Q2 acts as a variable resistor, stepping down the voltage from the power supply to match the need of the LED's. so Q2 will need a heatsink if there is a high LED current or if the power source voltage is a lot higher than the LED string voltage. with a large heatsink, this circuit can handle a LOT of power.

The Q2 transistor specified will work up to about 18V power supply. If you want more, look at my Instructable on LED circuits to see how the circuit needs to change.

With no heat sinks at all, Q2 can only dissipate about 1/2 watt before getting really hot - that's enough for a 200mA current with up to 3-volt difference between power supply and LED.

Circuit function:

- Q2 is used as a variable resistor. Q2 starts out turned on by R1.

- Q1 is used as an over-current sensing switch, and R3 is the "sense resistor" or "set resistor" that triggers Q1 when too much current is flowing.

- The main current flow is through the LED's, through Q2, and through R3. When too much current flows through R3, Q1 will start to turn on, which starts turning off Q2. Turning off Q2 reduces the current through the LED's and R3. So we've created a "feedback loop", which continuously tracks the current and keeps it exactly at the set point at all times.

Step 3: Wire the LED

Picture of Wire the LED

connect leads to the LED

Step 4: Start Building the Circuit!

Picture of Start Building the Circuit!

this circuit is so simple, i'm going to build it without a circuit board. i'll just connect the leads of the parts in mid-air! but you can use a small proto-board if you want (see photos at the end for an example).

first, identify the pins on Q1 and Q2. laying the parts in front of you with the labels up and the pins down, pin 1 is on the left and pin 3 is on the right.

comparing to the schematic:
G = pin 1
D = pin 2
S = pin 3

E = pin 1
B = pin 2
C = pin 3

so: start by connecting the wire from the LED-negative to pin 2 of Q2

Step 5: Keep Building

Picture of Keep Building

now we'll start connecting Q1.

first, glue Q1 upside-down to the front of Q2 so that it is easier to work with. this has the added benefit that if Q2 gets very hot, it will cause Q1 to reduce the current limit - a safety feature!

- connect pin 3 of Q1 to pin 1 of Q2.

- connect pin 2 of Q1 to pin 3 of Q2.

Step 6: Add a Resistor

Picture of Add a Resistor

- solder resistor one leg of resistor R1 to that dangling LED-plus wire

- solder the other leg of R1 to pin 1 of Q2.

- attach the positive wire from the battery or power source to the LED-plus wire. it probably would have been easier to do that first actually.

Step 7: Add the Other Resistor

Picture of Add the Other Resistor

- glue R3 to the side of Q2 so it stays in place.

- connect one lead of R3 to pin 3 of Q2

- connect the other lead of R3 to pin 1 of Q1

Step 8: Finish the Circuit!

Picture of Finish the Circuit!

now connect the negative wire from the power source to pin 1 of Q1.

you're done! we'll make it less flimsy in the next step.

Step 9: Permanant-ize It

Picture of Permanant-ize It

now test the circuit by applying power. assuming it works, we just need to make it durable. an easy way is to put a large blob of silicone glue all over the circuit. this will make it mechanically strong and waterproof. just glob on the silicone, and make an effort to get rid of any air bubbles. i call this method: "BLOB-TRONICS". it doen't look like much, but it works really well and is cheap and easy.

also, tying the two wires together helps reduce strain on the wires also.

i've also added a photo of the same circuit, but on a proto-board (this one is "Capital US-1008", available at digikey), and with a 0.47-ohm R3.


satheeshbabu (author)2015-10-22

In calculations, you mentioned LED current is 0.5/R3. How did you arrive at the number 0.5? similarly, the next line how did you arrive 0.25/R3 as power?

KirkB8 (author)satheeshbabu2016-01-25

Given that the voltage drop across the current sense resistor (R3) is 0.5v - then using Power = IV = VI = V x V/R = 0.5 x 0.5/R3 = 0.25/R3.

I've built the circuit and the voltage drop across Vbe for Q1 = voltage drop for R3 - is indeed 0.5v. But how could that be predicted from the 2N5088 data sheet? What are the design steps?

(it is a great tutorial BTW - just I'm frustrated by my lack of understanding - even if the light does come on!)

jwhendy (author)KirkB82016-10-31

I was curious about this too, and if I'm understanding right has to do with the qualities of a transistor, but could be wrong. Here's why I think that:


Throughout my reading, I've seen folks using 0.5 - 0.7 for various calculations; thankfully some named it as V_BE, so I googled around leading me to the above. It sounds like an inherent property of the materials used. Here's two examples that spell it out a little more than this tutorial:


So, if we know that junction right above R3 is at 0.7V, I think that's how we get I = 0.7/R3 (though this tutorial uses 0.5). And power is V^2/R, so you get 0.25 (0.5^2). In the first link, his equation Vm = Vdd - Vled - 0.7 suggests that since the voltage will drop across the LED, and end up at 0.7 at the transistor, the MOSFET will have the balance.

Anyway, you had that already but felt like commenting since it perplexed me as well. I'm still not positive, but think he's using 0.5 instead of 0.7 and the value comes from a property of transistors. At least that's as close as I've gotten to something sane-ish!

JaneZ8 (author)jwhendy2017-12-26

The Voltage drop when used Germanium made components is around 0.5-0.6 but usually 0.5. If it's Silicium it's 0.7. That is as far as I remember from my lessons in high school :D
Maybe it's irelevant but i thought it might help :)

ElvinBurnett. (author)2017-10-28

This is a really cool circuit. Thank you for sharing. I was wondering, would this circuit work on a long strand of led lights to combat voltage drop over the length of the run?

Chloe_Lemaire (author)2017-06-08

Will battery be okay as my power supply? Since it is not a constant voltage will it affect the way the circuit works?

PirateKittyK made it! (author)2017-05-22

You can use pretty much any N-mosfet and NPN-transistor.
I used an IRF44Z and a 2N2222. My R3 is 2 parallel 2.8 Ohms. Mess with different values to get what you want.
A 1.4 Ohm, in my case gives about 375mA @12V which is what I want.
My LED is mounted in a junk heat sink with thermal compound and its black heat sink (only way to read correct with IR thermometer), reads 61ºC at 25ºC room temperature. (you want =< 80ºC)
The Mosfet reads 48ºC. Not putting a heat sink on that. Not hot enough.
R3 (1.4 Ohm) doesn't even get warm.

Thanks for sharing and gl to all.

chinmoy1955 (author)2016-05-11

Congratulations on a simple but effective constant current circuit.

The schematic symbol of the MOSFET shown in the circuit diagram is that of a P-Channel MOSFET, not N-Channel. Moreover, it is not a simple FET but a MOSFET.

The power adapter voltage can be calculated as follows:

V supply = 3.2 X n + 1V

Where 3.2 is the voltage drop across each led, n is the number of LED's in series.

The 1V added in the end is to take care of any voltage drop that may occur in the power supply due to regulation errors. So a 6V supply can only feed one LED. A 12V supply can feed a maximum of 3 LED's. From the above rough formula we can say that 3 LED's in series will require:

V = 3.2X3+1 = 10.6V

So a 12V supply will also work with 3 LED's, but to keep the power dissipation of the MOSFET at a minimum, it would be advisable to keep the power supply voltage as close to the calculated value as possible, thus 11V would dissipate less heat than a 12V supply. So if you have a 24V supply available, you can safely connect 7 LED's in series.

chinmoy1955 (author)simpletronic2017-01-15

You are wrong, the symbol is that of a P-Channel MOSFET. Below is the actual symbol of the Fairchild FQP50N06L MOSFET taken from their data sheet. Please be informed before commenting:

simpletronic (author)chinmoy19552017-01-15

Both symbols ( yours and the author´s ) are correct and correspond to an NMOSFET. The authors´s symbol is a simplified symbol which does not show the internal bulk-source connection. IEEE(Institute of Electrical and Electronics Engineers) accepts this symbol as a validNMOSFET symbol. Nothing more to say.

You are ALL wrong. That is the symbol for a exhaust bearing.

Look folks, you all know what it is supposed to be regardless of your view, why argue the point?

chinmoy1955 (author)2017-01-15

I had posted here about 8 months back about a serious flaw in the circuit diagram and related text. I am surprised to find that the original circuit is still being published without correcting the flaw. It seems the author does not bother to read the comments and the people following the post are also as much blind and misinformed as the author.

I would like to once again point out the mistake. The circuit symbol of the MOSFET shown is that of a P-Channel Mosfet, and not N-Channel as written in the article. But funnily, the part number given is that of a N-Channel MOSFET. The proper symbol for the MOSFET mentioned in the article is given below:

dfwdavid (author)chinmoy19552017-02-12

as simpletronic said, BOTH your symbol AND the author's symbol are correct and are both acceptable ways of drawing an N-channel MOSFET.

I have no illusions, however, that my picture will change your mind either.

I worked at my parents' hardware store in college. one time a man came in with a "blown fuse". I suggested we check it first with a multimeter. the fuse showed 0 Ohms, so I said the fuse was still good and hadn't been blown. the man said 0 Ohms meant open circuit. I politely restated it was a short circuit. he restated his assertion that it was open. I tried to diplomatically explain I was currently studying EE and showed him that the fuses on our shelves also measured 0 Ohms. I even touched the tips of the meter together and showed him the display now read 0 Ohms. he left without buying a replacement fuse because he was certain he was correct and that all of the fuses in our stock were just as bad as the one he had.

some people just "know" they are correct and no amount of evidence to the contrary will convince them otherwise.

liam2799 (author)2017-02-07

Does anyone have a diagram on how to make this on breadboard? please.

Edward Sun (author)2016-10-14

Could I use the IRF510 instead of the FQP50N06L?

psycobunny12 (author)2016-08-19

will you make it so i can buy this kit?

bigbuglamer made it! (author)2016-06-01

I made similar circuit many times, last one:

irlml2502 nfet, 4.7ohm resistor, KT315B legendary soviet union BJT (made in december 1981, exactly one year older than i'm ;D ).

LED is 300ma 940nm IR.

sola0005 (author)2016-01-18

Dan, can you please tell me what you refer to with "3 x 3W"? Elsewhere, I've read "power source: 3 x 3W". Wish I knew what that means!

DonCesare (author)sola00052016-02-20

Hi! I guess you might know what it means by now but in case you were still looking for the info, 3x3w simply means 3 leds of 3w. Usually they are wired in series.

BerndH4 (author)2015-11-18

The schematic shows 3x power leds and is usable for my design. Howerver:

I would like to use 2x 3 power leds in serie in parallel, each led with If = 350mA en Vf =3.2-3.5 V

Can I just use the 2x parallel 3 leds with R3 set to 0.71 ohm?

Using a 12Volt power supply.


sola0005 (author)BerndH42016-01-18

Maybe you can answer my question: Can you please tell me what Dan refers to with "3 x 3W"? Elsewhere, I've read "power source: 3 x 3W". Wish I knew what that means! Thanks.

b.orka (author)sola00052016-01-22

Not sure exactly where you saw those terms here -- but in electronics W represents "watts", so 3 x 3W may mean three 3-watt LEDs -- perhaps the capacity of the power supply?

DonCesare (author)2016-01-20

Awesome! But can I use a rfp30n06le n channel mosfet? thank you for your time and dedication

leetheguy (author)2015-12-06

Works like a charm! Thanks so much for this and for making it so easy for novices like me. :D

Fezder (author)2015-11-30

Thanks for sharing, this helped! Escpecially those calculations, but where those 0.5 and 0.25 came from?

awilliams1 (author)2015-11-04

I made a 1 amp version like the one you have on the Veroboard to power some white LED strips. Works awesome.

Mine is also blobtronics <-(love that term : )


MarcR15 (author)2015-10-15

Really great tutorial and ideas. Just wish I understood the theory better.

Maybe you could delve into that some day? I don't understand how the FET can be an adjustable resistor. Also, I was confused by the terms like power NFET, small transistor, etc. When I looked up the model you posted, it stated that it's a MOSFET. When googling for FETs, one usually gets results for MOSFETs. I suppose because google tries to include similar results.

mohamed uvaiz (author)2015-06-03

Sir.... I am not expert this area. How can I connect more LEDs, any modify this circuit..... Please help me......

ReneJ2 (author)mohamed uvaiz2015-07-16


ReneJ2 (author)mohamed uvaiz2015-07-15

choose the DC adapter with higer voltage & use the higher wattage R1 if necessary (the (!!!apropriate number of) LEDs'll keep the FET´s wattage the same = the excess voltage (ADAPTER DC OUT minus 3.99V per led)will dissipate as !!! heat on R1, Q2 , the led should convert 50% of it's energy to light the rest 50% heats and gradually degrades the LED junction)

Clark57 (author)2015-05-02


I have a Constant Current LED driver module, rated 80-100V @ 600ma. It currently runs a 45 LED string.

What I want it to do is to run a 23 string @ the same rating. The PROBLEM is that when I do, the supply kicks the total voltage down to equal the same voltage across each LED. Basically, I am trying to double the voltage at the same current, but it won't let me. (The LEDs are rated at over double their current power so I am not worried about blowing them.)

Real world:

I measure across the LED while 45 of them are on and the voltage is 2.225VDC @ 600ma. I then short the string to 23 and measure and it is still 2.225VDC@600ma, but I want 4.45VDC@600ma across each LED.

There's a micro blue potentiometer on this module and I don't know what it does (nor do I have a schematic) but I need to know what to do to adjust it up to 4.45VDC.

I need this, like, 8 years ago so please help!


Clark57 (author)Clark572015-05-02

PS If it can't do this, then how can I get it to give me 1200ma instead at the same volts?

d___b (author)Clark572015-05-31

That depends on the model of LED driver. What is the manufacturer's name and/or part number for the thing?

d___b (author)Clark572015-05-31

Your LED driver is only providing a constant current to the string of LEDs. The voltage across each LED is therefore set by the LEDs themselves [See the LED I-V Curve picture attached].

RonaldSaraswat (author)2015-04-03

How can I use this circuit in an array of, 20 10W power LEDS with 900ma current nd 9v forvard voltage ?

Can the same circuit be used or do I need to tweak it ?

and how many LED's can be put in series or parallel with a 12v dc power supply ?

ozdenakca (author)2015-03-18


how can i drive more power to the led cuz its not shining bright enough?

pgrferrari (author)2015-03-02

Hi, do you have a solution to wire a power led to a 6VAC power source?

rtorres20 (author)2015-01-27

If I wanted to make the same circuit but couple it to my bike's hub dynamo, what components would need to change? (Of course, aside from having a full wave rectifier.) I want to drive six 1w LEDs with an IF of 320mAh in two series/parallel branches. The hub provides a peak power of around 14.2 v at around 50 Km/h. Some insight would be amazing. Thanks!

sandoreduardo.pereyra (author)2015-01-18

Hola mi montaje es este , con un circuito que encontré en un sitio. Material un regulador 7805 y una R de 15 ohmios y led como muestra la imagen Ya que son de 4 Watts El inconveniente que calientan demasiado los reguladores 7805 y la Resistencia. No se si esto es normal dependiendo del diseño del circuito. Gracias por alguna sugerencia para este diseño. Ya que lo necesito para colocar en la moto con 12 volts 5 Amper

shijilt (author)2014-04-26

I want run 10 1 watt leds from a 3.7v 2000mAh battery for max time ... If I connect them directly , I will bet less than 30 minutes total ..

How to get more work time ? if I use a dc stepup circuit , is it gonna help ?

discostu956 (author)shijilt2015-01-17

The only way you can get more run time from this set up is to either limit power to the LED's so they are running under 1w each, or to increase your battery capacity. Can't get something from nothing. Also, connecting them in series, your going to need higher voltage, as the voltage adds up with series. At around 3.2v per LED, your going to need around 32v to run them properly, unless you go in parallel (which I have no experience with, only series)

exhornet (author)2015-01-05


Can you please tell me, can I drive a 3W Led with this circuit by modifiying some of the parts?

fmarquis made it! (author)2015-01-03

Easy and useful built. I used what I had on hand, namely a BC237B as the NPN Transistor and a FQP30N06L for the N-MOSFET. After 10min, with the small heat sink, the MOSFET is only warm to the touch. Resistor does not really heat. However, I should spend a few minutes to make a heat sink for the LED...

awesomelumens made it! (author)2014-10-25

nicely described I was unsure about this method. Than one day I found this page.thanks

EmmettO (author)2010-05-12

 Can this circuit be used to drive multiple LEDs? If so, would they be wired in parallel or series?

SvenM1 (author)EmmettO2014-09-10

Yes. In series. Always in series with LEDs. Know that it is possible to arrange parallel LEDs (and, more commonly, branches of LEDs in series) but it is of no concern to you at this moment. Because in this context, any engineer who asks that question is definitely not ready to balance parallel current draws, and the other complexities inherent with parallel LED drive circuits.

vanmankline (author)EmmettO2010-10-28

I'm not an expert on this topic, but I'm learning.... From what I know:

You would need to switch out the resistors and ensure the other components can handle the current you require. The way these work you will get better results if you wire the LEDs in series.

EmmettO (author)vanmankline2010-10-28

Thanks, I eventually figured it out.

sbadgujar (author)2014-07-23

How to connect 5 to 10 1W LED to this circuit. What changes I have to make in this circuit.

About This Instructable




Bio: Dan Goldwater is a co-founder of Instructables. Currently he operates MonkeyLectric where he develops revolutionary bike lighting products.
More by dan:Giant Xylophone made from Bed SlatsEasy Mothers Day Fudge (with small child)Mosaic Tile Pixel Art Car
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