Introduction: Adjustable Constant Current Source 4mA to 3A

Hello, in this project i want to show you aadjustable constand current source, wich maximal current is 3A by 39V. This is a very usefull circuit, to light up power LEDs or heater, also it is a very cheap and tiny circuit, which is easy to build. For the project you need basic soldering skills and basic elektronic skills.

Features:

Current range: 4mA to 3A
Voltage range: 1,2V to 35V
Temperature range: +-0°C to +125°C
Size: 4cm x 3cm x 5cm
Costs: 1$ to 5$


Step 1: Part List

Parts:

LM350 (or Equal)
Heatsink for the LM350
Terminals
circuit board
Potentiometer 15OHM  1W, 2W or 5W
Wire
Unisolatet wire

Tools:

Soldering coil
Solder
Screwdriver

Other:

power Supply
maybe a 10W LED or a 5W 2,7OHM Resistor for testing
a multimeter

Step 2: Circuit

The circuid is veeery simple, you only have to wire 2 electronic components and  2 terminals. be carefull with the LM350, dont confuse the pins!

Step 3: Assembling

Now assemble it, make sure that you got thick wire for the circuit board and the potentiometer because there can be currents about 3A. Also make sure that everything is good soldered and the heatsink is good connected to the lm350! also dont forget to mark the terminals with "+ - input and output"

Step 4: Testing (yay :D)

For testing you need a power supply (i prefer for that 12V), a multimeter, a resistor 2,7OHM 5W (!!!), and wires.

now connect the positive output of the constant current circuit to the 2,7OHm 5W Resistor, switch your multimeter to ampere measurement and connect the positive probe to the other side of the resistor.

Next connect the negative probe to the negative output of the constant current circuit.

!!! DONT short circuit parts of the circuit and DONT short circuit it with your FINGERS!!!

Regulations the potentiometer to the middle position!!!

now connect the power supply to the input of the circuit, turn it on and regulate it to 12V.

If you twist the potentiometer the multimeter should show the floating current, which should be from 0mA to 1,3A - SUCESS!


But why 1,3A i told it could reach 3A... it can you just have to regulate up the voltage, then the current should rise.


The output Voltage should be 0,7V lesser the the Input voltage.

Step 5: Final Step!

Troubleshooting:

errors could be:

confused the pins of the LM350
short circuits
electronic components burned out


Uses:

Heaters
Heating pad for 3D printers
High current LEDs
current limit




Thank you for reading! :)
i hope you like it, feel free to comment

Comments

author
MinaAminifar (author)2017-07-19

Hi, i want to make a source current 1nA-10mA, can i use this circuit??
If not, may you help me how can i do that?

author
Fragmaster (author)2015-03-26

I'm thinking of using this method to controll an LM338. Does the potentiometer need to be rated for 1W or more? Is it dissipating power? Power potentiometers are a bit expensive when buying one at a time, so I was wondering if a 150 ohm trimpot would work as well. I want to be able to adjust from 1A to 4.5A at 12V to a peltier cooler I'm using.

author
Jan_Henrik (author)Fragmaster2015-03-27

The pot doesn`t have to be rated for high currents

author
stefan-p (author)Jan_Henrik2016-09-04

are you sure , since following your diagram it seems all the current from the LDO's output will go straight through the pot , most likely burning it up

author
rlogiacco (author)stefan-p2016-09-05

The pot MUST sustain the whole sourced current, meaning for 3A at 10V you need a 30W potentiometer... not a very cheap component....

author
stefan-p (author)rlogiacco2017-04-30

https://www.amazon.com/Uxcell-Ceramic-Variable-Resistor-Rheostat/dp/B008LT4FL2 , not very expensive , but add the other components and you get the same price of a switch mode circuit of the same power rating as this

author
rlogiacco (author)stefan-p2017-04-30

Looking at the Amazon description I see no reason why that potentiometer should be capable to withstand the load power.
You must specifically search for a power potentiometer with a 30+W power rating.
On top of that, I would say 8$ for a pot is definitely expensive.

author
stefan-p (author)rlogiacco2017-04-30

It says 150 ohm and 50 watt . But yea still too expensive

author
rlogiacco (author)stefan-p2017-04-30

Actually I only see "power rating: 50"... Are you sure it does imply 50 watt?

author
WesleyV7 (author)2016-07-11

Can i use an LM338 instead of an LM350?

author
stefan-p (author)WesleyV72017-04-30

yes , it's just a more expensive and more powerful linear regulator ic , 3 amp on the 350 and 5 on the 338 , tho if you plan on getting the full 5 amps then you're going to put more power through the pot too , so take that into consideration

author
rlogiacco (author)2017-04-20

Sorry Sebastian, but you didn't read the datasheet properly. The adjustment pin is used to adjust voltage to use the LM350 as a Voltage Regulator, in which case the current toward the load is not flowing through the resistor/potentiometer.
Using the LM350 as a constant current source all the current going into your load is also flowing through the ADJ resistor (not the ADJ pin though), meaning you need a power potentiometer.

Look at figures 31,32 and 34 in the datasheet you linked: in all three of them current coming out of the LM350 is passing through R1, being it a resistor or a potentiometer. Current going into the ADJ pin is still minimal, but the resistor has also to sustain current going into the load.

author
SebastianS29 (author)rlogiacco2017-04-21

You are right! The schematic in the instructable was confusing, I now see that when in Constant current configuration, the resistor/POT will dissipate significant power. The LM317 works the same, except with only about 1.5A current limit. Probably a better part to use anyways. Thanks for the reply.

author
ngpha (author)2015-09-10

I cant change the value of current in the output

author
Jan_Henrik (author)ngpha2015-09-12

Hi, is everything soldered right? and how do you measure it?

author
Shubhz (author)2015-07-01

I am making one project to control stepper motor. I am using Arduino Mega, Nema 17 Stepper motor, adafruit TFTLCD, adafruit motor shield V1 and DIY Matrix Keypad. Actually the problem is, when I am using Stepper motor individually, it's working FINE. But when I am embedding my whole system and then after taking input from Matrix Keypad the motor was not working properly and slowly slowly my LCD Screen is also dimming. I think this is happening due to lack of current supply. So do the above current source can help me out? Right Now I am using Variable DC Power Supply having capacity of 500mA. Please Help me out...!

author
Jan_Henrik (author)Shubhz2015-07-09

You may want to add capacitors at the input of your circuit to stabilize the Supply voltage/Current

author
turbiny (author)2015-04-07

thnx for the circuit i inted to use 3 of these in my car to pwer a 10w RGB LED
should i add a capacitor and where?

author
Jan_Henrik (author)turbiny2015-04-08

you can add caps at the out and input of the circuit, a smaller one at the output and a greater one at the input

author
kainxavier (author)2013-12-17

Wait.... so this is basically all a LED driver is...?

author
Orngrimm (author)kainxavier2013-12-24

Yes and no... :)
A LED driver IS a constant current source. Thats also what this circuit does.
BUT: A real LED-driver doesnt waste this much energy as soon as it has to lower the voltage. Thats where more complicated switching constant Current sources come in.
Check http://parametric.linear.com/internal_power_switch_buck if you want a few good ones... Remember: You have to reconfigure them to accept the shuntvoltage as Uadjust and not a voltage from a voltagedivider. Otherwhise you have a constant voltage source and not constant current source.

author
Jan_Henrik (author)kainxavier2013-12-17

Yes you can use this circuit good to light up high current LEDs :)
but dont circuit them paralel, only in a row!!

author
lunakid (author)2013-12-18

Any idea about efficiency? I mean, the big heatsink is there for a reason, and it would be nice to know when to use this lovely simple circuit, or even just one simple serial resistor instead. (I did that for 3 serial 1W LEDs from a 12V car battery, with fairly reasonable losses).

author
Orngrimm (author)lunakid2013-12-24

The eff is quite low since what you do with the FET is you mimic a linear-regulator.
You can calculate the power used and the power wasted very easy:

Pused = Ioutput * Uoutput
Pwasted = Ioutput * (Uinput - Uoutput)
The efficiency is now simply
eff = (Pused + Pwasted) / Pused
(1 = 100%, 0.1 = 10% and so on)
Ay sou can see, the more you have to drop the voltage from the input to the output, the more power you waste.
If you want more effiiency, you have to go woth a switching regulator. They are a bit more expensive and may need 2-3 external components for that.

Also, you may add a capacitor to the input to prevent the LM350 or LM317 from starting to oscillate...

author
Jan_Henrik (author)lunakid2013-12-19

That is a realy good question, which i can´t awnser.
I don´t know how to calculate the efficiency, if you know how please tell me! :)
But i think its a good or high efficiency :)

author
padbravo (author)2013-12-19

tks for this circuit... I will use it with some hi power led's... tks!

author
Jan_Henrik (author)padbravo2013-12-20

Thank you for using it ;)

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