DIY Constant Current Load

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Posted in TechnologyElectronics

Introduction: DIY Constant Current Load

About: Awesome Electronics Tutorials, Projects and How To´s

In this small project I will show you how to make a simple adjustable constant current load. Such a gadget is useful if you want to measure the capacity of chinese Li-Ion batteries. Or you can test how stable your power supply is with a certain load. Let's get started !

Step 1: Watch the Video!

The video gives you all the information you need to build a constant current load. But I will present you some extra help in the following steps

Step 2: Order Your Parts!

Here is the small list of parts that you will need:

Ebay:

1x Vero board: http://rover.ebay.com/rover/1/711-53200-19255-0/1?...

1x 1Ω / 5W resistor: http://rover.ebay.com/rover/1/711-53200-19255-0/1?...

1x LM358: http://rover.ebay.com/rover/1/711-53200-19255-0/1?...

2x PCB terminals: http://rover.ebay.com/rover/1/711-53200-19255-0/1?...

1x IRLZ44N N-channel MOSFET: http://rover.ebay.com/rover/1/711-53200-19255-0/1?...

1x 500k potentiometer: http://rover.ebay.com/rover/1/711-53200-19255-0/1?...

Amazon.com:

1x Vero board: http://amzn.to/1OeN41p

1x 1Ω / 5W resistor:http://amzn.to/1FFJNEk

1x LM358: http://amzn.to/1OeMz7o

2x PCB terminals:http://amzn.to/1CuBQgM

1x IRLZ44N N-channel MOSFET: http://amzn.to/1H8jr1l

1x 500k potentiometer: http://amzn.to/1FFJxoT

Amazon.de:

1x Vero board: http://amzn.to/1yZ4k1J

1x 1Ω / 5W resistor:http://amzn.to/1yZ4u9x

1x LM358: http://amzn.to/1GtRhNH

2x PCB terminals:http://amzn.to/1GtRU9R

1x IRLZ44N N-channel MOSFET: -

1x 500k potentiometer: http://amzn.to/1yZ4Vkb

Step 3: Build the Circuit!

Here you can find the schematic for the build and the board design that I created. Make sure to interrupt the copper traces underneath the LM358.

Step 4: Success!

Now you should be able to build your own constant current load.
Feel free to check out my Youtube channel for more awesome projects:

http://www.youtube.com/user/greatscottlab

You can also follow me on Facebook, Twitter and Google+ for news about upcoming projects and behind the scenes information:

https://twitter.com/GreatScottLab

https://www.facebook.com/greatscottlab

13 People Made This Project!

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Please be positive and constructive.

Tips

Hello GreatScott, I have added a current cut-off feature on the constant current discharge circuit. The schematic is attached.

schematic_constant_current_load.png

5 Questions

i made the circuit but the current that flows from the battery is very low !!!

with 1 ohm resistor and 1 V Vref

i connected a 4 ohm resistor parallel eith the 1 ohm

and the current increases to 1 amp

what is the problem ????

Hi, i've made a mistake on my order and i bought a bunch of 100k potentiometers can i use them instead of the 500k ?

Yeah should not be a problem, as you can see the inputs on the potentiometer are the 5V and GND, this means that twisting the potentiometer changes the voltage, on the output of the potentiometer. This is called a resistor bridge, voltage divider etc. The higher ohm of the potentiometer is chosen to save energy.

Thanks for providing this tutorial. I have two questions though:
1. Would it be possible to expand this idea in a way that current load disconnects completely after battery voltage falls below certain value? This is AFAIK essential for battery health.
2. There are many current load circuits on the net, mostly much more complex, usually having another op amp at the reference voltage supply side and couple of resistors and capacitors elsewhere. I know just enough electronics to understand your circuit, but I have no idea if other more complex circuits have any serious advantages over yours. Do they, and if yes, in what way?

1. It does that. When VBAT is lower than output of RV1, U1B always outputs GND to the gate of Q2. Q2 becomes an open circuit. I have tested with a cut-off of 3V, and the current between J2 terminals is lower than 10mA.
2. It depends on what you want to do. If you need really precise discharge current, thermal cut-off, high discharge currents, you may need to modify the circuit.

2 more answers

I have replied to your questions ;)

1 - Take a look at my tip on how to add cut-off voltage feature on the circuit.

1. It does that. When VBAT is lower than output of RV1, U1B always outputs GND to the gate of Q2. Q2 becomes an open circuit. I have tested with a cut-off of 3V, and the current between J2 terminals is lower than 10mA.
2. It depends on what you want to do. If you need really precise discharge current, thermal cut-off, high discharge currents, you may need to modify the circuit.

Take a look at my tip on how to add cut-off voltage feature on the circuit.

You can cheaply obtain these transistors by ordering them from China. But yes, it takes few weeks to obtain them.

If I am right, that the battery still discharges after its voltage hits cut-off voltage, but at lower, unknown currents. Yes, I agree that the battery is protected, but further discharge is unnecessary and cannot be added to capacity as you don't know the current value.

The obvious solution is following the voltage by microprocessor like Raspberry Pi and then turn off the discharge by GPIO; I wonder if there is any purely electronic solution.

3 more answers

1. You are right, but they didn't have the IRLZ44N at my local electronics shop.
2. I don't know what you mean. You want to discharge up to cut-off voltage to evaluate its capacity and at the same time, don't overdischarge and damage your cell. I am attaching a typical discharging curve of a 18650 cell.

18650b_discharge.png

Two comments though.
1. IMHO, with IRLZ44N for Q2, as in original circuit, VCC=5V would suffice.
2. Isn't that after voltage of battery hits cut-off value, battery keeps emptying, but this time at constant cut-off voltage? Because in batteries lower emptying currents mean larger voltages.

98 Comments

hi sir great scott...may i have a question? since the op amp LM358 have 2 comparator...can i use them both??

2pcs. IRFZ44N, 2pcs. 500k potencio meter, 2pcs 1ohm resistor but 1pcs LM358....

thanks for your time to reply...

3 replies

Hello! I did exactly what you said, but as i need to have the same current in the second battery, I've use the same feedback of potentiometer on pin 5 (pin 3 and 5 shorted) in that way you can use only one potentiometer for the same load on both

But if you want to do 2 different loads, so you can use 2 potentiometers instead.

I had to power with 10v to get the IRFZ44 to work 5v would not turn it on enough.

The IRFZ44N is not a logic level, so yeah, you would need more voltage, the IRLZ is the logic level version, but as long as you can supply the turn on voltage I don't see why either would not work. There are also other mosfets that are logic level that should work.

Hey guys,

Wondering if someone could help me. I got the circuit working, but the project I'm working on at school requires the discharge current to be at minimum 4A. Does anyone know a way I could alter the circuit to allow greater discharge currents? I tried using a 12V source and reducing the resistor value, but it only makes small changes, and seems like I am capping at 2.5A.

Thanks

1 reply

You are probably limited by VDS. Your VGS is quite high even with a 1Ohm resistor, at 11-4 = 7V. I am taking 11 instead of 12V, since the LM358 should have a 1V drop at output not being rail-to-rail.
You need to keep reducing the resistor. One thing you did not tell though is the battery voltage and your cut-off requirements.

I was thinking about trying to use the second comparater to measure the voltage of the battery without a voltage divider and then use that to somehow trigger the device to stop. I built it and love it, and have used it a lot, I put USB capabilities on it for super easy battery bank testing, but also use it for cells, but since you have to watch cells it would be nice if I could basically make it shut off when the battery reaches the desired voltage. Dunno, pretty new but maybe I could us the second one to test voltage and as soon as it drops to the desired end voltage, it could trigger a flip flop to shut off the circuit.

Any ideas would be awesome.

Hi,

I have tried to recreate this exact circuit, but have always failed. What do we attach at "load", like a resistor or what?

1 reply

The "load" is the battery that you want to discharge.

Hi Guys,
I want to build this Project but wich mosfet do i have to buy ? I found those 2 at the local electronics shop IRLZ44PBF 1 N-Kanal 150 W TO-220AB or IRLZ44NPBF 1 HEXFET 110 W TO-220
Thanks for your help.

Today I ran some tests.

2A 12v for 40 minutes with no fan and pcb sitting on bench so no air flow underneath.

The heat sink had reaches 107 deg C which I think is the safe limit for my setup.

After standing pcb off of the bench and applying a small fan , the heat sink dropped to 40deg C

So at this point I turned the power up to 4A @ 12V. The heat sink is about 54deg and the resistor is around 200 deg C ( the coating is good to 350 deg C)

All looks great.

see my post

I have just ordered the parts, I calculate with a 3c/w heat sink I can take 30w.

So I think 2.5A @ 12v.

Hoping to build this in the next 3 days , so will find out if I'm completely wrong or not !

I will post the results good or bad!

Hi I'm going to try and build this in a few days time.

I hope it will be good for 30w

2.5A @ 12v

I will be using a 3C/W heat sink to get rid of the extra power dissipated by the MOSFET.

I am new to this so will post the results.

I found these which may help.

http://www.ko4bb.com/Test_Equipment/DynamicLoad/

Info on constant current load

http://mustcalculate.com/electronics/heatsink.php

Heat sink calculator

I don't see why not! just replace the resisto with your led! but I have to remind you I'm a noobie just like you c:

Well the thing is I found this circuit used as a constant current discharge for a battery thus I'm not really familiar with it. Also what do you mean by replacing the resistor with an LED? :3 Can i just simply put a 10w LED as a load?

Yeah. I actually tried it on a breadboard ^_^ it does work if you replace the Resistor with your LED (or whatever you want to power with a constant current).
not sure though, because I didn't have a logic level mosfet and used 12v for the op amp!

the load+ mosfet works as a voltage divider afaik!

So what you mean is that I cam simply change the 1ohm 5w Resistor with a 10w LED and power up the circuit with +12v through the op amp, after that what do I do with the load?
~I'm way too noob for this circuit sorry :3