DIY Constant Current Load

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Introduction: DIY Constant Current Load

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.

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.

Comments

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 ????