What is Constant Current Source – Facts Explained

In this post we try to analyze what is constant current source and how it affects a load, or how it may be used with a load correctly for achieving the most efficient results.

The following discussion between me and Mr. Girish will clearly explain what is CC or how constant current operates.

How a Constant Current Source Works.

Question put by Mr. Girish.

I am trying to build a Arduino based Li-ion charger with a display, but I am with loads of confusions, if possible try to correct my puzzlement.

I have attached a diagram it is similar to which I am working with.

The LM317 in CC and CV mode, I have limited the voltage to 4.20V and current to 800mA (for 2AH battery) with 1.5ohm 1 watt resistor.

I am getting exactly 4.20V at output (open circuit) and short circuit current of exactly 0.80A.

But when I connect a  Li-ion battery (with half charge which is old batteries from laptop) the current consumption is just  0.10A, and almost discharged battery not consuming not more than 0.20A.

If charging is done at this rate it might take 10 hrs or more for reaching full battery, which is not feasible.

Is it possible to force the current to flow through the battery at 0.80A rate?

As far as i know the batteries are in good condition.

Will the Current be Forced into the Load

My second question is: Does constant current source pump current into a load or is it just a maximum current limiter?

Answer

If you are supplying 4.2V and 800mA to a 3.7V/800mAH or to a 2AH  cell, then everything is correct and nothing should be changed, because your charging specifications are perfect.

If the battery is not charging at the given full rate then the problem has to be with the battery not with the charging procedure.

You can try confirming the results with another meter if possible, to be entirely sure.

By the way a good battery should have accepted the 0.8 mAH charging rate and should have shown an immediate rise in its body temperature...if that is not happening then I guess the problem has to be with the battery.

You can also try another Li-ion battery and check whether it behaves in the same way or not. or you can try raising the current to full 1.5 amps, and check the response, but make sure to mount the ICs on a good heatsink, otherwise they will shut off.

Constant current source will not pump current, its job is is restricted to not to allow the load to consume current above the CC's specified value at any circumstances. However ultimately it is the load which decides how much current it is supposed to consume. Current limiter will only work to stop the consumption if it reaches over the  specified rating, and nothing more.

Feedback from Mr.Girish

Exactly, what I discovered too, but in YouTube, I have seen many people saying it "pumps" the current through the load. They limited the current to 12.6 mA with 100 ohm resistor and I am getting short circuit current of around 12.6 mA, they connected number of LEDs in series and took reading, current flow remains the same 12.6mA. The input volt is hiked to 24V, but the LED remains without any harm.

link: www.youtube.com/watch?v= iuMngik0GR8

I too replicated the experiment and got the same result. I think this may look like current "pumping" but obviously not "pumping".

I think this video conclusion cannot be applied to Li-ion batteries, since LEDs are current driven devices.

In case of Li-ion battery, if we connect two in series we have to increase the voltage to 8.4V and not keeping the same voltage or unconditionally higher voltage as LEDs.

I am assuming that my batteries are faulty.

Answer:

In the video the person says a 1amp constant current source will push 1 amp to a 1 ohm and also to a  100 ohms regardless of the resistance value? it implies that it will do the same to a 1K resistor?? that's grossly incorrect...just try it with a 1K resistance.

You can apply Ohm's law and get the results quickly.

Constant current simply means that the source will never allow the load to consume more than the specified rating of the source, this is the ultimate truth for any constant current source.

It is the load that ultimately decides how much current it will consume.... provided the load V specs matches the source V specs.

This is the reason why we use different resistors with different LEDs, because resistors do resist current depending on their values.

It may be any kind of load, whether battery or LED or bulb or SMPS, as long as the V spec matches the source V spec,  the current draw will be decided by the load.

The current source can do nothing but wait until the load tries to pull more than the rated value, and here the CC comes into action and stops the load from doing this.

Our mains input has around 50 amp current CC, does that mean it will push this current in our appliance, then we would see our appliances catching fire every now and then...;)

You can pump current by disturbing the voltage, that is by increasing the V beyond the load's V rating, which is technically wrong.

Feedback:

I too agree on this and I think the reason why LEDs able to lit without any harm at 24V because the current is limited to 12.6mA  which would also affect the voltage (V and I are proportional and no voltage regulator in it). since the current is constant, the terminal LED voltage must also stay fairly constant. I done the same experiment and got 2.5 to 3V across the LED at 17V input.

Reply:

Yes that's another aspect, if the current is below the load's max current specs then the voltage will drop to the load's rated V specs, regardless of the input voltage increase, .....but not if the current is more than the loads rating, then it will burn the load.

That's why when we use a low current capacitive power supply, even though the input conversion produces 310VDC across the LED, it quickly drops to the connected LED's fwd drop value, because the current is limited by the low value capacitor which may be rated lower than the loads's max amp rating.

In the above indicated capacitive power supply, the output from the bridge is around 310V DC, but yet it is quickly dropped at the zener diode's value without burning the zener diode. This happens because of low constant current from the capacitive supply which is unable to cause any harm to the zener diode, due to the much higher wattage of the zener diode.

Conclusion

From the above discussion we understand the following aspects regarding a constant current source:

• Constant Current Supply has only one job to do, stop the connected load from drawing more current than the input's CC rating.
• For example, a 7812 IC can be considered as a 1 amp 12V CC/CV regulator IC, because it will never allow the load to consume more than 1 amp and more tha 12V, regardless of the load rating.
• Alternatively,  as long as the load's voltage rating matches the Constant current supply's Voltage rating, it will consume a current as per its own specification.
• Suppose we have a 12V supply with a 50 amp CC, and we connect a load rated at 12V 1 amp, so what will be the consumption of the load.
• It will be 1 amp strictly, because the load's V spec is correctly matched with the supply's V specs.

What happens if the supply V increases.

It will be then devastating for the load, as it will be forced to consume dangerous higher levels of current than its 1 amp rating, and finally it will burn.

Simple Constant Current, Constant Voltage Circuit Using Transistors

The following image shows how a simple yet very reliable CC/CV regulator can be built using a couple of transistors or BJTs.

The 10K pot can be used for adjusting the required constant voltage output level, while Rx cab be set for fixing the constant current level at the output.

Rx may be calculated with the help of the following formula:

Rx = 0.7 / the Desired CC Level

About the Author

I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. I am also the founder of the website: https://www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials. If you have any circuit related query, you may interact through comments, I'll be most happy to help!