Low voltage constant current battery drain

[Vincevl]

Hi Folks,

I've been experimenting with building a Nickel Iron battery and I have my first cell working. I've been able to charge the cell using my bench power supply. However I want to drain the battery using a constant current load so I can test the cell performance.

Presently I've been draining the battery using a Joule thief LED circuit that draws about 25mA but the problem is that as the battery drains the load drops. The charged cell is 1.4V and I want to drain it down to 0.8V at about 50mA.

I searched the web and found LM317 circuits but they are unstable with Vin < 1.25V.

Any suggestions for a circuit would be appreciated. Please keep it simple I'm a newbie to electronics.

Many thanks,
Vince

 

[ronv]

constant current

The low voltage makes it harder, but something like this should work. It needs to be a rail to rail op amp.

 

[Vincevl]

Thank you very much! That was fast. I'm freaking amazed! I put your circuit into circuit wizard. They didn't have the LM6142A but the substitute I used seems to be working as you designed it. Varying the 1K pot changes the constant current, even if my cell voltage drops to 0.8V Brilliant!!

My next task is to incorporate a timer circuit to charge and then discharge the cell over and over to condition the cell.

Cheers,
Vince

View attachment 65607

 

[ronv]

Interesting battery. How do you charge it? Constant current, constant voltage? How do you determine it is fully charged? We could probably make a circuit to cycle it.

 

[MrAl]

Hi,

I wouldnt mind seeing some pics of the battery and more info on it too if you care to share. Will be interesting.

 

[Vincevl]

Edison Nickel-Iron NiFe Battery Cell

Hi All,

Here's a couple pictures of the NiFe cell driving a Joule thief circuit. The cell is made of perforated tubing with the positive electrode filled with alternating layers of Nickel(II) hydroxide and graphite and the negative tube is filled with steel wool. The electrolyte is Potassium Hydroxide in distilled water.

I just finished the cell and it holds a voltage of about 1.2 > 1.4V. I was hoping it would have more capacity, but the cell could still be forming oxides. I've been charging the cell with 50mA of current with my bench power supply. I don't really know when the cell is charged. I may have been over charging (I need to experiment). However, NiFe cells are able to withstand overcharging and also deep discharges.

I'd really like a circuit that I could charge the cell for a given time and then automatically discharge the cell to 0.8 volts, and then charge it up again, but this is beyond my electronics abilities (any ideas would be appreciated). Apparently this will help to increase the capacity of the cell.

Cheers,
Vince

View attachment 65639

View attachment 65640

 

[Vincevl]

Hi ronv,

I'm trying to understand your circuit but am having trouble. I could not find reference on the internet for the way you connected the op amp. Most examples I found have a feedback loop across the inverting pin. Your set up takes the inverting pin directly to the transistor collector. Are you using the op amp as a comparator or an amplifier? Could you describe how the circuit is working. I'm a electronics newbie so any help is appreciated.

 

[ronv]

It's not quite a comparator, but it is a very high gain amplifier. It just makes the voltage across the 10 ohm resistor match the voltage set by the pot. Since the resistor is always the same it makes the current constant.

 

[Vincevl]

Just so I can understand. Are you setting the gain by (5000+1000)/10 Ohms = 600 times or ???

 

[ronv]

It's called open loop gain since there is no feedback from the output to the - input. It is about 8000 for the op amp you used in the simulation. So a 1 mv change at the input would try to make the output change by 8 volts.

 

[dougy83]

The amplifier is working in a closed loop mode, wherein it adjusts the output to make the inverting (-) input the same voltage as the non-inverting (+) input.

The gain of the amplifier is roughly Iout = Vin/10. Vin is the voltage on the non-inverting input.

EDIT: where Iout is the transistor collector current

 

[ccurtis]

This circuit is somewhat simpler. Current draw is set to about 50mA, but can be adjusted by R2, or R3. Simple series voltage regulator across fixed load resistor (R3), thereby drawing constant current from battery. Works down to a least .8V, but somewhat less. Cheers.

 

[Vincevl]

Thanks! I'm going check this one out.
Cheers

This circuit is somewhat simpler. Current draw is set to about 50mA, but can be adjusted by R2, or R3. Simple series voltage regulator across fixed load resistor (R3), thereby drawing constant current from battery. Works down to a least .8V, but somewhat less. Cheers.