I want to discharge a battery at a constant C rate (0.2C) the battery is 3.7V and 380mAh. I did some calculations and found out that the current the circuit will need to take is 76mA. I am planning to build this with LED's. I am not sure but from what i know LED's take a constant current, i am not sure if this is 20ma or 15ma or something else. Also I am not sure if I should connect them in series or in parallel. Can someone help me come up with a circuit I could make that takes up a constant current of 76mA. Any suggestions of LED's i could buy off amazon do to this would also be helpfull
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\$\begingroup\$ Ah, what you need is a CCS - Constant Current Source. For LED newbies, I would recommend to start off reading lednique.com. Happy learning. Cheers. \$\endgroup\$– tlfong01Feb 28, 2021 at 1:32
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4\$\begingroup\$ You are confusing two things. LEDs and diodes aren't constant current. They are constant voltage which means you can drive them with constant current supplies to ensure they do not burn out, because their voltage drop is constant and will make no attempts to change to limit current through them. \$\endgroup\$– DKNguyenFeb 28, 2021 at 1:34
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\$\begingroup\$ A Lithium battery cell is 3.2V when almost completely discharged and is 4.2V when fully charged. 3.7V is half a full charge. Why so do want to discharge it more? If its voltage becomes less than about 3.0V then it is ruined and might explode or catch on fire if charging is attempted. Read about it at www.batteryuniversity.com . \$\endgroup\$– AudioguruFeb 28, 2021 at 1:55
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2\$\begingroup\$ This is an X-Y problem. It would be better to ask for advise about the original problem rather than asking how to implement a solution that may not solve your problem. \$\endgroup\$– ScienceGeyserFeb 28, 2021 at 2:05
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1\$\begingroup\$ LEDs should be operated by constant current, they do not provide a constant current characteristic by themselves \$\endgroup\$– Russell McMahon ♦Feb 28, 2021 at 3:50
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2 Answers
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The most you should discharge a Li-ion cell to is about 2.8 to 3.0V to prevent damage to the cell.
Here is a very simple circuit that maintains a fairly constant discharge rate from 2.8 to 4.2V (and beyond).
simulate this circuit – Schematic created using CircuitLab
R2 biases D1 (actually an IC shunt reference that regulates at 1.24V) and provides a small amount of base current for Q1. Q1 provides the remainder of the base current Q2 requires.
R1 is chosen so that 1.24V/R1 ~= 76mA. A bit of current that is not metered goes through R2 so the current is not precisely constant.
Here is a quick LTspice simulation:
The circuit will typically work well below 2.8V-3.0V so you need to provide some means of cutting off the current before the cell is damaged.
Just for comparison, as I already had it set up, here is @Michal's circuit simulated:
It's possible that he got different results depending on the models used.
In real life, it would be much nicer to make a circuit that operates from an independent regulated power supply such as 5.0V which would eliminate both the low voltage limit and most of the slope in the current-voltage line. It would also be easy to incorporate a low voltage cutoff.
answered Feb 28, 2021 at 2:45
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You must drain it to resistor because the Leds rapidly decrease conducting under 2,8V. I expect you want to go lower. Probably simplest constant current circuit:
R1 and R3 about 2Wats power rating