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I have spent many hours looking through the Internet and this forum but could not find an answer to my question.

Goal: I want to discharge a lithium cell from nominal voltage of 3.7V to 0V. Essentially, I want to build a discharge circuit without a cut-off voltage for discharge protection. I am aware, that this will irreversibly damage the cell. That is fine. I want to take it apart and avoid any chance of a short circuit.

Ideally, I can set a constant discharge current (e.g. 500mA) using a programmable controller and can monitor the voltage of the cell. Once the cell has reached a voltage of 0V, it should be automatically short-circuited to allow for relaxation of the cell.

Question: how do I construct such a circuit? What elements do I need and how do I wire them up?

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    \$\begingroup\$ You've stated your goal, but forgotten to actually ask a question. Stack Exchange sites are all about Q&A, but there's no Q here for anyone to A ... \$\endgroup\$
    – brhans
    Commented Dec 20, 2021 at 23:00
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    \$\begingroup\$ It almost sounds like you've already answered your own "question". Discharge at a safe constant current, monitor the voltage, then apply a short at 0V. What are you asking, exactly? (There's no question in the question.) \$\endgroup\$
    – JYelton
    Commented Dec 20, 2021 at 23:00
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    \$\begingroup\$ You will have to find an unprotected cell to do this of course. The simplest option is a resistor of a few ohms. Even before the nominal 0% voltage, the battery's output voltage will collapse rapidly once it is nearly discharged, so I don't see any benefit in a more complicated solution. \$\endgroup\$
    – user16324
    Commented Dec 20, 2021 at 23:01
  • \$\begingroup\$ Do be aware that even fully discharged, the chemicals in a lithium cell can be pretty nasty stuff. Wear lab gloves. \$\endgroup\$
    – Hearth
    Commented Dec 21, 2021 at 0:35
  • \$\begingroup\$ Thanks for the replies. I added the question to the original post. @user_1818839: how long will this take? As you said, the voltage drops for low SOC - a constant resistor e.g. 5 Ohm would mean the last 20% will take a long time, no? \$\endgroup\$
    – bron
    Commented Dec 21, 2021 at 6:27

4 Answers 4

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If you have an ordinary triple-output lab power supply handy, you could do something like this (outputs set to +/-10V at a few mA and +5V at sufficient current to discharge the cell):

enter image description here

The op-amps are halves of a dual LM358. Q2 can be something like a TIP32 or you can use a bipolar PNP darlington (eg. TIP122) and replace Q1/Q2

The TL431 produces a 2.5V reference voltage, which establishes the desired 'on' current in conjunction with R2 (in the example, it should be a 1W resistor).

The battery is simulated in the circuit as a capacitor in series with a voltage source. Note that it is connected so that Vbat is negative initially. The Darlington sees 2.5V + 3.6V initially, so power dissipation is 6.1V times the current. Use an appropriate heatsink.

When the initial -3.6V on the battery rises to 0V then the second op-amp output goes high and prevents any more current from flowing. That output going high can act as an indicator.


That said, for a one-off, personally I'd just put a suitable resistor or the initial discharge current across it and check the voltage periodically. If it's taking too long to get down to a very low voltage attach a lower value resistor. You could also use an incandescent lamp, which automatically drops in resistance as the voltage drops, but they're not so well stocked these days, especially in oddball voltages.

Edit: tip122 is a NPN Type darlington so it is the opposite/complementary to an PNP transistor and for me didn't work properly. For me it messed up the circuit and maybe two bc557/bc558. The PNP complementary to TIP122 would be TIP127. Maybe it would work with theese PNP Transistors. Now I have no PNP anymore of this type and the circuit didn't work as expected. It drained the cells but slowed extremely down at 20mah discharge current and about 0.8v. The cutoff had never been reached. I'll give it a try with some pnp 40v 0.5-0.8a Transistors.

https://www.el-component.com/bipolar-transistors/tip122

https://www.el-component.com/bipolar-transistors/tip127

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Here is an example.

R1 sets the current as its voltage is maintained at about -0.75V by the opamp that "compares" it to the R2/R3 divider. (Note that the voltage across R1 is about 0.25V).

You'll need to adjust for:

  1. Opamp current and Q1 limitations: you probably need to replace this with a darlington and possibly multiple power transistors, and you need to add a heat sink (500mA @ 5V is more than 2 W (5V because of the extra drop to R1)). You should be able to get rid of the overcurrent.
  2. Base voltage/Collector voltage: to ensure the required voltage drop, you can change the R2/R3 ration so that the reference voltage is -0.75V.
  3. You need to add protection(s) to avoid that the battery voltage goes negative. And/or you can add comparators to make the short circuit (preferably through a resistor) and disable the constant current circuit. R1 also dissipates quite some power (0.5W if it's a 1 Ohm resistor @500mA, 0.25W for a 0.510 Ohm resistor).

schematic

simulate this circuit – Schematic created using CircuitLab

The transistor needs to be an NPN here so that no gain is introduced in the loop, otherwise the stability at unity gain is no longer guaranteed (an opamp is generally tuned to have a stable closed loop at least above unity gain - unless you choose an uncompensated one).

EDIT: Extra explications following comment

How does this circuit work.

The resistor R1 has a voltage drop proportional to the current flowing through it. That current is essentially the same as the current coming from the battery (minus the base current).

The opamp is rigged such that the voltage drop across R1 is regulated to the voltage drop across R3.

To understand the feedback loop follow these steps:

  1. Suppose that the current from the battery gets lower.
  2. That results in a reduces voltage drop across R3.
  3. The lowers the voltage on the - input of OA1.
  4. So the voltage difference from + to - increases.
  5. An increasing differential input voltage on OA1 results in an increasing output voltage on OA1.
  6. The increasing output voltage increases the Base to Emitter voltage of Q1, and an increasing Base to Emitter current.
  7. The increasing Base to Emitter current allows for an increased Collector to Emitter current (if the voltage source can provide it and if Q1 is not saturated).
  8. The increased Collector to Emitter current, increases the voltage drop across R1, which increases the '-' input and counteracts the initial change on the '-' input (in step 3). So this is the closed negative feedback loop.

As the negative feedback loop ensures that the voltage drop across R1 is the same as the one across R3 and the latter is a fixed voltage because of the voltage divider accros a known source, the current in R1 is "known".
It is the voltage drop across R3 divided by the value of R1, so here it is $$\frac{ {1V}\frac{{R3}}{R2+R3}}{R1}=\frac{{1V}0,23}{0.51\Omega}=450mA$$ . I chose the component values approximately, so it's not exactly 500mA - it suffices to change one of the resistor values.

With regards to the protection in point "3." further above.

If the battery's terminal is pulled towards -1V, it may very well be that it still conducts (without the lithium battery protections). And I do not think that is a good idea.
I updated the voltage generator representing the battery so that it also generates a negative voltage. You can see that there is still current when the battery voltage is negative (supposing that the battery supports it, the internal resistance was not modeled).

So I suggest to protecting against this (for instance a diode with anode to the negative terminal and kathode to the positive terminal of the battery would already offer some protection as it would limit the battery voltage to about -0.7V).

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  • \$\begingroup\$ Thank you for this diagram! This is what I'm after. The simulation shows, that this circuit maintains a constant current of 500 mA - however, I'm having trouble understanding how it works. I don't understand how the opamp keeps the voltage across R1 at 0,75V. I also don't understand point 3. Could you take some time to explain the circuit to me? \$\endgroup\$
    – bron
    Commented Dec 21, 2021 at 7:28
  • \$\begingroup\$ I added extra explications to the answer. I hope it's more clear like that. Not that the voltage of R1 is maintaned at -0.75V, so the voltage across it is -1V - (-0.75V)=0.25V (more precisely computed it's about 0.23V), I did no write "across". \$\endgroup\$
    – le_top
    Commented Dec 21, 2021 at 10:50
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Concerning Spehro Pefhany's post in this thread:

The op-amps are halves of a dual LM358. Q2 can be something like a TIP32 or you can use a bipolar PNP darlington (eg. TIP122) and replace Q1/Q2

The Darlington pnp tip122 missleaded me. So the circuit broke two BC557. In Order to get the results mentioned by the author, I tried to get the bc557bc558 and a npn type transistor to drain the current from the cells. Now I read, that tip122 is not a PNP but a NPN type transistor. Propper PNP type complementary to the tip122 would be the TIP127

https://www.el-component.com/bipolar-transistors/tip122

https://www.el-component.com/bipolar-transistors/tip127

Unfortunatley the two bc557/558 burned because of over current/over Voltage or because of use of the TIP122 equivalent NPN transistor, some overvoltage or other configuration issues. So I wasn't able to try the circuit as written by the author(For some ohter the parts I had to try some equivalent circuitry).

So now for the faulty use with an npn darlington transistor like TIP122 it seems that the cells discharged rapidly, but at 20 mah discharge current and about 0.8v slowed down extremely. Maybe with an PNP 40v 500mah Transistor it will work. I'll give it a try as sson I'll get a PNP 40v 0.5a Transistor.

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    \$\begingroup\$ The order of the post may change over time, so refering to the "second post" maybe misleading soon. \$\endgroup\$
    – Jens
    Commented Jul 20, 2022 at 1:58
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Most cells contain internal protection circuits that disconnect the external terminals when the internal cell V becomes too low. Therefore your external circuit won't fully discharge the cell.

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  • \$\begingroup\$ I think this depends on the type of cell. If LiPo (secondary) I would agree they usually do have a protection circuit. But if LiSOCl2 for example (primary), none of the cells we typically use have such protection. \$\endgroup\$
    – JYelton
    Commented Dec 20, 2021 at 23:08
  • \$\begingroup\$ I am considering unprotected cells, but good point. Is there any way to safely discharge a protected cell to 0V? \$\endgroup\$
    – bron
    Commented Dec 21, 2021 at 6:32

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