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I built up a Rspberry Pi self-driving car and everything was fine while I collected data for the Neural Net. After NN finished its training, I want to test the system. At the beginning, everything was fine for like 5 mins. After that the left motors of the car won't work anymore. The first thing I thought about is that at least one of the 2 Li ion batteries I use was drained (I use 2 Li ion batteries connected in series for 2 motors and 2 more for the other motors). I measured the voltage of the batteries, and one of them had 0 voltage. I thought fine I just have to charge it...

Well ... it was not so easy. When i plugged the battery in the cell phone and connected it to the charger, the phone won't turn on (even after 10-15 minutes of charging). Read something about sleep mode of a battery and decided to apply some voltage to the battery (used a pack of 1.5V of batteries). After, I measured the voltage, it was around 2.7. I though good, it helped, let me try to verify the fix using the phone, but it still won't turn on ... Re-verified the voltage and it was 0.0 ...again ...

To summarize, if i connect the Li ion battery to a device that would use it's energy the voltage drops to 0.0, when I apply voltage to the battery, its voltage become something like 2.7V and this behavior is repetitive.

P.S. The other 2 battries seem fine. I have no idea what is the issue. The only way I can think of is to buy new ones, but maybe there is something I don't know. Esp that I need this system working on Monday (morning) and today is almost Sunday (the store I buy from is closed)...

I use this type of batteries: https://www.cnet.com/products/nokia-bl-4u-cellular-phone-battery-li-ion-series/specs/

Any help is highly appreciated.

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  • \$\begingroup\$ Please provide a link to the datasheet for the batteries. \$\endgroup\$ – Elliot Alderson May 26 '18 at 17:09
  • \$\begingroup\$ Lion batteries discharged that deep are just broken. Dispose them before they start to burn or explode \$\endgroup\$ – PlasmaHH May 26 '18 at 17:40
  • \$\begingroup\$ It sounds like you were drawing far too much current from a cellphone battery. Do you actually know the power requirements for your small car? Many of the cellphone batteries have internal battery management or fuses to protect the cells or minimize fire risk. If you have exceeded the rating the battery is dead. \$\endgroup\$ – Jack Creasey May 26 '18 at 17:58
  • \$\begingroup\$ @PlasmaHH, they are pretty new, like 1-2 weeks. After I purchased them, I used them till they discharged a bit (like 15-20 % left) and fully charged just once. Is it possible to kill them this fast? \$\endgroup\$ – Postica Ðenis May 26 '18 at 18:02
  • \$\begingroup\$ @JackCreasey, I'm not not sure. I use L298N motor driver to control the rotation direction of the motor (switching + with - and so on) and its speed (using pulse-width modulation). I use this car chassis alibaba.com/product-detail/… . I used like 50 % of the available power as at 100% the car was moving to fast to accurately move on a track made from A4 paper sheets. \$\endgroup\$ – Postica Ðenis May 26 '18 at 18:07
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The effect you see is the standard behavior of electronically protected battery when it reaches the protection over-discharge threshold.

When you load and drain the battery in a device, the voltage of internal cell will gradually drop down. When it drops below the threshold imposed by protective electronics (typically 2.5 - 2.7 V), the electronics disconnect the cell from external electrodes, to prevent overdischarge.

Now, the actual voltage on cell (internal) terminals depends on ESR and load current. When you disconnect your load, the cell voltage will instantly jump up by dV = ESR * Iload. The protection circuit usually has some hysteresis, so it won't immediately disengage, and the external battery terminals will be at zero output again.

Then two scenarios can happen, (a) the cell has high self-discharge rate, so the output will remain at zero, until you put the battery into a charger, and (b) the cell might self-recover a little bit, after some time, so the internal no-load voltage become above the upper threshold of the protection circuit, and external terminals will show the voltage (2.7V) again. Loading this discharged battery will result of a very short "working" time, then the voltage will again drop to zero, due to workings of battery protection.

Obviously this behavior will have many timing variants depending on hysteresis levels of the protection circuit, current ESR of the cell, and load current.

You should disregard this marginal behavior at the verge of nearly-discharged cell. Electronics in your device/car should stop (disconnect) using batteries if their (external terminal) voltage drops below 3 V, and re-charge the batteries immediately.

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