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Hello Guys!!
I play 1:10 R/C. In my chassis I have a pack of SUB-C batteries plug onto the electronics as Power Supply. There are several Cell Tray Discharger or Batteries Equalizer in the market but the price is kinda expensive for me. I looked over my friend's batteries and I found that in every side-by-side cell he has, he soldered a resistor and a diode, not sure what it is but I would like to know also how this thing works! Several facts regarding SUB-C type Ni-MH batteries are: they have what so called natural discharging, but it takes more than weeks and my friend told me that his packs are discharing themselves automatically by the resistive load of the resistor, and it stops (cut off) around 0.5V per cell. He used this method like 10 years and one of his pack is 12 years old, peak charge around 2200mAh but his cell is actually 2000mAh. What a trick!!! He had this method and his cells are damn healthy. I believe several ppl have their own opinion regarding this design. Please do comment if you like, thanks! This circuit should work like this: 1. Batteries work like normal when plugged and used. 2. When it is not used, resistor will discharge it slowly. Cut off determined by a diode. 3. When it reaches the cut off voltage, each cell stops discharging. 4. Batteries are ready to be charged! Anyone knows what type of diode he used and how it is connected? does he use any Zener? Care to comment? Thanks!
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Experienced Member
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I would guess he uses something like a 1N4148 and a series resistor wired across each cell. A normal diode will stop conducting around 0.5V at 25°C. The discharge current will vary as the cell voltage varies. At full charge it will be around (1.2-0.7)/R where R is the series resistor. So, if you want to initially discharge at 10mA you would need a resistor of (1.2-0.7)/0.010 = 50Ω.
Mike. |
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Experienced Member
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![]() Take a look, please do comment my schematics. These resistors are used to determine the discharging current. Hmm.. anyone knows the calculation of discharging? How about a 3300mAh Ni-MH sub C packs? How long does it takes to discharge it with 10mA ??? Is it 3300 : 10 = 330 Hours?
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Experienced Member
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All the resistors should have the same value. To discharge a 3300mAh pack at 10mA will take 330 hours but, I assume you discharge the pack when you use it and you just want to get rid of any remaining charge. Why don't you ask your friend what value resistors he uses.
Mike. |
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Experienced Member
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@Pommie: Well, he is not around, so I can't help myself to figure it out but this LTSpice program tells me nice numbers of mA discharging current. LOL. So now I'll take 50 Ohms for each cells.
I think the disadvantage of this circuit is only it keeps discharging no matter what, so playing time will be less, but it is for the performance sake of the batteries, it is wise... hmmm!
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Btw I have another question, perhaps u would like to answer it.. I saw several discharger tray or batts conditioner. They have like 20A discharging rate. Hmm.. don't you think it is too high? How about putting only 1 Ohm of resistor value. It shows me here: (with LTSPICE simulation) 235mA current thru the Diode and Resistor.
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I thought that the idea of these diode/resistor circuits was to balance the cells and the discharging was done by your R/C car. Battery conditioners are different, they cycle the batteries by charging and discharging them repeatedly.
For the resistors, I would start at around 25Ω (20mA). My logic for choosing this value is that the cells should be discharged (after use) to around 10% of their capacity and 20mA will mean they will be balanced the next day. BTW, in your simulation, gradually reduce the cell voltage and see at what voltage the current becomes negligible. It should be around 0.5V. Mike. |
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Moderator
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For my Cell discharger, I use a 10ohm Sand resistor (high wattage) at 10W and it gets slightly warm. It may be a bit too low, but for my 2500NIMH cells it serves the purpose.
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