Yes, that is correct. 'C' in battery nomenclature stands for capacity. In the case of the NCR1865B, 3.4 A/Hr (3400 mA/Hr).
The less the discharge current, the more capacity can be used and the longer the battery life.
spec
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Yes, that is correct. 'C' in battery nomenclature stands for capacity. In the case of the NCR1865B, 3.4 A/Hr (3400 mA/Hr).Is that the same as 6.8A?
2*3.4
Thank you very very much for all the research!
Did not expect this project to grow so much but it is very fun and Im learning a lot from it
The reason to cut off at 3.5v is because I have been told that it is not good for the batteries going lower than that. But if it is ok to go lower like 3.1 to 3.3v or something I'm more than happy. A second protection if the PCB fails seems to be good as well.
Temperature is not a problem. (Well it depends how you see it only having summer a day or two every year). Im only going to use it during summer days or inside.
I don't know if cut off circuit current consumption is imortant? Just thought it would be a good idea to cut everything at a certain voltage.
Approximately At the same time I joined here I found this circuit that should be doing the same thing like mikes. I dont know if they are equal or which one is better? https://3.bp.blogspot.com/-Y1MicUfYujY/UlTqUj6jwlI/AAAAAAAAFYw/ryuPSF4z1Mg/s1600/low+voltage+cutoff+circuit.png
A Lithium battery cell averages 3.7V during a discharge so four make 14.8V. How does an amplifier produce 40W from such a low power supply voltage? Oh, 40 Whats?it is a 40w class d amp
No probs. I always learn a lot from ETO questions and posts too.
I'm afraid posts tend to grow a bit when I am involved.
Sound very good, I will go for what you recommend. Can I go for 3.1*4=12.4vWhile a cut off of 3.5V would certainly be good for battery life, you would be losing an awful lot of battery capacity. The nominal voltage of the NCR18650B is 3.6V rather than the other common 3.7V, so for a good compromise I would recommend discharging down to 3.1V and charging up to 4.1V. It is all a matter of balance; remember you are hitting the battery quite hard with the discharge current of 2 to 3A with a max of 5A. This is around 1C which is not good for battery life, compared to C/10 for discharge and charge which would give optimum practical life.
Is it possible to say how much current mikes circuit consume?The cut off circuit (COC) always consumes current, but then so does the battery itself (you could say). A very good COC would consume 10% of the battery equivalent self discharge current so that, in practical terms, the COC would not affect the battery shelf life. Even when the COC turns off it will normally consume some current.
The self discharge of a LiIon battery is around 1.7% per month at 25 deg C (around 20% per year) (LiIon are the best common rechargeable battery in this respect). For the NCR18650C, this equates to an equivalent discharge current of 82uA, so a good COC would consume no more than 8.2uA.
I''m using s mono class d amp TPA3110 from 9-A Lithium battery cell averages 3.7V during a discharge so four make 14.8V. How does an amplifier produce 40W from such a low power supply voltage? Oh, 40 Whats?
Do your amplifiers use the Tripath TA2024 class-D amplifier IC? Tripath went bankrupt a number of years ago and a Chinese company copies their IC lately. Its maximum supply voltage is recommended to be 13.2V which is exceeded by your 16.8V fully charged battery. It produces only 5.6W into 8 ohms or 9.6W into 4 ohms at clipping per channel with a 12V supply. The datasheet does not show any other supply voltage.
That's good to hear.For me it is only interesting so that is absolutely not problem
Yes, no problem. We can even give you a simple formula so you can work out the resistor values for yourself.Sound very good, I will go for what you recommend. Can I go for 3.1*4=12.4v
With same setup that mike gave me?
I can't answer for Mike, but going by his many posts on ETO, he is always pleased to help.I feel bad about asking him change the resistors to different values if I want to go for something between 3.1-3.4.
That would be possibleWould be great if I could just trim it with the pot with same setup.
Yes, that can easily be calculated.Is it possible to say how much current Mike's circuit consumes?
Sounds great!
Then I only need to know a few things before buying components.
1. Formula how to calculate for that resistor (other things) depending on cut off. Would like to try between 3.1-3.5. Or for different packs in future.
2. Im not sure about "M1 that has a low Ron". A specific link would be helpful to one I can buy.
Would be great if it works on different setups?
3. Circuit consume is interesting
Thank you very much!
... Can I go for 3.1*4=12.4v
With same setup that mike gave me?
I feel bad about asking him change the resistors to different values if I want to go for something between 3.1-3.4. Would be great if I could just trim it with the pot with same setup.
Is it possible to say how much current mikes circuit consume?
...
Looking forward to see what you are working on. I will build both of themOK. I am still working on the circuit that I mentioned previously. If it turns out OK I will post that for your consideration, in addition to Mike's circuit, quite soon. If the circuit does not work out I will also tell you. I would not be surprised if other members didn't post circuits too.
I will let you know what bits to order for my circuit and perhaps Mike will advise you on the questions about his circuit.
spec
Thank you Mike! Don't think I understand how it was calculated but appreciate helping me.Make R8=42.2K (nearest 1% value). The trim pot makes it adjustable from ~12V to ~13V.
After it disconnects the load, the circuit draws ~170uA from the batteries. While the load is still connected, the battery current is the load current +~170uA.
Put the ON-OFF switch between the batteries and my circuit.
The datasheet from Texas Instruments for the TPA3110 amplifier IC shows that the output is 40 Whats at very high distortion into 4 ohms when it is overheating, has both of its bridged channels in parallel and uses an 18VDC power supply. With your 14.8V average supply it produces about 22 Watts with fairly low distortion. It will be hot but not overheating. Will you use two of these amplifiers for stereo?
I was going to add: you really need some kind of latching comparator circuit or it will just keep chattering on and off since the battery voltage will jump up when the load is removed.I have just been having a look at the NCR18650B datasheet. I don't like to be a kill-joy, but it appears that you would be lucky to get 500 mA/H from your batteries with a cutoff of 3.5V and that would be at a temperature of 25 deg C.
Also, I suspect that having a cut off voltage just 100mV below the battery nominal of 3.6V is going to cause all sorts of problems. For example when you consider the drop in terminal voltage due to the load current and the battery internal resistance, especially when switching on in a cold environment.
Not only that, but the terminal voltage drops with a decrease in temperature, so on a cold day the circuit may never turn on. I believe it gets a bit cold in Sweden from time to time.Battery rebound also makes the situation more critical. To further compound the problems, the datasheet applies to brand new batteries: with use, battery characteristics deteriorate.
spec
NCR1865B datasheet
**broken link removed**
The battery maker should know this data but it depends on load current.Thank you very very much for all the research!
The reason to cut off at 3.5v is because I have been told that it is not good for the batteries going lower than that. But if it is ok to go lower like 3.1-3.3v or something Im more than happy. A second protection if the PCB fails seems to be good as well.
I was going to add: you really need some kind of latching comparator circuit or it will just keep chattering on and off since the battery voltage will jump up when the load is removed....
A problem with the Chinese board are that they cannot explain in English if it is a charger or if it is a protection circuit because they say both.Does this look correct connecting batteries, pcb, charger, lvc and load?
Thank you! I will spend some time try understanding this as well.Here's one I used on my camper so I wouldn't run the battery too low to start the truck, it worked real good, except for the fact that I put the reset button in a very awkward place.
Because some batteries regenerate their voltage once the load is removed, I made this so it locks itself out until reset or bypassed, either one will reset it if the voltage is above cut off, bypass has to carry the full current of the load, reset button has minimal current.
I show several switches, but that is get different switching times for simulation, only one is needed for resetting the circuit once the battery is recharged.
Jeff
View attachment 99295
Wow, Impressive and easy to read. I´m overwhelmed by all help I´m having here. I really hope this thread will be an asset for other rookies like me.Here is another LiIon voltage cut off circuit (I haven't optimized the resistor values yet):
NOTES
(1) N1 is a low current (1uA) comparator with a CMOS push pull output
(2) ZD1 is a 2.5V low current (1uA) precision (+-0.2%) Zener diode
(3) R5/R4 provides hysteresis to take care of battery rebound.
(4) RV1 sets the cut off voltage to 3.1V per cell and can be adjusted to give +-400mV variation in cut off voltage
(5) The circuit layout and routing of lines is important and should be as shown on the schematic to maintain accuracy and reduce the likelihood of frequency instability.
(6) The capacitors are for decoupling to assist frequency stability and have no function in regard to the basic circuit operation. They should be physically as shown on the schematic.
(7) R1 is a gate stopper to prevent the PMOSFET from oscillating at a high frequency (not peculiar to this circuit) and should be mounted on the PMOSFET gate pin. It plays no part in the basic circuit function.
PARTS LIST
To Be Defined
A problem with the Chinese board are that they cannot explain in English if it is a charger or if it is a protection circuit because they say both.
Other problems are that its overvoltage limit is too high, its undervoltage limit is too low and its current is way too high. Therefore it will provide NO protection for your little battery.
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