Noob batteries in series question

Question

I was wondering, if I wire for example 2 2700ma batteries in series to get roughly 8.4 v cell fully charged but use a step down convertor to bring the voltage down to 5v do I gain back any of the capacity between the two cells? Ideally I would want to wire in parallel for the project but unfortunately I need to draw up to 3.7ish V which would become impossible as the batteries discharge. I'm new here so please go easy on me I'm trying to do as much reading as I can but would appreciate the help.

gecoz · Accepted Answer

Hi Steven, The short answer to your question is no, with the batteries in series the max capacity you can aim for is the original capacity of the single battery, if they are identical, otherwise the capacity of the resulting battery would be equal to the lowest capacity. Generally speaking, if you use higher voltage to power up your load you end up drawing more current from the battery, hence making it discharge faster. If you imagine the device you are powering up with the battery as a simple resistor (approximation), and think about Ohm's law ( V = R * I ), you can understand why this happens: your load would stay constant, so if you increase the voltage (V), the current draw increases too, so that R remain constant. If you are planning to use the battery in parallel instead, to increase the capacity, just make sure the 2 batteries are identical (or as identical as they can be), otherwise they won't be balanced, which will lead to one of them eventually failing. I hope it helps, Fabio

Jan Cumps · Answer

yes. at least with the buck converter way less energy will be wasted as heat compared to the linear FET. With a buck-boost converter, Steven would also be able to get some more energy out of the cell when the combined battery voltage drops below the required voltage.

gecoz · Answer

Hi Steven, If I understand your question correctly, you are asking why, if the nominal voltage of the battery if 4.2V, I read a voltage of 3-3.7V once the battery is connected to the circuit. The battery nominal voltage is defined as the voltage read at the terminals when the battery is disconnected, i.e. with no load attached to the battery. Ideally, the perfect battery could provide any amount of current, keeping its nominal voltage the same, i.e. the battery would have no internal resistance. The reality is quite different: batteries do have some internal resistance, and you can see its effects only when you connect the battery to a circuit. Once attached to a load, the current drawn from the battery causes a voltage drop on the internal resistance, and the result is a lower voltage read at the battery terminals, as the sum of the voltage drop on the internal resistance and the voltage drop on the load must still be equal to the nominal voltage of the battery. I hope I'm not making things more confusing with my explanation. Fabio

michaelkellett · Answer

If I understand your circuit correctly (a schematic would help !) you have a battery, a mosfet (controlled we don't know how) and a load. if the mosfet is operating in a linear way it drops the voltage between the battery and the load, and the wasted power (current x drop across mosfet) comes out in heat in the mosfet. If you use a switching controller (Google 'buck regulator') it acts differently and the current in the load will be greater than the current drawn from the battery but the voltage across the load will be less than that across the battery. The power into the load will always be less than the power drawn from the battery because no controller can ever be 100% efficient. MK

michaelkellett · Answer

We really need to see a schematic - I think what you have described is is PWM drive of a mainly inductive load, which will work rather like a buck regulator and can be efficient. There should be a diode somewhere (to help with efficiency and keep the MOSFET safe) - without a schematic I can't say where, and I need more information about the coil - part number, data sheet whatever. If you coil is really a 0.24 ohm load the current with 3.7V across it would be 15.4A - seems a lot for batteries. What does the coil do - in many cases it isn't necessary to put anything like as much power into a solenoid to keep it energised as there is to get it to pull in at first. MK

moderategamer · Answer

Here's a rudimentary diagram it's missing a few items but nothing that is important for demonstration purposes

moderategamer · Answer

No this is a one off making it as a present for my dad who is most likely never going to use it. I get the best batteries on the market all individually tested. I think it's entirely possible to do it with an Arduino safely and 75 watts is just the limit I'm setting actuall use will be 45 - 50 watts at short draw of around 2 seconds I have many high quality commercial mods that can draw 200 watts safely so I think 45 -50 is safe. Also 18650's can give out high amp loads well. I think there's a point where you can be overly safe and think mine will be 100x safer than the unregulated mods which are the ones that cause accidents. The actuall power delivery is on a closed circuit independant of the Arduino due to the mosfet so there's no real chance of any hardware fault on the Arduino side and with the correct resistors and fuse (which btw no commercial mods use) the circuit can be very safe. I understand if you don't feel comfortable helping though that said I would appreciate the help, I mean I'm going to do it anyway may as well get some sound advice xD If you do feel like advising me would love to hear your thoughts and knowledge about the buck convertor.

michaelkellett · Answer

For a heater load a buck converter has limited advantage, one needs to know a lot more about the load to decide if its worthwhile. Do you know the resistance of the heater cold and at working temperature (there is usually quite a difference). MK

michaelkellett · Answer

You can attach indirectly by adding a link, I use a free Dropbox account for such things. I'm interested in Shabaz's info because it shows that it is expected that battery pack designs would need two faults to fail, and the system would then need three faults to a fire. The only design I have with a biggish (2.6AH) lithium battery uses a protected battery and poly fuse protection against shorts. I used a complete charge/load management chip. In the vaping application I don't think you could use protected batteries because of the very high currents. MK

michaelkellett · Answer

At 15A the MOSFET will dissipate 15 * 15 * .047 = 10.6W

At 20A it will be 18.8W - it will get hot - at 20A about 25% of the power is going into the MOSFET - so it's not a good choice.

From the graph that fuse will take 1 minute to blow at a current of 50A.

From the Sony VTC5A battery data sheet:

If the over 35A discharge occur, the allowable time of operating over current protection

comply with the below table.

35Aを超える放電が発生する場合、過電流保護作動までの許容時間は下記表に従うこと

Discharge current

放電電流

35～40A

～60A

～80A

～100A

Time

時間

<78sec.

<26sec.

<14sec.

<6.7sec

Interpolating it looks as if the fuse will not save you if the current is around 50A - that fuse is no good for this application.

To get a current of 50A you would need the fault load on the battery to be about 0,064R (including its own internal resistance) - this is a figure that is quite likely to happen unless you very carefully design against it.

With regard to the TI regulator - do you have the experience to design a board to implement that circuit - it is a seriously non-trivial thing to do.

I hate to say it but I think you are way out of your depth here - if I were asked to design this thing and put my name on it being safe, I'd expect at least three board spins and a the job the take several months.

For example I would probably go for three levels of fusing - electronic current limiting/power monitoring (possibly dual system with fault tolerance), current sensitive fuse and a thermal fuse.

MK

gecoz · Answer

To try and grab this concept, lets think the mosfet just as a "black-box" converter, placed between the battery and the load (the battery is connected the converter input terminals, and the load is connected to the converter's output terminals). In your case, your converter should transform your voltage from let's say 6V down to 3V. If your transformer was ideal, this conversion would happen with no losses, i.e. the power (P = V * I) transferred from the converter to the load would be equal to the power transferred from the battery to the converter (no losses). Again, in the real world, there is no such thing as no losses, and the converters have a defined efficiency (i.e. the fraction/percentage of power effectively transferred from the input to the output of the converter), as Michael pointed out. Therefore, using higher voltages leads to more power wasted because of the losses. But I think what you are finding more confusing is the fact that battery capacity and run time are not the same thing. The capacity of a battery is a property of the battery, due to the physics of how it is built (the chemical reactions involved, etc...), and it is independent from the way the battery is used. The run rime of a battery depends entirely on the way it is used: if you have 2 identical batteries, and you connect them to 2 circuits that draw one double the amount of the current than the other, the run time of the battery attached to the higher-drawing circuit will last half the time compared to the other. Again, I'm trying to simplify things hoping that this doesn't confuse you more, as I know some concepts are hard to grasp when you are moving your first steps in the world of electronics. Fabio

moderategamer · Answer

I'm not able to measure the resistance at temperature but the maths say at 500 degrees Celsius the resistance would be roughly 0.58344 ohms which would bring the amp draw down to about 6-7 amperes. Cold temp is the first number I quoted at 0.24 ohms. Take these numbers with a grain of salt though as they're calculated with googled coefficients for stainless steel. sorry I made a mistake there got my Fahrenheit and Celsius mixed up at 500 degrees Fahrenheit resitance would be 0.38904 ohms and draw 9.95 (2dp) amperes.

gecoz · Answer

So, what you are trying to build is basically a PWM-controlled MOSFET driven heater, whose heating element is powered by a 2.7Ah battery. Have you already worked out the duty cycle for heating up the coil? This will determine how long one charge will last for but, more importantly, will determine if the battery is working within safe operating margins. This leads to my second question: do you know the maximum continuous discharge current for your battery? For example, a Sony 2.2Ah High Discharge Rate battery is only safe up to 10A max continuous current. Drawing 15A, if done continuously, will definitely put your battery under huge stress, with risks of over-heating. Assuming your coil resistance will vary between 0.24 ohm and 0.39 ohm, you definitely will also have to take into account the battery internal resistance in your calculations. For a good quality brand new battery, the internal resistance will be around 0.03 ohm (will increase as battery ages). Fabio

Noob batteries in series question

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