Effect of switching current on Li-ion battery packs - Page 1
Forum section: electronics / beginners

author: VEGETA
author_rank: Super Contributor
timestamp: February 07, 2020, 03:51:47 pm
content: Hello,I've nearly completed the design of a linear lab PSU with SEPIC switching pre-regulator which can deliver 30v 3 amps from 2S2P battery pack.The pack should be around 8.4v maximum, say 8v typical... simulation in LTSpice is good.However, the pack is about 5000 mAH or say 4500 mAH but the switching maximum current according to attached image is around 20-30 amps and zero amps.So I wonder if this could be done? looking forward to your answers.Note: green is final output current at 30v... and blue is the ripple switched current from the battery pack (nevermind the negative since it is probed that way).

author: fcb
author_rank: Super Contributor
timestamp: February 07, 2020, 04:29:46 pm
content: It will depend on the inductance of your battery pack and switching frequency.But I would expect significant heating from your pack - can you try reducing the ripple with capacitors on the input of your converter?

author: Rigolon
author_rank: Regular Contributor
timestamp: February 07, 2020, 05:08:29 pm
content: Batteries are really complex if you go deep in how they work, and i'm not an expert but here is my thoughts on the subject.In a 2S2P battery pack you have 4 batteries. 2 pairs connected in series, each pair consists in 2 batteries connected in parallel.So to have 5000mAh 2S2P battery pack, each battery has the capacity of2500mAh. This is important, so keep that number in mind.The capacity of the battery represents the value that would fully discharge a battery if draw constantly in 1 hour. (There are more variables to this but I will not get into that). Let's say a 2500mAh means you can supply 2.5A for an hour and then the battery dies, if you supply double (5A) the battery dies in half an hour.But here is the catch.The thing is that the capacity of the battery it's not a real value for every case. Usually lower consumption means even more time that can be used and higher means less time. You would expect that in a 2500mAh battery that supplies only 250mA you would get 10 hours, but in reality you can get more, lets say 11~12hours. The same applies for higher consumption. If you go 10 times the capacity current (25A) you'd believe that you can supply current for 1/10 of the time (6 minutes). But in reality will supply less, perhaps 3 to 4 minutes or so. All these values are only illustrative as this characteristics changes with the type of batteries and a lot more specs.Although when using high frequency the batteries will show behavior of capacitors that would allow higher current output, I wouldn't think it would be able to output 10 times higher the value specified in the capacity for much longer than in constant DC.Also there is a max current output that over this value will damage the battery, this are all specs to care for.I'm not saying that you wouldn't be able to do what you want. But as I'm not sure that you will, I'm sharing what I know about the subject so you can check a few things before risking blowing up your battery pack.EDIT: my knowledge is more related to lead-acid batteries and they are really different than Li-Ion batteries. As I checked a few information it seems they don't lose much capacity even when discharging with 10C rate (10x 2500mA). And since you are using pulsed I guess you won't have much problems.But on important thing to remember is that when using pulsed discharging the battery allows less cycles (life-time) than rather using constant low DC consumption.

author: VEGETA
author_rank: Super Contributor
timestamp: February 07, 2020, 05:49:10 pm
content: Quote from: fcb on February 07, 2020, 04:29:46 pmHmmm for the absolute state which is 30v 3a total output power would be 90 watts. The design is optimized to reduce heating by using the pre-regulator at 1v difference from linear one which is done using ltspice and all loops are compensated for stability.for the pack it would get 20 amps of pulses but also 0 amps within the same pulse... shouldn't this count? we know that in order to get overheating you need prolonged overload current but this is not the case here.Notice that this is the maximum case, normal uses are way below this. However, I wanted to make sure this works.QuoteYes, but that is the user's responsibility right? i mean he knows he will drain his batteries fast. However, my responsibility is to make it safe and working perfectly without errors.QuoteI put 2 in series and 2 in parallel to make it 5000 mah to allow more capability. Now each pair should get 10 amps in pulses.

author: fcb
author_rank: Super Contributor
timestamp: February 07, 2020, 06:19:50 pm
content: Quote from: VEGETA on February 07, 2020, 05:49:10 pmOK. I think you are missing the point of what I'm trying to say.If you look at your graph, the blue trace is solid - this represents a dynamic signal (what frequency?) - basically you are drawing an AC current from your battery. So what matters is how your battery is able to cope with an AC current! And this will depend on cell construction, chemistry, state of charge etc..If your battery has any inductance at all, then there will be an effect and you can't treat your load totally within the DC domain.With sufficient capacitance (and the right type of capacitors) between the battery and the converter you will 'smooth out' the AC ripple - reducing the AC component you are presenting to the battery and then you are treating the battery more as DC device.

author: VEGETA
author_rank: Super Contributor
timestamp: February 07, 2020, 06:28:24 pm
content: QuoteNo, the current is DC only. I told you the graph is just how I could probe in LTSpice. No AC current. The design is a bit unorthodox which uses a floating regulation with reference to the output... etc.So kindly assume all DC current and rethink about it. Maybe my problem of probing misguided you.

author: fcb
author_rank: Super Contributor
timestamp: February 07, 2020, 06:41:37 pm
content: Quote from: VEGETA on February 07, 2020, 06:28:24 pmYou've edited your original text. But still it doesn't seem clear to me that you are not presenting the battery with a large AC component - but if you say it's DC; then it's DC and will proceed as DC.The answer is 'it depends' - it depends on the type of cells in your battery pack as to whether or not they will stand the high current discharge. I've done tests on many types of Lithium cell in recent times and in general they are all capable of 5-10C discharge with some heating. Cells with built in protection tend to crap out at between 1-2C discharge rate.You'll just have to build the system and try the battery to see how the cell's react - you'd have no problem using some generic battery for a 1C discharge, but at the rates you want - you should defintly test/control what battery your clients use.

author: VEGETA
author_rank: Super Contributor
timestamp: February 07, 2020, 07:04:42 pm
content: Quoteyes it is. It is a floating regulator design with the output positive is "GND" of the battery. However the load gets DC current as well as from battery. However, the direction depends on where and how you probe... thus magnitude is the only important thing.QuoteI will use normal quality batteries assuming they are not protected, since there will be a charging and protection circuit available.However, shouldn't the C rating be about the continuous current? we are talking about instantaneous current which is averaged through inductors (DC-DC converter) to give the continuous maximum current of 3 amps which is totally ok. any battery will be able to give such a current but the point is the instantaneous current pulses as shown.In your tests, did you make 10C continuous current? what is the configuration (single or dual cells? parallel?)?QuoteI should do tests of course. The ultimate solution is to put in another 2 cells in parallel which keeps total voltage at 8.4 and gets more current which should be enough... still using the exact same circuit. This adds cost to those who wanna get or make the circuit but the performance is gonna be a monster. Or better yet, they could choose to put only 2 or 4 batteries according to their uses... with maximum of 6 batteries.

author: fcb
author_rank: Super Contributor
timestamp: February 07, 2020, 08:06:48 pm
content: Quote from: VEGETA on February 07, 2020, 07:04:42 pmsomeone else pls

author: VEGETA
author_rank: Super Contributor
timestamp: February 07, 2020, 08:13:47 pm
content: Quote from: fcb on February 07, 2020, 08:06:48 pmWhy are you saying this? I meant any li-ion battery (18650) can give 3 amps of current... is this weird??about the 20 amps inst. current... yes it goes through inductors to be smoothed to maximum of 3 amps.... is this wrong?please tell me if i made a big mistake.

author: fcb
author_rank: Super Contributor
timestamp: February 08, 2020, 01:02:29 pm
content: OK. Lets try again.Is this statement correct: 3A is your average (RMS) current draw from the battery and this is being drawn from the battery in short pulses (PEAK) of perhaps 30A.

author: fcb
author_rank: Super Contributor
timestamp: February 08, 2020, 01:11:23 pm
content: C is a ratio, and is calculated thus: (dis)charge divided by rated capacity of the battery.So 0.4A discharge current of a battery rated at 4Ah would be called a "0.1C discharge" and should exhaust the battery after 1/0.1C=10hours.Likewise charging a the same battery at 2C would mean charging the same battery at 8A and result in a 30 minute charge time.My tests where on various 18650 cells rated at 2 to 2.8Ah.

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 01:32:54 pm
content: Quote from: fcb on February 07, 2020, 06:19:50 pmExactly. Next thing to do - redo simulation. Introduce real capacitor(s) between battery and converter, use real components with real impedances everywhere including battery andwires.

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 03:53:01 pm
content: Quote from: fcb on February 08, 2020, 01:02:29 pmWhen I set the final voltage at 30v and final current limit to a little bit more than 3 amps (to allow 3 amps) + at 10 ohms of load... I get 30v and 3 amps of output through that 10 ohms load. this is supposed to be the absolute maximum rating ever of this design. Normal uses are way below this.At this statuation... switching current through the pre-regulator taken from the battery is about 20-30 amps of pulses as seen in the picture and attached complete ltspice simulation file.____Quoteso these 2000 mah batteries gave you +10 amps of continuous current? no severe voltage drop or something?QuoteI have attached the simulation so you can check yourself. it is very well done and compensated through help from a certain member in this forum long time ago... I spent sleepless nights to learn and modify it until we made it stable and working.kindly check it, it is slightly modified to use consolidated parts like 40k resistors (in real-life would be 4 of 10k in series) instead of 43.2k which gave 300khz switching frequency while 40k gives slightly more than 300khz. this has no major effect on the design.Also, the design uses not-orthodox type of grounding and referencing since it is a floating regulator... so when you probe, keep holding mouse left button and drag to get the drop voltage needed. You'll know when you see.thanks!

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 04:27:42 pm
content: Quote from: VEGETA on February 08, 2020, 03:53:01 pmYes indeed you have zero impedance battery, ideal (and small) capacitor after battery, no battery wires "modelled", L3 & L4 also are ideal. You would want to update circuit and run simulation again.That "not-orthodox type of grounding" is. What's the problem to make OUT- as simulation ground?

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 04:37:57 pm
content: Quote from: ogden on February 08, 2020, 04:27:42 pmSo what is your suggestion? i posted the spice simulation so you can modify and test your ideas to see if it fits. I am open to these modifications.QuoteThe circuit is indeed not so straightforward, if you are interested and have time to waste then checkthis threadwere this circuit was developed from scratch.The circuit makes the GND as the reference and as Vout+ itself. There is a DC-DC isolated module which acts as an isolated power source to power op-amps, MCU, LCD, and control circuit (DAC,ADC,etc..) which has its negative reference at the Vout+ itself... this is called the floating regulator. I learned this idea from Kleinstein here in this forum as he witnessed it in commercial supplies. That is all the trick in the circuit, the rest is normal.The topology of this circuit is not really the most important here but rather the pulsing current, whether 2s2p pack of 18650 can handle it or not.

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 04:55:42 pm
content: Quote from: VEGETA on February 08, 2020, 04:37:57 pmIt is up to you to find out & introduce impedance of your battery, wires, said capacitors & inductors. Don't ask me.QuoteI am afraid that you confuse real world ground with simulator ground. Those can be two different things. If your simulation ground (triangle symbol) placement actually breaks simulation - you are doing it wrong. Just introduce something like "GND1" for MCU, LCD / whatever (name that floating ground "GND1") but connect simulator ground to OUT- where it shall be. Whether 2s2p pack of 18650 can handle pulsing current you will ask *after* you correctly simulate that pulsing current.

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 05:32:05 pm
content: QuoteOk, I will but I wouldn't mind a hint.QuoteReal world ground (triangle symbol) is exactly where I want it to be. The ground of the design is the output positive itself.QuoteSimulation works and it is not broken. As I told you, the design is exactly as we wanted it to be.QuoteHmm what should be the benefit of changing the terminology if the final result in real world is the same and simulation works?QuoteI fail to understand why is reversing the direction of current or changing the probing method will affect this? the current pulsing is in the design itself as the LT3757A explains and we picked the 0.003R shunt resistor to allow such high current peaks in order to be able to output enough final output current.

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 05:39:38 pm
content: Quote from: VEGETA on February 08, 2020, 05:32:05 pmTriangle symbol is not real world ground in LTspice. It is simulation ground. There is no real world ground as such in LTspice. Decision of which wire is real world ground is up-to you.QuoteThere are two problems in your simulation: #1 use of ideal, non real components #2wronginconvenient for simulationconnection of LTspice ground. Problem #2 has nothing to do with current pulsing.QuoteChanging simulator ground changes ground reference of simulation voltages. Illogical simulation ground connection discouraged me to even press "run" button because I knew "measure" tool have wrong reference ground.

author: fcb
author_rank: Super Contributor
timestamp: February 08, 2020, 05:40:02 pm
content: Running your simulation (as supplied), if you look at the current draw on the power source (battery) at say 4.85ms, you get this:You'll see that the current is being drawn by your converter in pulses. Any inductance in the supply to this (e.g. battery chemistry, construction, leads, etc..) will have an effect on this ability to draw current and likely affect your stability/function of your converter.If I put in a small ESR for the battery (say 0.02ohms) and just look at the ripple current that C13 (which says 4.7uF x2 - but is actually set to 4.7uF not 9.6uF?):Peak currents of around 3.6A in C13 - that needs to be a pretty special 4.7uF capacitor to take that ripple. I would increase it and model is with at least it's ESR included.Here's what happens if you change:C13 to C1210C106K3RAC 10uF from KEMET (Mouser stocked part)L7 (wire), 100 mm of 4mm CSA wire (https://www.eeweb.com/tools/wire-inductance)You'll notice that the ripple current experienced by the battery is much lower. Perhaps adding a bit more inductance (a filter basically) will reduce this further (lower EMI, lower heating due to AC component in battery). I'd probably add a delibrate PI filter, or perhaps add a capacitor across the battery directly to create this filter with any wiring.You'll have to do some calculations with C13 to see if it will take the ripple (clue: Dissipation Factor!)

author: GeorgeOfTheJungle
author_rank: Super Contributor
timestamp: February 08, 2020, 06:12:37 pm
content: Quote from: VEGETA on February 08, 2020, 04:37:57 pmMany 18650s can handle 15..20 A max constant currentnowadays (e.g. LG INR18650HG2), but you've got to look that carefully in the cells' specs:Intended Use(power tools) andMaximum Continuous Current.In general the rule (for 18650s) has always been more current (max A) => less capacity (Ah/Wh), and in this sense 4S is better than 2S2P, and 2S2P better than 4P, because 4S lets you get the same watts with less amps.At those currents you won't get anywhere close to the nominal capacity (Ah/Wh) because the nominal capacity is usually calculated at less than 1C. For example, that INR18650HG2 has 3Ah @0.6A, but I2R losses at 20A areat leasta THOUSAND times more than at 0.6A... ((20/0.6)2)At those currents life expectancy is a disaster.

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 06:33:20 pm
content: Kindly see newer modified version in attachment, now total battery peak currents are about 1 amps of ripple (14.5A - 13.5A) instead of pulses of 20 amps and 0. Nice improvement right?what I did is adding around 50nH of inductance with 0.005R series resistance which is the PCB trace since I will surface mount the regulator very closely to the surface mounted batteries... so it would be even better than this.Combine that with now 3 of 10uF capacitor that you choose to get this result with about 3 amps of these capacitors ripple current current (assuming I could get cheaper ones from LCSC in a big reel with same specs or close). I guess this is what you meant previously right? but your capacitor ripple is lower.What I have mistaken hugely is the switching current at Isense and Rsense resistor of the LT3757A which is required... I assumed that the current drawn from the battery should be similar... but you guided me that it shouldn't.QuoteNow maximum battery pack current is about 15 amps continuous which means each battery gives 7.5 amps, this is much better. There is a very short (~3 ms) of more current at first but this is not continuous.QuoteI thought about this solution and maybe it is better overall? luckily the design allows this flexibility, any configuration works since it is a wide input SEPIC design... However I will need to adjust the charging and protection circuit for the complete circuit according to the final decision... which has nothing to affect this main circuit.Do you think I could pull maximum of 30 volts 3 amps out of say 4S1P cells of ~ 2000-2500 mah? the design allows this assuming the source is capable.

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 07:28:27 pm
content: As you see - capacitors indeed reduce power supply rippleYes, high current Li-Ion cells can sustain currents and discharge rate above 1C. There are even 10C cells/packs. Problem with such - they are more expensive than 1C rate cells, they degrade faster. For example 18650 cell at 4C discharge (2500mAh @ 10A) at the end will be around 50..60oC degree hot. Frequently overheated cells do not live long. (picture from internet search)

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 07:49:14 pm
content: Quote from: ogden on February 08, 2020, 07:28:27 pmWell, at 15A max we got 7.5A per cell which is doable even for Chinese 18650 batteries.. which is my aim. To be able to use this thing with any battery.I made another simulation with 16v input (4 batteries in series) and the total current from battery was around 7 amps with peaks between 6 amps and 7 amps total.So the question is: should I use 2S2P or 4P? since they are using same current per cell (~7A) -> at maximum ever state.

author: BravoV
author_rank: Super Contributor
timestamp: February 08, 2020, 08:08:50 pm
content: Quote from: VEGETA on February 08, 2020, 07:49:14 pmSeriously ? Especially using those "firey" brand 18650 cells like SureFire, MustFire and etc.

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 08:16:45 pm
content: Quote from: BravoV on February 08, 2020, 08:08:50 pmwe don't have these brands that you mentioned but rather other brands that performs good. They are cheap and good, not good as panasonic that is.I am working on implementing protection and charging circuitry to make sure even the unprotected cells can work without problems. Now the performance is another story since it depends on the batteries but my point is that the device should work with any battery. It is the user's fault for getting a cheap battery that lasts for a short time.

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 08:31:47 pm
content: Quote from: VEGETA on February 08, 2020, 07:49:14 pmThis part was already answered:Quote from: GeorgeOfTheJungle on February 08, 2020, 06:12:37 pmFor DC-DC converter also it is better to minimize input current because you minimize losses in the switches, rectifiers and inductors.

author: VEGETA
author_rank: Super Contributor
timestamp: February 08, 2020, 09:17:36 pm
content: QuoteI get it but it is as this:2S2P: max current from battery is ~15 amps but per cell battery is about 7A.4S1P: max current from battery is ~7 amps which is the same as per cell.What you mean is 15 amps of total current is worse since more current means more heat... thus per cell current is not the lone factor?Quotethis makes sense. I wonder if adding bigger caps will enhance the system... but without hindering the speed. I remember once I tried to add an inductor after the battery (not the 50nH wire one) but circuit didn't work.Concluding from all this that 4S is the correct choice here with making sure enough capacitors directly after the battery. Correct?

author: ogden
author_rank: Super Contributor
timestamp: February 08, 2020, 09:34:46 pm
content: Quote from: VEGETA on February 08, 2020, 09:17:36 pmSooner or later you get diminishing returns. You balance caps for ripple_voltage vs cost/size.

author: VEGETA
author_rank: Super Contributor
timestamp: February 09, 2020, 09:20:31 pm
content: Quote from: ogden on February 08, 2020, 09:34:46 pmI've modified the simulation a bit to consolidate some parts and so on... I also made ESR for each passive part as much as I could. Kindly see it.Is this ripple good enough?Also, see the very first +40A overshoot thing... I couldn't fix it even by putting 4.7u inductor.

author: ogden
author_rank: Super Contributor
timestamp: February 09, 2020, 10:08:48 pm
content: Quote from: VEGETA on February 09, 2020, 09:20:31 pmThat is current ripple of battery, right? Where is voltage ripple on input capacitors? You have to look at both.QuoteIgnore it. It is inrush current of input capacitor charge.

author: VEGETA
author_rank: Super Contributor
timestamp: February 10, 2020, 07:04:19 am
content: QuoteYes.QuoteI have attached photos for the voltage ripple, the 2nd photo is when output is stable. voltage ripple is about 0.15V, so is it good?QuoteI would ignore it but isn't it harmful to the batteries and supplies in general? It was oscillating and doing some overshoot then stabilize before adding caps though.

author: ogden
author_rank: Super Contributor
timestamp: February 10, 2020, 07:18:51 am
content: Quote from: VEGETA on February 10, 2020, 07:04:19 amYou mean 0.15V? - Yes, looks good. To me seems like input caps size is right. Why you are refusing to reconnect simulation ground to "OUT-" so you don't have to use math like in this case?QuoteThat "short circuit" is very short. In real world some spark will fly but nothing bad will happen. Obviously do not overdo capacity of input capacitors. The bigger they are, the louder bang will be when you connect battery

author: VEGETA
author_rank: Super Contributor
timestamp: February 10, 2020, 08:15:59 am
content: Quoteyes I modified it to be "v". I guess this is the best it gets.QuoteYes since increasing them will increase the inrush current plus will make it slower.Quotewhat math?This simulation is exactly as the one it should be in real-life, that is how this design is.If I connect ground to OUT-, then I will have to modify a lot of stuff like the isolated DC-DC ground and the ground at the output... since all referenced to it. It looks like negative current and so on but in real-life it is typical behavior. Check the thread where this thing got developed in for more info.Why do you differentiate between real-world ground and simulation ground?You can do this modification as a gift to you brother if you want to xD.

author: fcb
author_rank: Super Contributor
timestamp: February 10, 2020, 08:55:36 am
content: From your earlier description I’m assuming you are permenatly connecting the battery to the converter? So no switch?Back in the 90’s we built high power dc boost converters (250W+), these were connected to large NiMH packs as and when needed using expensive Fischer connectors. As the topology was basic ‘boost’ we needed to (1) protect against high in-inrush current when connected (2) disconnect under low voltage battery conditions and (3) disconnect under output short conditions.We used a big MOSFET in series with the battery and drove it from the microcontroller. It had no measurable impact on converter efficiency and allowed us to gently fill the front end filter capacitors and bring up the converter. We could detect shorted outputs, etc.. easily and protect battery and converter.

author: VEGETA
author_rank: Super Contributor
timestamp: February 10, 2020, 09:08:21 am
content: Quote from: fcb on February 10, 2020, 08:55:36 amYes, batteries are always connected. However, I could add a hardware switch to turn the entire device on or off. What is your recommendation?I assume you mean put the hardware switch after the input capacitors so that the caps are always charged so no inrush current will occur, right?

author: ogden
author_rank: Super Contributor
timestamp: February 10, 2020, 09:38:32 am
content: Quote from: VEGETA on February 10, 2020, 08:15:59 amBecause measure tool is referenced against simulation ground. Seems like you don't even try to comprehend what I am telling. So I give up.QuoteThanx, no. If you can't delete few triangles, name those busses as "VGND" or similar and put single triangle where it shall be - at battery "-" terminal, I am not going to.

author: fcb
author_rank: Super Contributor
timestamp: February 10, 2020, 10:10:50 am
content: Quote from: VEGETA on February 10, 2020, 09:08:21 amNo - you need to keep the path between the input capacitors and the SEPIC as short as possible.In your circumstance I would put a lowRDSon N-channel MOSFET in the negative of the battery and then control the gate with a switch. Probably also have some sort of soft-start (slug the rise time of the gate with a capacitor) - you won't have to slug it much to reduce the in-rush current massively. Probably make sure your controller IC is setup to come on after your input MOSFET is hard on.Your input MOSFET will cost alot less than a physical switch that can cope repeatedly with the in-rush current.

author: VEGETA
author_rank: Super Contributor
timestamp: February 10, 2020, 10:28:50 am
content: QuoteBut won't this make the sepic converter always on? I mean if I put a mosfet at the negative side of the battery, how can it be powered with the microcontroller? in this case, maybe you meant powering the mosfet with a physical switch to its gate connecting it to the battery?The LT3757A has a soft-start pin, going beyond the 10nF capacitor will hinder its working as I tested it in simulation. So if we go with this solution, we might put a 10nf at the gate of the mosfet to make it a bit slower. I guess this is better than using an analog switch.I tried playing around with soft-start pin of the sepic ic but couldn't do much thus i kept it as is (10nF).Let's say we went with this inrush current as is, is it gonna be a problem assuming physical or mosfet switch?Now with 4S pack the voltage is 16v, thus I need to lower it a bit to feed it into the isolated DC-DC converter module. Previously I used a boost to make it from 6-8.4v to 12v but now it should be from say 12-16.8v to 12v using a buck converter assuming 3v is the lowest voltage each cell can reach. If we go to say 2.5-2.7v, pack voltage is gonna be 10-10.8v so it is less than 12v! in this case the buck will output about 9v or even less. I need to solve this the cheapest and simplest way possible.

author: fcb
author_rank: Super Contributor
timestamp: February 10, 2020, 12:34:39 pm
content: Just put your on/off switch in line with R1.

author: VEGETA
author_rank: Super Contributor
timestamp: February 10, 2020, 01:22:36 pm
content: Quote from: fcb on February 10, 2020, 12:34:39 pmThanks for this circuit, I will try to modify it to consolidate parts and make it suitable for 16v input.The rest of my questions needs more investigation to be solved.regards

author: fcb
author_rank: Super Contributor
timestamp: February 10, 2020, 02:43:05 pm
content: Quote from: VEGETA on February 10, 2020, 01:22:36 pmThat circuit will work fine at Vbattery=16V. If you are concerned about exceeding VGSmax (+/-20V) stick a 10V zener across the gate and source.

author: VEGETA
author_rank: Super Contributor
timestamp: February 11, 2020, 09:58:15 am
content: Quote from: fcb on February 10, 2020, 02:43:05 pmI intend to have a 13v zener diode elsewhere (input of DC-DC module), so I will put another one for this circuit. Also, probably getting consolidated parts like 10k or 100k resistors only. putting a small cap on the on-off switch will help in bouncing effect if it happens with a traditional switch (not tactile button).

author: fcb
author_rank: Super Contributor
timestamp: February 11, 2020, 01:41:40 pm
content: Quote from: VEGETA on February 11, 2020, 09:58:15 amYou don't need to debounce the switch - the capacitor across the gate-source will do that for you.

author: VEGETA
author_rank: Super Contributor
timestamp: February 17, 2020, 09:39:17 pm
content: Nothing new here but re-arranged GND stuff to be referenced in a better way, still the circuit is the same.Now GND is the battery negative and I assigned "float GND" to the floating supply circuit.