Single NiMH over-discharge protection! HELP NEEDED - Page 1
Forum section: electronics / projects

author: soubitos
author_rank: Frequent Contributor
timestamp: June 19, 2018, 08:59:54 pm
content: Does anybody have any ideas on how to protect a NiMH a single NiMH cell from over discharging???I would be happy if i could "disconnect" load after cell drops to around 1V (+.1-0.2V) but i have found no way to do that ...

author: Gyro
author_rank: Super Contributor
timestamp: June 19, 2018, 09:12:37 pm
content: There's no way to over-discharge a single NiMH (or NiCd) cell, they can safely be taken down to and stored at 0V.Over-discharge only occurs in series multi-cell batteries, where the weakest cell gets reverse charged by the stronger ones when the battery runs low. It's reverse charging is what they can't tolerate.

author: splin
author_rank: Frequent Contributor
timestamp: June 19, 2018, 11:37:06 pm
content: Quote from: Gyro on June 19, 2018, 09:12:37 pmCycle life of NiMh cells suffers if you deep discharge them below 1V so disconnecting at 1V is probably a good idea, providing the protection circuitry doesn't draw significantly more than the self-discharge of the cell.There are several possible solutions but more information is needed - continuously powering a very low power device like a clock is a very different problem to a power tool which has high current drain but intermittent and possily infrequent use. Here are a few questions for starters:1) How much current does your application draw?2) How much current can you allow the protection circuit to consume?3) What is the duty cycle of the load - mostly on, mostly off or unpredictable?4) Is it cost sensitive?5) How many cells do you need to monitor/protect? Is it a single cell or a number of cells in series which you want to monitor indiviually, disconnecting the load if any cell drops below 1V?6) Is it space/volume/weight sensitive?7) Is the cell manually switched into the circuit or can it be switched on by some electronic signal?8 ) Does the application use a microcontroller that you can use for the monitoring and protection?9) What voltages are available in the circuit other than the single cell's 1V minimum?10) Do you have any electronics design experience or are you looking for a ready made solution?11) What size/capacity cells are you using?12) What temperature range will it experience?13) Is this a one off, or is to be put into production?A typical solution is to use a transistor, usually a mosfet which can have very low on resistance, requires almost no power to maintain in the on state and has almost no leakage current in the off state. If you only have 1V available though you might not be able to find a suitable Mosfet with a sufficiently low gate threshold voltage in whch a high gain bipolar might be suitable.Another possibility is to use a relay but a) you might not be able to find one that operates at 1V and b) will use a relatively large amount of power. You could avoid the latter by using a latching relay but the control circuit may be a bit more complex and you will need to have sufficient energy available to switch it. That could be stored in a capacitor so you are not reliant on the battery being able to provide it when almost exhausted. Latching relays probably wouldn't be suitable if the application may be subject to mechanical shock, vibration or strong magnetic fields which could flip the relay state.You then need a control circuit which drives the switch and monitors the voltage without using too much power. I don't know of any off the shelf device but a typical circuit may use 2 or 3 CMOS invertors as a latch to drive the switch. 74AXP series gates are very low power and specified to operate down to .8V; there are probably others.A comparator circuit and voltage reference to change the latch state is also needed and can be more of a problem if you need very low power (uAs.) operation or low cost. You could use a low power CMOS timer circuit that powers up the voltage monitor periodically to reduce power consumption. If you have a microcontroller with an ADC you use that to monitor the voltage and drive to latch circuit.If 1V minimum is too restrictive you could use a joule thief or charge pump to boost the voltage - several manufacturers make suitable low power, low input voltage devices intended for energy harvesting. You might even find one with a suitable voltage reference and comparator function.A more radical idea is to use a fuse to supply the circuit and short circuit the battery+fuse to blow the fuse, assuming there is enough energy to blow the fuse. A circuit breaker would make it reusable but perhaps a bit bulky.

author: Audioguru
author_rank: Super Contributor
timestamp: June 20, 2018, 02:05:02 am
content: In their Ni-MH battery manual, Energizer Battery Company recommends storing a Ni-MH battery fully charged with no load.

author: Gyro
author_rank: Super Contributor
timestamp: June 20, 2018, 09:57:50 am
content: Again, the OP is talking about a single cell. The minimum ~1V recommendations are to prevent the danger of cell reversal in multi-cell batteries. Single high capacity NiCd cells for aircraft and similar applications used to come from the manufacturer with shorting springs fitted. I doubt that NiMH is any different. Cell reversal is what kills them.Edit: From the Energizer NiMH Handbook and applications manual (attached)...

author: capt bullshot
author_rank: Super Contributor
timestamp: June 20, 2018, 10:37:02 am
content: Quote from: Gyro on June 20, 2018, 09:57:50 amYou can do that with NiCd cells of all flavours, that's indeed the best way to store them for long term.The usual (home and office use) NiMH (AA) cells don't like deep discharge, they tend to react with reduced capacity and / or high internal resistance and get unusable if you store them for longer time discharged.

author: capt bullshot
author_rank: Super Contributor
timestamp: June 20, 2018, 10:41:53 am
content: Quote from: soubitos on June 19, 2018, 08:59:54 pmDepends on your application. I might give it a try with a low voltage (3V or 5V coil) relay. Very simple circuit:Connect a N/O contact in series to the cell, the coil on the load side. You'll have to energize the relay by some other means (lab power supply), once it is energized, the coil gets supplied by the cell. Such small relays tend to release at maybe 20% of their nominal voltage, so if the cell gets discharged, the relay will open and disconnect the load. You should test some relays with a lab supply for its release voltage, one should be able to find one with a release voltage around 1V.

author: Gyro
author_rank: Super Contributor
timestamp: June 20, 2018, 10:58:08 am
content: Quote from: capt bullshot on June 20, 2018, 10:37:02 am@capt bullshot:I added a clip from the Energizer manual [EDIT: before] seeing your reply. I found a reference later in the manual to "creep leakage" past the seals when stored under load at 0V for prolonged periods, so it is clearly undesirable from that perspective, I stand corrected. I guess it depends whether the condition will persist for a significant period in the OP's setup.

author: soubitos
author_rank: Frequent Contributor
timestamp: June 20, 2018, 08:26:24 pm
content: I will try to answer here as it covers most if not all aspects and questions asked.Quote from: splin on June 19, 2018, 11:37:06 pmThe "load' is in fact a joule thief with minimal current drawn, I would say 5mA at most under certain conditions and changes to its output.Quote from: splin on June 19, 2018, 11:37:06 pmIt would be best to be as low as possible, certainly not in the range of 10's of mA ideally under 5mAQuote from: splin on June 19, 2018, 11:37:06 pmIts a joule thief which most of the time likes to switch at around 100Khz but it varies with voltage etc so i'd say unpredictable LOLQuote from: splin on June 19, 2018, 11:37:06 pmVERY!!!!! I am trying to find something as simple and cheap as a passibes around an under 10c ic/mosfetQuote from: splin on June 19, 2018, 11:37:06 pmThis is about a single cell application.. i could make it 2S if absolutely needed but i'd have to at least reduce them from AA to AAA or even 2/3AAA as this has to be as small and compact as possible too.Quote from: splin on June 19, 2018, 11:37:06 pmsensitive no.. but the smaller the betterQuote from: splin on June 19, 2018, 11:37:06 pmThe cell is on a charger IC but the load is always present.Quote from: splin on June 19, 2018, 11:37:06 pmNo.. even if it did, it would have to operate with a singe NiMH cell and i am not aware of any and even if i did, i am not able to program them (yes, i have no idea really!)Quote from: splin on June 19, 2018, 11:37:06 pm5V from a microUSB when charging, not always presentQuote from: splin on June 19, 2018, 11:37:06 pmI can follow guidelines, read datasheet and think i can design simple stuff (hence why i deal with simple chargers to begin with LOL)Quote from: splin on June 19, 2018, 11:37:06 pmLike mentioned, max size 1XAA capacity could be any available, i'd say in the range of 800-900mAh anything else would be lower than that and smaller size esp if the only way to offer some sort of protection to the cells is to go 2SQuote from: splin on June 19, 2018, 11:37:06 pmThe charger offers over-temp protection for the cells while charging but i would leave that one open as it could be used outdoors with -10 to +40 max range (there is no demand to work at this range as soon as it will not get damaged!)Quote from: splin on June 19, 2018, 11:37:06 pmI wouldnt dare to say production but i like to make 10pcbs for the prototypes and test the market with them if they work without changes needed. Then i might make 20-30-50pcs...Quote from: splin on June 19, 2018, 11:37:06 pmI have tried with a P channel (AO3401) with Gate to negative Source to positive and Drain as the positive terminal of the "pack" ... i found it searching around for some clue but it didnt work... i tried with a power supply, not battery to adjust voltages and the strange thing was that once voltage dropped under 0.5V i got negative readings on the DMM!!!!! I wouldnt say it works but if i have to search for a suitable mosfet, what do i look for ?Quote from: splin on June 19, 2018, 11:37:06 pmA relay of any kind would be out of the question for cost and size alone!Quote from: splin on June 19, 2018, 11:37:06 pmI haven't though of using the JT output to power a protection circuit to be honest.... that would be interesting!!I regulate the voltage to 5.1V at some point, that would be more than enough as soon as total current to operate the protection circuit is VERY low.. now that i think of it, i might have a few voltage monitoring ICs of very low cost and size (SOT23) that could possibly be used.. HMMMMMMMMMMMQuote from: splin on June 19, 2018, 11:37:06 pmI dont know what you have in mind here but... HMMMM use the JT output... yes, that MIGHT workTHANK YOU and thanks to all who took the time to post here, i really appreciate it!

author: IanB
author_rank: Super Contributor
timestamp: June 20, 2018, 09:01:32 pm
content: As others have said, you don't really need to protect a single cell. Taking it below 1.0 V will not harm it very much.If the voltage gets too low the JT will stop working, at which point you can presumably disconnect the battery? Only if it is unattended for long periods might you start to worry.

author: soubitos
author_rank: Frequent Contributor
timestamp: June 20, 2018, 09:12:49 pm
content: Quote from: IanB on June 20, 2018, 09:01:32 pmJT will take it under 0.6V if left working. I would like to protect the cell from premature aging and loss of capacity.My original choice was a chinese JT IC with NiMH "charger" or a "proper" charger taking care of the cell allowing for longer service life. So it there IS a low cost solution, i would like it in..

author: soubitos
author_rank: Frequent Contributor
timestamp: June 21, 2018, 09:06:51 pm
content: This is the solution i came up with and hope it works.I use CN305 which is a "Low power window detector"http://www.consonance-elec.com/pdf/datasheet/DSE-CN305.pdfWhat is happening isWhen an empty battery (under 0.8V) is connected to the circuit and no external (microUSB) power is present, CN305 and Mosfet are both OFF and nothing is working. To activate the Joule thief you need to apply power to the charger. At this point power reaches CN305 directly from microUSB via a diode (used shottky instead of 1N4004 seen under the arrow pointing right), CN305 is sensing battery voltage and if under 0.9V, its UV output is LOW mosfet stays OFF.When battery voltage raises above 0.9V, UV output goes HIGH, mosfet turns ON and powers the Joule Thief at the next stage.Joule Thief (JT) has a rectifier diode at its output plus a 3.3V zener diode and a 10uF capacitor to smoothen voltage out. CN305 consumes only a few uA and the total JT load should not exceed 10-20mA at any given point.Now, lets assume battery is full and external power disconnected. CN305 is now powered from the JT "regulated" output and as long as the battery is above 0.9V its UV output is HIGH.. once battery drops bellow 0.9V, UV goes LOW, mosfet goes OFF and since CN305 has open drain outputs, mosfet will stay OFF "biased" by the small 12K resistor at its Gate... To activate again, you need to recharge the battery...PS this is taking very little pcb space and total cost of parts is under 0.10$ i think (say 0.20$ including shipping/taxes etc hehehehe)I hope my english is good enough to explain how i think this will work, does anybody feel different?

author: IanB
author_rank: Super Contributor
timestamp: June 21, 2018, 11:05:27 pm
content: NiMH cells are funny things. If you disconnect a load from them their terminal voltage will always try to increase to 1.2 V. It is very, very difficult to discharge them enough so that their voltage will go below 1 V and stay there. Try it yourself. Discharge a cell with a resistor of, say, 100 Ω until the voltage goes below 0.6 V. Then disconnect the resistor and watch what happens. Most likely the cell voltage will climb back up close to 1.2 V.So any battery cut-off circuit you use will need logic to keep the power switched off until the battery voltage is higher than 1.2 V or so.

author: soubitos
author_rank: Frequent Contributor
timestamp: June 21, 2018, 11:13:43 pm
content: The load is minimal 10-20mA max i think so the battery will not dive noise down from high currents etc where voltage drop is expected even on a fully charged battery.. the idea of this circuit is to disconnect the load when under 0.9V and trust me, a JT if made properly and left alone can bring the battery down to 0.5-0.6V easily... From experience, when used with GOOD primary batteries they almost always go as low.. of course they bounce back like any battery i think.... I am not experienced with NiMH i admit it but the task is to NOT let the battery drop bellow 0.8-0.9V tops.. if it does it will disconnect and if i am correct it will not reconnect the battery even if it bounces back up unless you connect external power (to charge the battery)

author: IanB
author_rank: Super Contributor
timestamp: June 22, 2018, 12:01:40 am
content: I just recommend you do the test in order to find out what happens. Take a good NiMH cell like an Eneloop and discharge it down to 0.5 V. For 10-20 mA you can use say a 50 Ω resistor, or use your actual Joule thief. When the voltage gets down below 0.6 V disconnect the load and leave the battery to rest for a few hours. Watch what happens to the open circuit voltage over that time.

author: soubitos
author_rank: Frequent Contributor
timestamp: June 22, 2018, 08:45:37 am
content: That is exactly my problem.. i dont want to let the NiMH deeply discharge.. i want to shut it off when it reaches 0.8-0.9V and let it disconnected until the charger is powered up again. Charger will have a max charge current of 200mA a second stage timer of 3hrs and NTC control on the cells and the batteries will be 600-850mAh ...The question is.. will this way of disconnecting the cell work?I have a pcb ready to order for this but i'd like a second or third opinion!

author: IanB
author_rank: Super Contributor
timestamp: June 22, 2018, 03:34:21 pm
content: Quote from: soubitos on June 22, 2018, 08:45:37 amOpinions won't help you. Opinions are just that: opinions. What you need are facts!So again, I tell you, you need to do your own testing with the cells you plan to use. If you want to stop the discharge at 0.9 V that's fine. Do a test discharge down to 0.9 V, disconnect the load, and watch the cell voltage to see what happens. I predict the voltage will rebound back to 1.2 V. But don't take my word for it, test it.The thing is that NiMH cells really, really want to be at 1.2 V. You won't get the voltage to rest below 1.2 V unless you severely abuse the cell, or unless the cell is damaged.

author: soubitos
author_rank: Frequent Contributor
timestamp: June 22, 2018, 03:46:26 pm
content: My friend, I KNOW they will rebounce.. the idea is that without powering the CN305 again, it will keep the mosfet and JT off waiting for next recharge.....

author: splin
author_rank: Frequent Contributor
timestamp: June 23, 2018, 12:32:37 am
content: Quote from: soubitos on June 21, 2018, 09:06:51 pmYour English is excellent but there are a couple of problems with your solution.1) The UV/ output will go active, ie. will switch to 0V when the UV sense input drops below the threshold which will turn on your MOSFET. That's easily fixed by using the OV/ detector instead2) The threshold voltage for both the UV and OV detectors are between 1.09V and 1.15V (for a falling input voltage which is what you are looking for). That is a bit high for a NiMh, especially at low temperatures when the voltage drops, and would leave quite a lot of capacity unused but might be OK for you; however it means the battery must be connected directly to the OVIN pin, not via a resistive divider as shown in your circuit.A better alternative for your application may be to use a boost converter which will be quite a lot more efficient than an unregulated JT and are really cheap. There are a lot of cheap Chinese parts but you want one with a shutdown/chip enable (CE)/Enable (EN) pin and low current consumption, particularly when turned off.The NCP1402 may be suitable at $.09 each. It works from .8V, consumes < 1uA when off and has a chip enable:https://www.aliexpress.com/item/100pcs-lot-NCP1402SN33T1G-SOT23-5-NCP1402SN33-SOT23-NCP1402/32848848958.html?spm=2114.search0104.3.147.7925157fr3LOwq&ws_ab_test=searchweb0_0,searchweb201602_3_10152_10151_10065_10344_10068_10130_5722815_10324_10342_10547_10325_10343_10546_10340_5722915_10548_10341_5722615_10696_10190_10084_10083_10618_10307_10820_10303_5722715_10059_306_100031_10103_10624_10623_10622_5722515_10621_10620,searchweb201603_35,ppcSwitch_5&algo_expid=d2630408-bd98-4ab4-be6a-4b445ecb82be-23&algo_pvid=d2630408-bd98-4ab4-be6a-4b445ecb82be&priceBeautifyAB=0RICHTEK also have a selection of low cost converters.There are probably lots of cheap ICs for the undervoltage shutdown that can be used to switch at .9V but finding the cheap Chinese parts isn't easy. It can be achieved by using a couple of bipolar transistors as shown in the attached simulation. V1 is the boost converter output which would be 3.3V but is set to 1.6V to improve the scaling of the traces. C1 is there to briefly turn on Q1 when the battery is connected to ensure the boost converter starts up - but you could use a switch or connect C1 to the USB charger.This isn't a particulary good circuit as the switching threshold will depend on the characteristics of the particular transistors you use and it will vary with temperature (Vbe reduces by approx 2mV per degree C) but should be good enough for your application. It also stops working when the battery drops below 550mV but the boost converter IC isn't likely to start up again at that voltage, but it's operating current may increase somewhat. You'd have to experiment. It is cheap though.Finally, you could use a boost converter chip which wont operate from a single NiMh because you can use its own 3.3V output to power the IC. The MT3608 is even cheaper at around $.07 from Aliexpress. The problem is that it's not clear from the datasheet what it's behaviour is when the input voltage drops below 2V which will happen when the battery voltage drops below .9V It has an internal undervoltage lockout at 1.98V but what voltage does that contine to work down to? Again, you'd have to experiment.