Li-Ion Protection IC "Too Protective" - Always in Overcurrent Detect Mode - Page 1
Forum section: electronics / projects

author: Matthew Lind
author_rank: Contributor
timestamp: July 23, 2018, 07:21:22 pm
content: Hello folks,I am running into an issue with my battery protection circuit that I am having trouble making progress on, and I am hoping that some of you can give me new ideas for things to try.To start off with, my high-level circuit goal is to have a Li-Ion coin cell battery light up 4 (or more) LEDs. The coin cell needs to be recharged (I am using the Microchip MCP73831, which works beautifully) and protected from the standard undervoltage / overcurrent.The issue that I'm hitting involves the protection circuit not allowing full current. This circuit is based around the Diodes Inc. AP9101C, specifically the AP9101CK6-CQTRG1 with Vm sense = 0.15v (see page 14 on the datasheet, all the way at the bottom). Here is the datasheethttps://www.diodes.com/assets/Datasheets/AP9101C.pdf.My design goals for the protection circuit are:Undervoltage cutoff at 2.8vOvercurrent cutoff <= 200maNormal current supply ~35ma, perhaps more if I want to add a few more LEDsNOT as concerned about overvoltage, since the charger IC handles thisMy circuit is identical to the "Typical Applications Circuit" on page 2 of the datasheet. I'm using all of their recommended values as well for R1, C1, and R2. My load (4 LEDs nominal, or sometimes just a test resistor) is across P+ and P-.The current that I measure with just my LEDs and battery is 26mA. Then when I hook up the protection circuitry, and short P- to Vbat- to wake it up from power-down mode, I measure a current of only 17mA.To investigate further, I did the following tests (quoted from an email to Diodes Inc):QuoteOn the face of it, it makes sense that the current is being limited, because the current-sense pin, Vm, has a voltage higher than the overcurrent cutoff value of 0.15v in each case. But what I don't understand iswhythis is happening.From the best of my reasoning, the voltage on the Vm pin is dictated by the current, and the RdsON values of the MOSFETs. If this is true, with my tests I would expect a value for Vm to be, at most:Vm = 5.2mV = 26mA * 100mOhm/MOSFET * 2 MOSFETsI've tried 3 different MOSFETs with different RdsON values. Something like 80 mOhm, 100 mOhm, and 1 Ohm. Furthermore, each of these MOSFETs have Vthreshold values < 2.8v (undervoltage cutoff).Here are the MOSFETs I have tried:https://www.digikey.com/product-detail/en/on-semiconductor/NTR4003NT1G/NTR4003NT1GOSCT-ND/1967270https://www.digikey.com/product-detail/en/diodes-incorporated/ZXMN3B14FTA/ZXMN3B14FCT-ND/1211599https://www.digikey.com/product-detail/en/on-semiconductor/NTR4501NT1G/NTR4501NT1GOSCT-ND/687133I can hook up my LEDs and battery to the MOSFETs, and simply tie their gates to Vbat+, and the circuit performs as expected, full LED brightness.Also, I can take Vm pin and simply tie that to Vbat- and the circuit provides full current and undervoltage protection. This makes sense because Vm = 0v < 0.15v so it thinks there is no current flowing.I guess I'm just stumped here, because it seems like it should work, but it does not.Are there some other MOSFET properties that I am neglecting to factor in?Do ya'll have another battery protection circuit that you've gotten to work that I should look into instead?I'm doing this on a breadboard, could this be a factor?Anyway, any help is greatly appreciated. Please let me know what additional information you would like.

author: Matthew Lind
author_rank: Contributor
timestamp: July 23, 2018, 11:36:47 pm
content: Attaching the circuit schematic so you don't have to open the datasheet

author: Matthew Lind
author_rank: Contributor
timestamp: July 24, 2018, 02:15:37 am
content: In an act of desperation, I have tried a fresh AP9101C chip. No dice.Then I rotated the IC 180 degrees because maybe that dot next to that pin isn't the pin-1 marker? This produced bright LEDs, but it doesn't cut off at low voltage, whereas the correct orientation does.From what I can tell, the current that is lighting the LEDs isn't going through either MOSFET, but is actually going into the Vm pin. I think this because I see no difference in brightness when I disconnect Q2's source from P-.If anyone out there has any ideas, I would hugely appreciate hearing them. I'm pretty much at the end of my understanding at this point. Maybe someone could comment on how current could flow INTO the Vm pin given this functional block diagram?

author: Matthew Lind
author_rank: Contributor
timestamp: July 24, 2018, 02:27:33 am
content: Ok, here's another thought. The circuit relies on passing current through the body diode of Q2 for discharging. Could it be that the body diode's voltage drop is greater than 0.15v, which is why this thing is always on?If that is true, then I should expect that the circuit will work when I bypass Q2 and hook P- directly to Q1's drain.I'll test hat out.

author: Matthew Lind
author_rank: Contributor
timestamp: July 24, 2018, 02:33:31 am
content: It did not work. In fact, I get the same brightness when I completely disconnect the MOSFETs from everything. The entirety of the current is flowing through Vm. Bizarre.

author: Matthew Lind
author_rank: Contributor
timestamp: July 28, 2018, 11:04:08 pm
content: Alright, received new components.As suggested to me by Diodes Inc, I've ordered an "auto-wakeup" version of the AP9101C. I swapped it in and found absolutely no difference in the way the circuit performed.Then I also tried an ABLIC S-8261 IC that is a drop-in replacement, and again found absolutely no difference in how the circuit performs.So it must mean that something else in my circuit is wrong. It is either the MOSFETs, though I see no difference when I try different MOSFETs, or it could be....what? The fact that the load is a group of 4 parallel LEDs?If anyone out there has any ideas whatsoever, I would really appreciate them. I don't know what else to do at this point. Have any of you ever seen a working circuit using one of these Li-Ion protection ICs?

author: Matthew Lind
author_rank: Contributor
timestamp: July 29, 2018, 12:03:19 am
content: Maybe I've made a mistake in my circuit.Here is a drawing of what my circuit is theoretically like.

author: Matthew Lind
author_rank: Contributor
timestamp: July 29, 2018, 12:08:48 am
content: I made a mistake in my drawing.Here's the correction

author: Matthew Lind
author_rank: Contributor
timestamp: July 29, 2018, 12:34:45 am
content: I've been reading up more on MOSFETs. Perhaps the issue has something to do with the gate charge that is required to fully turn the MOSFET on.The way this failure would work is the DO pin would attempt to turn the gate on, but it would have too low of current to charge the gate before the voltage on the sense pin exceeded the overdischarge voltage.The reason I don't really think it is this is because when I ground Vm sense pin or short P- to gnd, the DO pin should be charging the MOSFET gate, so when I return the circuit to normal, it should be at "maximum chooch".

author: Matthew Lind
author_rank: Contributor
timestamp: July 29, 2018, 12:44:59 am
content: Ok let me ask ya'll this: if you were in my shoes, would you be using an oscilloscope on all of the pins? Or what troubleshooting techniques would you employ?I don't have a scope, but this might be the point where I get one.

author: sleemanj
author_rank: Super Contributor
timestamp: July 29, 2018, 12:59:07 am
content: #1 "uses the body diode", no t in normal operation. A MOSFET is bidirectional, as long as the Gate-Source voltage is in such a state to turn the MOSFET on, current can flow in either direction, once it is on you can ignore the body diode (it is effectively shorted out of circuit).#2 it is usual to use a dual-fet rather than two discrete devices, you probably want the fets to be fairly well matched, the classical chinese IC to use is the 8205A (with a DW01 protection IC)#3 the current limit is set by way of choosing fets with appropriate RDSon such that the voltage drop across the pair at the desired current limit (use Ohm's law) is some specific voltage (which depends on the precise part number of the protection IC, see page 12 of datasheet), you could of course fudge the RDSon by adding some additional series resistance in the right place#4 "just your leds and battery", how are you limiting current in this configuration?#5 measure the Vgs of your fets, looking just at the first link to ta fet you posted, the headline is "4v" for Vgs to get 1.5 Ohm RDSon, since you have two of these in series even if you could manage to get 4v Vgs, which you can't (your cell is only barely 4v when it's fully charged), you have at best 3 Ohm of extra series resistance there

author: Matthew Lind
author_rank: Contributor
timestamp: July 29, 2018, 01:27:31 am
content: Thank you for your response! I'm looking at each of your points.QuoteFrom what I can tell, because DO and CO are independent, sometimes one MOSFET is turned off while the other is turned on. For example, when the battery is fully charged, the CO MOSFET will be off, but the DO MOSFET will be on, so current will flow through the CO MOSFET body diode. Of course, in the nominal voltage range, both MOSFETs will be on.QuoteI will go ahead and try that dual-fet out. What do you mean by the fets being "fairly well matched"? Do you think small variations due to manufacturing tolerances might cause issues?QuoteYes, I thought originally that this was the source of my problem. RdsON being too high would cause the voltage at the VM pin to be higher. However, using Ohm's law, even a high value RdsON should provide enough current for my application. The 4 LEDs are only pulling <= 37ma.So that would give:Vvm = Iload * (RdsON*2)0.15 v = 37 ma * 2RdsONRdsON = 2.02I have tried MOSFETs with RdsON = 140 mOhmI've also simply tested the MOSFETs by connecting their gates to VBat+ to ensure that they can provide the necessary current.QuoteGood point, that was me being hasty. I have 59 Ohm resistors in series with the LEDs.QuoteI have been using lower RdsON MOSFETs than the first linked. I will absolutely measure the Vgs of the MOSFETs I am currently using and make a new post.Again, thank you so much for your response. I greatly appreciate it, and having the opportunity to respond to your points is helpful for me to continue to work through my reasoning on this.

author: sleemanj
author_rank: Super Contributor
timestamp: July 29, 2018, 01:52:58 am
content: Quote from: Matthew Lind on July 29, 2018, 01:27:31 amYes that is outside of normal operation though, ideally you don't want to trigger any of those conditions if you can avoid it, the protection is a "just in case something goes wrong". Although relying on it for low-voltage cutoff is generally OK, as soon as you start charging it the voltage will rise above and both fets will be on almost immediately anyway.QuoteMatched with respect to their characteristics (from manufacturing variance). Although it probably doesn't make much difference in this case.QuoteRdsON at what Vgs though.I would measure to determine what RDsON you are actually getting, ie, measure the current passing through the fets, measure the voltage, and calculate, if that does not explain the reduced current then you can start looking elsewhere.

author: EAvMx
author_rank: Newbie
timestamp: November 30, 2021, 09:15:55 am
content: Have you figured out how to make it work? I'm facing the same issues

author: thm_w
author_rank: Super Contributor
timestamp: November 30, 2021, 11:26:22 pm
content: Quote from: EAvMx on November 30, 2021, 09:15:55 amWhat FETs?

author: EAvMx
author_rank: Newbie
timestamp: December 01, 2021, 03:24:03 pm
content: Quote from: thm_w on November 30, 2021, 11:26:22 pmI'm using DMN2029USDhttps://www.diodes.com/assets/Datasheets/DMN2029USD.pdfEdit: I attached the schematic of the circuit

author: thm_w
author_rank: Super Contributor
timestamp: December 02, 2021, 12:17:46 am
content: Vdoc = 0.100VFET ~2x60mOhmSo around 1.6AWhy are you taking VBAT after the 330R resistor and using it elsewhere in the circuit? That line should only be going to the AP9101C.What current can you pass before the circuit trips?Is BAT- going directly to the battery and nowhere else right? VBAT- is not connected in any way to the battery?

author: EAvMx
author_rank: Newbie
timestamp: December 10, 2021, 10:09:29 am
content: Quote from: thm_w on December 02, 2021, 12:17:46 amVBAT goes to a PMOS for loadsharing.Correct, BAT- is connected exclusively to the battery and VBAT- is not connected at all to the battery.I wrote to DIODES INC and they replied this IC (AP9101CAK6-XXXXXX) only recovers from overdischarge status when load is disconnected and it surpases overdischarge release voltage or a charger is connected as shown in page 9 figure 1. I don't get why do they state (page 7, section 3: Overdischarge status) that recovery can be done:QuoteThanks everyone for your time invested!