Optimizing reverse polarity protection current consumption - Page 1
Forum section: electronics / projects

author: okw
author_rank: Frequent Contributor
timestamp: April 07, 2025, 03:28:54 pm
content: I have a question about sizing the two resistors for minimal quiescent current, but also ensure low enough resistance through the MOSFET.With the two 10k resistors, it consumes 745uA, which drains the battery (4.2V 750mA Li-Ion) in 35-40 days. If I up them to 100k or 1M, I reduce it to 79uA and 9uA respectively, but I'm not sure how high I can go, before affecting how the MOSFET turns on (saturates?). The product consumes around 50-120mA when turned on, but could have short 200-300mA bursts from time to time.I would be nice if someone could point to me how to do the calculations, so I could size the resistors properly, and maybe even replace the transistor and/or MOSFET to better ones. The MOSFET have really low Rds(on), but I might need to consider other specifications too(?).

author: Peabody
author_rank: Super Contributor
timestamp: April 07, 2025, 04:06:09 pm
content: Could you explain the function of Q1? Why not just connect the gate directly to BAT+?

author: okw
author_rank: Frequent Contributor
timestamp: April 07, 2025, 04:38:18 pm
content: Quote from: Peabody on April 07, 2025, 04:06:09 pmI got "inspiration" here:http://blog.deconinck.info/post/2017/12/22/18650-Battery-charger-reverse-polarity-protectionNow, the trick is to have the gate turning the switch On or Off at the right moment. The first thing is to put a resistor from the source to the gate to make sure the MOSFET is OFF "by default", and the second thing is to add a transistor to override that resistor and flip the voltage on the gate when the battery is correctly positioned.

author: Peabody
author_rank: Super Contributor
timestamp: April 07, 2025, 09:13:50 pm
content: Well, I've just not seen that before, and wonder why it's any better than the version in this other video:

author: okw
author_rank: Frequent Contributor
timestamp: April 07, 2025, 10:13:55 pm
content: That one uses a P-ch and breaks positive rail, mine uses N-ch and breaks ground. Not sure what is most efficient. And what works well with a TP4056+DW01 protection.Usually breaking the positive rail is best practice, especially with ICs, as you might have some artefacts when chips are powered, but ground is floating.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 07, 2025, 11:18:48 pm
content: Quote from: okw on April 07, 2025, 03:28:54 pmThat was designed for a charger, so they do not care about current drain.For low voltage battery operation, you can use a low threshold FET and connect the gate to B+, & via a resistor/zener if you are paranoid.There are parts like PMCB60XNE that are spec'd at 2.5V gate and have ESD diodes included, so you may be able to omit the zener.VGS = 2.5 V; ID = 2 A; Tj = 25 °C - 17 25 mΩIGSS gate leakage currentVGS = 4.5 V; VDS = 0 V; Tj = 25 °C - - 2 µAVGS = -4.5 V; VDS = 0 V; Tj = 25 °C - - -2 µA

author: okw
author_rank: Frequent Contributor
timestamp: April 07, 2025, 11:48:05 pm
content: Quote from: PCB.Wiz on April 07, 2025, 11:18:48 pmit seems the HL2312 I'm using has even lower VGS (0.45-1.0V) compared to PMCB60XNE which is 0.5-1.1V.And IGSS is 1/100th of that of PMCB60XNE.https://www.lcsc.com/datasheet/lcsc_datasheet_2501071555_hongjiacheng-HL2312_C28646272.pdfAm I missing something?

author: Peabody
author_rank: Super Contributor
timestamp: April 08, 2025, 01:36:31 am
content: The circuit in the video I linked to is only for the battery as a source of current. It won't work with a charger because current flows into the battery to charge it, and then out of the battery to power the load. Here's a Great Scott video that goes through his efforts, and ends up with a circuit that looks like yours, except he has a diode in there:Any way, as to your original question, it seems you should be able to use 100K resistors at least, and possibly even higher on the gate resistor.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 08, 2025, 02:28:12 am
content: Quote from: okw on April 07, 2025, 11:48:05 pmThe part I quoted has ESD diodes included.If you add an external zener to your part, they become similar.

author: Zero999
author_rank: Super Contributor
timestamp: April 08, 2025, 08:50:08 am
content: Just use a single MOSFET. There's no need for any other parts. The quiescent current is simply equal to the gate leakage.Either N-channel or P-channel can be used. N-channel devices are more efficient because they have a lower on resistance, compared to P-channel. P-channel has the advantage of the input and output being able to share a common negative rail.

author: okw
author_rank: Frequent Contributor
timestamp: April 08, 2025, 10:15:49 am
content: Quote from: Zero999 on April 08, 2025, 08:50:08 amThanks. I'll try without the BJT and resistors.Do you understand the reasoning behind putting a "resistor from the source to the gate to make sure the MOSFET is OFF by default"? Wouldn't the MOSFET either unsaturate by itself (battery change takes at least a couple of seconds), or when inserting it the wrong way, immediately block (even if the MOSFET is still saturated from the last correct insertion)? Is it to limit any initial surge that might escape through the MOSFET before it switches?The transistor is probably there to not "bleed" the MOSFET gate directly, which would result in a continuous 0.37mA leakage when polarity is correct.

author: Zero999
author_rank: Super Contributor
timestamp: April 08, 2025, 10:22:25 am
content: Quote from: okw on April 08, 2025, 10:15:49 amThe MOSFET is bypassed by its internal body diode, in the forward biased state.Regarding the speed, the MOSFET will turn off fairly quickly. If you're really concerned about it being too slow, then connect a diode in reverse parallel with the load to clamp any very short reverse polarity spikes.