Reverse polarity protection with minimal power loss - Page 1
Forum section: electronics / beginners

author: somuchtyler
author_rank: Contributor
timestamp: August 03, 2020, 02:44:52 am
content: Hey guys,I am trying to add reverse polarity protection to my project that is designed to run for months at a time.I really want to use these Keystone 55 AAA battery clips, since they are so cheap.However since there's going to be no physical way to block someone from inserting AAA batteries the wrong way I want to add reverse polarity protection.I want minimal power loss and I am running 4xAAA batteries that would put out 4-5V in series, I got a couple questions.1) Would it be better to use a diode with a resettable fuse? Would these two work?https://lcsc.com/product-detail/PTC-Resettable-Fuses_RUILON-Shenzhen-Ruilongyuan-Elec-SMD1206P010TF_C20981.htmlhttps://lcsc.com/product-detail/Switching-Diode_Nexperia-BAV99-215_C2500.html2) or would it be better to use a MOSFET with a resistor plus zener diose? like this

author: Dabbot
author_rank: Regular Contributor
timestamp: August 03, 2020, 03:11:10 am
content: I vote for a simple schottky reverse protection.Pick a nice low voltage one and you'll hardly notice the voltage drop. Especially when you're only drawing enough to keep it running for months.I'm assuming you're using NiMH, judging by the 4-5V remark? The voltage will remain level through the discharge cycle, so you won't need to worry about a schottky in the mix.

author: Nusa
author_rank: Super Contributor
timestamp: August 03, 2020, 05:00:00 am
content: Or you could price out a proper pc pin 4 AAA holders and compare it to the price of 8 clips plus whatever protection components you decide on. Are they really too expensive to consider?Or you could find some fiber washers of the right size and glue them on the positive contacts so that only the positive button on a battery can engage the contact.Another method would be to use a bridge rectifier, which means the circuit will work regardless of which way the batteries are inserted. Assuming the entire set is in the same direction, of course...that's another argument for a proper holder.

author: somuchtyler
author_rank: Contributor
timestamp: August 03, 2020, 06:09:27 am
content: Quote from: Dabbot on August 03, 2020, 03:11:10 amI am going to use either 3 or 4 AAA batteries. I don't think I want to use a schottky diode because they have a voltage drop under forward bias. say 0.25V. That means at 100mA, you're dissipating 25mW of power.Quote from: Nusa on August 03, 2020, 05:00:00 amWouldn't a bridge rectifier that wreck the power efficiency of my circuit?

author: Dabbot
author_rank: Regular Contributor
timestamp: August 03, 2020, 06:29:09 am
content: Quote from: somuchtyler on August 03, 2020, 02:44:52 amQuote from: somuchtyler on August 03, 2020, 06:09:27 amPerhaps there's been a misunderstanding here. What kind of power demand does your project have?You're not going to get 100mA out of AAAs for months at a time, unless that demand is intermittent. And even then, 25mW extra in those periods is meaningless.

author: somuchtyler
author_rank: Contributor
timestamp: August 03, 2020, 06:42:59 am
content: Quote from: Dabbot on August 03, 2020, 06:29:09 amSorry I am a bit of a noob to electronics, more of a software developer myself. The total power usage of my project is 0.06mA or 60μA.

author: Dabbot
author_rank: Regular Contributor
timestamp: August 03, 2020, 06:59:42 am
content: Quote from: somuchtyler on August 03, 2020, 06:42:59 amNo need to apologize, you're in the right place!A schottky will be fine in this scenario. Pick one rated for 1A at 20V or less. Lots of cheap choice here.Generally speaking, the lower the reverse voltage rating and the higher the forward current rating, the lower the forward voltage drop. Check the datasheets etc.

author: jpanhalt
author_rank: Super Contributor
timestamp: August 03, 2020, 07:20:57 am
content: Given reverse installation of the batteries, with the fuse, you have heavy current flow until the fuse opens. With the mosfet, there is no current flow.BTW, there is no need for the zener diode or gate resistor with a maximum of 6 V. The FQP47P06 may not be the best choice at 4.5 V, and it is large. A logic level device might be preferred.

author: TimFox
author_rank: Super Contributor
timestamp: August 03, 2020, 03:44:46 pm
content: A typical Schottky rectifier, 1N5817, is rated for 1 A and 20 V from ON (originally Motorola):The typical curve only goes down to 20 mA current, where the forward voltage is 0.23 V at 25 C (lower for 100 C) case temperature.Eyeballing the extrapolation, we expect well below 0.2 V below 1 mA.

author: Peabody
author_rank: Super Contributor
timestamp: August 03, 2020, 05:04:13 pm
content: Well I'm gonna vote for the mosfet. Since you don't need the zener or the gate resistor, it's one part either way, and the mosfet will have a much lower voltage drop, so it won't be dissipating any battery power as heat, and the cutoff voltage for your load won't be reached quite as soon. Of course at your expected current levels, it doesn't really matter, but generally speaking the mosfet is the more elegant solution. Something like an NPD6020P might be a better choice because of its low GS threshold voltage. But in an SOT-23 surface mount package, something like an Si2323 would be fine.There is also the issue of Schottky reverse leakage current, which I don't really know how to deal with, if that's even necessary.Anyway, just be sure to orient the mosfet the right way. The drain goes to the battery, and the source to the load. The gate is just tied directly to ground.You can also do this on the low side with an N-channel mosfet.

author: TimFox
author_rank: Super Contributor
timestamp: August 03, 2020, 05:14:15 pm
content: Typical reverse current (25 C) for the 1N5817 rectifier diode at 6 V reverse bias is 100 uA. The higher-voltage 1N5819 is 50 uA at 6 V.The general-purpose 1N5711 signal diode, rated for 70 V, has only 200 nA at 50 V reverse voltage, with somewhat higher forward voltage.What is the off-state leakage of the MOSFET?

author: Peabody
author_rank: Super Contributor
timestamp: August 04, 2020, 02:33:32 pm
content: The NDP6020P is 100nA if I'm reading the datasheet correctly. The thing about Schottky diodes is that their leakage current increases exponentially as they get warmer. Now of course at the OP's current levels, it wouldn't get warm, but in a high current circuit, leakage might need to be dealt with somehow.

author: radiolistener
author_rank: Super Contributor
timestamp: August 04, 2020, 02:48:17 pm
content: diode cannot protect from reverse polarity, because it has some voltage drop. So, some amount of current with reverse polarity can flow through your device.If your power supply can provide very high current, for example 50-1000 Amps, it can make fire and high damage to your device even with 0.5 V.It's better to use mechanic relay to protect your device from reverse polarity.Here is reverse polarity protection with overvoltage protection for 20-30 Amps power supply, used for transceiver for example.

author: TimFox
author_rank: Super Contributor
timestamp: August 04, 2020, 03:04:31 pm
content: The original question was about a very low current circuit, powered by AAA cells. There are two questions for the diode solution: forward voltage drop (his application could probably tolerate 300 mV) and leakage current when reverse-biased (is 100 nA dangerous to the low-power load?). The Schottky diode in series works if 200 mV is acceptable, and another diode (Schottky or PN) across the load could safely conduct the miniscule leakage current, allowing maybe 100 mV reverse voltage.The same questions apply to a MOSFET circuit.

author: Peabody
author_rank: Super Contributor
timestamp: August 04, 2020, 04:18:54 pm
content: In this case I don't think leakage matters either way, but more generally my understanding is that in high current situations in which the diode would get quite warm in normal operation, leakage could be a problem if a battery is replaced backwards while the diode is still warm. I'm not an engineer, but my understanding is that a mosfet's leakage is not affected by temperature to that extent. Plus, with a much lower voltage drop across the mosfet, it isn't going to get as warm in the first place.But for the OP, I think it's still true that the diode will drop about 300mV, which means the batteries will effectively wear out sooner. There's a lower supply voltage point at which the circuit will no longer operate properly and he will reach that point sooner with the diode because of the voltage drop. There would be a much lower drop across the mosfet.So I'm just saying that the better general solution for this function is the mosfet.

author: TimFox
author_rank: Super Contributor
timestamp: August 04, 2020, 04:35:30 pm
content: The MOSFET will work, but the forward drop for his circuit will be < 300 mV at his current, and actually falls if the diode warms up. I can find the leakage current for the Schottkys, but I don't know the leakage current for the MOSFET. Again, if leakage is a problem for safety of the device, a shunt diode can tame it for either series device. How sensitive is his circuit to 200 mV difference in the battery voltage?

author: Zero999
author_rank: Super Contributor
timestamp: August 04, 2020, 05:23:44 pm
content: Quote from: radiolistener on August 04, 2020, 02:48:17 pmWell that's not true. Diodes are frequently used for reverse polarity protection. In most applications, the voltage drop is not an issue.QuoteThat's excellant for high current applications and has the lowest loss possible. The only downside is the relay coil will use a considerable amount of power and will exceed that of the device itself, in a low power application.

author: radiolistener
author_rank: Super Contributor
timestamp: August 05, 2020, 01:34:42 pm
content: Quote from: Zero999 on August 04, 2020, 05:23:44 pmThis is not an issue for low power devices, just because you can simply put power through diode (in series).But voltage drop is a serious issue for powerful devices. For high power device you cannot simply put diode in series, because it will dissipate a lot of power due to high current.And you cannot simply put fuse and diode in parallel, because some amount of voltage with reverse polarity still applied to the device due to voltage drop on the diode.

author: JustMeHere
author_rank: Frequent Contributor
timestamp: August 05, 2020, 01:40:19 pm
content: Look at the schematic for Adafruit HUZZA32https://learn.adafruit.com/assets/41630

author: Zero999
author_rank: Super Contributor
timestamp: August 05, 2020, 01:43:33 pm
content: Quote from: radiolistener on August 05, 2020, 01:34:42 pmIt depends on what you mean by high power. High voltage or current? Diodes are better at higher voltages.I like the MOSFET solution myself. It has a much lower voltage drop, than a diode, without the static current, required by a relay coil.

author: Peabody
author_rank: Super Contributor
timestamp: August 05, 2020, 01:59:34 pm
content: Quote from: JustMeHere on August 05, 2020, 01:40:19 pmWell, that's a load sharing circuit, not really reverse polarity. If VBUS is high, that turns off the mosfet and lets VBUS power the regulator. If there's no VBUS power, the gate pulldown resistor turns on the mosfet, which lets the battery supply power. The diode has to be there to prevent the battery from turning its mosfet off. But this all could have been done just using another diode on the battery line - the source with the highest voltage would power the regulator. But Adafruit went to some trouble to use the mosfet instead. I think that's to avoid the voltage drop of a diode.

author: radiolistener
author_rank: Super Contributor
timestamp: August 05, 2020, 01:59:49 pm
content: Quote from: Zero999 on August 05, 2020, 01:43:33 pmUsual amateur electronics needs 12-24 V. When it consume 30-60 Amps, it's hard to protect it with diode.If you put diode in series with the load, the voltage drop about 0.5 V leads to 0.5*60 = 30W heat dissipation on the diode. Also it leads to lower voltage on the load.If you put diode in parallel to the load (to make short circuit for reverse polarity), 0.5V with more than 60A will be delivered to your device in reverse polarity. It may lead to serious damage, not only damage to the transistors and IC, but also damage of PCB.

author: JustMeHere
author_rank: Frequent Contributor
timestamp: August 08, 2020, 08:41:44 am
content: Quote from: Peabody on August 05, 2020, 01:59:34 pmThat's what I thought when I first saw it, but then I saw this circuit. They look the same if you remove vbus from the Adafruit one. I think it does both jobs.https://www.eenewsautomotive.com/content/power-supplies-automotive-start-stop-systems/page/0/2

author: Peabody
author_rank: Super Contributor
timestamp: August 08, 2020, 02:16:32 pm
content: Quote from: JustMeHere on August 08, 2020, 08:41:44 amThis circuit has a zener diode, not a schottky, and you don't need it or the resistor if your supply is 5V. So then you're just back to using a p-channel mosfet for reverse polarity protection. It's not at all clear to me that a load sharing circuit protects against reverse polarity, particularly given where it's located. It certainly wouldn't protect any charging circuit that comes before the battery.

author: FriedMule
author_rank: Frequent Contributor
timestamp: August 10, 2020, 09:38:40 am
content: Please do NOT follow any of my advices, before anyone has commented on them, I am properly even more noob than you, but here it goes. :-)Why not use a BJT that only activates if it gets positive voltage, and let that control a super low resistance relay?I have made a simulation here:http://tinyurl.com/yyk72yvzto see it in action, right click on the battery and chose "swap terminals" a relay with maybe 199mOhm may be possible?

author: Zero999
author_rank: Super Contributor
timestamp: August 10, 2020, 11:30:00 am
content: Quote from: FriedMule on August 10, 2020, 09:38:40 amThe transistor is being used as a diode. It would be better if it were replaced with a rectifier diode, which will have a much higher reverse voltage rating. Refer to the circuit posted by radiolistener.https://www.eevblog.com/forum/beginners/reverse-polarity-protection-with-minimal-power-loss/?action=dlattach;attach=1040668;image

author: FriedMule
author_rank: Frequent Contributor
timestamp: August 12, 2020, 02:01:28 pm
content: My idea was to keep the Voltage drop as low as possible and keep the components out of the normal flow. I could not find something other than placing a low distance relay-type in the main line. The resistance is properly a bad idea and could be replaced by a reverse diode, to block the flow.But I understand that it was not a great solution, so thanks for teaching:-)

author: Zero999
author_rank: Super Contributor
timestamp: August 12, 2020, 10:21:26 pm
content: Quote from: FriedMule on August 12, 2020, 02:01:28 pmYour circuit is the same as the one I linked to, only the transistor is being used as a diode, rather than a proper diode. The base-emitter junction of a BJT is a diode.

author: winniethepooh_icu
author_rank: Contributor
timestamp: August 14, 2020, 04:25:45 am
content: Looks like the blind leading the newbie here.The first question that should have been asked: How much current supplies the intended device?Now consider the forward drop of a diode you have chosen -> Is the voltage drop at this current, and associated power loss acceptable?If yes, use a diode.If no, use a MOSFET. TI makes some nice parts which integrate "ideal diode" type functionality and they are widely used in the industry.I suspect, for an AAA battery application, that the diode will be more than sufficient.Can't believe someone posted intent to use AAA batteries and people reply with solutions involving relays, mention 60A, 1000A, whatever. A nice laugh.