Fuse and Diode Reverse Polarity protection - Want to get it right - Page 1
Forum section: electronics / beginners

author: Veramacor
author_rank: Regular Contributor
timestamp: November 21, 2012, 08:13:13 pm
content: Hello all, I have a circuit I would like to protect from reverse polarity, or in other words, someone put the batteries in the wrong way!I have seen and tried a PTC fuse, but I'm not too thrilled about the heat it can generate, plus the leakage current.I have some expensive electronics I would like to reverse polarity protect. My circuit is ROUGHY equivalent to :3.6V BatteryPIC18F Microcontroller3.3 V RegulatorWIFI ModuleThe above circuit at most will draw about 250 milliamps of current. Typical is about 40 milliamps.Can anyone suggest a proper fuse and diode combination? I know that PTC fuses are resettable, but I prefer a traditional fuse. I'm a beginner so I do not know what type of diode is best.Thanks for your ideas!

author: tom66
author_rank: Super Contributor
timestamp: November 21, 2012, 08:24:31 pm
content: How about a PMOS reverse protection circuit?- very minimal voltage drop- no fuse required

author: ptricks
author_rank: Frequent Contributor
timestamp: November 21, 2012, 09:41:06 pm
content: Use a schottky diode with its anode attached to the + input and its cathode connect to power the rest of the circuit. A schottky has only a .1-.3V drop and will prevent the reverse power situation. Pick out a diode based on twice the reverse voltage you are using to power the circuit and at least 125% of the current requirement.

author: Jon Chandler
author_rank: Frequent Contributor
timestamp: November 21, 2012, 10:59:04 pm
content: Here are a few articles that may be of interest.Power Protection CircuitsPower Protection Circuits - A Closer Look at the Series DiodePower Protection Circuits - A Closer Look at the Crowbar CircuitPower Protection Circuits - A Closer Look at the P-Channel MOSFET CircuitThe P-channel MOSFET circuit works well with almost no voltage drop.

author: Veramacor
author_rank: Regular Contributor
timestamp: November 22, 2012, 12:35:15 am
content: Thanks for all the suggestions.My application I forgot to mention aims at being an ultra low power application. So the typical 30 milliamp current draw will only happen maybe a dozen times a day at the most. Say maybe 100 seconds per day at 30 milliampsMost of the time I intend to sleep both the PIC and the wifi module and will be using about 50 microamps of constant current.A .1 Volt drop the schottky suggestion may be too much power spent. The p-fet sounds interesting, but im not sure on its power consumption.Why wouldnt my original suggestion be optimal? A fuse to my knowledge poses no voltage drop or power consumption, the only thing i have to worry about is the reverse leakage current of the diode.Maybe if I re-ask the question: whats the best reverse polarity protection in an ultra-low power application?

author: JackOfVA
author_rank: Unknown
timestamp: November 22, 2012, 12:59:05 am
content: Fuses have DC resistance - after all, the heat that melts the fusible link has to come from I2R heating.You should check the data sheet for the fuse you have in mind as it will have a typical voltage drop at rated carry current spec.

author: Veramacor
author_rank: Regular Contributor
timestamp: November 22, 2012, 01:05:38 am
content: Thanks Jack,That is the root of the question right there. What in theory, would be the lowest current cost solution? My gut says that it has to be a fuse, if I get a fuse that burns at 500 milliamps, I would hade to think that the resistance of that fuse wire at 30 milliamps is close to zero?

author: ejeffrey
author_rank: Super Contributor
timestamp: November 22, 2012, 02:17:42 am
content: Quote from: Veramacor on November 22, 2012, 12:35:15 amYou are using a linear regulator, correct? If so, the voltage drop in the protection circuit will not affect the power draw at all as long as it doesn't violate the dropout limit of the regulator. A small voltage drop on a diode or fuse or PTC will only reduce the power dissipated by the regulator. The only thing that will cause lowered efficiency is leakage from the shunt element (i.e., the diode, or gate leakage in the FET). That will be a few microamps for a 1N4000 series at room temperature, but a lot more at high temperature. FET leakage will be zero.So what you really need to look at is whether the voltage drop of the fuse/fet/shottky/PTC will make your circuit cut out while there is still life left in the battery. For that, you need to figure out the minimum voltage your circuit will operate at and compare to the discharge curve for the battery and make sure that you are using (say) 90% of the available current.One thing to watch out for... Regulator dropout voltages and resistive/diode losses increase at high currents. A micropower circuit that occasionally draws 250 milliamps is quite a tricky problem. It could be running fine at 3.4 volts from the battery, then the radio kicks on and causes a brownout that resets everything.

author: Veramacor
author_rank: Regular Contributor
timestamp: November 22, 2012, 03:23:12 am
content: Thanks ejeffrey - good info.I'm not too concerned with brownouts as I am going to use either a Tadiran 19Ah at 3.6V or a 3 cell 4.5V AA battery pack.I think I may end up using what is attached as it looks like 3 milliohms is the efficiency I'm looking for.Found this on instructable btw:http://www.instructables.com/id/Reverse-polarity-protection-for-your-circuit-with/

author: robrenz
author_rank: Super Contributor
timestamp: November 22, 2012, 04:01:04 am
content: http://www.edn.com/design/analog/4368527/Simple-reverse-polarity-protection-circuit-has-no-voltage-drop

author: Veramacor
author_rank: Regular Contributor
timestamp: November 22, 2012, 05:24:50 am
content: I just so happen to have a dozen of these - forgot i had them...http://www.digikey.com/product-detail/en/2N7002K-TP/2N7002K-TPMSCT-ND/2642046I tested and it works beautifully. After about 1.2 Volts the power supply effectively matches the source drain voltage. I flipped the polarity, 0.000 volts on the Multi!Anyone see a problem using this MOSFET for 3.3V at 250ma Max?

author: JackOfVA
author_rank: Unknown
timestamp: November 22, 2012, 12:21:36 pm
content: Quote from: Veramacor on November 22, 2012, 01:05:38 amA 3AG type, fast blow, 250V 500 mA fuse has a cold resistance of about 0.5 ohms. (Littlefuse 312/318 Series) At 30 mA, the voltage drop will be 15mV; at 250 mA the drop will be 125mV more or less. (Some change will be seen in resistance as the current approaches a significant percentage of the rated value as the fusible link will heat and thus will change resistance.)Jack

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: November 22, 2012, 12:32:50 pm
content: You should be able to get an nchannel fet at less than 10mOhms.Something in this series might suit. PSMN012-60YSI used it in my latest project, but I haven't tested it yet.

author: arza.homes
author_rank: Newbie
timestamp: February 12, 2013, 11:07:52 am
content: Hello Guys,I am facing the exact same problem, worse - the current requirement is much lower. Attached is a picture of what my circuit looks like.I want to use a P MOSFET for reverse polarity protection which has a very low Rds. The supply voltage is 3V, followed by a mosfet and then a LDO (2.85V or 2.7V,100mA). This is extremely low power application, so i am looking for a P-channel mosfet which has a low Rds at load current ranging from 10mA to 100mA. The issue is, almost all the mosfet out there specify Rds(on) for currents ranging from 1A to 25A, there is no data sheet which gives me data at 10-100mA! Anyways, my search is still on...and i think i will end up buying a dozen different types and testing them all.PSMN012-60YS has a Rds of 8-11 milli ohms at VGS = 10 V; ID = 15 A; Tj = 25 °C;In between, is this issue closed at your end Veramacor?A 0.2V schottky or a PFET?Arza

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 12, 2013, 12:09:03 pm
content: Quote from: robrenz on November 22, 2012, 04:01:04 amnot just the relay may not be ativated by low voltage reversed application (just as in the reply posted), but i think even if there is enough voltage to activate the coil, before that time there will be reverse current flowing in the circuit for a moment possibly damaging the sensitive circuit. i think for low power app, a simple schottky diode in series will be acceptable, but for high powered application, we may reverse the relay setup (circuit normally open) and only engaged when there is forward (correct) voltage. power used by relay coil should be negligible compared to what is used by the circuit.

author: gcp6ca
author_rank: Contributor
timestamp: March 03, 2013, 11:46:21 pm
content: I had to use a Li-Ion battery that needed reverse battery protection so that it could power the electronics as well as be charged. It is a modified version of the P-Channel MOSFET reverse protection. I used PMV48XP which has an on resistance of 120mR if Vgs is 1.8V @ 100mA. That is a 1.2mW of power loss at an extreme condition. For a depleted Li-Ion (3V) the Rds would be about 80mR @ 2.4A/125C so if we assume that is is the same at 100mA then it would be a 0.8 mW power loss. The resistors surrounding it waste 0.18 mW but they can be made larger to further reduce that.I also used it with MCP1824 3.3V regulator which has a dropout characteristic of about: Vdrop = I * (0.0006 V/mA) and Pdroploss = I*I*(0.0000006 W/mA/mA). At a 100mA it would be about: Vdrop = 60 mV and Pdroploss = 6mW.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: March 04, 2013, 02:58:41 am
content: It might be better with a sharper cutoff and some hysteresis.I haven't got a circuit to hand but instead of q2 an op-amp and use a zener for one of its inputs.edit: sorry posted in wrong thread

author: gcp6ca
author_rank: Contributor
timestamp: March 04, 2013, 03:10:54 am
content: That was the simplest way I could think of since I did not have too much space to fit it in. For the fastest response wouldn't comparators be better? Also, it would take more power using an op-amp or comparator, especially with the zener.

author: arza.homes
author_rank: Newbie
timestamp: March 13, 2013, 10:47:58 am
content: I got a low Vgs(th) FET - NTR2101P from On-Semi. Got it tested as well, the voltage drop at 100mA is very less. Should fit my application need. Adding a 100k to the source pin makes sense? Just to stop the flow of any leakage current or a smaller value is better? Somewhere someplace i did hear "using huge resistor values is a bad design, huh"?ThanksArza

author: alanb
author_rank: Frequent Contributor
timestamp: March 13, 2013, 10:54:36 am
content: Could there be a simple solution by designing the battery connector so that the battery cannot be inserted the wrong way round.

author: fcb
author_rank: Super Contributor
timestamp: March 13, 2013, 11:59:15 am
content: Use a 1A size schottky diode, the drop will be very small, also, alot of people tend to design in voltage regulators were they don't actually need them.

author: gcp6ca
author_rank: Contributor
timestamp: March 13, 2013, 03:24:43 pm
content: Quote from: arza.homes on March 13, 2013, 10:47:58 amThat depends completely on what you are trying to do. Lowering R4 will turn off the P-Channel faster and draw more idle current. If you use large values on R5 & R6, it is usually okay since that 2N7002 switches really fast. To figure how long it will take to turn off the MOSFET: tf = Qg/I = Qg/Vdd*R4. For PMV48XP it would be 11nC/4.2V*100k = 262us.Quote from: alanb on March 13, 2013, 10:54:36 amI actually had it since I connected the negative end to a spring and the positive end to a flat surface. If it was reverse the plastic cover on protected 18650 batteries would prevent the negative end from making contact with the flat surface. For me the problem was during assembly, if the connector wire was installed backwards then it would damage the charging IC and replacing it would not be easy since the board it conformally coated.Quote from: fcb on March 13, 2013, 11:59:15 amUsing a diode only protects the circuit and you would need a separate one for the charger but that voltage drop (of even 0.2V) would be a fairly big drop in capacity. Of course it is not a problem if it is being charged separately. If you were to use PMEG3020ER,115 which has a voltage drop of about 0.21V at 100mA that is a power loss of 21mW and compared to the 0.8mW of using a P-Channel MOSFET, that can be a really big difference for low power applications.