Revers Polarity Protection

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ljbeng
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Revers Polarity Protection

Post by ljbeng »

How would you protect things in the picture below from the power supply being connected backwards? (non-mechanical solutions please) The serial port may or may not be connected to another device. The other device may or may not have a common earth ground connection back to the use supplied power. I need the protection all mounted on the "Electronics Micro Etc" pc board.

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Chris Smith
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Post by Chris Smith »

From what I can see in the photo, a simple bridge rectifier will steer the power in the right direction no matter what the input is.

The only draw back is will the diodes drop the voltage too much from the PS for the device to operate?
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haklesup
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Post by haklesup »

A single diode placed between the power and ground such that it is reverse biased when powered correctly will do the trick. When plugged in backward the diode will be forward biased and short the supply to ground.

For this to be totally successful, you need to consider a couple other things. First the supply. What is its maximum current. It should be selected to be only a little more than you need to power the circuit. This way when it is plugged in backward, the short caused by the protection diode will not carry too much current and the supply can collapse (reduce its voltage) due to current limiting. If the supply is too big, it will fry the protection diode after a few seconds and you will no longer have any protection. The second consideration is to select a rectifier diode with sufficient power rating to shunt the entire power supply current and a reverse voltage rating 5 to 10X more than the supply. so a 50V or 100V PIV diode will generally work. That 1n5400 will work with quite a short circuit load when required.

This solution is very common. In fact I have fixed a bunch of wall wart powered stuff which people had used the wrong wart on. Usually the protection diode and maybe a resistor is blown right at the power connector input.
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philba
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Post by philba »

why not just a diode in series with the +12V line? This is assuming that the +12V feeds a power supply that provides lower voltages to the micros and other electronics (5V, 3.3V, etc). If this is an automotive product, then that is the best solution.

If you need exactly +12V to the equipment then I would use a PMOSFET in series with the 12V line and connect the gate to gnd. when the gate is 0, it will conduct, when it's +12, it won't. Choose one with a low Rds(on). I'd put a moderate size resistor between gate and gnd.

I think a bridge would be a problem as it will lift the ground the electronics sees. You will effectively have 2 grounds, one .7V above the other. (actually floating based on total current). You will need to be careful as to which one you use. Your sensors will then need a separate gnd line to the equipment. Comm might be a bit tricky due to that as well.

The reverse diode across gnd and +12 seems like a problem to me - why should reverse power cause the product to require maintenance? Maybe a circuit breaker would work...

Also, any +12v sensor inputs need to be protected as well.
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Bob Scott
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Post by Bob Scott »

haklesup wrote:A single diode placed between the power and ground such that it is reverse biased when powered correctly will do the trick. When plugged in backward the diode will be forward biased and short the supply to ground.
I agree with this. Just add a suitably rated fuse in the 12V line before the diode. If the power is connected in reverse, the fuse will blow.

And the other way would be to add the diode in series with the 12V line only if the equipment can operate at all times with 0.7 volts less than 12V. This method does not need a fuse if the supply is reversed by accident.
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haklesup
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Post by haklesup »

The series diode would be suitable as long as the supply current to the load were sufficiently low so that you only saw 1V or less of drop across the diode. The 1N5400 will have 8V across it at 5A for example.

For devices that run on just a few 1.5V batteries (or similar) this should work just fine since they already should be tolerant of low supply voltage (as batteries do while they drain) and generally have low current to begin with.

Philba's suggestion of using a diode connected FET sounds like it would work too. I've just not seen that yet and don't have a good feel for voltage drop vs supply current.

The parallel diode does not interact with the load (unless it is leaky) and is only doing something during a fault situation.

Circuit breakers generally protect the source not the load. By the time they trip, the load (application circuit) has already failed. The CB may prevent further damage but often does not prevent the initial overstress caused by excess or reverse bias voltage.
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philba
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Post by philba »

Putting in a fuse to protect against battery reversal just seems like a problem. Consider cars - some joker puts the battery in backwards and you have to get the all the electronics serviced? Seems extreme.


As to the P-MOSFET approach. It's pretty simple.
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With correct polarity, the gate is at 0V and the MOSFET turns on. With the input voltage reversed, the gate is at 12V and the MOSFET never turns on. In the case I show here, the MOSFET has an Rds(on) of a hair less than .1 Ohm. With a 1 A load, this translates into a bit less than 80 mV voltage drop. I used switchercad to verify this. The MOSFET might get a little warm - 80 mW dissipation. The only issue is that this isn't a cheap solution like the series rectifier.
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Chris Smith
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Post by Chris Smith »

absolutely not

I’ve seen this misconception before from others. The assumption that a diode only drops its rated Vf (forward voltage) value without regard for the current flowing through it.

A diode only has this rated voltage drop just after it turns on and with the current through it generally less than 1mA. Look at a datasheet, Vf is spec'd at a particular forward current. After that the diode acts like a resistor of 5 to 50 ohms. The OP's application would put up to 7.5A through the diode making for much much more than 0.7V across it. Even at 1A, several volts is not unrealistic and that's why its easy to burn up a diode.

A series diode could balance the load simply by virtue of the resistive portion of its characteristic but at more of a power loss that is necessary. If operated below a minimum current load, the non linear portion of the diode would be active and any difference between the two diodes would cause a load imbalance which would quickly run away to one supply with the lower Vf diode (derated due to proir heating). For 7.5A you would need a heat sink mountable schotkey diode. Far more expensive than a 5W resistor.

BTW, you would want the load sharing resistors to be non inductive. Avoid wire wound. Carbon comp is actually a good choice for material. Low inductance is indicated in the catalog for applicable resistors.05
We cant conceive that a diode drop in one circumstane is negative, then tht next its ok?>


We cant conceive that a diode drop in one circumstance is negative, then the next its ok?>

What is needed is the experimentation of all the proper situations, for your self ,...is the proper situation.
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philba
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Post by philba »

Please note the comment "actually floating based on total current" in my post. However, it doesn't change the point. Note that the OP said "may or may not have a common ground".
ljbeng
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Post by ljbeng »

One of the points I wanted everyone to see was that if the 12 volts is connected backwards, the 12 volt line is now connected to the other devices ground through the serial port connection. I have considered all these ideas to protect the main electronics, but how do I protect from the reverse polarity now connected to another systems ground?
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Chris Smith
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Post by Chris Smith »

A rectifier bridge will sort out the correct polarity, no matter what direction the pole are, as long as the voltage is high enough to over come the slight drop.

Even AC will be corrected into DC but you dont have any filters so ripple will come through.

I placed one on my Timming light that was 12 volts, and never needed to worry which clip was black or white, or missing its color.
ljbeng
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Post by ljbeng »

chris, I like that idea and the board is usally powered by 24vdc and I only need 5v and 3.3v on the board so the bridge should not be problem.
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Post by dyarker »

philba,

Would a N-MOSFET in the negative supply lead solve ljbeng's problem?????

haklesup,

This is a DC situation, so current should not exceed the diode's continous current rating. The 1N5400 is 3A diode, and the spec sheet says Vf is 1V max at 3A. In a power supply rectifying AC into a large filter cap, with a 3A load; the peak current through the diode will be higher. Then you do have to use the diode's higher Vf in the calculations. And if ambient temperature will higher than 25C, you have derate the max current. Unless you're making a lot of power supplies, it is easier not to exceed 3/4 of the diodes continuous rating. ie loads upto 750mA, use 1A diodes; upto 3A load use 5A diodes; etc.

With lots of spare voltage, the bridge is probably best.
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haklesup
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Post by haklesup »

Hopefully that diode will never see continuous duty, that a user would realize the polarity swap within a few seconds. If the supplies were to come up with a transient, that situation should also settle within a second. However, as I said, I have fixed a number of devices whit this diode blown, it is not foolproof and sometimes acts like a fuse itself. It does tend to limit damage further in even when some stuff gets blown.

Though the 1n4001 is a 1A diode, I was able to apply 5A to it for over 20 seconds without damage (it did get hella hot though). If you add to that bob scotts idea of a fuse, when the power is reversed, the diode conducts enough to blow the fuse. A diode would only need to shunt the overstress just long enough to blow the fuse. The spec sheet gives safe design values which will result in the stated reliability. My curve tracer was testing a real device. I got about 2.3V at 1A on that actual device.

Recall mey comment about the current rating of the power supply. say the application circuit operated at 12V and 1.0A nominally. Power that project with a 12V, 1.2A supply. Now if you get a short or reverse polarity, the diode needs to shunt only 1.2A and the supply should collapse to about 2.5V (acording to the curve trace) until the fault is cleared. Only if you put a very powerful supply on there would you fry the diode.

A bridge rectifier is nothing more than 4 diodes in one package. You can use that if you find it convenient or you can use seperate diodes. At that level it's simply a construction issue. If you have more than just one supply and ground, that may be useful. Discrete diodes are available in the same ratings as bridge rectifiers in general but many tend to have heat sinks so they can have higher current ratings. BRs do not necessarily have better characteristics than discrete diodes but they can when you compare two specific parts.
We cant conceive that a diode drop in one circumstance is negative, then the next its ok?>

What is needed is the experimentation of all the proper situations, for your self ,...is the proper situation
I don't understand this comment or I might reply. By negative do you mean the "current" or "not desireable".

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philba
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Post by philba »

dyarker wrote:philba,

Would a N-MOSFET in the negative supply lead solve ljbeng's problem?????
I assume it could be made to work but the common ground issue would be a worry.

I would also be worried about the common ground issue with the bridge - won't that effectively short out the two low side diodes?

By common ground problem - I am referring to this
The OP wrote:The serial port may or may not be connected to another device. The other device may or may not have a common earth ground connection back to the use supplied power. I need the protection all mounted on the "Electronics Micro Etc" pc board.
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