Reverse Polarity Protection Circuits

Introduction: Reverse Polarity Protection Circuits

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It is often useful to provide protection against accidental reverse polarity for your circuits. This brief review will explore three simple methods for adding this protection to your projects. For a more in-depth tutorial, see this article.

Diode
Simply using a diode as shown in the first circuit is often a good approach. The advantages are simplicity and cost. The disadvantages are larger power loss for larger circuit loads and a substantial voltage drop. Normal rectifier diodes typically drop around 0.8 volts. If your circuit draws little power and can handle such a drop, then the blocking diode will work for you.

You can improve this circuit somewhat by using a Schottky diode. It has a lower voltage drop - usually about 0.6 volts, but you can get some lower than that. There is one potential problem with using Schottky's though. They have more reverse current leakage, so they may not offer sufficient protection. I suggest avoiding using Schottky diodes for reverse protection.

PNP Transistor
A greatly improved protection circuit can be provided by using a pnp transistor as a high-side switch as shown in the second circuit. The saturated voltage is much lower than it is with diodes, so the voltage drop and power loss are much lower.

The limitations of this approach is the fact that there is some power loss from the base current, and that loss is constant regardless of the circuit's current power draw. In circuits where a very low quiescent current is typical, this approach could greatly increase it. For circuits which are usually active and draw modest amounts of power, this simple type of protection is hard to beat.

P-Channel FET
For the ultimate in low voltage drop and high current capability, replacing the PNP transistor with a P-channel FET as shown in the third circuit can't be beat. Please note that the FET is actually installed in the reverse orientation as it would normally be used. This direction is so that the slight leakage current through the FET's intrinsic body diode will bias the FET on when the polarity is correct and block current when reversed, thus shutting off the FET.

If the supply voltage is less than the FETs maximum gate to source voltage, you only need the FET, without the diode or resistor. Just connect the gate directly to ground. If after checking your FET's spec sheet, you find that Vcc could exceed the maximum Vgs, then you must drop the voltage between the gate and the source. The circuit shown does exactly that. By inserting a zener diode with a voltage less than the maximum Vgs, it limits the voltage to a safe level. Calculate the resistor value to provide enough current to properly bias the zener diode chosen.

Choosing
Each of these circuits offer a different set of advantages and disadvantages. I have listed them in order of increasing complexity and cost. In choosing what is best for your circuit, examine what your voltage and power needs are. Then use the simplest circuit that will suffice for those needs.

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    15 Discussions

    0
    nwsayer
    nwsayer

    1 year ago

    I have traditionally used a reverse biased rectifier diode across the input. It presents a short-circuit to reversed input voltages, but more often than not the source supply either has a fuse or is a modern switching supply with short circuit protection (hiccup mode). The advantage this scheme offers is that there is no voltage drop or impedance in the power path at all (normally). In some cases, you can substitute the rectifier diode for a TVS diode to get some measure of transient voltage excursion protection as well.

    0
    AdrianF2
    AdrianF2

    Reply 1 year ago

    Yes, where the droput is not an issue. You lose more than 1V with a rectifier bridge

    0
    domints
    domints

    Reply 7 years ago on Introduction

    Bridge rectifier looks like the best solution because no matter how power source is connected, the output will always have the same polarity. But there's one big disadvantage. High voltage drop, twice as in first method, between 1,2 - 1,8V, sometimes more... So if you supply 12 volt, your device will get a bit more 10 volts. But if you supply 1V, you won't get anything a output, because diodes will 'eat' everything.
    Of couse sometimes it doesn't matter, but often does :)

    0
    OkiA4
    OkiA4

    3 years ago

    i use battery li-ion 18650(3.7 v) to arduino pro mini 3.3v.

    pnp or mosfet?

    0
    AdrianF2
    AdrianF2

    Reply 1 year ago

    MosFet due to low droput.

    0
    Prasanna12525
    Prasanna12525

    2 years ago

    Dear sir,
    I have small problem with one circut the problem is wen relay short condition negative voltage flows from short relay throw positive terminel so that fuse will dorn so many times so my reqairment is any voltage from relay (+) terminol to source shuld not be go . can you say solution for thet. relay source voltage is 110vDC current : 1A plz give solution this is my email id is: prasanna12525@gmail.com

    0
    AdrianF2
    AdrianF2

    Reply 1 year ago

    IRF5210 will do the job.

    0
    mrmonteith
    mrmonteith

    2 years ago

    Using the FET I wasn't sure on selection for a circuit that draws around 2-3amps? Judging from the datasheets when they're turned on the resistance is for all practical purposes zero. Is that right?

    0
    zaherdirkey
    zaherdirkey

    3 years ago

    Thank you, Any good PNP transistor example to protect Ateml328 with 5v?

    0
    farmerkeith
    farmerkeith

    Reply 3 years ago

    For Atmel328 at 5V, I think you can use any general purpose PNP transistor. The Atmel328 itself only draws about 20 mA. However your circuit may have peripherals that take more current. Still, a PNP like the 2N3906 has a max current of 200 mA so it should be plenty - although I think if you actually need that much current then a) you should use either a transistor with more current capacity (eg TIP125) or go to the FET solution which is better anyway.

    For FET I would look at IRF4905 for a high power solution, or BS250 for lower currents. Both these FETs have a +-20V Gate - Source rating so you don't need the resistor and zener diode in circuits running at or below 12V, even with unregulated power supplies.

    1
    FilipW
    FilipW

    4 years ago

    Don't forget that you can just as well use an NPN or N-FET transistor, and place it on the negative side of the input (mirror the circuit). N-polarity transistors usually have better characteristics. But keep in mind that this cuts off the ground, and the ground can sometimes find undesired paths. On isolated circuits, this is not an issue.

    0
    tesla man
    tesla man

    7 years ago on Introduction

    Could you use an LED in the place of the diode?

    0
    emajko159
    emajko159

    Reply 4 years ago on Introduction

    You can. But classic signalization LEDs have max current 20mA. Of couse you can use High power LEDs, but if you do not need strong light source, you get needless strong light and heat.