To narrow the focus a bit from an earlier question:

Is a Schottky diode appropriate for reverse polarity protection?

I'd like to prevent mishaps from a user connecting DC power in reverse, but I'd also like as low a voltage drop as possible. Can you explain what the reverse leakage current is and whether it would be a concern or not in this scenario?

The application is a small device that operates on 9-12 volts DC at less than 100 mA.


Just as an example, I am expecting users to be able to use 6 AA cell batteries in series, either alkaline or NiMH. In the latter case, the batteries are 1.2V, so the total voltage is only 7.2V. I am using a 5V voltage regulator with a dropout voltage of 1.3V, so therefore my minimum operating voltage is 6.3V. A bias protection diode with 0.7V drop is going to raise that minimum to 7.0V. As the batteries are drained, I expect to dip below the 7.0V requirement very quickly, and therefore not use the full capacity of the batteries efficiently.

If a 0.3V diode is used, the minimum requirement is lowered to 6.6V, which I feel is a better fit for use with NiMH batteries.


3 Answers 3


The bigger problem you're likely to run into is operation under forward bias conditions. Schottky diodes still have a voltage drop under forward bias, say 0.25V.

That means at 100mA, you're dissipating 25mW of power. Better than a standard silicon diode, but not great especially for a battery constrained device.

A better way to get reverse bias protection is to use a P-Channel MOSFET. MOSFET's act more like a resistor when saturated, and it's possible to get MOSFETs with low on resistances.

Let's assume we have a 1 ohm on resistance. At 100mA, that's a 0.1 V drop across the MOSFET and 10mW dissipation. 1 ohm on resistance is kind of lousy for a MOSFET, you can get some which have significantly less on resistance. I'm not entirely sure about the leakage current through MOSFET's, but I seem to remember it being quite small.

To hook up the mosfet:

Connect the drain to the positive battery terminal, connect the gate to the negative terminal, and connect your load to the source. For added protection you can add a zener diode and a resistor across the source/gate.

A more complete explanation can be found here.

  • \$\begingroup\$ Thanks for this suggestion. I wasn't aware of using a MOSFET in this way (I normally use them to drive solenoids and motors!). I am planning on also using a zener diode plus fuse for overvoltage protection. I am experimenting with zener diodes versus transient voltage suppressors (TVS). \$\endgroup\$
    – JYelton
    Commented Jan 10, 2013 at 3:40
  • 3
    \$\begingroup\$ You can also do the same with an N-channel MOSFET on the low side, which can be helpful because it's often easier to find N-channel with lower RDS(on). See TI SLVA139: Reverse Current/Battery Protection Circuits: ti.com/lit/an/slva139/slva139.pdf \$\endgroup\$
    – Joe Baker
    Commented Jan 10, 2013 at 10:14
  • \$\begingroup\$ I've tried to use Schottky diodes to separate voltage sources. I have 2 solar panels that (should be) connected in parallel. Only one panel is lit at the time. So I wanted to avoid current leakage from active panel to another one. I have found that Schottky diodes don't help me to separate the sources due to high reverse current leakage. The solution with MOSFET looks the best approach to the problem. Thank you for the idea! \$\endgroup\$ Commented Jun 4, 2016 at 10:59

I'd say that a Schottky is an excellent choice as its forward voltage is less than what the ESD (Si) diodes on your semiconductors probably have. You are then diverting this mistake, and its current away from more delicate parts into something that is beefier / more robust.

Reverse leakage can be an issue with battery-powered devices. Schottky's do have higher reverse leakage current (due in part to their lower forward voltage). This spec will be in the datasheet and increases with temperature.

Reverse leakage current is simply the deviation away from the ideal case in which there would be zero current flow. Having reverse leakage means it will eat some of your battery power. At 100 mA it will be unnoticeable, but it is the offstate leakage that you care about. Just make sure that the Schottky is AFTER the power switch and its leakage will be moot at that point.

  • 1
    \$\begingroup\$ When you say it will eat some of the battery power, are you are referring to the state in which the circuit is hooked up in reverse bias? The diode will definitely be after a power cutoff switch. \$\endgroup\$
    – JYelton
    Commented Jan 10, 2013 at 3:37
  • \$\begingroup\$ I'm just sayin that the leakage current is a liitle bit more consumption. that's all. I should have said REVERSE leakage. I'll fix. \$\endgroup\$ Commented Jan 10, 2013 at 3:39
  • 2
    \$\begingroup\$ How can it leak current if you've got a power switch (off)? Regardless of it's infront or behind the switch, current can't flow through? \$\endgroup\$
    – aaa
    Commented May 10, 2016 at 11:43

Reverse leakage current would only apply in the opposite polarity case. This is current that is leaking through the semiconductor material when the diode is in the off condition because it is incorrectly biased. Thus in your application, it is not an issue.

However, you are right to be wary of the forward voltage drop. Only you can decide what an acceptable voltage drop is based on what your power source is and the voltage needs of your regulator or power supply are. Since you are buffering your power supply using a voltage regulator, I would guess that a small voltage drop (less than 0.7 V) would be acceptable. As an example, such a voltage drop might be important if the circuit protection was for a analog voltage sensing application instead.

  • 1
    \$\begingroup\$ I've added some additional information pertaining to what's acceptable and why. \$\endgroup\$
    – JYelton
    Commented Jan 10, 2013 at 3:21

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