Need help understanding reverse voltage protection using a diode in parallel

An article I ran across in an issue of Make had a BB Arduino using a 1N4005 diode in parallel for reverse voltage protection to avoid voltage drop. Trying to learn more about this method I started searching around for some info and everything I've found indicates a fuse should be used between Vcc and the cathode of the diode. Can anyone take a look at this circuit and help me understand why a fuse isn't needed?

Link to article (fig 2.5 & 2.6 specifically): http://makezine.com/projects/make-36-boards/make-your-own-dmn-board-part-1-layout/

My fallback is to just use a 4001 in series between Vcc and the voltage regulator and use a 9v battery where the drop won't be noticed, but as the article states, you have more powering options if you can get rid of the drop. Any help would be greatly appreciated.

Cost?... diodes are cheaper.

I prefer the voltage drop method in series over this method ..

This is used because the battery voltage is 6v to begin with, if the battery is reversed the diode shorts killing itself... a resettable fuse would be better imo.

Basically, that rather than blocking the reverse voltage like the series diode, you create a crowbar to shunt the current through the diode. Assuming that the device has a higher forward voltage than the diode it should work fine. But if the diode fails, if you have a high current power supply, for example, then all the reverse voltage will be applied to your device.

In the series case, the diode is unlikely to fail.

Also, at the instant the reverse voltage is applied it is a race to see if your device can draw enough current to destroy itself before the diode pulls the voltage down to a safe level.

Frankly, I don't think this is a good practice.

The parallel diode to ground is popular in ham radio higher power mobile radios where any voltage drop decreases the maximum RF output power possible a little. But if has to be coupled with a fuse or being fed with a voltage source that has some kind of automatic voltage fold-back. For typical arduino type circuits the series diode is the better way to go, as no current will flow if input polarity is reversed.

Series diode, use a low Vf diode, like a schottky: http://www.digikey.com/product-detail/en/1N5817-TP/1N5817-TPCT-ND/950586 0.45V @ 1A about the best you can do for low-cost thru hole

0.36V if you want to pay a little more http://www.digikey.com/product-detail/en/RA%2013V1/RA%2013V1CT-ND/4572566

0.22V if you want to pay a litte more and can go SMD http://www.digikey.com/product-detail/en/LSM115JE3%2FTR13/LSM115JE3%2FTR13CT-ND/1634547

Or better still a fuse!

A polyfuse is my way :)

Polyfuse does not protect against reverse voltage.

Poly fuses take time to blow so they tend to be useless at protecting semiconductors.

If you use a reverse diode across the supply you are actually hoping that the power supplie's over current limit circuit kicks in before the diode melts, which is a good bet for most normal curmstances.

The circuit shown would protect against a reverse-connected power supply when the power supply has built-in current limits (either because of short-circuit protection in its regulator, or inherent in the power supply design.) This would be true of most wall-warts, probably. It would be scarily NOT true of, say, LiPo or Car batteries. The usual solution is a fuse or polyfuse in-line with the power supply; if the power is connected backward, the diode is forward-biased and conducts enough current to blow the fuse (much more than the correctly running circuit.) Such a fuse need not be on the board itself (fuses can be annoyingly large!)