I fail to see how the design presented on that website would do anything other than burn out the MOSFET if you tried to pass much current through it. If you look at the next few posts in the thread, it's people telling the OP that it won't work.
In order to turn on the MOSFET, you need Vgs to be high enough (datasheet will tell you what this needs to be; it starts turning on an Vgs(th), but to be fully on, it should be at least the lowest voltage for which Rds(on) is specified 4.5v for a typical logic level MOSFET). But if the FET is on, then resistance between drain and source would be very low, so Vgs wouldn't be high enough to turn the FET on.
What would happen with that circuit is that the voltage across the FET would settle to a voltage where the FET is part-way on, and the high Rds would result in the MOSFET heating up and burning out.
A MOSFET can be used as an ideal diode, but that circuit will not achieve it - and you need active control driving the gate. You can't just wire it up as simply as you would wire a diode (think about it, if this worked, who would use a diode?)