Author of this idea should be sued !
EDIT: actually not... read below.
You should not use this circuit !!!
Let me write a general description why. (This text below is applicable to most chips -> read as: DO NOT use this "solution" also in any other circuit, with any processor/microcontroller/logic, etc.)
This is a large subject I'm diving in now, so I'm going to simplify it a bit and exclude several details to keep the explanation straight-forward.
Chips have some protection elements attached internally to each input pin. Let's suppose that these protection elements are simple diodes, one is connected to the internal positive supply rail (Vdd, Vcc, whatever you name it), and the other is connected to the chip ground (GND, Vss). Their purpose is to protect the chip against some shocks (eg. static electricity). This "resistance to shocks" is also listed in datasheet for each chip, one of the measuring method is with so-called Human-Body Model (representing a person touching the chip pin)...
Ok, so, these protection elements are not needed for normal operation (the reason that there is not much discussion about them in general discussions), but are crucial in several other occasions that might quickly occur.
We know that diodes have some voltage drop when they are conducting (read as: they are burning some energy, so they are heating more or less; depending on their construction and current,... and also some other parameters).
(By the way: If you ever wondered why chips' maximum allowed voltages range is slightly above Vcc and slightly below GND, you've just got the answer.
Ok, let's take a look what is going on when you connect 5V to 3.3V input:
Since the voltage diff is 1.7V, the internal protection diode between input pin and Vcc internal rail starts to conduct (diode's anode is connected to input pin and its cathode to Vcc). As a result:
- complete chip is now supplied with 5V (slightly less due to diode voltage drop) and
- path for complete chip's supply current is established to flow through this protection diode.
Since this current is usually quite significant (according to diode's structure), it starts to overheat, the rest of the chip is also not very glad to have higher voltage then it should be.
Most likely, the protection diode will blow quickly. It can blow into two forms, either to conducted ("wire") or to disconnected state (like blown fuse).
In first case, you got input pin permanently connected to a Vcc (so the complete chip voltage got increased since there is no drop on a diode anymore).
In second case, you "just" lost proper protection (but fatal when you next time grab the circuit with your hands, for example) - remaining part of the chip might remain alive... until other elements on the input (transistors) don't start to slowly gave up due to permanent overvoltage.
Going back to this particular circuit - the resistor DOES NOT decrease voltage to 3.3V, but limits the current through the protection diode mentioned above. In other words, this resistor might save protection diode not to blow, but DOES NOT prevent chip core to be stressed with 5V as long as you provide logic "1" on the input.
Simple solution: If you add one resistor, you can get proper voltage converter. Connect additional resistor between input pin and ground. This is called "voltage divider" (resistive version, to be precise). Check Google for further details.
For digital circuits we're dealing with here, values of these resistors should be few kilo ohms (but keep their ratio, of course).
Best wishes, Ivan Zilic.