eccdreded es 'bastante' recomendable conectar ambos VSS y ambos VDD. Conectando solo un par el pic anda, pero no es recomendable.
Te paso un par de links por si te interesa el tema, estan en ingles, si no entiendes algo por favor preguntame que te lo puedo traducir.
http://forum.microchip.com/tm.aspx?m=146914&mpage=1&key=Vss%2cpower𤘧http://forum.microchip.com/tm.aspx?m=116773&mpage=1&key=VSS𜤇http://forum.microchip.com/tm.aspx?m=116773&mpage=1&key=VSS𜣊Destaco las siguientes frases,
1) Many fast chips have more than one set of supply pins. The problem is that each and every logic operation takes a gulp of current from the power supply. You need a decoupling capacitor to provide this current without the power rails dropping so far that the chip might become unreliable. The longer the wires from the capacitor to the logic gates, the more series inductance there is, and so the more the voltage drop.
You should feed power in at every Vdd and Vss pin.
Capacitors built inside chips are very expensive. It is often cheaper to have multiple external capacitors and multiple power pins. Even when you factor in the cost of each and every hand-soldered capacitor on a discrete board.
There are exceptions, where convenience is of paramount importance. For example some members of the MAX232 family do not need as many external capacitors. But these parts are more expensive.
2) It is a must that you connect every VDD/VSS pair of pins. In some smaller chips, the multiple VDD/VSS pins connect at the same metallization, i.e., the power bus inside the chip is electrically connected via metal runs to all VDD/VSS pins, and in these chips the need to connect all power pins is to distribute current density and power dissipation. Even for those chips you should not leave unconnected power pins.
Since you never know how the internal connections are to the chip power grid, you must always connect all VDD/VSS pairs. On most VLSI chips, the different VDD/VSS pairs will power different parts of the chip. For current density and joule effect issues, it makes more sense to distribute the power through the chip area.
That's why you should decouple every pair of VDD/VSS pins as if they were independent.
The board layout needs to guarantee that all VDD/VSS pins will be at the same voltage, with a fraction of milivolt from one another, or a very nasty effect will result: on-chip current from one VDD pin to another VDD pin.
3) If you don't connect one pair, your board might work, but is not guaranteed to be reliable.
Saludos
