I need a 1.95v for a VDDCORE of a microprocessor I'm using, what would be the best way to get 1.95V from 3.3v? or is a regulator the best way.
It's not necessary to provide the
VDDCORE voltage externally.
VDDOUT can provide the typical 1.8VDC for the core, so you can just connect this to
VDDPLL with external decoupling caps (described in section 5.3).
Section 5, and specifically 5.3 provides the info you're looking for:
VDDCORE pins. They power the logic of the device; voltage ranges from 1.65V to 1.95V, 1.8V typical. It can be connected to the VDDOUT pin with decoupling capacitor. VDDCORE is required for the device, including its embedded Flash, to operate correctly.
Voltage Regulator The SAM7X512/256/128 embeds a voltage regulator that is managed by the System Controller. In Normal Mode, the voltage regulator consumes less than 100 μA static current and draws 100 mA of output current. The voltage regulator also has a Low-power Mode. In this mode, it consumes less than 25 μA static current and draws 1 mA of output current. Adequate output supply decoupling is mandatory for VDDOUT to reduce ripple and avoid oscillations. The best way to achieve this is to use two capacitors in parallel: one external 470 pF (or 1 nF) NPO capacitor should be connected between VDDOUT and GND as close to the chip as possible. One external 2.2 μF (or 3.3 μF) X7R capacitor should be connected between VDDOUT and GND. Adequate input supply decoupling is mandatory for VDDIN in order to improve startup stability and reduce source voltage drop. The input decoupling capacitor should be placed close to the chip. For example, two capacitors can be used in parallel: 100 nF NPO and 4.7 μF X7R
And Section 5.4 shows a typical powering scheme.
A regulator is pretty much the only way. Depending on how much current you need you could use an LDO or a switching converter. Any other approach would probably not give you a stable and accurate enough voltage.