Input Leakage Current
To determine your resistors voltage drop from the gate you need to use the leakage current from the datasheet. Microchip specifies an "Input Leakage Current" on their datasheets. The [datasheet that I have looked up][1] specifies an input leakage current of 1uA. This could cause a .1V or 100mV, which is only double what Robert calculated, probably not a problem on your signal.
Now remember, if you are dividing a 30V signal down to 30/11 (2.7v) volts full read then the 100mV is added to this, causing up to 3% error on your 30V signal.
If you need a resolution of 1V, divide that by 11 and then add the 100mV. This 100mV could be larger than the 1V signal.
Input Capacitance
Robert is correct, there will be a capacitance, but this really specifies an amount of time that is needed to take the ADC measurement. This also, combined with the input resistance you chose, creates a low pass filter; if you were wanting to measure signals with a higher frequency, you are not going to be able to capture them.
Reducing the error
The easiest way is to either reduce your resistance on your divider, or to buffer your signal. When you buffer the signal you will replace the PIC's leakage current with your op-amps leakage current which you can get quite low.
This 1uA is a worst case, unless it costs you a large amount to make minor changes to the design, fab your design and test how bad it is for you.
Please let me know if there is anything I can do to make this easier to read.
Yes, from the datasheet:
Due to the analog input resistive divider formed by R1
and R2 in Figure 5, any significant analog input source
resistance (R SOURCE) results in gain error. Furthermore, R SOURCE causes distortion due to nonlinear
analog input currents. Limit RSOURCE to a maximum
of 100Ω.
So practically, this means you need to drive the input with an amplifier. Although, a resistor divider made with a 192 Ω and 200 Ω resistor will meet specs if your signal has a low output impedance.
Best Answer
Nothing to do with aliasing. I won't go into what aliasing is but it is so far removed from the issue that you should go do some reading to understand what aliasing actually is before ever suspecting or worrying about aliasing in anything.
It's telling you how much the input will load down a signal source you feed to it, and adding a 300K effectively increases the impedance of the signal source and would utterly demolish the ADC's reading of the signal.
simulate this circuit – Schematic created using CircuitLab
Voltage divider giver: \$V_{load} = \frac{R_{load}}{R_{load}+R_{source}}V_{source}\$
Obviously you want \$V_{load} = V_{source}\$ which requires
\$input.impedance.of.load >> output.impedance.of.source\$
so as much voltage as possible appears across load and not lost in the source via voltage divider.
So do you see the problem if you make \$R_{source}\$ significant relative to \$R_{load}\$? It would result in something akin to drinking a sizeable portion of the wine barrel in order to sample the taste.
If you want to protect the ADC with a large series resistance, put a buffer in front of the ADC and diode clamp the input to the buffer. The buffer will have much higher input resistance so can tolerate a larger (but still small relative to the buffer) input resistance and not load down the signal as much. Simultaneously, the buffer has a low output impedance to work well with your ADC's input impedance.
simulate this circuit
The diodes, whether external or internal (for ESD protection), clamp the voltage to within on forward diode voltage drop of the power rails and are what actually protect the pin. The resistor in turn protects the diodes from frying by limiting the current through the diodes and dropping the extra voltage across the resistor rather than the diodes (which are basically short circuits when conducting).