filterimpedance-matchinglow pass
I am building a low pass filter. This is the equation I'm using to calculate what frequencies I want to pass though, but I have a couple questions on details.
$$
f_c = \frac{1}{2\pi RC}
$$
What goes into R? Does the resistance of the load (R2) affect R? Say R2 is fixed, are there any rules for selectring R1, like load matching?
simulate this circuit – Schematic created using CircuitLab
The cap near the power pin is not to protect the part from noise, but to keep the part from generating noise as the logic switching causes rapid changes in supply current. Ideally the cap would supply instantaneous demands for more current without increasing current all the way back to the power source.
The sum of the impedances on the PSU side of the circuit - the internal impedance of the PSU plus the inductance, resistance, and capacitance of the traces or planes - is enough to give some low pass filtering on the input side of the cap. I think of the cap as a tiny a power supply that can respond to demands with a bandwidth in the multi-MHz range. The larger regulators that supply a full circuit react far too slowly and the cap is a temporary source of power that replaces or bypasses (or decouples) the PSU. Placing the cap close to the power pin on a chip minimizes resistance and inductance that would slow the response.
CMOS parts consume most of their power while switching state. For microprocessors this means on clock edges and the current draw is in little fast spikes. The size of the spikes varies as fast as the clock as every instruction uses different combinations of internal circuits. Imagine the circuitry used in checking a register for zero versus fetching data from RAM. The power needed fluctuates at the clock rate. The greater the current changes, the bigger the cap. Calculating the right size is a matter of estimating for most of us and the 0.1uF ceramic cap is so common that it is very low cost. Capacitor construction is also a concern as well as change with temperature. Some can respond quicker than others and some vary by 80% over the commercial temperature range.
They are also called bypass caps because: 1) They can "bypass" (short) high frequency PSU noise to ground. 2) They can "bypass" the PSU and respond to high frequency demands for power.
Also called "decoupling caps", a more accurate term for high frequencies as they "decouple" the power demand between the part and the PSU.
Your low pass filter is loading your high pass filter. Think of a simple voltage divider: 
. In general, if you have 2 circuits connected together, the first one's output to the second one's input, you do not want the second one to attenuate the signal that passes through the first one. Therefore, you must make sure that the second circuit's input impedance (represented by R2 in the voltage divider) is larger than your first circuit's output impedance. How much larger? The rule of thumb is 10x larger.
The output impedance of your high pass is roughly the value of R1, or 10 kOhm. The input impedance of your second circuit is also roughly R2 = 10 kOhm. So you need to decrease R1 by a factor of 10 or increase R2 by a factor of 10 (or some other combination such that R2 = 10R1) and adjust C1 and C2 accordingly to obtain the same cutoff frequencies. Once you do, you will see that your bandpass filter behaves much better.
Along the same lines, you don't want to increase R2 beyond 500 kOhm because your load is 5 MOhm. The same rule applies!
Best Answer
You can use the Thevenin equivalent circuit to see what your capacitor "sees".
Your capacitor will see an input voltage of Vin*R2/(R1+R2) in series with a resistor equal to (R1||R2). Hence your load will affect your voltage amplitude and also your cut-off frequency. If it is a problem, you may use an op-amp configured as a voltage follower between the output of your filter and your load.