Electrical – Topology for 1st order active low-pass filter

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.

The problem arises from adding the \$7.5 \Omega\$ at the output. If you calculate the Thevenin Equivalent you'll get the following circuit, where the \$5 \Omega\$ resistor is the parallel equivalent of \$7.5 \Omega\$ and \$15.5 \Omega\$.

^{simulate this circuit – Schematic created using CircuitLab}

You can see that the output voltage has an insertion loss from the factor \$\frac {7.5} {7.5+15.5}\$ and it is where the 9.6 dB comes from.

The \$5 \Omega\$ equivalent resistor explains why the frequency cutoff changes to 2100 Hz.