Electrical – Pi filter design for SMPS output ripple

Let's get you started on just the first filter for now.

The first filter is just a simple inverting op amp amplifier. For such a circuit with input impedance \$Z_{I}\$ and feedback impedance \$Z_{F}\$ the transfer function is simply $$A(j\omega) = \frac{v_{O}}{v_{I}} = -\frac{Z_{F}}{Z_{I}}$$

An ideal op amp can force the output to any voltage necessary regardless of the load so \$R_{L}\$ does not affect the transfer function.

In this case you simply have $$Z_{I} = R_{1}$$ and $$Z_{F} = R_{2}||\frac{1}{j\omega C}$$

Simply plug these impedances into the transfer function above to get the transfer function for your filter (I'll leave that as an exercise to you).

For the DC gain you simply set \$\omega = 0\$ in the transfer function and solve for the resistance values that give you \$|A(j\omega)| = |A(0)| = 1\$. Note that the capacitor does not affect the DC gain since it is an open circuit at DC. Mathematically

$$\lim_{\omega \to 0}\frac{1}{j\omega C} = \infty$$

and infinite impedance is simply an open circuit. The above limit should also provide some insight as to why \$\omega = 0\$ corresponds to DC. As \$\omega \to 0\$ the frequency gets lower and lower until it's just a constant value -- i.e. DC.

Recognizing that the capacitor is an open circuit at DC you can ignore it and simply set \$R_1\$ and \$R_2\$ to give you a DC gain of \$1\$ (by inspection, this is \$R_1 = R_2\$). The resistors aren't fully determined by the DC gain requirement, they just have to be equal. You can choose a reasonable value like \$1\$k\$\Omega\$ for them. (I'm assuming the specified DC gain of \$1\$ is an absolute value since you have an inverting topology -- there's no way to make it positive.)

The cutoff frequency \$\omega = 1000\$ is used to set the value of the capacitor. Set \$|A(j\omega)| = 0.7\$ and \$\omega = 1000\$ and solve for \$C\$. The fact that \$R_1 = R_2\$ makes this easier.

The last part about "physically realistic" values means that you can't use very specific values for your resistors and capacitors like \$1074.23\Omega\$. Choose standard resistor and capacitor values which come the closest to your desired cutoff frequency \$\omega = 1000\$. Use series and parallel combinations of resistors and capacitors for more accurate values -- e.g. you can form \$500\Omega\$ out of two standard \$1\$k\$\Omega\$ resistors in parallel.

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!