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The following is the diagram of interest:

enter image description here

The impedance of the filter is:

$$Z_o=\frac{R_f L_f s}{R_f L_f C_f s^2+L_f s+R_f}$$

and the bode plot with all different components and the total impedance should look like this: enter image description here

The way I intrupruted the frequency response is that the impedance follows the inductor at low frequency and it follows the capacitor at high frequencies and the resistor damps the resonance at the resonance frequency, which implies that the total impedance follows the resistor at the resonance frequency.

Now, what confuses me is that the instructor of the course I am taking said that this filter design is not good as it experiences resistor losses at low frequencies when the capacitor acts as an open circuit, which is also mention in Erickson and Maksimovic's book. enter image description here

But as far as I can tell, at low frequencies the resistor will not introduce losses because the current will flow therough the inductor which looks like a short circuit at low frequency and this is exacly what the bode plot shows.

So, what does the instructor mean when he said the resistor will create losses at low frequency?

The instructor proposed the following alternative design to fix "the loss issue".

enter image description here

P.s. After giving the problem some though and more research I think what the instructor was referring to is the power dissipation from the input side. The parallel RC design acts as intended if we look at its impedance from the output side. however, if we look at it from the input side we will see that at low frequency, The impedance is purly DC and this is what the instructor meant by resistive power loss at DC.

if we derive the impedance and plot its Bode we get:

$$Z_{in}=\frac{R_fC_fL_fs^2+L_fs+R_f}{R_fC_fs+1}$$ enter image description here

On the other hand, the alternative design will have the additional series capacitance C_2 dominating at low frequencies, which prevents the resistance from disspiating DC power coming from the input side $$Z_{in}=\frac{R_fL_fC_{f1}C_{f2}s^3+(C_{f1}+C_{f2})s^2+R_fC_{f2}s+1}{R_fC_{f1}C_{f2}s^2+(C_{f1}+C_{f2})s}$$

enter image description here

Does my interpretation make sense? please let me know if you think otherwise.

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  • \$\begingroup\$ Try it out in the "Falstad simulator", the oscilloscope settings have various options to plot all kinds of data, very good. \$\endgroup\$
    – NaturalDemon
    Feb 10 at 11:45
  • \$\begingroup\$ Some words/senetences in your contribution sound as if it would be a "bad" thing to have "losses" in a filter circuit. I think - in contrast, losses are necessary and desired because the have damping characteristics and are necessary to select a certain wanted filter response. Otherwise we have - in your example - steep resonance effects which cannot fulfill typical lowpass requirements. \$\endgroup\$
    – LvW
    Feb 10 at 12:10
  • \$\begingroup\$ Please don't listen to simulator recommendations unless supported by engineers with recognizably vast experience of using sim tools. The undamped and damped responses in your bode diagram appear to have their names swapped. Are you talking input impedance here? \$\endgroup\$
    – Andy aka
    Feb 10 at 12:11
  • \$\begingroup\$ Without R, at low frequency, the current wouldn't flow almost at all because the cap acts as an open. The resistor in parallel allows for current to bypass the high impedance cap, so I guess this is what your instructor interpret as loss. \$\endgroup\$
    – Sredni Vashtar
    Feb 10 at 12:46
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    \$\begingroup\$ A few things, please use \$s\$ in your Laplace expression and not a capital \$S\$. Second, the losses your instructor refers to are ohmic losses in the inductor (\$r_L\$) and the capacitor (\$r_C\$). Check the term equivalent series resistance or ESR in the web and you will find more information about these parasitics. If you now consider the ohmic loss of the inductor, in dc, when \$s=0\$, then the attenuation is simply \$H_0=\frac{R}{R+r_L}\$. Adding the extra capacitor may damp the filter - make it more lossy in ac - but it will not "fix" the dc loss : ) \$\endgroup\$
    – Verbal Kint
    Feb 10 at 12:57

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