Energy dissipation in the resistor of a passive lowpass filter

[pwood]

Hello everyone, this is my first post, and I hope I have put this in the right area. I am looking to understand something about the behavior of a low pass filter. When an alternating current is passed through a low pass filter, the higher the frequency, the lower the voltage output of the filter. Therefore, the voltage drop across the resistor must become larger with higher frequencies. Is this because the capacitor becomes less of an interruption in the current flow, with respect to time? That is, because current is flowing more often (probably the wrong word to use), the resistor dissipates more energy? Thanks in advance.
-PWood

 

[yungman]

It depends on the circuit the LPF drive also. If you are talking about a simple RC LPF driving into a high impedance load, yes, more power is wasted on the resistor at high frequency beyond the cut off frequency.

that's the reason a lot of LPF don't use RC, they use LC instead where the series element is/are L and the impedance goes up with frequency and draw less current. Case in point, look at the passive Bessel, Butterworth filters etc.

 

[marcusl]

It depends on the circuit the LPF drive also. If you are talking about a simple RC LPF driving into a high impedance load, yes, more power is wasted on the resistor at high frequency beyond the cut off frequency.

that's the reason a lot of LPF don't use RC, they use LC instead where the series element is/are L and the impedance goes up with frequency and draw less current. Case in point, look at the passive Bessel, Butterworth filters etc.

Ok up to last sentence. Butterworth filters can be dissipative and, in fact, a one pole RC filter has a Butterworth response.

 

[yungman]

Ok up to last sentence. Butterworth filters can be dissipative and, in fact, a one pole RC filter has a Butterworth response.

I mean not as bad, if you have a source resistor and termination resistor, you always have dissipation. Most LC filter should be an improvement to RC filter.

 

[alphysicist]

I can provide an explanation for the energy dissipation in the resistor of a passive lowpass filter. The behavior described in your post can be attributed to the relationship between the capacitor and the resistor in the filter circuit.

In a lowpass filter, the capacitor acts as a frequency-dependent impedance, allowing lower frequencies to pass through while blocking higher frequencies. As the frequency of the alternating current increases, the capacitor's impedance decreases, allowing more current to flow through it. This results in a larger voltage drop across the capacitor and a smaller voltage output from the filter.

The resistor, on the other hand, has a constant impedance regardless of frequency. As the current passing through the resistor increases with higher frequencies, more energy is dissipated in the form of heat through the resistor. This is because the resistor converts electrical energy into heat, according to Ohm's Law (V = IR).

Therefore, the larger voltage drop across the resistor at higher frequencies is due to the increased current flow and subsequent energy dissipation. This can also be thought of as the resistor "absorbing" more energy from the circuit as the frequency increases.

I hope this explanation helps clarify the behavior of a lowpass filter and the role of the resistor in energy dissipation.