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Consider the following circuit:

schematic

This is a simple first-order low-pass RC filter. The cutoff frequency is:

\$ f_c = \dfrac{1}{2 \pi (1k\Omega) (1\mu F)} \approx 159 Hz \$

The reactance of C1 at \$f_c\$ is:

\$ X_{C1} = \dfrac{1}{2 \pi (159 Hz) (1uF)} = 1k\Omega \$

So what if we pretend that C1 was just a resistor, with a resistance that's a function of frequency? If you are a 159 Hz sinusoid, then the circuit might as well be this (I think...):

schematic

At least, if we only care about the frequency response, and not the phase response, of the circuit. The voltage between T1 and T2 is half of V1, and this is consistent with the calculation of the cutoff frequency above.

I think with this circuit, we can pick any frequency, calculate the reactance, replace C1 with a resistor of that value, and get an equivalent circuit for the purposes of calculating the frequency response.

But what about this circuit?

schematic

Maybe not so much. Under what circumstances can we simplify impedances this way?

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    \$\begingroup\$ I wouldn't ever. What you're missing is Z = jX. The reactance gives the magnitude of the impedance, but it's at "right angles" to a resistive impedance. And, as your last example shows, ignoring this can make you miss big effects. \$\endgroup\$
    – The Photon
    Jan 23, 2013 at 23:39
  • \$\begingroup\$ @ThePhoton doh...drawn incorrectly. I'll fix. \$\endgroup\$
    – Phil Frost
    Jan 24, 2013 at 0:54
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    \$\begingroup\$ @ThePhoton anyway, I fully understand that ignoring the complex part of impedances leads to errors in some (maybe most) circumstances. The appeal of converting reactances to impedances is that it's more intuitive to think about frequency dependant resistors than complex numbers. Complex numbers are mathematically elegant, but not intuitive at all. I'm just looking to better understand the bounds of this simplification, and maybe get some insight into why it works sometimes. \$\endgroup\$
    – Phil Frost
    Jan 24, 2013 at 1:03

2 Answers 2

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Unfortunately, you cannot simply replace reactances by resistances because you are then ignoring the phase shift. In your first example, the cutoff frequency of a low-pass filter is defined as the frequency for which the output voltage is 3 dB down (0.707) from its maximum, not half. So replacing the capacitive reactance with an equal value resistance does not yield the correct voltage output. This is because the voltage and current in the real circuit with the capacitor are not in phase, but you are forcing them to be in phase by replacing capacitive reactance with a pure resistance. I really don't think frequency dependent resistors (which are not available as passive components) are more intuitive than capacitors and inductors which are easily constructed as passive components. In your second circuit, you completely lose the concept of resonant frequency because, unlike capacitive reactance and inductive reactance, resistances are always positive. Thus R3 and R4 can never cancel each other out. I think you need to stick with reactances and learn to apply them to real circuits. It will be worth the effort.

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  • \$\begingroup\$ Well, you don't have to tell me that the 2nd circuit doesn't work. That's why I gave it as an example. But you have corrected me on one point, and that is the first circuit doesn't work, either. I guess I forgot to consider that "-3dB" means not \$V_1/V_2 = 0.5 \$ but rather \$ V_1^2/V_2^2 = 0.5 \$ or \$ P_1 / P_2 = 0.5 \$. I guess I never noticed the error since I've never had to apply this analysis to anything more demanding than an audio AC coupling circuit. \$\endgroup\$
    – Phil Frost
    Jan 24, 2013 at 2:55
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It fundamentally misses the energy storage aspect of the components, which will manifest itself a a phase delay or resonances depending upon the order of the system.

But you know that already. So I will extend the thoughts to two further areas.

Case 1): Thought experiment: in the case where you are measuring the time averaged power of the system at a single frequency in a down stream detector or circuit. You might do as a A-B comparison for noise for example.

but there is a real world example where this is done all the time -> replacing resistors with capacitors.

Case 2): Switch cap circuits: where in the combination of a capacitor and a sampling/shorting switch gets you the equivalent resistance in sampled systems. This is done because the area and power consumed is far less, it is compatible to standard CMOS processes and the matching is better.

\$\ R_{equ}=1/{4Cf_{clk}}\$ is the formulae used for the translation from C and F to R.

This is how almost all CMOS ADC's DAC's etc. are designed.

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