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Capacitive reactance in ohms is defined as:

$$X_C(\omega) = \frac{1}{2\pi f C}$$

where the unit of C is farad.

Focusing on the frequency \$f\$ in the equation:

If the input signal to the capacitor is a single current pulse with 50 ps rise time, 10 ps ON time and 100 ps fall time, how can we roughly quantify \$X_C\$?

Null
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user1245
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    You can also use I = C dV/dt (easily derived) and derive an equivalent "R" via Ohm's Law (cancelling out the V, since dV/dt is proportional to peak V for the described pulse) –  Oct 22 '21 at 12:59
  • What are you trying to do? You seem to be stepping off on the wrong foot. – stretch Oct 22 '21 at 14:49
  • user1245, I think @user_1818839 pretty much provides the shortest, simplest way to look at your question. Two answers, though. One from rise and one from fall times. Apparently, you already know there is some C that is being driven and you know the rise and fall times, too. So you already know your 1/f must be long enough to accommodate those and also the OFF time which wasn't mentioned. So a separate approach would be to use a numerical FFT (and windowing function) to confirm if there is a dominant 'f'. (Seems mostly triangular/sawtooth from what you write, though.) – jonk Oct 22 '21 at 19:27
  • Xc is defined for a sinewave. Xc is not reshaped by a time domain signal. It is what it is. – Andy aka Oct 23 '21 at 08:57
  • Strictly speaking, you've given $X_C(f)$, not $X_C(\omega)$ since there's no $\omega$ on the RHS. The only difference is a factor of $2\pi$, though. – Null Oct 25 '21 at 12:07

1 Answers1

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There are two main ways to approach this problem, namely (1) time domain and (2) frequency domain. You are thinking about the signal in the time domain, and you are trying to understand the capacitor impedance (resistance) in the frequency domain. This is actually the main source of confusion.

(1) Think about the signal in frequency domain. Since signal seems to be aperiodic, you would use Fourier transform to find how signal "looks like" in the frequency domain. Aperiodic signals, in general, consist of all frequencies (from 0 Hz to infinity). The Fourier transform would give you sine-wave amplitudes (and phase, but this is irrelevant here) for each frequency - this is called a spectrum. The equation you gave tells you how capacitor behaves for each of these frequencies. With this information, you can find the current spectrum.

(2) Think about the circuit in the time domain. In that case you do not use equation for capacitor impedance but rather time-domain description of the capacitor: \$i = \mathrm{C\ dv/dt}\$. Since current is a derivative of the voltage (times the capacitance), the slower the rise time the smaller the derivative, which means less current. The worst case is when you have step-wise change in voltage, in which case gives you infinite current. This is of course not possible in the real-world because you always have some resistance which limits the current. But, the resistance is not included in this analysis, so theoretically, for step-wise voltage change on the capacitor you get infinite current at that moment!

Note that the approach (2) will give you exact voltages and currents in the circuit, and the approach (1) is usually used to get "a feeling" about how circuit behaves for different types of signals. Faster the rise and fall times means more high-order frequencies. Since capacitor "amplifies" high-frequency components (higher the frequency higher the amplification gain), voltage signal with more high-frequency components will give you current with even more high-frequency components (read this as glitches), which is something you want to avoid.

Null
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Marko Gulin
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  • Why dont you say capacitor resists less in high frequencies instead of saying amplifies? – user1245 Oct 22 '21 at 13:30
  • Because in frequency representation we usually think in terms of "gain". When gain is in range (0,1) we say that the system attenuates (in your terminology this means "resists more"), and when the gain is greater than 1 the system amplifies (or "resists less"). But this "more" and "less" implies that you are comparing it to some other frequency, whereas attenuates/amplifies can be used with a single frequency. – Marko Gulin Oct 22 '21 at 13:36