2
\$\begingroup\$

The current through a capacitor is given by \$i_c=C\cdot \frac{dv_c}{dt} \$.

Let's say the voltage across the capacitor is a cosine wave. \$v_c=\cos(\omega t) \$.

Due to the complex exponential function we can write this as \$v_c=\Re(e^{j\omega t}) \$.

Let's calculate the current $$i_c=C \cdot \frac{d}{dt}\Re(e^{j\omega t}) $$ $$i_c=C \cdot j\omega \Re(e^{j\omega t}) $$ $$i_c=j\omega C\cdot v_c $$

The impedance is defined as \$Z_c=\frac{v_c}{i_c} \$.

Which finally makes us arrive at $$Z_c=\bigg(\frac{i_c}{v_c}\bigg)^{-1}=\frac{v_c}{i_c}=\frac{1}{j\omega C} $$

A similar thing can be done for an inductor.

My question is, do these formulas always hold? In the derivation I assumed that the voltage was a sinusoid (well, a phase shifted sinusoid) but this is not always the case.

What if the voltage across the capacitor is a sawtooth function or maybe a triangle wave? Then the derivation above wouldn't work at all.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ I assumed that the voltage was a sinusoid ...but this is not always the case. A sinusoid is the "base" of all waveforms. Any waveform, whatever the shape can be constructed by a sum of multiple sinusoids (Fourier sum), for each of those sinusoids, the above applies and all can be summed (super position). \$\endgroup\$ – Bimpelrekkie Feb 16 at 17:09
  • 4
    \$\begingroup\$ @Bimpelrekkie To be pedantic, a sinusoid is one of infinitely many possible bases of all waveforms. \$\endgroup\$ – Hearth Feb 17 at 1:51
  • \$\begingroup\$ Just so that you do not get surprised later: for real capacitors this only holds approx up to a certain frequency. \$\endgroup\$ – lalala Feb 17 at 8:40
  • \$\begingroup\$ @Bimpelrekkie You might be interested in the Walsh-Hadamard transform. Or wavelets. \$\endgroup\$ – user253751 Feb 17 at 10:49
7
\$\begingroup\$

The differential equations which use \$\frac{di}{dt}\$ and \$\frac{dV}{dt}\$ are more fundamental. They do not do not care about any abstractions such as "frequency", "sinusoids", or "canned" waveforms which, in a sense, need you to know what is going to happen in the future. As a result, you can always use the differential equations.

The impedance equations which use \$\omega\$ are derived from the differential equations using sinusoids as inputs. If you choose to work with the impedance equations instead of the differential equations then you need to break the input down into component sinusoids using Fourier analysis, perform the analysis for each sinusoid than add them up again at the end via super position. Don't forget to account for the phase shifts.

\$\endgroup\$
3
  • \$\begingroup\$ The only problem with dI/dt , dV/dt is measuring it beyond the BW of your fastest DSO, whereas impedance or network analyzers are needed to do this. \$\endgroup\$ – Tony Stewart EE75 Feb 18 at 1:14
  • \$\begingroup\$ @DKNguyen does the signal have to be periodic? That's what I'm thinking, because you can't Fourier transform a non-periodic function, right? \$\endgroup\$ – Carl Feb 18 at 15:07
  • \$\begingroup\$ @carl No real signal is truly periodic because it starts and ends rather than having always existed and existing infinitely into the future. You can Fourier a non-periodic signal just fine but it will have infinite frequency components because that is what is required to make it perfectly not exist in the past and perfectly stop in the future. That's why you get other frequencies in the spectra on a real sinewave signal if you Fourier it from start to finish. \$\endgroup\$ – DKNguyen Feb 18 at 15:14
3
\$\begingroup\$

My question is, do these formulas always hold?

Apart from the physical limit cases when transmission line theory take over, the formulas always hold irrespective of waveform shape: -

$$I = C\cdot \dfrac{dv}{dt}$$ $$V = L\cdot\dfrac{di}{dt}$$

\$\endgroup\$
2
  • \$\begingroup\$ When you refer to transmission theory, I assume you mean distributed vs lumped models. \$\endgroup\$ – DKNguyen Feb 16 at 17:24
  • \$\begingroup\$ I'm referring to components whose physical size make it impossible to treat them with the standard formulas due to the frequency being applied having a wavelength that is significant. \$\endgroup\$ – Andy aka Feb 16 at 17:27
1
\$\begingroup\$

The impedance formulae always hold true (within specs) but the spectrum of the input signals can be varied from sinusoidal so the response depends on the circuit transfer function. s domain plots or Smith charts or Bode amplitude and phase plots will demonstrate this.

\$\endgroup\$
1
\$\begingroup\$

My question is, do these formulas always hold?

The answer is really "yes" and "no". Other answers have explained the "yes" answer, but they all depend upon a capacitor or inductor being "ideal". Real capacitors and inductors have "stray reactances" and resistances. But even if we ignore these, real capacitors have dielectrics which are not a vacuum (although air comes close). Real inductors have cores which are not a vacuum (although air again comes close).

The significance of these facts is this:

The reactance of a real capacitor will deviate from that of an ideal capacitor, and that deviation will depend upon both frequency and upon amplitude. The permittivity of every real non-vacuum dielectric is non-linear, (although air comes close).

Similarly, the reactance of a real inductor will deviate from that of an ideal inductor, and that deviation will depend upon both frequency and amplitude. The permeability of every real, non-vacuum core is non-linear (although air comes close).

Everything that is said above about the reactances deviating from ideal applies also to the differential equations that govern ideal capacitors and inductors. Real components will behave differently from the differential equations for ideal capacitors and inductors

\$I = -CV'\$

\$V = -LI'\$

even if stray inductance, capacitance and resistance are accounted for.

Design objectives in practical power circuits generally include minimizing bulk, weight, and cost. Unfortunately, these objectives conflict with linearity of components. Inductors with magnetic cores are highly non-linear, but are used in power circuits because they are smaller, lighter and cheaper than their more linear equivalent counterparts. Similarly with capacitors. In practical circuits where these components are used near their voltage or current limits, their reactance may differ quite significantly from the values obtained with small signals. Power supply engineers generally need to take the non-linearity of their reactive components into account.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.