7
\$\begingroup\$

Why is a capacitor a linear device?

One property for linearity is that the capacitance or some such parameter must not change with voltage or current. Is this enough to make a device linear?

A few sources say that the \$Q=CU\$ has a linear characteristic with voltage and so it is a linear device but wouldn't there be at least one such parameter in a MOSFET/diode that does change with respect to voltage or current in a linear manner - for example the voltage of a diode decreases linearly with the temperature.

So what should I exactly consider for linearity?

\$\endgroup\$
  • 2
    \$\begingroup\$ All calculations that we do are oversimplification of reality. If it gets you the numbers you need to do some decisions, then we use those calculations. Thats what physics is about. \$\endgroup\$ – PlasmaHH Jun 7 '16 at 11:43
  • 7
    \$\begingroup\$ It has non linear I-V characteristics NO it does not You start talking about "a" capacitor, you do not specify what type. Until you do we talk about an ideal capacitor which is a linear device. You should reformulate your question because now it looks like you don't have a clue. \$\endgroup\$ – Bimpelrekkie Jun 7 '16 at 11:47
  • 3
    \$\begingroup\$ The definition of linearity, and it's a necessary and sufficient definition to use superposition, is that if f(a+b) = f(a) + f(b), then the function f is linear, its response to the sum of two stimulii is equal to the sum of the responses to those stimulii individually. \$\endgroup\$ – Neil_UK Jun 7 '16 at 12:34
  • \$\begingroup\$ @FakeMoustache Yes you are right, IV is linear for an ideal. Was thinking something different and wrote it wrong, I have removed it now. \$\endgroup\$ – Bhuvanesh Narayanan Jun 7 '16 at 13:19
  • 1
    \$\begingroup\$ The term linearity has different meanings. For two variables x and y linearity means that their relation is y = const * x. But in this context we are looking at functions of time i(t) and u(t). For functions linearity has a more "extended" meaning, namely u(t) = F[ a * i1(t) + b*i2(t)] = a * F[i1(t)] + b * F[i2(t)], i.e. F[..] is a linear operator. See my answer below. \$\endgroup\$ – Curd Jun 7 '16 at 19:37
22
\$\begingroup\$

First of all, an I-V curve does not make any sense for a capacitor. This is because a capacitor follows the following equation: $$i = C \frac{dV}{dt}$$

Note that the current depends on the rate of change of voltage. So you can have the same current at two different voltages, if the rate of change is the same.

The reason a capacitor is a linear device is because differentiation is linear. Superposition becomes: $$i_1 +i_2 = \frac{d}{dt}(v_1 + v_2) = \frac{dv_1}{dt} + \frac{dv_2}{dt}$$

\$\endgroup\$
  • \$\begingroup\$ Thank you for your quick answer , yes that was the part I missed out to consider ! I was thinking too much but did not consider that differentiation also accounts for linearity. \$\endgroup\$ – Bhuvanesh Narayanan Jun 7 '16 at 20:33
5
\$\begingroup\$

Your assumption is wrong:

It has non linear I-V characteristics

An ideal capacitor, just like an ideal resistor, has linear I/V characteristics.

Since you're obviously learning linear circuit analysis (judging by your knowledge of the superposition principle), I'm absolutely certain you've learned (or will very soon learn, by reading your course's material) about representing harmonic currents by complex currents.

With complex currents and voltage representations, it's really easy to see that a capacitor is a linear device.

\begin{align} I(t) &= C\frac {dU(t)}{dt} & \text{the elementary capacitor formula}\\ &\text{hinting at linearity}\\ I_\text{sum}(t) &= C\frac{dU_\text{sum}}{dt}&U_\text{sum} = U_1+U_2\\ &\overset!= C\frac {dU_1(t)}{dt}+C\frac {dU_1(t)}{dt}\\ \end{align} which is the case because the differentiation \$\frac d{dt}\$ is linear.

You might really just be confused by "linear" as term.

\$\endgroup\$
  • 1
    \$\begingroup\$ by the way, if you look at this in the purely DC static case: no matter what constant voltage you apply to a capacitor, after everything has settled, the current going through the cap is 0; N times 0 is still zero, so linearity isn't really broken in any case. \$\endgroup\$ – Marcus Müller Jun 7 '16 at 11:50
  • \$\begingroup\$ +1 for the last line. Linear function and linear operation are completely different things. \$\endgroup\$ – Crowley Jun 7 '16 at 15:43
  • \$\begingroup\$ @MarcusMüller Sure, I was really confused with the differentiation as whether that would account to linearity but I get it now. And also to the fact that the voltage saturates after a while and then it does not allow any further increase but even then as you have mentioned the current is still 0 which I did not thin of but is true and makes things even more clear. \$\endgroup\$ – Bhuvanesh Narayanan Jun 7 '16 at 20:28
4
\$\begingroup\$

The formal definition of a "linear" function, as in linear system, is that if you scale the input of the function by some amount, the output is scaled by that same amount:

  y = f(x)
  f(Ax) = Ay

Note that F() adding a offset inside violates this.

As you said, one way to describe a capacitor is V = Q / C. This says that the voltage on a capacitor is proportional to the charge it is holding, and that proportionality constant is the inverse of the capacitance. In the parlance of a linear equation as above, V = f(Q). Since f(Q) = Q/C, it should be clear that this equation is linear because:

  A * Q / C = A * V

for arbitrary values of A.

\$\endgroup\$
  • \$\begingroup\$ Yes I thought the same but I also felt that every component such as a diode might have some parameter that varies linearly with voltage for example in a diode the Vd decreases proportionally to temperature. But here in this case its Q but the current is proportional to the rate of change of Q and thus allowing superposition possible in circuit analysis. \$\endgroup\$ – Bhuvanesh Narayanan Jun 7 '16 at 20:24
1
\$\begingroup\$

A capacitor is a linear component because voltage and current as functions of time depend in a linear way on each other.

In the context of relations of two functions (of time) to each other (and not just values at one instance of time) linearity means that the principle of superposition holds (as Neil_UK has pointed out). The principle of superposition says that the function of a linear combination equals the linear combination of the functions, i.e. \$f(ax + by) = a f(x) + b f(y)\$.
This is the case not only for multiplication by a constant but also for the differentiation operator and integration operator.

I.e.

Not only muliplication by a constant like
\$u(t) = R i(t)\$
is a linear operation upon a function but also integeration and differentiation:

\$u(t) = \frac{1}{C} \int i(t) dt\$ and

\$u(t) = L \frac{d}{dt} i(t)\$.

Therefore not only resistors but also (ideal) capacitors and inductors are linear components.

\$\endgroup\$
  • \$\begingroup\$ Thank you! Yes I missed out on that part to consider that even differentiation and integration also are linear operations. \$\endgroup\$ – Bhuvanesh Narayanan Jun 7 '16 at 20:20
0
\$\begingroup\$

If we look to the capacitor when connected across a AC supply, then it can be easily said that it can be treated as a linear element. Linear elements are those which current voltage relationship is linear. V is proportional to I.

For an AC supply:
v=v'e^jwt (here v'=amplitude of the applied ac voltage).

Now for a capacitor:
q=cv,
i=dq/dt=c(dv/dt),
i=c.jw.v'e^jwt,
i=jwc.v,
v=i/(jwc)=i/z

z= impedance of the capacitor. Hence linear

\$\endgroup\$
  • 2
    \$\begingroup\$ Welcome to EE.SE. We use standard English on the site and that includes proper capitalisation and punctuation. It makes the post more legible and creates a better impression of the author. This site uses MathJAX as is evident in the other posts. See suluclac.com/Wiki+MathJax+Syntax Use \$ for in-line equations. Use $$ for standalone equations. \$\endgroup\$ – Transistor Aug 6 '16 at 14:10

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.