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In Charge Tunneling Rates in Ultrasmall Junctions section 2.1, the authors consider the problem of charge relaxation in a simple circuit shown in Figure A. The implicitly use an assumption made about the respresentation of voltage in the Laplace domain that I do not understand. To pose the question, let us first lay out the point of the calculation.

enter image description here Figure A: A capacitor coupled to an arbitrary impedance $\tilde{Z}(\omega)$ and a dc voltage source $V$.

Denote the equilibrium charge across the capacitor is $Q_e$. At time $t=0$ we drop some extra charge on the capacitor, making the total capacitor charge $Q_0$. We'd like to solve for the time dependent charge $Q(t)$ on the capacitor.

From the definition of impedance we have $$\hat{V}(p) = \hat{Z}(p) \hat{I}(p)$$ where $V$ is voltage, $Z$ is impedance, and $I$ is current. Using the derivative rule for Laplace transforms (straightforward algebra/calculus), we find $$\hat{I}(p) = p \hat{Q}(p) - Q_0$$ which gives us $$\hat{V}(p) = \hat{Z}(p)(p\hat{Q}(p) - Q_0) \, .$$

To finish solving the problem, we need to express $\hat{V}$ in terms of $\hat{Q}$. In the paper Equation (9), the authors simply write $$\frac{Q_e}{pC} = \frac{\hat{Q}(p)}{C} + \hat{Z}(p)(p \hat{Q}(p) - Q_0)\ \, .$$ which implies that $$\hat{V}(p) = \frac{Q_e}{pC} - \frac{\hat{Q}(p)}{C} \, , \tag{$\star$}$$ but I don't understand why that's the case.

Why is equation $(\star)$ correct? Does it come from some considerations of the boundary conditions or some other physical reasoning, or from a simple mathematical consideration?


In this question, $\tilde{x}$ indicates a Fourier transform defined as $$\tilde{x}(\omega) \equiv \int_{-\infty}^\infty x(t) \exp(-i \omega t) \, dt$$ and $\hat{x}(p)$ indicates a Laplace transform defined as $$\hat{x}(p) \equiv \int_0^\infty x(t) \exp(-pt) \, dt \, .$$

With this Fourier convention:

  • A causal linear response function $Z$ has $Z(t<0)=0$.

  • The Fourier and Laplace transforms are related by $\hat{x}(p) = \tilde{x}(-ip)$ (again assuming causal functions).

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    $\begingroup$ This seems to be an exercise so I have added the appropriate tag. But I am wondering how you justify this as a conceptual question, ie not off topic? It seems to me that you are asking for someone to point out the error in your calculation, which I understand to make the question off topic. $\endgroup$ – sammy gerbil Mar 2 '17 at 2:45
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    $\begingroup$ Just because you add the word conceptually to a question does not make it a conceptual question. The answer given by Skyler is not conceptual, it is a calculation. He has interpreted your question as "How do I obtain equation $(*)$?" ... What kind of conceptual answer would satisfy you? $\endgroup$ – sammy gerbil Mar 2 '17 at 2:52
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    $\begingroup$ This question is discussed on Meta. $\endgroup$ – rob Mar 2 '17 at 4:27
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    $\begingroup$ @sammygerbil You should probably read my response to the meta post. The gist is that I think your edit is inappropriate. Specifically, two issues: (1) this question doesn't deserve the tag, which is discussed by the meta post, and (2) you applied quote formatting to something which is not a quote. $\endgroup$ – David Z Mar 2 '17 at 4:56
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    $\begingroup$ I apologise for incorrectly using the quote format. I did not realise that it is intended only for quotes rather than to emphasize the problem that you are attempting to solve. But I still disagree about the tag, and will respond in detail to DavidZ's answer in Meta. $\endgroup$ – sammy gerbil Mar 2 '17 at 5:51
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There are two possibilities here:

  1. Your $Q_0$ is not the total charge on the capacitor, but the added charge to $Q_e$. The voltage $V(p)$ is the voltage across the rest of the circuit. We can see this from your equation $$ \hat{I}(p) = p\hat{Q}(p) - Q_0$$ and putting in the equilibrium configuration $Q(t) = Q_e\theta(t)$ where $\theta(t)$ is the Heaviside function that turns the equilibrium charge on at $t=0$. The transforms are $\hat{I}(p) = Q_e$ and $\hat{Q}(p) = Q_e/p$, implying $Q_0 = 0$ in equilibrium. This happens because the value $\theta(0)$ is a convention that can be anything from $0$ to $1$, and the implicit convention of this Laplace transform is that it is $0$.

    Note that the current computed as $\hat{I}(p)$ has a sign problem: You can't exactly put a charge of $Q_0$ onto a capacitor, you must decide on which of the plates you put it. If you increase the charge on the plate that is already charged in the equilibrium configuration, then the current will flow opposite to the current when the capacitor was charged. So, if $Q_0$ denotes putting more positive charge on the already positively charged plate, the voltage across the rest of the circuit will be $V(t) = \frac{Q_e}{C}\theta(t) - \frac{Q(t)}{C}$, where $Q(0) = Q_0$.

  2. Your $Q_0$ is the total charge on the capacitor, and the voltage you are computing is the voltage across the impedance. Then, by Kirchhoff's laws, we have $$ V+V_C+V_0 = 0,$$ where $V_C = Q(t)/C$ is the voltage on the capacitor and $V_0 = -Q_e/C$ is the contant voltage (not causally turned on this time!) across the DC voltage source. The latter equation only holds if the impedance has neglegible capacitance, since then we have $V = 0$ in equilibrium where $I=0$ and $Z$ is non-singular.

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  • $\begingroup$ Honestly, to me it seems that the authors just made a big mess with the Laplace transform. They want to arrive at the equation (10), $Q(t) = Q_e+(Q_0-Q_e)R(t)$, where the charge is expressed as a function of the initial charge $Q_0$ on the capacitor and on the equilibrium value $Q_e$, which is the final charge on the capacitor. But since the Laplace transform does not allow an immediate writing of an equation as a function of the final value, they wrote a wrong argument. There's also no need to consider $V$ and $Q_e$ at all in the analysis of the relaxation time: it's a linear circuit! $\endgroup$ – Massimo Ortolano Mar 4 '17 at 21:58
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Starting with: $$\hat{V}(p) = \hat{Z}(p)(p\hat{Q}(p) - Q_0) \, .$$

and using our knowledge of ideal capacitors undergoing reactive impedance (a purely imaginary term for impedence) such that:

$$\hat{Z}(\omega)=\frac{1}{i\omega C}=\hat{Z}(p)=\frac{1}{pC}$$

if we distribute the impedence we get $$\hat{V}(p) = (\frac{p\hat{Q}(p)}{pC} - \frac{Q_0}{pC}) \, .$$

but to do this division we are dividing by a complex number (namely $0 + i\omega t$), meaning we should multiply both terms by it's complex conjugate $0 - i\omega t$ or $p*=-p$ leading to

$$\hat{V}(p) = (\frac{|p^2|\hat{Q}(p)}{|p^2|C} - \frac{p^*Q_0}{|p^2|C})$$

With $\frac{p^*}{|p^2|} being \frac{i}{i^2}\frac{|\omega|}{|\omega|^2}=\frac{-i}{w}$ Which leads to the result: $$\hat{V}(p) = \frac{\hat{Q}(p)}{C} \ - \frac{Q_e}{|p|C} \tag{$\star$}$$

Which could be written as $$\hat{V}(p) = \frac{\hat{Q}(p)}{C} \ + \frac{Q_e}{pC} \tag{$\star$}$$

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    $\begingroup$ See diagram. $Z$ is not the impedance of the capacitor. $\endgroup$ – DanielSank Mar 2 '17 at 1:42
  • $\begingroup$ hmm...noted, what is known about $Z$ then? $\endgroup$ – Skyler Mar 2 '17 at 2:50
  • $\begingroup$ It's an impedance, but that is wholly irrelevant to understanding why $\hat{V}(p)=Q_e/pC - \hat{Q}(p)/C$. $\endgroup$ – DanielSank Mar 2 '17 at 3:19
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    $\begingroup$ How does the initial condition $Q_0$ morph into the circuit property $Q_e$ in this derivation? $\endgroup$ – ACuriousMind Mar 2 '17 at 16:54

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