# LC circuit with series diode (finding the linear average)

I've the following series LC circuit:

Find the (linear) average of the current trough the diode. The ideal is a ideal component. The initial conditions in the circuit are equal to zero.

MY WORK

I first shorted the diode (so pretend the ideal diode is not in the circuit). Using Faraday's law I can write:

$$\text{V}_{\space\text{in}}\left(t\right)+\text{V}_{\space\text{C}}\left(t\right)=-\text{L}\cdot\text{I}_{\space\text{in}}'\left(t\right)\tag1$$

Now, I also know that $\text{V}_{\space\text{in}}\left(t\right)=\sin\left(2\pi\cdot t\right)$, $\text{I}_{\space\text{C}}\left(t\right)=\text{V}_{\space\text{C}}'\left(t\right)\cdot\text{C}$ and $\text{I}_{\space\text{C}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)$. Where $\text{L}$ is the inductor value (in Henry) and $\text{C}$ is the capacitor value (in Farad).

So, I can also write:

$$\begin{cases} \frac{\text{d}}{\text{d}t}\left(\sin\left(2\pi\cdot t\right)\right)+\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}=-\text{L}\cdot\text{I}_{\space\text{in}}''\left(t\right)\\ \\ \text{I}_{\space\text{in}}\left(0\right)=0\\ \\ \text{I}_{\space\text{in}}'\left(t\right)=0 \end{cases}\tag2$$

Which gives:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{2\pi\cdot\text{C}\cdot\left(\cos\left(2\pi\cdot t\right)-\cos\left(\frac{t}{\sqrt{\text{C}\cdot\text{L}}}\right)\right)}{4\pi^2\cdot\text{C}\cdot\text{L}-1}\tag3$$

Using the given values:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{6\pi\cdot\left(\cos\left(2\pi\cdot t\right)-\cos\left(\frac{t}{\sqrt{3}}\right)\right)}{12\pi^2-1}\tag4$$

Plotting function $\left(4\right)$, gives:

(I used Mathematica to make the plot)

Now I put the diode back into the circuit and it will cut every negative part of the current. So every negative part of the current will become zero and only the positive parts of the current are flowing in the circuit.

To find the (linear) average I need to find:

$$\overline{\text{I}}_{\space\text{D}}=\overline{\text{I}}_{\space\text{in}}=\lim_{\text{n}\to\infty}\frac{1}{\text{n}}\int_0^\text{n}\text{I}_{\space\text{D}}\left(t\right)\space\text{d}t\tag5$$

Where $\text{I}_{\space\text{D}}\left(t\right)$ is the input current ($\text{I}_{\space\text{in}}\left(t\right)$) with all the negative parts blocked.

Now, I tried to find the values for $t$ when $\text{I}_{\space\text{in}}\left(t\right)$ equals zero:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{6\pi\cdot\left(\cos\left(2\pi\cdot t\right)-\cos\left(\frac{t}{\sqrt{3}}\right)\right)}{12\pi^2-1}=0\space\Longleftrightarrow\space t=\frac{6\pi\cdot\text{k}}{\sqrt{3}\pm6\pi}\tag6$$

Where $\text{k}\in\mathbb{Z}$

From now on I do not know how to continue.

• This is a trick question. try again. What is the steady state current? – Tony Stewart Sunnyskyguy EE75 Jan 3 '18 at 17:45
• It is not a linear circuit. Am unsure what "(linear) average current" might mean. Do you know? – glen_geek Jan 3 '18 at 17:47