# Circuit Analysis - Diode Small Signal Voltage

I'm struggling with this circuit analysis about a pn-junction diode.

I'm given I_D = 1.066 * 10^-7 A when V_D = 0.6V and I_D = 2.2 * 10^-5 A when V_D = 0.8V (all at room temp).

My calculated values for the diode are I_S = 1.2124 * 10^-14 and n = 1.5010.

1.5 seems rather big for the n value, but I pressed on.

I'm then given a circuit involving the same diode above where V_DD = 1V and small signal v_dd(t) = 0.01cos(ωt), with V_D = 0.7V.

I'm to find the diode small-signal resistance r_d, and the maximum voltage across the diode v_D dude to v_dd(t) and V_DD.

My calculated r_d is huge and I have no idea why! Here are my workings for reference.

Well, the Shockley diode equation, gives the relation between the voltage across and the current through a diode:

$$\text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)\space\Longleftrightarrow\space\text{V}_\text{D}=\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(1+\frac{\text{I}_\text{D}}{\text{I}_\text{S}}\right)\tag1$$

Where $$\\text{I}_\text{D}\$$ is the diode current, $$\\text{I}_\text{S}\$$ is the reverse bias saturation current, $$\\text{V}_\text{D}\$$ is the voltage across the diode, $$\\text{q}\$$ is the electron charge, $$\\text{k}\$$ is the Boltzmann constant, $$\\text{T}\$$ is the temperature and $$\\eta\$$ is the ideality factor.

\begin{cases} \begin{alignat*}{1} \displaystyle\frac{533}{500}\cdot10^{-7}&=\displaystyle\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\cdot\frac{3}{5}}{\displaystyle\eta\text{k}\text{T}}\right)-1\right)\\ \\ \displaystyle\frac{11}{5}\cdot10^{-5}&=\displaystyle\text{I}_\text{S}\left(\exp\left(\frac{\displaystyle\text{q}\cdot\frac{4}{5}}{\displaystyle\eta\text{k}\text{T}}\right)-1\right) \end{alignat*}\end{cases}\tag2

Which, when solved gives:

\begin{cases} \begin{alignat*}{1} \text{I}_\text{S}&\approx1.405247060\cdot10^{-14}\space\text{A}\\ \\ \frac{\displaystyle\eta\text{k}\text{T}}{\displaystyle\text{q}}&\approx0.03778662918\space\text{V} \end{alignat*} \end{cases}\tag3

Notice that you can find the input current $$\\text{I}_\text{input}\$$ to this circuit by solving:

$$\text{V}_\text{DD}+\text{V}_\text{dd}\left(t\right)=\text{I}_\text{input}\text{R}+\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(1+\frac{\text{I}_\text{input}}{\text{I}_\text{S}}\right)\tag4$$

And, so the voltage across the diode is given by:

$$\text{V}_\text{D}=\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(1+\frac{\text{I}_\text{input}}{\text{I}_\text{S}}\right)\tag5$$

Solving $$\(4)\$$ for the input current gives:

$$\text{I}_\text{input}=\frac{\displaystyle\eta\text{k}\text{T}}{\displaystyle\text{q}\text{R}}\cdot\mathscr{W}\left(\frac{\displaystyle\text{q}}{\displaystyle\eta\text{k}\text{T}}\cdot\left(\text{I}_\text{S}\text{R}\exp\left(\frac{\displaystyle\text{q}\left(\text{V}_\text{DD}+\text{V}_\text{dd}\left(t\right)+\text{I}_\text{S}\text{R}\right)}{\displaystyle\eta\text{k}\text{T}}\right)\right)\right)-\text{I}_\text{S}\tag6$$

WHere $$\\mathscr{W}\left(\cdot\right)\$$ is the Lambert W-function.