# Diode-Resistor-Capacitor Circuit Equations

So I took the time to measure the current dependency on voltage of a diode I have. I applied an exponential fit to it, and have a pretty reliable equation (within 1%).

I'm interested in how a diode-resistor-capacitor series circuit response to different signals. Naturally, I'm starting with just DC voltage.

The equation that I have for the voltage/current dependency for the diode is of the form

$$I=ae^{bV_D} \tag{1}$$

where $$V_D$$ is the voltage across the diode.

Using Kirchhoff's law, I get the following differential equation with an initial condition:

$$V = RQ' + \frac{1}c Q + \frac{1}b \ln\left(\frac{Q'}a\right)$$

$$Q(0)=0$$

where $$R$$ is the resistance of the resistor, $$c$$ is the capacitance of the capacitor, $$a$$ and $$b$$ are the exponential regression constants from equation (1), and $$V$$ is the applied DC voltage.

Does anyone know if it's possible to analytically solve this equation?

• If you put a capacitor in series with anything, than at DC the current will be 0. – The Photon Jul 11 '17 at 3:56
• After enough time, yes that's correct. But not initially. I'm interested in modeling what happens in that first second. Presumably the current follows a type of exponential decay. – Ben Jul 11 '17 at 4:08
• When we talk about analyzing a circuit at DC, we usually mean DC steady state. If you want to talk about what happens when the input voltage is changed, we usually call that a transient analysis. – The Photon Jul 11 '17 at 4:48
• AFAIK, there is no analytical solution to this circuit. However there are dozens of different computer programs available that can provide a numerical solution as accurate as could possibly be useful. LTSpice is a well known free (as in beer) one. – The Photon Jul 11 '17 at 4:49
• Any non-linear term is usually a pain in the neck. Diodes are usually linealized because of that. – FGSUZ Apr 8 '19 at 22:06

Well, we know that:

$$\text{V}_\text{in}\left(t\right)=\text{V}_\text{D}\left(t\right)+\text{V}_\text{R}\left(t\right)+\text{V}_\text{C}\left(t\right)\tag1$$

And we also know that:

• $$\text{I}_\text{D}\left(t\right)=\text{I}_\text{S}\cdot\left(\exp\left(\frac{\epsilon\cdot\text{V}_\text{D}\left(t\right)}{\eta\cdot\text{k}\cdot\text{T}}\right)-1\right)\tag2$$

Where $\text{I}_\text{S}$ is the reverse saturation current, $\epsilon$ is the electronic charge, $\text{k}$ is the Boltzmann's constant and $\text{T}$ is the absolute temperature and $1\le\eta\le2$.

• $$\text{V}_\text{R}\left(t\right)=\text{I}_\text{R}\left(t\right)\cdot\text{R}\tag3$$
• $$\text{I}_\text{C}\left(t\right)=\text{V}_\text{C}'\left(t\right)\cdot\text{C}\tag4$$
• $$\text{I}_\text{in}\left(t\right)=\text{I}_\text{D}\left(t\right)=\text{I}_\text{R}\left(t\right)=\text{I}_\text{C}\left(t\right)\tag5$$

So, we get:

$$\text{V}_\text{in}'\left(t\right)=\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}\cdot\frac{\text{I}_\text{in}'\left(t\right)}{\text{I}_\text{S}+\text{I}_\text{in}\left(t\right)}+\text{I}_\text{in}'\left(t\right)\cdot\text{R}+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag6$$

For example, when the input voltage is constant we get:

$$\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}\cdot\frac{\text{I}_\text{in}'\left(t\right)}{\text{I}_\text{S}+\text{I}_\text{in}\left(t\right)}+\text{I}_\text{in}'\left(t\right)\cdot\text{R}+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}=0\space\Longleftrightarrow$$ $$\int\text{I}_\text{in}'\left(t\right)\cdot\frac{\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}\cdot\frac{1}{\text{I}_\text{S}+\text{I}_\text{in}\left(t\right)}+\text{R}}{\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}}\space\text{d}t=\int-1\space\text{d}t\tag6$$

Substitute $\text{u}:=\text{I}_\text{in}\left(t\right)$:

$$\int\frac{\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}\cdot\frac{1}{\text{I}_\text{S}+\text{u}}+\text{R}}{\text{u}\cdot\frac{1}{\text{C}}}\space\text{d}\text{u}=\text{C}\cdot\left\{\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}\int\frac{1}{\text{I}_\text{S}+\text{u}}\cdot\frac{1}{\text{u}}\space\text{d}\text{u}+\text{R}\int\frac{1}{\text{u}}\space\text{d}\text{u}\right\}=\mathcal{C}-t\tag7$$

So, we get:

$$\text{C}\cdot\left\{\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}\cdot\frac{1}{\text{I}_\text{S}}\cdot\ln\left|\frac{\text{I}_\text{in}\left(t\right)}{\text{I}_\text{in}\left(t\right)+\text{I}_\text{S}}\right|+\text{R}\cdot\ln\left|\text{I}_\text{in}\left(t\right)\right|\right\}=\mathcal{C}-t\tag8$$

Now, we can write:

$$0.02353823794935365<\frac{\eta\cdot\text{k}\cdot\text{T}}{\epsilon}<0.05569380628470534\tag9$$

When $0 ^\circ\text{C}=\frac{5463}{20}\space\text{K}\le\text{T}\le\frac{5463}{20}\space\text{K}=50 ^\circ\text{C}$