Diode-Resistor-Capacitor Circuit Equations

Question

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:

$$V = RQ' + frac{1}c Q + frac{1}b lnleft(frac{Q'}aright)$$

$$Q(0)=0$$

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?

Jan · Answer

Well, we know that:

$$text{V}_text{in}left(tright)=text{V}_text{D}left(tright)+text{V}_text{R}left(tright)+text{V}_text{C}left(tright)tag1$$

And we also know that:

$$text{I}_text{D}left(tright)=text{I}_text{S}cdotleft(expleft(frac{epsiloncdottext{V}_text{D}left(tright)}{etacdottext{k}cdottext{T}}right)-1right)tag2$$

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 $1leetale2$.

$$text{V}_text{R}left(tright)=text{I}_text{R}left(tright)cdottext{R}tag3$$
$$text{I}_text{C}left(tright)=text{V}_text{C}'left(tright)cdottext{C}tag4$$
$$text{I}_text{in}left(tright)=text{I}_text{D}left(tright)=text{I}_text{R}left(tright)=text{I}_text{C}left(tright)tag5$$

So, we get:

$$text{V}_text{in}'left(tright)=frac{etacdottext{k}cdottext{T}}{epsilon}cdotfrac{text{I}_text{in}'left(tright)}{text{I}_text{S}+text{I}_text{in}left(tright)}+text{I}_text{in}'left(tright)cdottext{R}+text{I}_text{in}left(tright)cdotfrac{1}{text{C}}tag6$$

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

$$frac{etacdottext{k}cdottext{T}}{epsilon}cdotfrac{text{I}_text{in}'left(tright)}{text{I}_text{S}+text{I}_text{in}left(tright)}+text{I}_text{in}'left(tright)cdottext{R}+text{I}_text{in}left(tright)cdotfrac{1}{text{C}}=0spaceLongleftrightarrow$$
$$inttext{I}_text{in}'left(tright)cdotfrac{frac{etacdottext{k}cdottext{T}}{epsilon}cdotfrac{1}{text{I}_text{S}+text{I}_text{in}left(tright)}+text{R}}{text{I}_text{in}left(tright)cdotfrac{1}{text{C}}}spacetext{d}t=int-1spacetext{d}ttag6$$

Substitute $text{u}:=text{I}_text{in}left(tright)$:

$$intfrac{frac{etacdottext{k}cdottext{T}}{epsilon}cdotfrac{1}{text{I}_text{S}+text{u}}+text{R}}{text{u}cdotfrac{1}{text{C}}}spacetext{d}text{u}=text{C}cdotleft{frac{etacdottext{k}cdottext{T}}{epsilon}intfrac{1}{text{I}_text{S}+text{u}}cdotfrac{1}{text{u}}spacetext{d}text{u}+text{R}intfrac{1}{text{u}}spacetext{d}text{u}right}=mathcal{C}-ttag7$$

So, we get:

$$text{C}cdotleft{frac{etacdottext{k}cdottext{T}}{epsilon}cdotfrac{1}{text{I}_text{S}}cdotlnleft|frac{text{I}_text{in}left(tright)}{text{I}_text{in}left(tright)+text{I}_text{S}}right|+text{R}cdotlnleft|text{I}_text{in}left(tright)right|right}=mathcal{C}-ttag8$$

Now, we can write:

$$0.02353823794935365<frac{etacdottext{k}cdottext{T}}{epsilon}<0.05569380628470534tag9$$

When $0 ^circtext{C}=frac{5463}{20}spacetext{K}letext{T}lefrac{5463}{20}spacetext{K}=50 ^circtext{C}$

Diode-Resistor-Capacitor Circuit Equations

One Answer

Add your own answers!

Ask a Question