# Finding an expression for charge (Q) given an I-V equation

1. ### VinnyCee

492
An ideal semiconductor diode is a nonlinear element that obeys the following I-V equation:

$$I\,=\,I_s\,\left(\,e^{\frac{V}{V_{th}}}\,-\,1\right)$$

where $I_s$ is a constant (saturation current) and $V_{th}$ is a constant (thermal voltage, $V_{th}\,=\,\frac{k_B\,T}{q}$).

Assuming the applied voltage is given by

$$\begin{displaymath} V\,=\,\left\{ \begin{array}{ll} 0 & for\,t\,<\,0 \\ Bt & for\,t\,\geq\,0 \\ \end{array} \right. \end{displaymath}$$

where B is a known constant.

Find an analytic expression for the charge Q(t) that has passed through the diode over the period from 0 to t. Also find the analytic expressions for the power dissipated by the diode p(t) and for the total energy dissipated by the diode w(t) over the period from 0 to t.

Now assuming $I_s\,=\,1\,\times\,10^{-14},\,V_{th}\,=\,25.85\,mV$ and $B\,=\,90\,\frac{mV}{s}$ use MATLAB to plot I(t), Q(t), p(t), and w(t). Do your plots for t = 0 to 10s.

Find the time $\tau$ at which a total of 1 C of charge has passed through the diode ($Q(\tau)\,=\,1\,C$) and find the values of $p(\tau)$ and $w(\tau)$.

MY WORK SO FAR:

$$Q\,=\,\int_0^t\,i\,dt\,=\,\int_0^t\,\left(I_s\,e^{\frac{V}{V_{th}}}\,-\,I_s\right)\,dt$$

$$Q(t)\,=\,\left[I_s\,e^{\frac{V}{V_{th}}}\,t\,-\,I_s\,t\right]_0^t\,=\,\left(I_s\,e^{\frac{V}{V_{th}}}\,-\,I_s\right)\,t$$

Do I have the first part (equation for Q) of the question right?

Last edited: Sep 12, 2006
2. ### VinnyCee

492
Mabye the integral is wrong? Is this better?:

$$Q\,(t)\,=\,\int_0^t\,\left(I_s\,e^{\frac{B\,t}{V_{th}}}\,-\,I_s\right)\,dt$$

$$Q\,(t)\,=\,\left[I_s\,e^{\frac{B\,t}{V_{th}}}\,t\,-\,I_s\,t\right]_0^t\,=\,\left(I_s\,e^{\frac{B\,t}{V_{th}}}\,-\,I_s\right)\,t$$

Last edited: Sep 12, 2006