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

[VinnyCee]

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?

 

[VinnyCee]

Mabye the integral is wrong? Is this better?:

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

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

 

[piercas]

Yes, your expression for Q(t) is correct. It represents the total charge that has passed through the diode over the time period from 0 to t. To find the analytic expressions for power dissipated (p(t)) and total energy dissipated (w(t)), we can use the equations P = VI and W = ∫Pdt, where P is power, V is voltage, I is current, and W is energy.

To find p(t), we can substitute the given equation for V(t) into the I-V equation for the diode to get the expression for I(t). Then, we can multiply I(t) by V(t) to get the power dissipated by the diode at each time step. This can be written as:

p(t) = I(t) * V(t) = I_s * (e^(V(t)/V_th) - 1) * V(t)

Similarly, we can find the expression for w(t) by integrating p(t) over the time period from 0 to t. This can be written as:

w(t) = ∫p(t)dt = ∫[I_s * (e^(V(t)/V_th) - 1) * V(t)]dt

To plot these functions in MATLAB, we can use the given values for I_s, V_th, and B to create a vector of time values from 0 to 10 seconds. Then, we can use the equations for I(t), Q(t), p(t), and w(t) to calculate the corresponding values at each time step and plot them using the "plot" function.

To find the time at which a total of 1 C of charge has passed through the diode, we can use the equation Q(t) = 1 and solve for t. This will give us the time \tau at which Q(\tau) = 1 C. We can then substitute this value of \tau into the equations for p(t) and w(t) to find their corresponding values at that time step.