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

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SUMMARY

The discussion focuses on deriving an expression for charge (Q) in an ideal semiconductor diode governed by the I-V equation I = I_s(e^(V/V_th) - 1). The applied voltage is defined as V = Bt for t ≥ 0, where B is a constant. The participant successfully integrates the current to find Q(t) and seeks confirmation on the correctness of their integral approach. The values used for saturation current (I_s = 1 × 10^-14 A) and thermal voltage (V_th = 25.85 mV) are specified, along with the requirement to plot I(t), Q(t), power p(t), and energy w(t) using MATLAB for the time interval of 0 to 10 seconds.

PREREQUISITES
  • Understanding of semiconductor physics and diode operation
  • Familiarity with the I-V characteristics of diodes
  • Proficiency in MATLAB for plotting and numerical analysis
  • Knowledge of calculus, specifically integration techniques
NEXT STEPS
  • Learn how to implement numerical integration in MATLAB for charge calculations
  • Study the derivation of power and energy expressions from current and voltage
  • Explore the effects of varying saturation current (I_s) and thermal voltage (V_th) on diode performance
  • Investigate the transient response of diodes under different voltage applications
USEFUL FOR

Electrical engineers, semiconductor physicists, and students studying diode behavior and circuit analysis will benefit from this discussion, particularly those interested in charge dynamics and energy dissipation in semiconductor devices.

VinnyCee
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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}<br /> V\,=\,\left\{ \begin{array}{ll}<br /> 0 &amp; for\,t\,&lt;\,0 \\<br /> Bt &amp; for\,t\,\geq\,0 \\<br /> \end{array} \right.<br /> \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:
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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
 
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