# Ideal Diode Voltage Drop

I am a little unsure about why the ideal diode can't have positive voltage drop... Looking back at my notes, it just simply says that "It's not a feasible solution" and that only when voltage drop is zero or negative can there be a solution.

This is because you're using the completely ideal model for a diode.

Shockley's diode equation for the static behavior of a PN junction is given by

$$I=I_\mathrm{S} \left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right),\,$$

If you graph this for normal values of the parameters, you'll see that it's practically 0 for all values of the voltage drop less than 0, and it starts to have a very high slope for a very small voltage drop and the curve is practically vertical.

One of the main reasons this is done is because this equation is highly non-linear, and even a simple, one mesh circuit will yield an algebraic equation which cannot be solved in terms of elementary functions, so an approximate solution to a circuit is obtained by idealizing this non-linear equation.

Another step to make the model more accurate is to incorporate a voltage source when the diode is on. The value of this voltage source depends highly on what semiconductor the diode is made of. I will list some standard values:

Si: 0.7 V
GaAs: 1.2 V
Ge: 0.25 V

Yet another step taken to get closer to the Shockley equation is to add a resistor in series with this voltage source to account for the non-infinite slope of the curve.

Hope this helps. :)