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This is the Shockley equation for the current of the diode:

$$I_D = I_S \left( e^{\frac{V_D}{nV_T}} - 1 \right)$$

It is also valid for $V_D < 0$, when this equation tends to be

$$I_D = - I_S$$

Wikipedia specifies that the equation fits for a "moderate" reverse bias.

But which is the equation describing the diode current in breakdown region? Is it still an exponential current?

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The Shockley equation assumes that the breakdown wasn't produced yet. If you are in the breakdown region things get complicated. You need the ionization rates because the impact ionization generation is what predominates. Read chapter 7.4 from Donald Neamen's - Semiconductor Physics and Devices, but the formula you would use:

$$\frac{d I_n(x)}{dx} = I_n(x) \alpha_n + I_p(x) \alpha_p \approx I \alpha$$

Where I simplified the ionization constants of electrons and holes as equal $\alpha_n \approx \alpha_p = \alpha$ and the total current is $I = I_n(x) + I_p(x)$. Now, for a junction of width W:

$$I_n(W) - I_n(0) = I \int_0^W \alpha(E) dx $$

On avalanche $I_n(W) >> I_n(0)$ and $I_n(W) \approx I$, so, by definition, to find the breakdown voltage you have to find the electric field that produce $\int_0^W \alpha(E) dx = 1$. If you want to find the current at a given point $x_0$ for a given voltage other than the breakdown, first you have to find $\alpha(E)$, then $I_n(0)$ (I guess this last from Shockley equation) and then solve the above equation.

Is it still an exponential current?

From this I found that:

$$\alpha(E) = A \exp\left(-\frac{B}{E}^{\beta}\right)$$

With $E$ the electric field, I assume proportional to the voltage and $A$, $B$, $\beta$ constants. The graph of $I(V)$ is more weird than an exponential, obviously will decrease faster than an inverted exponential.

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