Current-Voltage Characterics

Under forward-bias, minority carrier injections occur, that is, electrons are injected into the p-side, whereas holes are injected into the n-side.

Ideal Characteristics

Ideal current-voltage characteristics based on the following assumptions:
(1) the depletion region has abrupt boundaries and, outside the boundaries, the semiconductor is assumed to be neutral;
(2)  the carrier densities at the boundaries are related by the electrostatic potential difference across the junction;
(3) the low-injection condition, that is, the injected minority carrier densities are small compared with the majority carrier densities (in other words, the majority carrier densities are changed negligibly at the boundaries of neutral regions by the applied bias);

(4) neither generation nor recombination current exists in the depletion region and the electron and hole currents are constant throughout the depletion region.

From the section of thermal equilibrium, we know \(V_{\mathrm{bi}}=\frac{k T}{q} \ln \left(\frac{N_{\mathrm{A}} N_{\mathrm{D}}}{n_{\mathrm{i}}^{2}}\right)\) (At thermal equilibrium, the majority carrier density in the neutral regions is essentially equal to the doping concentration). Therefore, here we have $$V_{b i}= \frac{k T}{q} \ln \frac{p_{p o} n_{n o}}{n_{i}^{2}} = \frac{k T}{q} \ln \frac{n_{n o}}{n_{p o}}$$Rearrange this equation gives $$n_{n o} = n_{p o} \text{exp}({q V_{b i} / k T})$$When a bias is applied, we have$$n_{n} = n_{p} \text{exp}({q(V_{b i}+V) / k T})$$with \(V\) positive for forward bias and negative for reverse bias.  For the low-injection condition, the approximation \(n_{n} \cong n_{n o}  \) holds true. Therefore \(n_{p o} \exp \left(q V_{b i} / k T\right)=n_{p} \exp \left(q\left(V_{b i}+V\right) / k T\right)  \), which can be simplified to $$n_{p}= n_{p o} \text{exp}(e^{q V / k T})$$ or $$n_{p}-n_{po}= n_{p o} \text{exp}(e^{q V / k T}-1)$$Similarily, we have $$p_n- p_{no}= p_{no}(\text{exp}(q V / k T)-1)$$Under our idealized assumptions, no current is generated within the depletion region; all currents come from the neutral regions. In the neutral n-region, there is no electric field, thus the steady-state continuity equation reduces to$$\frac{d^{2} p_{n}}{d x^{2}} -\frac{p_{n} -p_{\text {no }}}{D_{p} \tau_{p}}=0$$With the boundary condition \(p_{n}-p_{n o}=p_{n o}(\exp (q V / k T)-1)\) and \( p_{n}(x=\infty)=p_{n o}\), the solution gives$$p_{\mathrm{n}}-p_{\mathrm{n} 0}=p_{\mathrm{n} 0}\left[\exp \left(\frac{q V}{k T}\right)-1\right] \exp \left[\frac{-\left(x-x_{\mathrm{n}}\right)}{L_{\mathrm{p}}}\right]$$where \(L_p =\sqrt{D_{p} \tau_{p}}\)  is the diffusion length of holes (minority carriers) in the n-region. Because the current is constant throughout the device, we just look at the current at \( x=x_{\mathrm{n}} \), then we get $$J_{\mathrm{p}}\left(x_{\mathrm{n}}\right)=-\left.q D_{\mathrm{p}} \frac{\mathrm{d} p_{\mathrm{n}}}{\mathrm{d} x}\right|_{x_{n}}=\frac{q D_{\mathrm{p}} p_{\mathrm{n} 0}}{L_{\mathrm{p}}}\left[\exp \left(\frac{q V}{k T}\right)-1\right]$$ Similarly, we obtain for the neutral p-region, the diffusion electron density is $$J_{\mathrm{n}}\left(x_{\mathrm{p}}\right)=\frac{q D_{\mathrm{n}} n_{\mathrm{p} 0}}{L_{\mathrm{n}}}\left[\exp \left(\frac{q V}{k T}\right)-1\right]$$ The total current density is ( the ideal diode equation)$$\begin{array}{c}
J=J_{\mathrm{p}}\left(x_{\mathrm{n}}\right)+J_{\mathrm{n}}\left(-x_{\mathrm{p}}\right)=J_s\left[\exp \left(\frac{q V}{k T}\right)-1\right] \\
J_{\mathrm{s}} \equiv \frac{q D_{\mathrm{p}} p_{\mathrm{n} 0}}{L_{\mathrm{p}}}+\frac{q D_{\mathrm{n}} n_{p 0}}{L_{\mathrm{n}}}
\end{array}$$where \(J_{s}\) is the saturation current density.

(1)  In the forward direction with positive bias on the p-side, for \(V \geq 3 k T l q\), the rate of current increase is constant
(2)  At \(300 K \) for every decade change of current, the voltage change for an ideal diode is \(60 \mathrm{mV}(=2.3 \mathrm{kT} / q)\). In the reverse direction, the current density saturates at \( -J_s \)

The total current for \(p^{+}-n\) junction(one-side abrupt junction) is $$J=\frac{q D_{p} p_{n o}}{L_{p}}(\text{exp}({q V / k T})-1)=\frac{q D_{p}}{L_{p}} N_{V}(\text{exp}({[q V-(E_{F}-E_{V})] / k T})-1)$$


 

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