Question

For an ideal p-njunction rectifier with a sharp boundary between its two semiconducting sides, the current Iis related to the potential difference Vacross the rectifier by $\mathit{l}\mathbf{=}{\mathit{l}}_{\mathbf{0}}\left(e^{eV/kT}-1\right)$, where ${\mathit{l}}_{\mathbf{0}}$, which depends on the materials but not on Ior V, is called the reverse saturation current.The potential difference Vis positive if the rectifier is forward-biased and negative if it is back-biased. (a) Verify that this expression predicts the behavior of a junction rectifier by graphing Iversus Vfrom to $\mathbf{-}\mathbf{12}\mathbf{}\mathbf{V}\mathbf{}\mathbf{to}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{V}$. Take $\mathit{T}\mathbf{=}\mathbf{300}\mathit{K}$and ${\mathit{l}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{n}\mathit{A}$. (b) For the same temperature, calculate the ratio of the current for a 0.50 V forward bias to the current for a 0.50 V back bias.

Short answer

1. The graph plotted for the given expression predicts the behavior of a junction rectifier.
2. The ratio of the current for a forward bias to the current for a 0.50 V backward bias is $2.5\times {10}^8$.

Step-by-step solution

The given data

• The graph voltage values range from $-0.12V\mathrm{to}+0.12V,\mathrm{for}T=300K\mathrm{and}{l}_0=5nA$
• Voltage of the forward bias,$V=+5.0V$
• Voltage of the backward bias,$V=-5.0V$

Understanding the concept of current in a p-n junction rectifier

When the positive terminal of the battery is connected to the p-type and the negative terminal of the battery to the n-type semiconductor, the junction is said to be forward biased. If negative is connected with p and positive with n-type semiconductor, then the junction is said to be reverse biased. The current flows through the p-n junction when it is forward biased.

Formula:

The given current equation,$l=l_0\left(e^{eV/kT}-1\right)$ (i)

Here e is the elementary charge, V is the volume, T is the absolute temperature and $l_0$is the initial value of current.

a) Calculation to plot the current graph

The vertical axis represents the current in nanoamperes and the horizontal axis represents the voltage.

Hence, the graph plotted for the given expression predicts the behavior of a junction rectifier.

b) Calculation of the ratio of the current in the forward bias to the backward bias

Using the given data and equation (i), the ratio of the current for a 0.50V forward bias to the current for a backward bias is given by:

$$\begin{gathered} \frac{l{|}_{V=+0.50V}}{l{|}_{{}_{V=-0.50V}}}=\frac{l_{0}exp\left[\left({\displaystyle \frac{\left(+0.50eV\right)}{\left(8.62\times {10}^{-5}eV/K\right)\left(300K\right)}}\right)-1\right]}{l_{0}exp\left[\left({\displaystyle \frac{\left(-0.50eV\right)}{\left(8.62\times {10}^{-5}eV/K\right)\left(300K\right)}}\right)-1\right]} \\ =2.5\times {10}^8 \end{gathered}$$

Hence, the ratio value is $2.5\times {10}^8$.