Suggested languages for you:

Americas

Europe

Q40P

Expert-verifiedFound in: Page 1274

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**For an ideal p-n junction rectifier with a sharp boundary between its two semiconducting sides, the current I is related to the potential difference V across 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 I or V, is called the reverse saturation current. The potential difference V is 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 I versus V from 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.**

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

- 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, $V=-5.0V$

**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.

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.

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:

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

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

94% of StudySmarter users get better grades.

Sign up for free