## Related questions with answers

A diode has $w_{d o}=0.4 \mu \mathrm{m}$ and $\phi_{j}=0.85 V$. (a) What reverse bias is required to triple the depletion-layer width? (b) What is the depletion region width if a reverse bias of 7 V is applied to the diode?

Solution

VerifiedConsider:

$\begin{aligned} w_{d 0}&=0.4 \ \mathrm{\mu m} \\ \phi_{j}&=0.85 \ \mathrm{~V} \end{aligned}$

a) From Eq.(3.18) of the textbook we obtain the expression for the depletion-layer width for an applied reverse-bias voltage:

$\begin{aligned} w_{d}&=w_{d0} \sqrt{1+\cfrac{v_{R}}{\phi_{j}}} \end{aligned}$

To triple the depletion-layer width is required a reverse bias voltage of:

$\begin{aligned} 3 w_{d0} &=w_{d 0} \sqrt{1+\cfrac{v_{R}}{\phi_{j}}} \\ 9 w_{d0}^{2} &=\left(w_{d 0} \sqrt{1+\cfrac{v_{R}}{\phi_{j}}}\right)^{2} \\ 9 w_{d0}^{2} &=w_{d0}^{2}\left(1+\cfrac{V_{R}}{\phi_{j}}\right) \\ 8 &=\cfrac{v_{R}}{\phi_{j}} \\ v_{R} &=8 \phi_{j}\\ &=8(0.85 \ \mathrm{V}) \\ &=6.8 \ \mathrm{~V} \end{aligned}$

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