If doping concentration increases in n and p side of pn diode. Then depletion layer thickness become
Question
If doping concentration increases in n and p side of pn diode. Then depletion layer thickness become
Solution
The thickness of the depletion layer in a pn diode is inversely proportional to the doping concentration. This means that if the doping concentration on the n and p sides of the diode increases, the thickness of the depletion layer will decrease.
Here are the steps to understand this:
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The depletion region of a pn diode is formed when the diode is manufactured, and n-type and p-type materials come into contact. At this junction, free electrons from the n-type material recombine with holes from the p-type material, creating a region devoid of charge carriers called the depletion region.
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The width of this depletion region is dependent on the doping concentration of the n and p materials. The higher the doping concentration, the more charge carriers are available to recombine and the narrower the depletion region becomes.
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Therefore, if the doping concentration on both the n and p sides of the diode increases, the number of charge carriers available for recombination increases. This results in a decrease in the thickness of the depletion layer.
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This relationship is governed by the formula W = sqrt((2εVbi)/(q*(Na+Nd))), where W is the width of the depletion region, ε is the permittivity of the semiconductor material, Vbi is the built-in voltage, q is the charge of an electron, and Na and Nd are the doping concentrations of the p-type and n-type materials, respectively. As Na and Nd increase, W decreases.
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However, it's important to note that while increasing the doping concentration decreases the depletion width, it also increases the junction capacitance and reduces the breakdown voltage of the diode. Therefore, in practical applications, a balance must be struck between these factors.
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