For an ideal p - n junction rectifier with a sharpboundary between its two semic

Question

For an ideal p-n junction rectifier with a sharpboundary between its two semiconducting sides, the currentI is related to the potential difference V acrossthe rectifier by

I =I0(eeV/kT -1),

where I0, which depends on the materials butnot on I or V, is called the reversesaturation current. The potential difference V ispositive if the rectifier is forward-biased and negative if it isback-biased. For T = 330 K, calculate the ratio of thecurrent for a 0.31 V forward bias to the current for a 0.31 V backbias.

Explanation / Answer

I(V= 0.31) = I0*(exp(e*0.31V/kT)-1) I(V= -0.31) = I0*(exp(e*-0.31V/kT)-1) Ratio = I(V=0.31)/I(V=-0.31) =I0*(exp(e*0.31V/kT)-1)/I0*(exp(e*-0.31V/kT)-1) Ratio = (exp(e*0.31V/kT)-1)/*(exp(e*-0.31V/kT)-1) e = 1.602e-19 C k = 1.38e-23 J/K T = 330 K Ratio =(exp(1.602e-19*0.31/(1.38e-23*300))-1)/(exp(1.602e-19*-0.31/(1.38e-23*300))-1) Ratio = 1.62 e5 (absolute value) EDIT: Links removed for Cramster's policy

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