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 = 280 K, calculate the ratio of thecurrent for a 0.87 V forward bias to the current for a 0.87 V backbias.

Explanation / Answer

T = 280 K, V1 = +0.87 V, V2 = -0.87V = -V1 I1 = Io(eeV1/kT - 1) I2 = Io(e-eV1/kT - 1) =e-eV1/kT Io(1 - eeV1/kT) =-e-eV1/kT Io(eeV1/kT - 1) =-e-eV1/kTI1I1/I2 =-eeV1/kT = -4.42*1015

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