For an ideal p - n junction rectifier with a sharp boundary between its two semi

Question

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

I = I0(eeV/kT - 1),

where I0, 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. For T = 270 K, calculate the ratio of the current for a 0.26 V forward bias to the current for a 0.26 V back bias.

Please show step by step solutions

Explanation / Answer

k*T/e =1.38*10^-23*270/1.6*10^-19 =23.2875 meV   (mili electron-volts)

k is boltzmann constant , e is the electron charge.

I(+0.26) / I(-0.26) = [exp(+0.26/0.0232875) -1] /[exp(-0.26/0.0232875) -1] = -70600

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