Wksheet #3 Problem 1 barrier height of etBn 0.67 eV and effective Richardson con

Question

Wksheet #3 Problem 1 barrier height of etBn 0.67 eV and effective Richardson constant is A* 114 A/K2- cm2 at 300K. A Si pn junction has Na-1018 cm-3, Nd-1016 cm-3, Dp 10 cm2/s, Dn-25 cm2/s, tp0-n0-10-7 s. From Example 9.5, the reverse-saturation current density for the schottky barrier diode is JST- 5.98 x 10-5 A/cm2 and reverse-saturation current density of the pn junction is Js - 3.66 x 10-11 A/cm2. Using these parameters, determine the forward-bias voltages required to produce a current of 10uA in each diode. Assume each cross-sectional area is 10-4 cm2. SHOW YOUR WORK. Consider a tungsten barrier on silicon with a measured

Explanation / Answer

For a Schottky barrier diode, we have -

J = JST [ exp(eVa/KT) - 1 ]

Neglecting the (-1) term, we can solve for the forward-bias voltage.

Va = (KT/e) ln (J/JST)

Va = VT ln (J/JST)

VT = KT/e = 0.0259 V

J = 10 x 10-6/ 10-4 = 0.1 A/cm2

JST = 5.98 x 10-5 A/cm2

JS = 3.66 x 10-11 A/cm2

Hence,

Va = (0.0259) ln (0.1/5.98 x10-5)

Va = 0.192 V

For the pn junction diode, we have -

Va = VT ln (J/JS)

Va = (0.0259) ln( 0.1/3.66 x10-11)

Va = 0.563 V

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