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Gate 2019: Electronic Devices and Circuits Quiz 3 (App update required to attempt this test)
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Question 1
In an abrupt p-n junction, doping on p-side is NA = 1015/cm3 and doping on n-side is ND = 1013/cm3, then
Question 2
Calculate the junction built-in voltage of a diode (in mV), given that its intrinsic concentration is 1.5x1010/cm3; doping concentration of ‘p’ and ‘n’ side are 1017/cm3 and 1016/cm3 respectively. [Assume room temperature conditions – VT = 26mV.]
Question 3
Consider a region of silicon device of electrons and holes, with an ionized donor density of. The electric field at x = 0 is 0 V/cm and the electric field at x = L is 50 kV/cm in the positive x direction. Assume that the electric field is zero in the y and z directions at all points.
Given q =1.6 x10-19 coulomb, for silicon, the value of L in nm is_____
Given q =1.6 x10-19 coulomb, for silicon, the value of L in nm is_____
Question 4
Two ideal pn junction have exactly the same electrical and physical parameters except for the band gap of the semiconductor materials. The first has a bandgap energy of 0.525 eV and a forward-bias current of 10 mA with Va =0.255 V. The second pn junction diode is to be designed such that the diode current I=10 at a forward-bias voltage of Va =0.32 V. The bandgap energy of second diode would be_____ (eV)
Question 5
An ideal one-sided silicon n+p junction has uniform doping on both sides of the abrupt junction. The doping relation is The built-in potential barrier is . The applied reverse bias voltage is .The space charge width is:-
(ni= 1.5 x 1010 cm-3,εr=11.7)
(ni= 1.5 x 1010 cm-3,εr=11.7)
Question 6
A silicon diode, with NA=1017 cm-3 and ND = 5 × 1017 cm-3 , is forward biased with Va = 0.05 V. Assume that the intrinsic concentration of silicon is ni = 1.08 ×1010 cm-3. What are the minority carrier concentrations ∆np(xp) and ∆ pn(xn) at the edge of transition region?
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