1
GATE ECE 2016 Set 2
Numerical
+2
-0
Consider a region of silicon devoid of electrons and holes, with an ionized donor density of $${\mathrm N}_\mathrm d^+=10^{17}\;\mathrm{cm}^{-3}$$. 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. GATE ECE 2016 Set 2 Electronic Devices and VLSI - PN Junction Question 4 English

Given q = 1.6 × 10−19 coulomb, $$\varepsilon$$0 = 8.85 × 10−14 F/cm, $$\varepsilon$$r = 11.7 for silicon, the value of L in nm is ________.

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2
GATE ECE 2016 Set 3
MCQ (Single Correct Answer)
+2
-0.6
Consider the charge profile shown in the figure. The resultant potential distribution is best described by GATE ECE 2016 Set 3 Electronic Devices and VLSI - PN Junction Question 3 English
A
GATE ECE 2016 Set 3 Electronic Devices and VLSI - PN Junction Question 3 English Option 1
B
GATE ECE 2016 Set 3 Electronic Devices and VLSI - PN Junction Question 3 English Option 2
C
GATE ECE 2016 Set 3 Electronic Devices and VLSI - PN Junction Question 3 English Option 3
D
GATE ECE 2016 Set 3 Electronic Devices and VLSI - PN Junction Question 3 English Option 4
3
GATE ECE 2015 Set 3
MCQ (Single Correct Answer)
+2
-0.6
The electric field profile in the depletion region of a p-n junction in equilibrium is shown in the figure. Which one of the following statements is NOT TRUE? GATE ECE 2015 Set 3 Electronic Devices and VLSI - PN Junction Question 5 English
A
The left side of the junction is n-type and the right side is p-type
B
Both the n-type and p-type depletion regions are uniformly doped
C
The potential difference across the depletion region is 700 mV
D
If the p-type region has a doping concentration of 1015 cm–3 , then the doping concentration in the n-type region will be 1016 cm–3
4
GATE ECE 2015 Set 1
Numerical
+2
-0
The built-in potential of an abrupt p-n junction is 0.75V. If its junction capacitance (CJ) at a reverse bias (VR) of 1.25V is 5pF, the value of CJ (in pF) when VR = 7.25V is ___________.
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