GATE 2014 EC (EC2) – Question 47
The diode shown has V_on = 0.7 V and is otherwise ideal. If Vi = 5 sin(ωt) V, find minimum and maximum Vo (V), respectively.

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Correct answer: (C) −5 and 3.85
Explanation
The diode begins conducting when the output exceeds2+0.7=2.7V. Below that threshold the branch is open, so vo=vi and the minimum is−5V. At vi=5V the diode is on and equal1kΩ resistors divide between5V and2.7V, giving vo=(5+2.7)/2=3.85V. Thus **−5V,3.85V,C**.