JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 13)
Two Zener diodes (A and B) having breakdown
voltages of 6 V and 4 V respectively, are
connected as shown in the circuit below. The
output voltage V0 variation with input voltage
linearly increasing with time, is given by :
(Vinput = 0 V at t = 0)
(figures are qualitative)_5th_September_Evening_Slot_en_13_1.png)
(Vinput = 0 V at t = 0)
(figures are qualitative)
_5th_September_Evening_Slot_en_13_2.png)
_5th_September_Evening_Slot_en_13_3.png)
_5th_September_Evening_Slot_en_13_4.png)
_5th_September_Evening_Slot_en_13_5.png)
Explanation
Till input voltage reaches 4V, both diode will stay deactvated so output voltage will be same as input voltage.
When input voltage reach 4V, then 4V diode will get activated and it will transfer only 4V to the output but as a rasistance is present in series with the 4V diode so some voltage drop will happen across the resistor. So output
voltage(Vo) will be = 4V + voltage drop.
When input voltage reach 6V, then 6V diode will get activated and it will transfer only 6V to the output no matter how much input voltage become.
When input voltage reach 4V, then 4V diode will get activated and it will transfer only 4V to the output but as a rasistance is present in series with the 4V diode so some voltage drop will happen across the resistor. So output
voltage(Vo) will be = 4V + voltage drop.
When input voltage reach 6V, then 6V diode will get activated and it will transfer only 6V to the output no matter how much input voltage become.
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