JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Semiconductors

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Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of

V_{0}

changes by: (assume that the Ge diode has large breakdown voltage)

(A)0.8 V
(B)0.6 V
(C)0.2 V
(D)0.4 V✓ Correct

Answer: (D)

Solution

Both diodes face the supply, but only the one that turns on at the lower voltage conducts, and it clamps the drop across itself. With the Ge diode forward biased it conducts first at

0.3 V

, so the output is

V_{0} = 12 - 0.3 = 11.7 V

. Reverse the Ge diode and it no longer conducts (its breakdown voltage is large), so now the Si diode carries the current and drops

0.7 V

, giving

V_{0} = 12 - 0.7 = 11.3 V

. The output therefore shifts by

11.7 - 11.3 = 0.4 V

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Semiconductors

Solution

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