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\mathrm{Ge} and \mathrm{Si} diodes conduct at \mathrm{0.3\ V} and \mathrm{0.7\ V} respectively. In the following figure if \mathrm{Ge} diode connection is reversed, the value of \mathrm{V_{0}} changes by

Option: 1

\mathrm{0.2\ V}


Option: 2

\mathrm{0.4\ V}


Option: 3

\mathrm{0.6\ V}


Option: 4

\mathrm{0.8\ V}


Answers (1)

best_answer

Consider the case when \mathrm{Ge} and \mathrm{Si} diodes are connected as show in the given figure.
Equivalent voltage drop across the combination \mathrm{Ge} and \mathrm{Si} diode \mathrm{= 0.3\ V}
\mathrm{\Rightarrow \text { Current } i=\frac{12-0.3}{5 \mathrm{k} \Omega}=2.34 \mathrm{~mA}}
\mathrm{\therefore \text { Out put voltage } V_0=R i=5 \mathrm{k} \Omega \times 2.34 \mathrm{~mA}=11.7 \mathrm{~V}}
Now consider the case when diode connection are reversed. In this case voltage drop
across the diode’s combination \mathrm{=0.7\ V}
\mathrm{\Rightarrow \text { Current } i=\frac{12-0.7}{5 \mathrm{k} \Omega}=2.26 \mathrm{~mA}}
\mathrm{\therefore V_0=i R=2.26 \mathrm{~mA} \times 5 \mathrm{k} \Omega=11.3 \mathrm{~V}}
Hence charge in the value of \mathrm{V_0=11.7-11.3=0.4 \mathrm{~V}}

Posted by

Irshad Anwar

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