Question

(a) Calculate the value of $V_0$ ​ and $I$ if the $Si$ diode and the $Ge$ diode start conducting at $0.7\text{V}$ and $0.3\text{V}$ respectively, in the given circuit.
(b) If the $Ge$ diode connection be reversed. What will be the new values of $V_0$ ​ and $I?$

• A. $\left(a\right)2.34\text{mA},11.7\text{V}$
• B. $\left(a\right)2.26\text{mA},11.3\text{V}$
• C. $\left(b\right)2.26\text{mA},11.3\text{V}$
• D. $\left(b\right)2.34\text{mA},11.7\text{V}$

Answer 1

Answer: (A), (C)

$\left(b\right)2.26\text{mA},11.3\text{V}$

As given in the circuit,

Voltage, $V=12\text{V}$ and Resistance, $R=5\text{k}\Omega$

$\left(a\right)$ In the initial connection, the current will flow through the $Ge$ diode, so

$I=\frac{12-0.3}{5\times {10}^3}=\frac{11.7}{5\times {10}^3}$

$=2.34\times {10}^{-3}\text{A}=2.34\text{mA}$

Output voltage,

$V_0=IR=V-V_{Ge}=12-0.3=11.7\text{V}$

$\left(b\right)$ When the connections of Ge diode are reversed, then current will through silicon.

$\therefore I=\frac{12-0.7}{5\times {10}^3}=2.26\times {10}^{-3}\text{A}=2.26\text{mA}$

Output voltage, $V_0=V-V_{Si}=12-0.7=11.3\text{V}$

Hence, $\left(A\right)$ and $\left(C\right)$ are the correct answers.