Question

$\text{Ge}$ and $\text{Si}$ diodes start conducting at $0.3\text{V}$ and $0.7\text{V}$ respectively. In the following figure if $\text{Ge}$ diode connection are reversed, the value of $V_0$ changes by :
(assume that the $\text{Ge}$ diode has large breakdown voltage)

• A. $\text{0.4 V}$
• B. $\text{0.6 V}$
• C. $\text{0.2 V}$
• D. $\text{0.8 V}$

Answer 1

The correct option is A $\text{0.4 V}$

Given that,
$\text{Si}$ threshold voltage $=0.7\text{V}$
$\text{Ge}$ threshold voltage $=0.3\text{V}$

Initially $\text{Ge}$ and $\text{Si}$ are both forward biased, so current will effectively pass through $\text{Ge}$ diode
$\therefore (V_0{)}_1=12-0.3=11.7\text{V}$

And if $\text{Ge}$ is reversed then current will flow through $\text{Si}$ diode

$\therefore (V_0{)}_2=12-0.7=11.3\text{V}$
So,
$|\Delta V_0|=|(V_0{)}_2-(V_0{)}_1|$

$|\Delta V_0|=|11.3-11.7|=0.4\text{V}$

Hence, option $\left(D\right)$ is correct.