Question

Ge and Si diodes start conducting at $0.3V$ and $0.7V$ respectively. In the following figure if the Ge diode connection is reversed, the value of ${{V}_{0}}$ changes by (assume that the Ge diode has large breakdown voltage)

A) 0.6V

B) 0.8V  

C) 0.4V 

D) 0.2V 

Answer 1

Hint: We must know that if two semiconductors, which are connected parallel, the current will tend to flow through the semiconductor which needs less voltage for starting conduction. Also, in a reverse-biased condition, the semiconductor will not conduct through major charge carriers. Here, we will consider two cases as given in the question and obtain the output voltage. Then we will find the difference in output voltage ${{V}_{0}}$.

Complete solution:

We will consider the first connection as given in the figure, where both the semiconductors are connected parallel and forward biased. We will draw a diagram indicating the connection and voltage drop in each semiconductor.        

Here, the current will flow through the Ge diode because the voltage needed for starting the conduction through Ge is $0.3V$ which is lesser than the Si diode. So current will be passing only through Ge diodes. Then the output voltage ${{V}_{0}}$ can be obtained as,

${{V}_{0}}=V-{{V}_{Ge}}$

Where, $V$ is the input voltage and ${{V}_{Ge}}$ is the voltage drop across the Ge diode.

So, ${{V}_{0}}=12-0.3=11.7V$

Therefore, the output voltage in the first condition is found to be $11.7V$.

Now, in the second case, where both the semiconductors are connected parallel and Ge diode is connected in reverse biased condition and the Si diode is connected in forward biased condition. We will draw a diagram indicating the connection and voltage drop in each semiconductor.            

Here, the current will flow through the Si diode because the Ge diode is connected in reverse biased condition and it will not conduct. So the current will be passing only through the Si diode. Then the output voltage ${{V}_{0}}$ can be obtained as,

${{V}_{0}}=V-{{V}_{Si}}$

Where, $V$is the input voltage and ${{V}_{Si}}$ is the voltage drop across the Si diode.

${{V}_{0}}=12-0.7=11.3V$

Therefore, the output voltage in the second condition is found to be $11.3V$.

So, the difference between these two voltages will give us the change in ${{V}_{0}}$.

$\Rightarrow \Delta {{V}_{0}}=11.7-11.3=0.4V$

That is the change in ${{V}_{0}}$ is $0.4V$. So, option c is the correct answer.

Note: If a diode is reverse biased, it will not conduct. This is due to the voltage at the cathode being higher than that of the anode. So no movement of major charge carriers will be happening. Hence it will be acting as high resistance. So, no current will flow through it until the electric field is so high that the diode breaks down.