(a) Calculate the value of V₀ and i, if the silicon and germanium diode start conducting at 0.7 V and 0.3 V, respectively

(b) If the Ge diode connection is now reversed, what will be the new values of V₀ and i?

 OR

Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction. 

(a) Ge diode will start conducting before the silicon diode does so. The effective forward voltage across Ge diode is

(12 - 0.3) V =11.7 V.

This will appear as the output voltage across the load, i.e.,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{o}}==11.7 \mathrm{V} \\ \mathrm{i}==\frac{11.7}{5\times {10}^3} \mathrm{A}=2.34 \mathrm{mA} \end{gathered}$$

 (b) On reversing the connection of Ge diode, it will be reverse biased and conduct no current. Only Si diode will conduct. Therefore,

$V_o=(12-0.7)\mathrm{V}=11.3 \mathrm{V}$

and current,  $i=\frac{11.3}{5\times {10}^3} \mathrm{A}=2.26 \mathrm{mA}$

OR

The small region in the vicinity of the junction which is depleted of free charge carriers and has only immobile ions is called depletion region.

The accumulation of negative charges in the p-region and positive charge in the n-region sets up a potential difference across the junction. This acts as a barrier and is called barrier potential V_B