The resistivity of a pure semiconductor is 0.5 Ωm. If the electron and hole mobility be 0.39 m²/V-s and 0.19 m²/V-s respectively then calculate the intrinsic carrier concentration.
A Ge specimen is doped with Al. The concentration of acceptor atoms is \[\sim 10^{21} atom/m^{3}\]. Given that the intrinsic concentration of electron hole pairs is \[\sim 10^{19}/m^{3}\], the concentration of electrons in the specimen is :
According to mass action law
\[ n_{e}\times n_{h}= n_{i}^{2} \]
\[ n_{h}=10^{21} ; n_{i}= 10^{19}; n_{e}=n_{i}^{2}/n_{h} = n_{e}= 10^{38}/10^{21}= 10^{17} \]
The contribution in the total current flowing through a semiconductor due to electrons and holes are 3/4 and 1/4 respectively. If the drift velocity of electrons is 5/2 times that of holes at this temperature, then the ratio of concentration of electrons and holes is :
Current i = neAv
\[ \frac{I_{1}}{I_{2}}= \frac{n_{1}}{n_{2}} * \frac{v_{d1}}{v_{d2}} \]
\[ \frac{3}{1}= \frac{n_{1}}{n_{2}} * \frac{5}{2} \]
\[ \frac{n_{1}}{n_{2}} = \frac{6}{5} \]
Two PN-junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be :
In circuit 1. First Diode is forward biased and second diode is reverse biased. So, potential across 1st diode is less than 2nd diode.
In circuit 2: Both the diodes are forward biased so potential difference across them is equal
In circuit 3: Both the diodes are reverse biased so potential difference across them is equal to V/2.