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The conduction band edge in the p material is not at the same level to that of conduction band edge in the n material. Is it true or false?
In a pn junction diode, the energy levels of the p material and n material will not be at same level. They will be different. So, the conduction band edge as well as the valence band edge of the p material will not be same to that of the n material.
Which of the following equations represent the correct expression for the shift in the energy levels for the pn junction?
The shift in the energy of the energy level will be the difference of the conduction band edge of the p material and conduction band edge of n material. In the energy level diagram, the conduction band edge of p material is higher than that of the n material.
Calculate the E_{o} given that N_{d}=1.5*10^{10}cm^{3}, N_{a}=1.5*10^{10}cm^{3} at temperature 300K?
E_{o}=kTln((N_{d}*N_{a})/(n_{i})^{2})
Substituting k=1.38*10^{23}/K, T=300k and the values ofN_{d},N_{a}and ni,
We get
E_{o}=0eV.
In a pn junction, the valence band edge of the p material is greater than which of the following band?
When the pn junction is formed, the energy levels of the p material go higher than the n material. That’s why the valence band of the p material will be greater than that of the n material.
Which of the following equations represent the correct expression for the band diagram of the pn junction? (E1=difference between the fermi level of material and conduction band of n material and E2=difference between the conduction band of n material and fermi level of n material)
From the energy band diagram of the pn junction, the option ‘a’ satisfies that band diagram.
Calculate the value of E_{o} when p_{no}=10^{4}cm^{3} and p_{po}=10^{16}cm^{3} at T=300K.
E_{o}=kTln(p_{po}/p_{no})
Substituting the values, we get
E_{o}=0.7eV.
Calculate the value of Dp when µp=400cm/s and V_{T}=25mV.
Dp= µp*V_{T}
=400*10^{2}*25*10^{3}
=0.1.
What is the value of kT at room temperature?
kT=1.38*10^{23}*300K
=4.14*10^{21}/ (1.6*10^{19})
=0.0256eV.
Is Vo depends only on the equilibrium concentrations. Is it true or false?
Vo is the contact potential of the junction when the junction is in equilibrium. If, the junction is not in the equilibrium, Vo can’t be calculated.
Calculate Vo when p_{po}=10^{16}cm^{3}, p_{no}=10^{4}cm^{3} and Vt=25mV.
V_{o}=V_{T} ln(p_{po}/p_{no} )
=25*10^{3}*ln(10^{16}/10^{4})
=0.69V.
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