Space Charge Region MCQ Quiz in मल्याळम - Objective Question with Answer for Space Charge Region - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 16, 2025
Latest Space Charge Region MCQ Objective Questions
Top Space Charge Region MCQ Objective Questions
Space Charge Region Question 1:
A silicon PN junction at T = 300 K with a p-type doping concentration of \({{\rm{N}}_{\rm{a}}} = {10^{18}}{\rm{c}}{{\rm{m}}^{ - 3}}\). For a maximum electric field of \(\left| {{{\rm{E}}_{{\rm{max}}}}} \right| = 3 \times {10^5}\frac{{\rm{V}}}{{{\rm{cm}}}}\) at a reverse bias \({{\rm{V}}_{\rm{R}}} = 25{\rm{V}}\), the n-type doping concentration \(\left( {{\rm{in\;}}{{10}^{16}}{\rm{c}}{{\rm{m}}^{ - 3}}} \right)\) is approximately ___________.
Given, \({\epsilon_{\rm{s}}} = 11.7 \times 8.854 \times {10^{ - 14}}\frac{{\rm{F}}}{{{\rm{cm}}}}\)
Answer (Detailed Solution Below) 1 - 1.3
Space Charge Region Question 1 Detailed Solution
The maximum electric field is given by
\(\begin{array}{l} {{\rm{E}}_{{\rm{max}}}} = \frac{{ - 2\left( {{{\rm{V}}_{{\rm{bi}}}} + {{\rm{V}}_{\rm{R}}}} \right)}}{{\rm{W}}}\\ \Rightarrow {{\rm{E}}_{{\rm{max}}}} = - {\left\{ {\frac{{2{\rm{q}}\left( {{{\rm{V}}_{{\rm{bi}}}} + {{\rm{V}}_{\rm{R}}}} \right)}}{{{\epsilon_{\rm{s}}}}}\left( {\frac{{{{\rm{N}}_{\rm{a}}}{{\rm{N}}_{\rm{d}}}}}{{{{\rm{N}}_{\rm{a}}} + {{\rm{N}}_{\rm{d}}}}}} \right)} \right\}^{\frac{1}{2}}} \end{array}\)
Now, \({{\rm{V}}_{{\rm{bi}}}}\) is usually less than \(1{\rm{\;and\;}}{{\rm{V}}_{\rm{R}}} = 25\)
So, we may approximate \({{\rm{E}}_{{\rm{max}}}}\) as
\(\begin{array}{l} {{\rm{E}}_{{\rm{max}}}} \approx - {\left\{ {\frac{{2{\rm{q}}{{\rm{V}}_{\rm{R}}}}}{{{\epsilon_{\rm{s}}}}}\left( {\frac{{{{\rm{N}}_{\rm{a}}}{{\rm{N}}_{\rm{d}}}}}{{{{\rm{N}}_{\rm{a}}} + {{\rm{N}}_{\rm{d}}}}}} \right)} \right\}^{\frac{1}{2}}}\\ \Rightarrow \left| {{{\rm{E}}_{{\rm{max}}}}} \right| = {\left\{ {\frac{{2{\rm{q}}{{\rm{V}}_{\rm{R}}}}}{{{\epsilon_{\rm{s}}}}}\left( {\frac{{{{\rm{N}}_{\rm{a}}}{{\rm{N}}_{\rm{d}}}}}{{{{\rm{N}}_{\rm{a}}} + {{\rm{N}}_{\rm{d}}}}}} \right)} \right\}^{\frac{1}{2}}}\\ \Rightarrow 3 \times {10^5} = {\left\{ {\frac{{2\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {25} \right)}}{{\left( {11.7} \right)\left( {8.854 \times {{10}^{ - 14}}} \right)}}\left( {\frac{{{{10}^{18}} \cdot {{\rm{N}}_{\rm{d}}}}}{{{{10}^{18}} + {{\rm{N}}_{\rm{d}}}}}} \right)} \right\}^{\frac{1}{2}}}\\ \Rightarrow 9 \times {10^{10}} = 7.723 \times {10^{ - 6}}\left( {\frac{{{{10}^{18}}{{\rm{N}}_{\rm{d}}}}}{{{{10}^{18}} + {{\rm{N}}_{\rm{d}}}}}} \right)\\ = 9 \times {10^{28}} + 9 \times {10^{10}}{{\rm{N}}_{\rm{d}}} = 7.723 \times {10^{12}}{{\rm{N}}_{\rm{d}}}\\ \Rightarrow {{\rm{N}}_{\rm{d}}} = \frac{{9 \times {{10}^{28}}}}{{\left( {7.723 - 0.09} \right) \times {{10}^{12}}}} = 1.18 \times {10^{16}}{\rm{c}}{{\rm{m}}^{ - 3}} \end{array}\)
Space Charge Region Question 2:
Comprehension:
Common Data for Questions 19 and 20 :
Consider a silicon p – n junction at room temperature having the following parameters :
Doping of the n – side = 1 × 1017 cm-3
Depletion width on the n – side = 0.1 μm
Depletion width on the p – side = 1.0 μm
Intrinsic carrier concentration = 1.4 × 1010 cm-3
Thermal voltage = 26 mV
Permittivity of free space = 8.85 × 10-14F ⋅ cm-1
Dielectric constant of silicon = 12
The peak electric field in the device is:
Answer (Detailed Solution Below)
0.15 MV. cm-1, directed from n-region to p – region
Space Charge Region Question 2 Detailed Solution
\({E_{max}} = \frac{{e{N_A}}}{{{\epsilon_{si}}}}{x_p} = \frac{{1.6 \times {{10}^{ - 19}} \times {{10}^{16}} \times 1 \times {{10}^{ - 4}}}}{{12 \times 8.85 \times {{10}^{ - 14}}}} = 0.15\;MV/c{m^{ - 1}}\)
The direction of electric field in a PN junction is always from N side to P side.
Space Charge Region Question 3:
Comprehension:
Common Data for Questions 19 and 20 :
Consider a silicon p – n junction at room temperature having the following parameters :
Doping of the n – side = 1 × 1017 cm-3
Depletion width on the n – side = 0.1 μm
Depletion width on the p – side = 1.0 μm
Intrinsic carrier concentration = 1.4 × 1010 cm-3
Thermal voltage = 26 mV
Permittivity of free space = 8.85 × 10-14F ⋅ cm-1
Dielectric constant of silicon = 12
The built–in potential of the junction
Answer (Detailed Solution Below)
Space Charge Region Question 3 Detailed Solution
From charge neutrality we have,
\({N_A}{x_p} = {N_D}{x_n}\)
or, \({N_A} = {N_D}.\frac{{{x_n}}}{{{x_p}}}\)
or, \({N_A} = {10^{17}}.\frac{{0.1}}{{1.0}}\)
∴\({N_A} = {10^{16}}\)
∴\({V_{bi}} = {V_T}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)
\( = 26 \times {10^{ - 3}} \times \ln \left[ {\frac{{{{10}^{17}} \times {{10}^{16}}}}{{{{\left( {1.4 \times {{10}^{10}}} \right)}^2}}}} \right]\)
= 0.761 V