Electric Field in a BJT MCQ Quiz - Objective Question with Answer for Electric Field in a BJT - Download Free PDF
Last updated on Jun 12, 2025
Latest Electric Field in a BJT MCQ Objective Questions
Electric Field in a BJT Question 1:
Operating Frequency limitation for a Bipolar semi-conductor device is due to
Answer (Detailed Solution Below)
Electric Field in a BJT Question 1 Detailed Solution
The correct answer is: 3) both (1) and (2)
Explanation:
-
Junction Parasitics (like capacitance and resistance at the PN junctions) slow down the switching speed of the device, especially at high frequencies.
-
Electron Mobility affects how quickly charge carriers can move through the semiconductor. Lower mobility results in slower device response, thus limiting high-frequency operation.
Both factors contribute to the frequency limitation of bipolar semiconductor devices.
Electric Field in a BJT Question 2:
Consider the npn transistor shown in figure.
Assume that one-half of the base current enters from each side of the emitter strip and flows uniformly to the center of the emitter. Assume the following parameters for the transistor:
NB = 1016 cm-3, xB = 0.70 μm
μp = 400 cm2/V-s, S = 8 μm
Emitter length L = 100 μm
The resistance between x = 0 and x = S/2 for the flow of base current (IB) will be ______ Ω.
Answer (Detailed Solution Below) 885 - 900
Electric Field in a BJT Question 2 Detailed Solution
The structure of the base region of the given NPN transistor is redrawn as:
The cross-sectional area of the base will be:
A = L × xB
The resistance is given by:
\(R = \frac{{\rho \ell }}{A} = \frac{\ell }{{\sigma A}}\)
σ conducting of the material given as:
σ = e μ N
μ = Mobility of the carriers
N = Number of carrier concentrations
Since the Base is of p-type, the conduction carriers will be holes, i.e.
σ = e μp NB
Now, the resistance between x = 0 to x = S/2 will be:
\(R = \frac{{S/2}}{{e{\mu _p}\;{N_B}\left( {L{x_B}} \right)}}\)
Putting the respective values, we get:
\(R = \frac{{4 \times {{10}^{ - 4}}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {400} \right)\left( {{{10}^{16}}} \right)\left( {100 \times {{10}^{ - 4}}} \right)\left( {0.7 \times {{10}^{ - 4}}} \right)}}\)
R = 893 ΩElectric Field in a BJT Question 3:
In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector doping in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close to unity, which one of the following conditions is TRUE?
Answer (Detailed Solution Below)
Electric Field in a BJT Question 3 Detailed Solution
The emitter injection efficiency is a measure of how efficient the emitter is in injecting electrons into the base.
It is important because the entire transistor action is due to the electrons injected into the base.
The holes, which are simultaneously injected into the emitter, are useless from this point of view.
In BJT the doping profile are related as-
NE> NC> NB
Now, as emitter injection efficiency of the BJT is close to unity we can say
NE≫ NB
Thus, NE≫ NB and NB < NC
Top Electric Field in a BJT MCQ Objective Questions
In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector doping in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close to unity, which one of the following conditions is TRUE?
Answer (Detailed Solution Below)
Electric Field in a BJT Question 4 Detailed Solution
Download Solution PDFThe emitter injection efficiency is a measure of how efficient the emitter is in injecting electrons into the base.
It is important because the entire transistor action is due to the electrons injected into the base.
The holes, which are simultaneously injected into the emitter, are useless from this point of view.
In BJT the doping profile are related as-
NE> NC> NB
Now, as emitter injection efficiency of the BJT is close to unity we can say
NE≫ NB
Thus, NE≫ NB and NB < NC
Operating Frequency limitation for a Bipolar semi-conductor device is due to
Answer (Detailed Solution Below)
Electric Field in a BJT Question 5 Detailed Solution
Download Solution PDFThe correct answer is: 3) both (1) and (2)
Explanation:
-
Junction Parasitics (like capacitance and resistance at the PN junctions) slow down the switching speed of the device, especially at high frequencies.
-
Electron Mobility affects how quickly charge carriers can move through the semiconductor. Lower mobility results in slower device response, thus limiting high-frequency operation.
Both factors contribute to the frequency limitation of bipolar semiconductor devices.
Electric Field in a BJT Question 6:
In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector doping in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close to unity, which one of the following conditions is TRUE?
Answer (Detailed Solution Below)
Electric Field in a BJT Question 6 Detailed Solution
The emitter injection efficiency is a measure of how efficient the emitter is in injecting electrons into the base.
It is important because the entire transistor action is due to the electrons injected into the base.
The holes, which are simultaneously injected into the emitter, are useless from this point of view.
In BJT the doping profile are related as-
NE> NC> NB
Now, as emitter injection efficiency of the BJT is close to unity we can say
NE≫ NB
Thus, NE≫ NB and NB < NC
Electric Field in a BJT Question 7:
Consider the npn transistor shown in figure.
Assume that one-half of the base current enters from each side of the emitter strip and flows uniformly to the center of the emitter. Assume the following parameters for the transistor:
NB = 1016 cm-3, xB = 0.70 μm
μp = 400 cm2/V-s, S = 8 μm
Emitter length L = 100 μm
The resistance between x = 0 and x = S/2 for the flow of base current (IB) will be ______ Ω.
Answer (Detailed Solution Below) 885 - 900
Electric Field in a BJT Question 7 Detailed Solution
The structure of the base region of the given NPN transistor is redrawn as:
The cross-sectional area of the base will be:
A = L × xB
The resistance is given by:
\(R = \frac{{\rho \ell }}{A} = \frac{\ell }{{\sigma A}}\)
σ conducting of the material given as:
σ = e μ N
μ = Mobility of the carriers
N = Number of carrier concentrations
Since the Base is of p-type, the conduction carriers will be holes, i.e.
σ = e μp NB
Now, the resistance between x = 0 to x = S/2 will be:
\(R = \frac{{S/2}}{{e{\mu _p}\;{N_B}\left( {L{x_B}} \right)}}\)
Putting the respective values, we get:
\(R = \frac{{4 \times {{10}^{ - 4}}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {400} \right)\left( {{{10}^{16}}} \right)\left( {100 \times {{10}^{ - 4}}} \right)\left( {0.7 \times {{10}^{ - 4}}} \right)}}\)
R = 893 ΩElectric Field in a BJT Question 8:
Operating Frequency limitation for a Bipolar semi-conductor device is due to
Answer (Detailed Solution Below)
Electric Field in a BJT Question 8 Detailed Solution
The correct answer is: 3) both (1) and (2)
Explanation:
-
Junction Parasitics (like capacitance and resistance at the PN junctions) slow down the switching speed of the device, especially at high frequencies.
-
Electron Mobility affects how quickly charge carriers can move through the semiconductor. Lower mobility results in slower device response, thus limiting high-frequency operation.
Both factors contribute to the frequency limitation of bipolar semiconductor devices.