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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)
- a)1.61 x 10-19 C
- b)1.13 x 10-15 C
- c)1.13 x 10-19 C
- d)1.61 x 10-15 C
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A junction diode is fabricated in which the p and n regions are doped ...
We have the formula to find out Charge stored on either side on a junction given as follows,
We don’t have the value of W with us, it needs to be calculated. This can be easily calculated using the following formula,
is the applied Voltage and V0 is the barrier potential.here V = 0 as no external bias is applied.
therefore,VB = V0 = 0.78V
Then the value of W is 2.01 x 10-6 cm.
So the value of QJ is found out to be 1.61 x 10-15 C
Most Upvoted Answer
A junction diode is fabricated in which the p and n regions are doped ...
X 10^18/cm^3 atoms of phosphorus and boron, respectively. The diode is then forward biased with a voltage of 0.7V. Given that the electron and hole mobilities are 0.1 m^2/Vs and 0.05 m^2/Vs, respectively, calculate the current flowing through the diode.
First, we need to calculate the minority carrier concentrations in the p and n regions. Since the p and n regions are doped equally, the minority carrier concentrations can be calculated as follows:
np0 = ni^2 / N_A = (1.5 x 10^10)^2 / 5 x 10^18 = 4.5 x 10^1 cm^-3
nn0 = ni^2 / N_D = (1.5 x 10^10)^2 / 5 x 10^18 = 4.5 x 10^1 cm^-3
Next, we calculate the excess minority carrier concentrations in the p and n regions:
Δpp = ni^2 / N_A exp(qV / kT) = (1.5 x 10^10)^2 / 5 x 10^18 exp(0.7 / (0.0259)) = 3.57 x 10^8 cm^-3
Δnn = ni^2 / N_D exp(qV / kT) = (1.5 x 10^10)^2 / 5 x 10^18 exp(0.7 / (0.0259)) = 3.57 x 10^8 cm^-3
The total minority carrier concentrations are then:
pp = np0 + Δpp = 4.5 x 10^1 + 3.57 x 10^8 = 3.57 x 10^8 cm^-3
nn = nn0 + Δnn = 4.5 x 10^1 + 3.57 x 10^8 = 3.57 x 10^8 cm^-3
Finally, we can calculate the current flowing through the diode using the equation:
I = qA(Dn*pp + Dp*nn)*V/L
Where:
q = electronic charge = 1.6 x 10^-19 C
A = cross-sectional area of the diode
Dn = electron mobility = 0.1 m^2/Vs
Dp = hole mobility = 0.05 m^2/Vs
V = applied voltage = 0.7 V
L = length of the diode
Assuming a cross-sectional area of 1 cm^2 and a length of 0.1 cm, we can substitute the values into the equation:
I = (1.6 x 10^-19)(1)(0.1)(0.1)(3.57 x 10^8 + 3.57 x 10^8)(0.7)/(0.1) = 3.2 x 10^-3 A
Therefore, the current flowing through the diode is 3.2 mA.
First, we need to calculate the minority carrier concentrations in the p and n regions. Since the p and n regions are doped equally, the minority carrier concentrations can be calculated as follows:
np0 = ni^2 / N_A = (1.5 x 10^10)^2 / 5 x 10^18 = 4.5 x 10^1 cm^-3
nn0 = ni^2 / N_D = (1.5 x 10^10)^2 / 5 x 10^18 = 4.5 x 10^1 cm^-3
Next, we calculate the excess minority carrier concentrations in the p and n regions:
Δpp = ni^2 / N_A exp(qV / kT) = (1.5 x 10^10)^2 / 5 x 10^18 exp(0.7 / (0.0259)) = 3.57 x 10^8 cm^-3
Δnn = ni^2 / N_D exp(qV / kT) = (1.5 x 10^10)^2 / 5 x 10^18 exp(0.7 / (0.0259)) = 3.57 x 10^8 cm^-3
The total minority carrier concentrations are then:
pp = np0 + Δpp = 4.5 x 10^1 + 3.57 x 10^8 = 3.57 x 10^8 cm^-3
nn = nn0 + Δnn = 4.5 x 10^1 + 3.57 x 10^8 = 3.57 x 10^8 cm^-3
Finally, we can calculate the current flowing through the diode using the equation:
I = qA(Dn*pp + Dp*nn)*V/L
Where:
q = electronic charge = 1.6 x 10^-19 C
A = cross-sectional area of the diode
Dn = electron mobility = 0.1 m^2/Vs
Dp = hole mobility = 0.05 m^2/Vs
V = applied voltage = 0.7 V
L = length of the diode
Assuming a cross-sectional area of 1 cm^2 and a length of 0.1 cm, we can substitute the values into the equation:
I = (1.6 x 10^-19)(1)(0.1)(0.1)(3.57 x 10^8 + 3.57 x 10^8)(0.7)/(0.1) = 3.2 x 10^-3 A
Therefore, the current flowing through the diode is 3.2 mA.
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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?
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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?.
A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?.
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A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)a)1.61 x 10-19 Cb)1.13 x 10-15 Cc)1.13 x 10-19 Cd)1.61 x 10-15 CCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice GATE tests.
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