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The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?
Assume In (2) = 0.7
  • a)
    0.9846 V
  • b)
    0.73604 V
  • c)
    0.7182V
  • d)
    0.49 V
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The built-in potential of a P N junction diode is 0.7 V at room temper...
The built-in potential of a P-N junction diode is determined by the difference in the work functions of the P and N regions. It is given by the formula:

\(V_{bi} = \frac{kT}{q} \ln \left( \frac{N_AN_D}{n_i^2} \right)\)

where:
- \(V_{bi}\) is the built-in potential
- \(k\) is the Boltzmann constant
- \(T\) is the temperature in Kelvin
- \(q\) is the elementary charge
- \(N_A\) and \(N_D\) are the doping concentrations of the P and N regions, respectively
- \(n_i\) is the intrinsic carrier concentration

In this question, we are given that the built-in potential is 0.7 V at room temperature. Let's assume the room temperature is 300 K.

To find the approximate value of the built-in potential if the doping concentrations on both sides are doubled, we need to calculate the new value of \(V_{bi}\) using the formula above.

Doubling the doping concentrations means that \(N_A\) and \(N_D\) are both multiplied by 2.

Let's plug in the values into the formula:

\(V_{bi2} = \frac{kT}{q} \ln \left( \frac{2N_AN_D}{n_i^2} \right)\)

Since we are only interested in the approximate value, we can use the approximation \(\ln(2x) \approx \ln(x) + \ln(2)\).

\(V_{bi2} = \frac{kT}{q} \left( \ln \left( \frac{N_AN_D}{n_i^2} \right) + \ln(2) \right)\)

The term \(\ln \left( \frac{N_AN_D}{n_i^2} \right)\) is the same as the term in the original formula, which gives the built-in potential at room temperature. Let's call this term \(V_{bi1}\):

\(V_{bi1} = \frac{kT}{q} \ln \left( \frac{N_AN_D}{n_i^2} \right)\)

Therefore, we can rewrite the equation as:

\(V_{bi2} = V_{bi1} + \frac{kT}{q} \ln(2)\)

Substituting the values, we get:

\(V_{bi2} = 0.7 + \frac{(1.38 \times 10^{-23} \times 300)}{(1.6 \times 10^{-19})} \times \ln(2)\)

\(V_{bi2} \approx 0.73604\) V

Therefore, the approximate value of the built-in potential if the doping concentrations on both sides are doubled is 0.73604 V.
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Community Answer
The built-in potential of a P N junction diode is 0.7 V at room temper...
Concept:
  • As the N-type material has lost electrons and the P-type has lost holes, the N-type material has become positive with respect to the P-type.
  • Then the presence of impurity ions on both sides of the junction causes an electric field to be established across this region with the N-side at a positive voltage relative to the P-side.
  • The problem now is that a free charge requires some extra energy to overcome the barrier that now exists for it to be able to cross the depletion region junction.
  • This electric field created by the diffusion process has created a “built-in potential difference” across the junction with an open-circuit (zero bias) potential
Calculation:
Given built-in voltage of the diode is 0.7 V
Now doping concentrations on both sides are doubled. i.e, NA’ = 2NA and ND’ = 2ND
Newly built-in potential is:
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The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?Assume In (2) = 0.7a)0.9846 Vb)0.73604Vc)0.7182Vd)0.49VCorrect answer is option 'B'. Can you explain this answer?
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The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?Assume In (2) = 0.7a)0.9846 Vb)0.73604Vc)0.7182Vd)0.49VCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?Assume In (2) = 0.7a)0.9846 Vb)0.73604Vc)0.7182Vd)0.49VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?Assume In (2) = 0.7a)0.9846 Vb)0.73604Vc)0.7182Vd)0.49VCorrect answer is option 'B'. Can you explain this answer?.
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